Class 7 Maths

Algebra Chapter 7 Equations Exercise 7 Solved Example Problems

Equations Introduction

An equation is a statement of equality that involves one or more literal numbers. The literal numbers are called unknowns or variables.

Any value of the variables that satisfy the given equation is called a solution or root of the equation. Many problems of arithmetic may be solved easily by forming equations properly and finding their solutions.

At this early stage, our aim is to form linear equations of one variable and to obtain their solutions.

Math Solution Of Class 7 Wbbse

Read and Learn More WBBSE Solutions For Class 7 Maths

Formation of equation

Suppose, we do not know the marks obtained by Ram in the examination.

But it is known that, if we add 15 marks with the marks obtained by Ram then the sum will be 90 marks.

Since, in algebra, the unknown number may be expressed by an alphabetic symbol, the marks obtained by Ram may be taken as x. Then we may express the previous statement as x + 15 = 90.

This process of equating different things is known as ‘an equation’.

The specific value of the unknown thing which satisfies the equation is known as the root of the equation.

Some useful information about equation

1. If same number is added to both sides of an equation, its sides remain equal.
2. If same number is subtracted from both sides of an equation, its sides remain equal.
3. If both sides of an equation is multiplied by the same number, its sides remain equal.
4. If both sides of an equation is divided by the same number, its sides remain equal.

Equation and identity

In an equation, two sets of expressions are equalised by a sign.

In each of these sets the known and unknown expressions are connected by the signs of addition, subtraction, multiplication or division.

Both sides become equal for a specific value of the unknown quantity.

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Example:

x + 2 = 5 is an equation. The two sides become equal for only one value of x, namely 3. Both sides do not become equal for any value of x other than 3.

An identity is expressed in a similar manner, but its difference with the equation is that, its both sides are equal for all the values of the unknown quantity.

Example:

x+5=x+ 4+ 1 is an identity because whatever may be the value of x, its sides are always equal.

Rules for solving an equation

The specific value of the unknown quantity which satisfies an equation is called the root of the equation. The determination of the root is called the solution of the equation. In order to solve an equation some rules should be kept in mind.

1. If a positive number changes side, it becomes negative.

For Example: If x+ 5 = 2x + 10 then x= 2x + 10 – 5.

In this case, + 5 of the left-hand side has become – 5 after going to the right-hand side.

Again, if 3.v + 2 = x + 15 then 3x + 2 – 15 = x.

In this case, + 15 of the right-hand side has become – 15 after coming to the left-hand side.

2. If a negative number changes side, it becomes positive.

For Example: If 5x + 25 = 2x – 10 then 5x + 25 + 10 = 2x

Again, if 7x – 2 = 3x + 5 then 7x = 3x + 5 + 2.

3. If a number is a multiplier of the left-hand side of an equation then the right-hand side has to be divided by it while transferring it to the right-hand side.

For Example: If 3x = 27, then x = \(\frac{27}{3}\) = 9.

In this case, 3 is a multiplier of the left-hand side, so 27 on the right-hand side has to be divided by 3.

4. If a number is a divisor of the left-hand side of an equation then the right-hand side is to be multiplied by it while transferring it to the right-hand side.

For Example: If \(\frac{x}{5}\) = 7, then x= 7 x 5 = 35.

In this case, 5 is a divisor of the left-hand side, so 7 on the right-hand side has to be multiplied by 5.

Method of solution of an equation when both sides of the sign of equality contain expressions involving known and unknown quantities

1. At first, each side has to be simplified.
2. Then the terms involving the unknown quantity are to be kept in the left-hand side and the other terms should be kept on the right-hand side.
3. Now each side has to be simplified again.
4. At last, by dividing the right-hand side by the co-efficient of the left-hand side the required root of the equation may be obtained.

 

Algebra Chapter 7 Equations Exercise 7 Some Examples Of The Formation Of Equations

Example 1. Adding 16 with 8 times a number, it becomes 40. Form an equation to find the number.

Solution:

Given:

Adding 16 with 8 times a number, it becomes 40.

Let, the number be x. Its 8 times is 8x. Adding 16 with it we get 8x + 16.

Hence, from the given condition 8x + 16 = 40; it is the required equation.

Example 2. Rupees 140 was distributed among Ram, Shyam and Jadu in such a way that Shyam got, half the money of Ram and Jadu got half the money of Shyam. Form an equation to find their shares.

Solution:

Given:

Rupees 140 was distributed among Ram, Shyam and Jadu in such a way that Shyam got, half the money of Ram and Jadu got half the money of Shyam.

Let, Ram got ₹ x.

Then Shyam got ₹ \(\frac{x}{2}\)

and Jadu got ₹ \(\frac{x}{2}\) X \(\frac{1}{2}\) = ₹ \(\frac{x}{4}\)

Hence, from the given condition,

x + \(\frac{x}{2}\) +\(\frac{x}{4}\) = 140 ; it is the required equation.

WBBSE Class 7 Algebra Equations Examples

Example 3. The present age of the father is 7 times that of the son. After 10 years, age of the father will be 3 times that of the son. Form an equation to find their present ages.

Solution:

Given:

The present age of the father is 7 times that of the son. After 10 years, age of the father will be 3 times that of the son.

Let, the present age of the son be x years.

Then the present age of the father is 7x years.

After 10 years—age of the son will be (x + 10) years and age of the father will be (7x +10) years

Hence, from the given condition,

7x +10 = 3 (x +10); it is the required equation.

Example 4. The half of a number is greater than \(\frac{1}{5}\) th of it by 6. Form an equation to find the number.

Solution:

Given:

The half of a number is greater than \(\frac{1}{5}\) th of it by 6.

Let, the number be x.

Half of the number is \(\frac{x}{2}\) and\(\frac{1}{5}\) th of the number is \(\frac{x}{5}\)

Hence, from the given condition, \(\frac{x}{2}\)= \(\frac{x}{5}\) + 6; it is the required equation.

Example 5. Of the total number of fruits with a fruit vendor, \(\frac{1}{5}\) was mango, \(\frac{1}{4}\) was apple, \(\frac{2}{5}\) was lichi and the rest 60 were oranges. Form an equation to find the total number of fruits with the vendor.

Solution:

Given:

Of the total number of fruits with a fruit vendor, \(\frac{1}{5}\) was mango, \(\frac{1}{4}\) was apple, \(\frac{2}{5}\) was lichi and the rest 60 were oranges.

Let, the total number of fruits be x.

Hence, from the given condition,

\(\frac{x}{5}\) +\(\frac{x}{4}\)+\(\frac{2x}{5}\)+60 = x; it is the required equation.

Solved Problems for Class 7 Equations

Example 6. Solve : \(\frac{x}{a}\) + b = \(\frac{x}{b}\)+ a .

Example 7. Solve: 50 + 3(4x – 5) + 8(x + 3) = x + 2.

Solution:

∴ -3.

Example 8. Solve: \(\frac{1}{2}\) (x +1) + (x + 2) + \(\frac{1}{4}\) (x + 3) = 16.

Solution:

∴ x= 13.

Class 7 Maths Equation Exercise Solutions

Example 9. \(\frac{1}{(x +1)(x +2)}\) + \(\frac{1}{x +2)(x +3)}\) + \(\frac{1}{(x +3)(x +4)}\) = \(\frac{1}{(x +1)(x +6)}\)

Solution:

∴ x = -7

Example 10. Solve \(\frac{1}{(x+1)(2)}\) + \(\frac{1}{(x+3)(3)}\) = 4

∴x= 3.

West Bengal Board Class 7 Algebra Assistance

Example 11. Solve: 16-5 (7x-2) = 13 (x-2) + 4 (13 – x)

∴ x = 0.

16-5 (7x-2) = 13 (x-2) + 4 (13 – x) = 0

Example 12. Solve \(\frac{x}{2}\) + \(\frac{1}{3}\) = \(\frac{x}{3}\) + \(\frac{1}{2}\)

∴ x = 1

Example 13. Solve: 3 (x – 1) – (x + 2) = x + 2 (x – 1)

∴x = – 3.

3 (x – 1) – (x + 2) = x + 2 (x – 1) = – 3.

Wbbse Class 7 Maths Solutions

Example 14. Solve: \(\frac{3x+1}{16}\) + \(\frac{2x-3}{17}\) =\(\frac{x+3}{8}\) + \(\frac{3x-1}{14}\)

or, 30x = 150 or, x = \(\frac{150}{30}\) = 5

Example 15. If x = 2t and y = \(\frac{3t}{5}\) – 1, then find the value of t for which x = 5y.

Solution:

Given:

x = 2t and y = \(\frac{3t}{5}\) – 1

The relation, x = 5y will be satisfied when

2t = 5(\(\frac{3t}{5}\) – 1) or, 2t = 5(\(\frac{3t-5}{5}\))

or, 2t = 3t – 5 or, 3t – 2t = 5 or, t = 5

∴ t = 5

Example 16. If x = at² and y = 2at, find the relation between x and y.

Solution:

Given:

x = at² and y = 2at

x = at² or, at² = x or, t² =\(\frac{x}{a}\)…… (1)

Again, y = 2at or, 2at = y or, t = \(\frac{y}{2a}\)

Wbbse Class 7 Maths Solutions

Algebra Chapter 7 Equations Exercise 7 Some Problems On Equation

Example 1. If the sum of three consecutive numbers be 60. then find the numbers.

Solution:

Given:

The sum of three consecutive numbers be 60.

Let, the three consecutive numbers be x, x + 1 and x + 2

∴ According to the question, x + x+1+ x + 2 = 60

or, 3x + 3 = 60

or, 3x = 60 – 3

or, 3x = 57

or, x = \(\frac{57}{3}\) = 19

∴ The numbers are : 19, 19 + 1, 19 + 2, or, 19, 20, 21

∴19, 20, 21.

Example 2. If the sum of two numbers be 60 and their difference be 40 then find the numbers.

Solution:

Given:

The sum of two numbers be 60 and their difference be 40

Let, the greater number be x. Then the smaller number is 60 – x

According to the question, x – (60 – x) = 40

or, x – 60 + x = 40 or, 2x = 40 + 60 or, 2x = 100

or, x = \(\frac{100}{2}\) = 50

∴ The greater number = 50 and the smaller number = 60 – 50 = 10

∴ 50, 10

.’. The greater number = 50 and the smaller number = 60-50 = 10 Ans. 50,10.

Example 3. In a number of two digits, the digit in the tens’ place is greater than that of the units’ place by 3 and the sum of the digits is 11. Find the number.

Solution:

Given:

a number of two digits, the digit in the tens’ place is greater than that of the units’ place by 3 and the sum of the digits is 11.

Let, the digit in the units’ place be x.

Then the digit in the tens’ place is x + 3.

According to the question,

x + x + 3 = 11 or, 2x + 3 = 11 or, 2x = 11 – 3 or, 2x = 8

or, x = \(\frac{8}{2}\) = 4

∴ The digit in the units’ place = 4 and the digit in the tens’ place = 4 + 3 = 7

∴ The required number = 74

∴ 74.

WBBSE Class 7 Chapter 7 Equations Guide

Example 4. The present age of the father is 6 times that of the son. After 20 years, age of the father will be twice that of the son. Find the present age of the father.

Solution:

Given:

The present age of the father is 6 times that of the son. After 20 years, age of the father will be twice that of the son.

Let, the present age of the son be x years. Then the present age of the father is 6x years.

After 20 years— age of the son will be (x + 20) years

and age of the father will be (6x + 20) years

∴ According to the question,

6x + 20 = 2(x + 20)

or, 6x + 20 = 2x + 40

or, 6x – 2x = 40 – 20

or, 4x =20 or, x = \(\frac{20}{4}\) = 5

∴ The present age of the father is 6 x 5 years = 30 years.

∴ 30 years.

The present age of the father is 30 years.

Example 5. 1/3rd of a bamboo is within the mud, 1/4th of it  is in water and 5 metres above water. What is the length of the bamboo?

Solution:

Given:

1/3rd of a bamboo is within the mud, 1/4th of it  is in water and 5 metres above water.

Let, the length of the bamboo be x metres.

Length of the bamboo within mud = 7

metres and length of the bamboo within water = \(\frac{x}{4}\) metres

∴ According to the question, x – (\(\frac{x}{3}\) + \(\frac{x}{4}\)) = 5

or, x – \(\frac{4x+3x}{12}\) = 5

or, x – \(\frac{7x}{12}\) = 5

or, \(\frac{5x}{12}\) = 5 or, x = 5 x \(\frac{12}{5}\) = 12

∴ The length of the bamboo is 12 metres.

Example 6. The ratio of monthly incomes of two persons is 4: 5 and the ratio of their expenditures is 7: 9. If each of them saves ₹50 per month, then find the monthly income of each of them.

Solution:

Given:

The ratio of monthly incomes of two persons is 4: 5 and the ratio of their expenditures is 7: 9. If each of them saves ₹50 per month,

Let, the monthly income of  first person be ₹ 4x

and the monthly income of the second person be₹ 5x

Since each of them saves ₹50 per month

∴ Monthly expenditure of the first person = ₹(4x- 50)

and monthly expenditure of the second person = ₹ (5x – 50)

∴ According to the question,

\(\frac{4x-50}{5x-50}\) = \(\frac{7}{9}\)

or, 9(4x-50) = 7(5x-50) or, 36x – 450 = 35x- 350

or, 36x – 35x = 450 – 350

or, x = 100

∴ Monthly income of the first person = ₹ 4 x 100 =₹ 400

Monthly income of the second person = ₹ 5 x 100 =₹ 500

∴ ₹ 400 and ₹ 500.

Understanding Equations for Class 7 Students

Example 7. A man encashed a cheque of ₹2000 from bank. He received some five rupee notes and some ten rupee notes. If he received altogether 300 notes then what was the number of five rupee notes?

Solution:

Given:

A man encashed a cheque of ₹2000 from the bank. He received some five rupee notes and some ten rupee notes. If he received altogether 300 notes

Let, the man receive x number of five rupee notes and (300 – x) number of ten rupee notes.

∴He received 200 five rupee notes.

Class Vii Math Solution Wbbse

Example 8. Total marks scored by Ram, Shyam and Jadu in an examination was 156. Shyam scored 9 marks more than Ram and Jadu scored 9 marks less than Ram. What were the marks scored by each?

Solution:

Given:

Total marks scored by Ram, Shyam and Jadu in an examination was 156. Shyam scored 9 marks more than Ram and Jadu scored 9 marks less than Ram.

Let, Ram scored x marks.

Then Shyam scored (x + 9) marks and Jadu scored (x – 9) marks

∴ According to the question, x+x + 9 + x- 9 = 156

or, 3x = 156

or, x = \(\frac{156}{3}\) = 52

Hence, Ram scored 52 marks, Shyam scored (52 + 9) marks

or 61 marks and Jadu scored (52-9) marks or 43 marks

∴ Ram scored 52 marks, Shyam scored 61 marks and Jadu scored 43 marks.

Example 9. The sum of the present ages of the father and his son is 65 years. 10 years ago, the ratio of their ages was 7: 2. Find the present ages of the father and his son.

Solution:

Given:

The sum of the present ages of the father and his son is 65 years. 10 years ago, the ratio of their ages was 7: 2.

Let, the present age of the father = x years and the present age of the son = (65 – x) years.

Then from the given condition,

\(\frac{x-10}{65-x-10}\) = \(\frac{7}{2}\)

or, 2(x – 10) = 7(55 – x) or, 2x – 20 = 385 – 7x

or, 2x + 7x = 385 + 20 or, 9x = 405

or, x = \(\frac{405}{9}\) = 45

The present age of the father = 45 years and the present age of the son = (65 – 45) years = 20 years.

∴ At present age of the father is 45 years and age of the son is 20 years.

Class 7 Maths Exercise 7 Solved Examples

Example 10. There were one rupee and five rupee coins in a money bag and the total amount was ₹170. If half of one rupee coin is replaced by five rupee coins, the amount becomes ₹210. How many one-rupee and five-rupee coins were there in the money bag at first?

Solution:

Given:

There were one rupee and five rupee coins in a money bag and the total amount was ₹170. If half of one rupee coin is replaced by five rupee coins, the amount becomes ₹210.

Let, at first there were x numbers of 1 rupee coins and y number of 5 rupee coins in the money bag.

According to the question,

x+ 5y = 170 …… (1)

If half of 1 rupee coins is replaced by 5 rupee coins, then number of

1 rupee coins  becomes x-\(\frac{x}{2}\) =\(\frac{x}{2}\) and the number of

5 rupee coins becomes Hence (y + \(\frac{x}{2}\))

∴ A number of 1 rupee coins is 20 and that of 5 rupee coins is 30.

Step-by-Step Solutions for Class 7 Algebra Problems

Example 11. A train crosses a bridge 264 m long in 20 sec, but it takes 8 sec to cross a signal post by the side of the rail line. Find the length of the train and its speed.

Solution:

Given:

A train crosses a bridge 264 m long in 20 sec, but it takes 8 sec to cross a signal post by the side of the rail line.

Let, the length of the train be x m.

Then, the train takes 8 sec to cross x m and it takes 20 sec to cross (x + 264) m.

Hence, \(\frac{8}{20}\) = \(\frac{x}{x + 264}\)

or, 5x = 2x + 528 or, 5x – 2x = 528 or, 3x = 528

or, x = \(\frac{528}{3}\) = 176

∴ Length of the train is 176 m.

Also, in 8 sec the train crosses \(\frac{176}{8}\) m = 22 m

In 1 sec the train crosses 176 m

In 60×60 sec the train crosses = \(\frac{22 X 60 X 60}{1000}\) km = 79.2 km

∴ The length of the train is 176 m and its speed is 79.2 km/hour.

Example 12. A fruit seller sold some bananas at ₹10 per dozen and some guavas at ₹4 per pair and thereby got ₹368. If the number of bananas exceeds that of guava by 20, how many dozens of banana did he have?

Solution:

Given:

A fruit seller sold some bananas at ₹10 per dozen and some guavas at ₹4 per pair and thereby got ₹368. If the number of bananas exceeds that of guava by 20,

Let, that fruit seller had x guavas and (20 + x) bananas.

price of 12 bananas is ₹ 10

price of 1 banana is ₹ \(\frac{10}{12}\)

price of (20=x) banans is ₹ \(\frac{5}{6}\) (20 + x)

price of 2 guavas is ₹ 4

price of 1 guava is ₹ \(\frac{4}{2}\)

price of x guava is ₹ 2x

According to the question,

\(\frac{5}{6}\) (20 + x) + 2x = 368

Algebra Chapter 6 Factorisation Exercise 6 Solved Example Problems

Factorisation Introduction

A very important topic in algebra is to resolve an expression into factors. The idea of factors in algebra is similar to that of in arithmetic.

You have already learned about the multiple and factor of a given number in arithmetic.

For example, 35 = 5×7. Hence, factors of 35 are 1, 5, 7, and 35.

In a similar way, the factors of ab are 1, a, b, and ab. Let us now define a factor.

WBBSE Class 7 Factorisation Solutions

Factor

Read and Learn More WBBSE Solutions For Class 7 Maths

If the product of two or more expressions is equal to another expression then those expressions are called the factors of the product.

Example: If p x q x r = x then the expressions p, q, and r are called the factors of x.

Therefore, by factorization of the expression x, three factors p, q, and r are obtained.

Math Solution Of Class 7 Wbbse Factorisation by a selection of common factors

If a polynomial contains one or more common factors in each of its terms then the common factor (or factors) are taken outside the bracket according to the distributive law and the remaining portion is kept inside the bracket.

Example:

⇒ \(a^2 b+a b+a b^2=a b(a+1+b)=a \times b \times(a+1+b)\)

⇒ \(x^3 y^2+x^2 y^3=x^2 y^2(x+y)=x \times x \times y x y x(x+y)\)

Factorization with the help of the formula of the square of a binomial

We know that, (a + b)² = (a + b)(a + b) and (a -b)² = (a-b) (a- b).

Some expressions can be factorised with the help of these two formulae.

Example: \(x^2+4 x y z+4 y^2 z^2\)

= \((x)^2+2 \cdot x \cdot 2 y z+(2 y z)^2=(x+2 y z)^2\)

Again, \(4 a^2-12 a b+9 b^2=(2 a)^2-2 \cdot 2 a \cdot 3 b+(3 b)^2\) 4a² – 12ab + 9b² = (2a)² – 2.2a.3b + (3b)²

= \((2 a-3 b)^2=(2 a-3 b)(2 a-3 b)\)

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Factorisation with the help of the formula of the difference of two squares

Applying the formula a²-b² = (a + b) (a-b), some expressions can be factorised.

Example : \(25 x^2-81 y^2=(5 x)^2-(9 y)^2=(5 x+9 y)(5 x-9 y) .\)

Factorisation by the simultaneous application of the formulae of the square of a binomial and the difference of two squares

We know that, a² + 2ab + b² = (a + b)² and a² -b² = (a + b) (a-b).

Some expressions may be factorised by applying these two formulae simultaneously.

Example: \(a^4+4\)

= \(\left(a^2\right)^2+(2)^2\)

= \(\left(a^2\right)^2+2 \cdot a^2 \cdot 2+(2)^2-4 a^2\)

= \(\left(a^2+2\right)^2-(2 a)^2=\left(a^2+2+2 a\right)\left(a^2+2-2 a\right)\)

= \(\left(a^2+2 a+2\right)\left(a^2-2 a+2\right)\)

 Algebra Chapter 6 Factorisation Exercise 6 Some Examples On Factorisation

Example 1. Factorise : 3x + 12.

Solution:

Given:

3x + 12

= 3.x + 3.4 = 3 (x + 4)

∴ 3 (x +4).

3x + 12 = 3 (x +4).

Example 2. Factorise : 25m – 35n.

Solution:

Given: 25m – 35n

25m – 35n = 5 x 5m – 5 x 7n

∴ 5 (5m- 7n)

25m – 35n = 5 (5m- 7n)

Solved Problems for Class 7 Factorisation

Example 3. Factorise : \(p^2 q r+p q^2 r+p q r^2\)

Solution:

Given: \(p^2 q r+p q^2 r+p q r^2\)

⇒ \(p^2 q r+p q^2 r+p q r^2\) = pqr (p + q + r)

∴pqr (p + q + r).

⇒ \(p^2 q r+p q^2 r+p q r^2\) = pqr (p + q + r).

Wbbse Class 7 Maths Solutions

Example 4. Factorise : \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\)

Solution:

Given: \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\)

⇒ \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\)

∴ \(2 a^2 b\left(2 a^2-3 a b+6 b^2\right)\)

⇒ \(4 a^4 b-6 a^3 b^2+12 a^2 b^3\) = \(2 a^2 b\left(2 a^2-3 a b+6 b^2\right)\)

Example 5. Resolve into factors (ax + by) (p – q) – (ax + by) (q – r).

Solution:

Given:

(ax + by) (p-q)-(ax + by) (q – r)

= (ax +by) {(p-q)-(q- r)}

= (ax + by) (p-q-q + r)

= (ax + by) (p-2q + r)

∴ (ax +by) (p – 2q + r).

(ax + by) (p – q) – (ax + by) (q – r) = (ax +by) (p – 2q + r).

Example 6. Resolve into factors: \(x^2-(a+b) x+a b\)

Solution:

Given:

⇒ \(x^2-(a+b) x+a b\)

⇒ \(x^2-(a+b) x+a b=x^2-a x-b x+a b=x(x-a)-b(x-a)\)

∴ (x-a) (x- b)

⇒ \(x^2-(a+b) x+a b\) = (x-a) (x- b)

Example 7. Resolve into factors : (P + q) (a + b) + r(a + b) + c(p + q + r).

Solution:

Given:

(P + q) (a + b) + r(a + b) + c(p + q + r)

(P + q) (a + b) + r(a + b) + c{p + q + r)

= (a + b) (p + q + r) + c(p + q + r)

∴ (p + q + r) (a + b + c)

(P + q) (a + b) + r(a + b) + c(p + q + r) = (p + q + r) (a + b + c)

Class 7 Maths Factorisation Exercise Solutions

Example 8. Factorise : \(\text { 2ay }-x^2-y^2+z^2 \text {. }\)

Solution:

Given:

⇒ \(\text { 2ay }-x^2-y^2+z^2 \text {. }\)

= \((z)^2-(x-y)^2\)

= {z + (x – y)} {z – (x – y)}

= (z + x – y) (z – x + y)

∴ (z + x-y)(z-x + y).

⇒ \(\text { 2ay }-x^2-y^2+z^2 \text {. }\) = (z + x-y)(z-x + y).

Wbbse Class 7 Maths Solutions

Example 9. Resolve into factors \(9(x-y)^2-25(y-z)^2\)

Solution:

Given:

⇒ \(\left.9(x-y)^2-25(y-z)^2=\{3(x-y)\}^2-\{50-z)\right\}^2\)

= \((3 x-3 y)^2-(5 y-5 z)^2\)

= {(3x – 3y) + (5y – 5z)} {(3x – zy) – (5y – 5z)}

= (3x – 3y + 5y – 5z) (3x – 3y – 5y + 5z)

= (3x + 2y – 5z) (3x – 8y + 5z)

∴ (3x + 2y – 5z) (3x -8y+ 5z).

⇒ \(9(x-y)^2-25(y-z)^2\) = (3x + 2y – 5z) (3x -8y+ 5z).

Example 10. Resolve into factors: \(x^4+x^2 y^2+y^4\)

Solution:

Given:

⇒ \(x^4+x^2 y^2+y^4\)

= \(\left(x^2\right)^2+2 \cdot x^2 \cdot y^2+\left(y^2\right)^2-x^2 y^2\)

= \(\left(x^2+y^2\right)^2-(x y)^2\)

= \(\left(x^2+y^2+x y\right)\left(x^2+y^2-x y\right)\)

= \(\left(x^2+x y+y^2\right)\left(x^2-x y+y^2\right)\)

∴ \(\left(x^2+x y+y^2\right)\left(x^2-x y+y^2\right)\)

⇒ \(x^4+x^2 y^2+y^4\) = \(\left(x^2+x y+y^2\right)\left(x^2-x y+y^2\right)\)

West Bengal Board Class 7 Algebra Assistance

Example 11. Resolve into factors : \(x^2-y^2-6 x a+2 y a+8 a^2\)

Solution:

Given:

⇒ \(x^2-y^2-6 x a+2 y a+8 a^2\)

= \(x^2-6 x a+9 a^2-y^2+2 y a-a^2\)

= \((x)^2-2 \cdot x \cdot 3 a+(3 a)^2-\left(y^2-2 y a+a^2\right)\)

= \((x-3 a)^2-(y-a)^2\)

= {(x – 3a) + (y – a)} {(x – 3a) – (y – a)}

= (x – 3a + y – a) (x – 3a – y + a)

= (x + y-4a) (x-y – 2a)

∴ (x + y- 4a) (x-y – 2a).

⇒ \(x^2-y^2-6 x a+2 y a+8 a^2\) = (x + y- 4a) (x-y – 2a).

Wbbse Class 7 Maths Solutions

Example 12. Factorise: \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\)

Solution:

Given:

⇒ \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\)

⇒ \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\)

= \(4 a^2+12 a b+9 b^2-25 c^2-10 c-1\)

= \(\left(4 a^2+12 a b+9 b^2\right)-\left(25 c^2+10 c+1\right)\)

= \(\left\{(2 a)^2+2.2 a \cdot 3 b+(3 b)^2\right\}-\left\{(5 c)^2+2.5 c \cdot 1+(1)^2\right\}\)

= \((2 a+3 b)^2-(5 c+1)^2=\{(2 a+3 b)+(5 c+1)\}\{(2 a+3 b)-(5 c+1)\}\)

= (2a + 3b +5c + 1) (2a + 3b – 5c – 1)

∴ (2a + 3b + 5c + 1) (2a + 3b – 5c – 1).

⇒ \(4 a^2-1+9 b^2-25 c^2+12 a b-10 c\) = (2a + 3b + 5c + 1) (2a + 3b – 5c – 1).

Example 13. Factorise: \(9(5 c+6 d)^2-16(3 c-2 d)^2\)

Solution:

Given:

⇒ \(9(5 c+6 d)^2-16(3 c-2 d)^2\)

= \(\{3(5 c+6 d)\}^2-\{4(3 c-2 d)\}^2\)

= \((15 c+18 d)^2-(12 c-8 d)^2\)

= {(15c + 18d) + (12c – 8d)}{(15c + 18d) – (12c – 8d)}

= (15c+18d+12c-8d)(15c+18d-12c+ 8d)

∴ (27c + 10d)(3c + 26d)

⇒ \(9(5 c+6 d)^2-16(3 c-2 d)^2\) = (27c + 10d)(3c + 26d)

WBBSE Class 7 Chapter 6 Factorisation Guide

Example 14. Factorise : \(x^4+6 x^2 y^2+8 y^4\)

Solution:

Given:

⇒ \(x^4+6 x^2 y^2+8 y^4\)

= \(\left(x^2\right)^2+2 \cdot x^2 \cdot 3 y^2+\left(3 y^2\right)^2-y^4\)

= \(\left.\left(x^2+3 y^2\right)^2-y^2\right)^2\)

= \(\left(x^2+3 y^2+y^2\right)\left(x^2+3 y^2-y^2\right)\)

= \(\left(x^2+4 y^2\right)\left(x^2+2 y^2\right)\)

∴ \(\left(x^2+4 y^2\right)\left(x^2+2 y^2\right)\)

⇒ \(x^4+6 x^2 y^2+8 y^4\) = \(\left(x^2+4 y^2\right)\left(x^2+2 y^2\right)\)

Example 15. Factorise: \(3 x^2-y^2+z^2-2 x y-4 x z\)

Solution:

Given:

⇒ \(3 x^2-y^2+z^2-2 x y-4 x z\)

= \(4 x^2-4 x z+z^2-x^2-2 x y-y^2\)

= \((2 x)^2-2 \cdot 2 x \cdot z+(z)^2-\left(x^2+2 x y+y^2\right)\)

= \((2 x-z)^2-(x+y)^2\)

= {(2x – z) + (x + y)}{(2x – z) – (x + y)}

= (2x – z + x + y){2x – z – x- y)

= (3x + y – z)(x -y-z)

= (x-y- z)(3x + y-z)

∴ (x – y – z){3x + y – z).

⇒ \(3 x^2-y^2+z^2-2 x y-4 x z\) = (x – y – z){3x + y – z).

Example 16. Factorise : \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2\)

Solution:

Given:

⇒ \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2\)

⇒ \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2=(2 b c)^2-\left(b^2+c^2-a^2\right)^2\)

= \(\left\{(2 b c)+\left(b^2+c^2-a^2\right)\right\}\left\{(2 b c)-\left(b^2+c^2-a^2\right)\right\}\)

= \(\left(2 b c+b^2+c^2-a^2\right)\left(2 b c-b^2-c^2+a^2\right)\)

= \(\left\{(b+c)^2-(a)^2\right\}\left\{(a)^2-(b-c)^2\right\}\)

= (b+ c + a)(b + c – a)(a + b – c)(a – b + c)

∴ (a + b + c)(a + b – c)(b + c – a)(c + a – b)

⇒ \(4 b^2 c^2-\left(b^2+c^2-a^2\right)^2\) = (a + b + c)(a + b – c)(b + c – a)(c + a – b)

Factorisation Techniques for Class 7 Students

Example 17. Resolve \(4 a^2-12 a b+9 b^2\) into two linear factors and find their sum.

Solution:

Given:

⇒ \(4 a^2-12 a b+9 b^2\)

= \((2 a)^2-2 \cdot 2 a \cdot 3 b+(3 b)^2\)

= \((2 a-3 b)^2=(2 a-3 b)(2 a-3 b)\)

Now, sum of the linear factors = 2a – 3b + 2a – 3b

∴ 4a – 6b

⇒ \(4 a^2-12 a b+9 b^2\) = 4a – 6b

Example 18. Resolve \(\frac{p}{q}\) – \(\frac{q}{p}\) into factors.

Solution:

⇒ \(\frac{p}{q}\) – \(\frac{q}{p}\)

= \(\frac{p^2-q^2}{p q}=\frac{(p+q)(p-q)}{p q}\)

= \(\left(\frac{p+q}{p}\right)\left(\frac{p-q}{q}\right)=\left(1+\frac{q}{p}\right)\left(\frac{p}{q}-1\right)\)

⇒ \(\frac{p}{q}\) – \(\frac{q}{p}\) = \(\left(\frac{p+q}{p}\right)\left(\frac{p-q}{q}\right)=\left(1+\frac{q}{p}\right)\left(\frac{p}{q}-1\right)\)

Example 19. Express a – 6 as the difference of two squares and factories.

Math Solution Of Class 7 Wbbse

Solution:

a – b

= \(\left(a^{\frac{1}{2}}\right)^2-\left(b^{\frac{1}{2}}\right)^2=\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)

Example 20. Three factors of an expression are a, a + \(\frac{1}{a}\) and a – \(\frac{1}{a}\) find the expression.

Solution:

The required express

= \(a\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right)=a\left(a^2-\frac{1}{a^2}\right)=a^3-\frac{1}{a}\)

Example  21. Factorise : \(x^4-3 x^2 y^2+9 y^4\)

Solution:

Given:

⇒ \(x^4-3 x^2 y^2+9 y^4\)

= \(\left(x^2\right)^2+2 \cdot x^2 \cdot 3 y^2+\left(3 y^2\right)^2-9 x^2 y^2\)

= \(\left(x^2+3 y^2\right)^2-(3 x y)^2=\left(x^2+3 y^2+3 x y\right)\left(x^2+3 y^2-3 x y\right)\)

∴ \(\left(x^2+3 y^2+3 x y\right)\left(x^2+3 y^2-3 x y\right)\)

⇒ \(x^4-3 x^2 y^2+9 y^4\) =  \(\left(x^2+3 y^2+3 x y\right)\left(x^2+3 y^2-3 x y\right)\)

Class 7 Maths Exercise 6 Solved Examples

Example 22.  Factorise: \(\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d\)

Solution:

Given:

⇒ \(\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d\)

= \(a^2 c^2-a^2 d^2-b^2 c^2+b^2 d^2-4 a b c d\)

= \(a^2 c^2+b^2 d^2-2 a b c d-a^2 d^2-b^2 c^2-2 a b c d\)

= \(\left(a^2 c^2+b^2 d^2-2 a b c d\right)-\left(a^2 d^2+b^2 c^2+2 a b c d\right)\)

= \((a c-b d)^2-(a d+b c)^2\)

= (ac -bd + ad + bc) (ac – bd – ad – bc)

∴ (ac-bd + ad + bc) (ac-bd-ad-bc)

∴ \(\left(a^2-b^2\right)\left(c^2-d^2\right)-4 a b c d\) = (ac-bd + ad + bc) (ac-bd-ad-bc)

Step-by-Step Solutions for Class 7 Factorisation

Example 23. Resolve into factors \(3 x^2-y^2-z^2-2 x y-4 x z\)

Solution:

Given:

⇒ \(3 x^2-y^2-z^2-2 x y-4 x z\)

= \(4 x^2-x^2-y^2+z^2-2 x y-4 x z=4 x^2+z^2-4 x z-x^2-y^2-2 x y\)

= \((2 x-z)^2-(x+y)^2=(2 x-z+x+y)(2 x-z-x-y)\)

= (3x+y-z) (x-y-z)

∴ (3x + y-z) (x-y – z).

∴ \(3 x^2-y^2-z^2-2 x y-4 x z\)  = (3x + y-z) (x-y – z).

Example 24. Factorise : 6xy – 9y + 4x – 6

Solution:

Given:

6xy – 9y + 4x – 6

6xy – 9y + 4x – 6 = 3y (2x – 3) + 2 (2x – 3)

∴ (2x – 3) ( 3y + 2)

6xy – 9y + 4x – 6 = (2x – 3) ( 3y + 2)

Example 25. Factorise: \(3 x^4+2 x^2 y^2-y^4\)

Solution:

Given:

⇒ \(3 x^4+2 x^2 y^2-y^4\)

= \(3 x^4+3 x^2 y^2-x^2 y^2-y^4\)

= \(3 x^2\left(x^2+y^2\right)-y^2\left(x^2+y^2\right)\)

∴ \(\left(x^2+y^2\right)\left(3 x^2-y^2\right)\)

∴ \(3 x^4+2 x^2 y^2-y^4\) = \(\left(x^2+y^2\right)\left(3 x^2-y^2\right)\)

Algebra Chapter 5 formulae And Their Applications Exercise 5 Solved Example Problems

Formulae and Their Applications Introduction

The entire structure of algebra is based on some formulae. We can form some easy secting AD at G and BC at H where GD = 6. formulae by simple algebraic multiplication and we may also verify them geometrically.

Actually, every formula is an identity that is true for all values of the literals present in that identity.

The fundamental formulae are of great importance because all the harder formulae are derived from the fundamental formulae. We may define a formula as follows:

A general relation, expressed in terms of algebraic symbols is called an algebraic formula or in brief a formula.

Math Solution Of Class 7 Wbbse

Read and Learn More WBBSE Solutions For Class 7 Maths

Formula for the square of the sum of two terms

(a + b)² = (a + b) (a + b)

= a (a + b) + b (a + b)

= a² + ab + ab + b² = a² + 2 ab + b².

Thus we see that,

The square of the sum of two numbers is equal to the square of the first number plus twice the product of the two numbers plus the square of the second number.

Geometrical representation of the formula (a + b)²  = a² + 2ab + b².

In the diagram, ABCD is a square. The length of each side of ABCD = a + b.

Hence, the area of the square ABCD = (a + b)².

Now, draw a vertical line EF perpendicular to AB and intersecting AB at E and CD at F where BE = b.

Again draw a horizontal line GH perpendicular to AD and intersecting AD at G BC and H where GD = b.

Let EF and GH intersect at P.

WBBSE Class 7 Algebra Formulae Solutions

Now, area of AEPG = \(a^2\), Area of EBHP = ab, area of GPFD = ab, and area of PHCF = \(b^2\).

Thus, the sum of these four areas = \(a^2+a b+a b+b^2=a^2+2 a b+b^2\)

Since, the area of ABCD = area of (AEPG + EBHP + GPFD + PHCF)

Therefore, \((a+b)^2=a^2+2 a b+b^2\) (Proved).

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

Formula for the square of the sum of three terms \((a+b+c)^2=\{(a+b)+c\}^2\)

= \((a+b)^2+2(a+b) c+(c)^2\)

= \(a^2+2 a b+b^2+2 a c+2 b c+c^2\)

= \(a^2+b^2+c^2+2 a b+2 b c+2 c a\)

Formula for the square of the difference of two terms

\((a-b)^2=(a-b)(a-b)\)

= a (a – b) -b (a – b)

= \(a^2-a b-a b+b^2=a^2-2 a b+b^2\)

Thus we see that,

The square of the difference of two numbers is equal to the square of the first number minus twice the product of the two numbers plus the square of the second number.

Geometrical representation of the formula \((a-b)^2=a^2-2 a b+b^2\)

Let ABCD be a square of side a.

Draw a horizontal line EF perpendicular to AD and intersecting AD at E and BC at F, where AE = b.

Next draw GH perpendicular to EF, where GF = b, and draw HP perpendicular to BC produced, where CP = b.

Let GH and DC intersect at Q.

Now, area of ABCD = \(a^2\), area of EGQD = \(\left(a-b^2\right)\),

area of ABFE = ab, area of GFPH = ab, area of QCPH = \(b^2\),

Now, area of EGQD = area of (ABCD + QCPH – ABFE – GFPH)

or, \((a-b)^2=a^2+b^2-a b-a b\)

i.e., \((a-b)^2=a^2-2 a b+b^2\) (Proved).

Some relations based on the square of the sum of two terms and the square of the difference of two terms.

1. Since \((a+b)^2=a^2+2 a b+b^2\) therefore, \(a^2+b^2=(a+b)^2-2\)
2. Since \((a-b)^2=a^2-2 a b+b^2\) therefore, \(a^2+b^2=(a-b)^2+2 a b\)
3. \((a+b)^2=a^2+2 a b+b^2=a^2-l a b+b^2+4 a b-(a-b)^2+4 a b\) Hence, \((a+b)^2=(a-b)^2+4\)
4. \((a-b)^2=a^2-2 a b+b^2=a^2+2 a b+b^2-4 a b=(a+b)^2-4 a b\) Hence, \((a-b)^2=(a+b)^2-4 a b\)
5. \((a+b)^2+(a-b)^2=a^2+2 a b+b^2+a^2-2 a b+b^2=a^2+a^2+b^2+b^2\)
   \(=2 a^2+2 b^2=2\left(a^2+b^2\right)\)
   Hence, \((a+b)^2+(a-b)^2=2\left(a^2+b^2\right) .\)
6. \((a+b)^2-(a-b)^2=\left(a^2+2 a b+b^2\right)-\left(a^2-2 a b+6^2\right)\)
   \(=a^2+2 a b+b^2-a^2+2 a b-b^2=4 a b\)
   Hence, \((a+b)^2-(a-b)^2=4 a b .\)

The formula for the product of the sum and the difference of two terms.

(a +b) (a – b) = (a – b) + 6 (a – b) = a² – ab + ab – b² = a² – b².

Thus we see that, The product of the sum and the difference of two numbers is equal to the difference of the squares of those two numbers.

Geometrical representation of the formula (a + b) (a – b) = a² – b².

Let ABCD be square of side a. Draw a vertical line EF perpendicular to AB intersecting AB at E where AE = b.

Next draw a horizontal line GH perpendicular to AD and intersecting AD at G where AG = b.

Now GH intersects EF at F, BC at Q.

Here QH = b, HP is drawn perpendicular from H on DC produced.

According to the construction, the area of EBQF = area of QHPC

Now, Area of GHPD = Area of (GQCD + QHPC)

= Area of (GQCD + EBQF)

= Area of (ABCD – AEFG)

or, (a +b) (a – b) = a² – b² (Proved).

Solved Examples for Class 7 Algebra Chapter 5

To express the product of two terms as the difference of two perfect squares,

⇒ \(a^2+2 a b+b^2=(a+b)^2\)

⇒ \(a^2-2 a b+b^2=(a-b)^2\)

By subtraction, \(4 a b=(a+b)^2-(a-b)^2\)

⇒ \(a b=\frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}\)

⇒ \(a b=\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\)

Algebra Chapter 5 formulae And Their Applications Exercise 5 Some Examples On (a ± b)² = a² ± 2ab + b².

Example 1. Find the square of (x + 2y).

Solution:

Given:

(x + 2y)

Square of (x + 2y)

= \((x+2 y)^2\)

= \((x)^2+2 \cdot x \cdot 2 y+(2 y)^2=x^2+4 x y+4 y^2\)

∴ \(x^2+4 x y+4 y^2\).

The square of (x + 2y) =  \(x^2+4 x y+4 y^2\).

Example 2. Find the square of 3ab + 2bc.

Solution:

Given:

3ab + 2bc

Square of (3ab + 2bc) = \((3 a b+2 b c)^2\)

⇒ \((3 a b)^2+2 \cdot 3 a b \cdot 2 b c+(2 b c)^2=9 a^2 b^2+12 a b^2 c+4 b^2 c^2 9 a^2 b^2+12 a \cdot b^2 e+4 b^2 c^2\).

The square of 3ab + 2bc = \((3 a b)^2+2 \cdot 3 a b \cdot 2 b c+(2 b c)^2=9 a^2 b^2+12 a b^2 c+4 b^2 c^2 9 a^2 b^2+12 a \cdot b^2 e+4 b^2 c^2\).

Example 3. Find the square of 101.

Solution:

Given:

101

Square of 101

= \((101)^2=(100+1)^2\)

= \((100)^2+2 \times 100 \times 1+(1)^2=10000+200+1=10201\)

∴ 10201.

The square of 101 is 10201.

Class 7 Maths Formulae Exercise Solutions

Example 4. Find the square of \(2 x^2-3 y^2\).

Solution:

Given:

⇒ \(2 x^2-3 y^2\)

Square of \(\left(2 x^2-3 y^2\right)\)

= \(\left(2 x^2-3 y^2\right)^2\)

= \(\left(2 x^2\right)^2-2 x 2 x^2 x 3 y^2+\left(3 y^2\right)^2=4 x^4-12 x^2 y^2+9 y^4\)

∴ \(4 x^4-12 x^2 y^2+9 y^4\)

The square of \(2 x^2-3 y^2\) = \(4 x^4-12 x^2 y^2+9 y^4\)

Example 5. Find the square of 498.

Solution:

Given: 498

Square of 498

= \((498)^2=(500-2)^2=(500)^2-2 \times 500 \times 2+(2)^2\)

= 250000 – 2000 + 4 = 250004 – 2000 =248004

∴ 248004.

The square of 498 is 248004.

Class Vii Math Solution Wbbse

Example 6. Simplify : \((5 a+3 b)^2+2 \cdot(5 a+3 b)(5 a-3 b)+(5 a-3 b)^2\). 

Solution :

Given:

⇒ \((5 a+3 b)^2+2 \cdot(5 a+3 b)(5 a-3 b)+(5 a-3 b)^2\)

Let 5a + 3b = x and 5a – 3b = y

Hence, the given expression = \(x^2+2 x y+y^2=(x+y)^2\)

= \((5 a+3 b+5 a-3 b)^2=(10 a)^2=100 a^2\)

∴ \(100 a^2\).

⇒ \((5 a+3 b)^2+2 \cdot(5 a+3 b)(5 a-3 b)+(5 a-3 b)^2\) = \(100 a^2\).

Example 7. Simplify : \((5 x-7 y)^2+2(5 x-7 y)(9 y-4 x)+(9 y-4 x)^2\)(5x – 7y)² + 2(5x – 7y)(9y – 4x) + (9y – 4x)².

Solution:

Given:

⇒ \((5 x-7 y)^2+2(5 x-7 y)(9 y-4 x)+(9 y-4 x)^2\)

Let, 5x -7y- a and 9y – 4x = 6

Then the given expression = \(a^2+2 a b+b^2=(a+b)^2\)

= \((5 x-7 y+9 y-4 x)^2=(x+2 y)^2=(x)^2+2 \cdot x \cdot 2 y+(2 y)^2\)

= \(x^2+4 x y+4 y^2\)

∴ \(x^2+4 x y+4 y^2\).

∴ \((5 x-7 y)^2+2(5 x-7 y)(9 y-4 x)+(9 y-4 x)^2\)(5x – 7y)² + 2(5x – 7y)(9y – 4x) + (9y – 4x)² = \(x^2+4 x y+4 y^2\).

Example 8. Simplify \((3 a+4 b)^2-2(3 a+46)(a+4 b)+(a+4 b)^2\).

Solution:

Given:

\((3 a+4 b)^2-2(3 a+46)(a+4 b)+(a+4 b)^2\)

Let, 3a + 4b = x and a + 4b = y

Then the given expression = \(x^2-2 x y+y^2\)

= \((x-y)^2\)

= \(\{(3 a+4 b)-(a+4 b)\}^2\)

= \((3 a+4 b-a-4 b)^2=(2 a)^2=4 a^2\)

∴ \(4 a^2\)4a²

\((3 a+4 b)^2-2(3 a+46)(a+4 b)+(a+4 b)^2\) = \(4 a^2\)4a²

Example 9. Simplify \((7 x+4 y)^2-2 \cdot(7 x+4 y) x(7 x-4 y)+(7 x-A y)^2\).

Solution:

Given:

⇒ \((7 x+4 y)^2-2 \cdot(7 x+4 y) x(7 x-4 y)+(7 x-A y)^2\)

Let 7x + 4y = a and 7x – 4y = b

Hence, the given expression = \(a^2-2 a b+b^2=(a-b)^2\)

= \(\{(7 x+4 y)-(7 x-4 y))^2\)

= \(\{7 x+4 y-7 x+4 y)^2\)

= \((8 y)^2=64 y^2\)

∴ \(64 y^2\)

⇒ \((7 x+4 y)^2-2 \cdot(7 x+4 y) x(7 x-4 y)+(7 x-A y)^2\) = \(64 y^2\)

West Bengal Board Class 7 Algebra Assistance

Example 10.  If x =-l then find the value of \(81 x^2-18 x+1\).

Solution:

Given:

x =-1 And \(81 x^2-18 x+1\)

⇒ \(81 x^2-18 x+1=(9 x)^2-2 \times 9 x \times 1+(1)^2\)81x² – 18x + 1 = (9x)² – 2 X 9x X 1 + (1)²

= \((9 x-1)^2=\{9(-1)-1\}^2=(-9-1)^2=(-10)^2=100\)

∴ 100.

Example 11. If p = 2, q = -1 and r = 3 then find the value of \(p^2 q^2+10 p q r+25 r^2\).

Solution:

Given:

p = 2, q = -1 and r = 3 And \(p^2 q^2+10 p q r+25 r^2\)

\(p^2 q^2+10 p q r+25 r^2\)

= \((\mathrm{pq})^2+2 \times \mathrm{pq} \times 5 r+(5 \mathrm{r})^2\)

=\((\mathrm{pq}+5 \mathrm{r})^2=\{2(-1)+5 \times 3\}^2=(-2+15)^2=(13)^2=169\)

∴ 169.

Example 12. If a = 1, b = 2 and c = -1 then find the value of \(4 a^2 b^2-20 a b c+25 c^2\).

Solution:

Given: 

a = 1, b = 2 and c = -1 And \(4 a^2 b^2-20 a b c+25 c^2\)

⇒ \(4 a^2 b^2-20 a b c+25 c^2\)

= \((2 a b)^2-2 \times 2 a b \times 5 c+(5 c)^2\)

= \((2 a b-5 c)^2=\{2 \times 1 \times 2-5(-1)\}^2\)

= \(\{4+5\}^2=(9)^2=81\)

∴ 81.

∴ \(4 a^2 b^2-20 a b c+25 c^2\) = 81

Example 13. Find the product 201 x 201 with the help of the formula.

Solution:

⇒ \(201 \times 201=(201)^2=(200+1)^2\)

= \((200)^2+2 \times 200 \times 1+(1)^2\) = 40000 + 400 + 1 =40401

∴ 40401.

201 x 201 = 40401.

Example 14. Simplify 0.82 x 0.82 + 2 x 0.82 x 0.18 + 0.18 x 0.18.

Solution:

Let a = 0.82 and b =0.18

Then the given expression =a x a+2 x a x b+b x 6

= \(a^2+2 a b+b^2=(a+b)^2=(0.82+0.18)^2\)

= \((1.00)^2=(1)^2=1\)

∴ 1.

0.82 x 0.82 + 2 x 0.82 x 0.18 + 0.18 x 0.18 = 1.

WBBSE Class 7 Chapter 5 Formula Guide

Example 15. What is to be added with x² + 4x +\(\frac{3}{2}\) so that the sum is a perfect square?

Solution:

x² + 4x +\(\frac{3}{2}\)

= (x)² + 2.x.2 + (2)²  – (2)² +\(\frac{3}{2}\)

= (x + 2)² – 4 + \(\frac{3}{2}\)

= (x + 2)²– \(\frac{5}{2}\)

Hence, to make it a perfect square \(\frac{3}{2}\) is to be added.

Example 16. If x² – 2x + 1 = 0, then find the values of  \(x+\frac{1}{x} \quad x^{10}+\frac{1}{x^{10}} \quad x^n+\frac{1}{x^n}\)

Solution: \(x^2-2 x+1=0\)

or, \((x-1)^2=0\)

or, x – 1 = 0

∴ x = 1

⇒ \(x+\frac{1}{x}=1+\frac{1}{1}=1+1=2\)

⇒ \(x^{10}+\frac{1}{x^{10}}=(1)^{10}+\frac{1}{(1)^{10}}=1+1=2\)

⇒ \(x^n+\frac{1}{x^n}=(1)^n+\frac{1}{(1)^n}=1+1=2\)

∴ 1. 2, 2. 2, 3. 2.

⇒ \(x+\frac{1}{x} \quad x^{10}+\frac{1}{x^{10}} \quad x^n+\frac{1}{x^n}\) = 1. 2, 2. 2, 3. 2.

Chapter 5 formulae And Their Applications Exercise 5 Some Problems On Various Formulae Derived From (a ± b)² = a² ± 2ab + b².

Example 1. Find the square of x + 2y – 3z.

Solution:

Given:

x + 2y – 3z

Square of x + 2y – 3z.

= \((x+2 y-3 z)^2\)

= \((x+2 y)^2-2 \cdot(x+2 y) \cdot 3 z+(3 z)^2\)

= \((x)^2+2 \cdot x \cdot 2 y+(2 y)^2-6 z(x+2 y)+9 z^2\)

= \(x^2+4 x y+4 y^2-6 z x-12 y z+9 z^2\)

= \(x^2+4 y^2+9 z^2+4 x y-12 y z-6 z x\)

∴ \(x^2+4 y^2+9 z^2+4 x y-12 y z-6 z x\)

The square of x + 2y – 3z = \(x^2+4 y^2+9 z^2+4 x y-12 y z-6 z x\)

Applications of Algebraic Formulae for Class 7

Example 2. If a + 6 = 5 and ab = 5, find the value of a² + b².

Solution :

Given:

a + 6 = 5 and ab = 5

⇒ \(a^2+b^2=(a+b)^2-2 a b\)

= (5)²– 2.5 = 25- 10= 15

∴ 15.

∴ a² + b²= 15.

Example 3. If a – 6 = 5 and ab = 14, find the value of a² + b².

Solution:

Given:

a – 6 = 5 and ab = 14

⇒ \(a^2+b^2=(a-b)^2+2 a b=(5)^2+2.14=25+28=53\)

∴ 53.

a² + b² = 53.

Example 4. If x + y =2 and x-y = 1, then find the value of \(8 x y\left(x^2+y^2\right)\).

Solution:

Given:

x + y =2 and x-y = 1

⇒ \(8 x y\left(x^2+y^2\right)\)

= \(4 x y \cdot 2\left(x^2+y^2\right)\)

= \(\left\{(x+y)^2-(x-y)^2\right\}\left\{(x+y)^2+(x-y)^2\right\}\)

= \(\left\{(2)^2-(1)^2\right\}\left\{(2)^2+(1)^2\right\}\)

= (4-1) (4+1) = 3 x 5 = 15

∴15.

∴ \(8 x y\left(x^2+y^2\right)\) = 15.

Wbbse Class 7 Maths Solutions

Example 5. If 2x + 3y = 9 and xy = 3, find the value of \(4 x^2+9 y^2\).

Solution:

Given:

2x + 3y = 9 and xy = 3

⇒ \(4 x^2+9 y^2\).

= \((2 \cdot x)^2+(3 y)^2=(2 x+3 y)^2-2 \cdot 2 x \cdot 3 y\)

=\((2 x+3 y)^2-12 x y=(9)^2-12 \cdot 3=81-36=45\)

∴ 45.

∴ \(4 x^2+9 y^2\)= 45.

Example 6. Express 35 as the difference of two squares.

Solution:

35 = 7 x 5

⇒ \(\left(\frac{7+5}{2}\right)^2-\left(\frac{7-5}{2}\right)^2\)

⇒ \(\left(\frac{12}{2}\right)^2-\left(\frac{2}{2}\right)^2\)

= \((6)^2-(1)^2\)

∴ \(35=(6)^2-(1)^2\).

Example 7. Express 4xy as the difference of two perfect squares.

Solution:

4xy = 2.x.2.y

⇒ \(\left(\frac{2 x+2 y}{2}\right)^2-\left(\frac{2 x-2 y}{2}\right)^2\)

⇒ \(\left\{\frac{2(x+y)}{2}\right\}^2-\left\{\frac{2(x-y)}{2}\right\}^2\)

= \((x+y)^2-(x-y)^2\)

∴ \(4 x y=(x+y)^2-(x-y)^2\).

Example 8. Express \(2\left(a^2+b^2\right)\) as the sum of two perfect squares.

Solution:

⇒ \(2\left(a^2+b^2\right)\)

= \(2 a^2+2 b^2\)

= \(a^2+a^2+b^2+b^2\)

= \(a^2+2 a b+b^2+a^2-2 a b+b^2\)

= \((a+b)^2+(a-b)^2\)

∴ \(2\left(a^2+b^2\right)=(a+b)^2+(a-b)^2\).

Wbbse Class 7 Maths Solutions

Example 9. Express 4a² + 1/4a² +1 as the difference of two perfect squares.

Solution:

⇒ \(4 a^2+\frac{1}{4 a^2}+1\)

= \((2 a)^2+2 \cdot 2 a \cdot \frac{1}{2 a}+\left(\frac{1}{2 a}\right)^2+1-2\)

= \(\left(2 a+\frac{1}{2 a}\right)^2-(1)^2\)

Example 10. If x = a + \(\frac{1}{a}\) and y = a – \(\frac{1}{a}\) find the value  of \(x^4+y^4-2 x^2 y^2\).

Solution:

⇒ \(x^4+y^4-2 x^2 y^2\)

= \(\left(x^2\right)^2+\left(y^2\right)^2-2 \cdot x^2 \cdot y^2=\left(x^2-y^2\right)^2\)

= \(\{(x+y)(x-y)\}^2=(x+y)^2(x-y)^2\)

⇒ \(\left\{\left(a+\frac{1}{a}\right)+\left(a-\frac{1}{a}\right)\right\}^2\left\{\left(a+\frac{1}{a}\right)-\left(a-\frac{1}{a}\right)\right\}^2\)

⇒ \(\left\{a+\frac{1}{a}+a-\frac{1}{a}\right\}^2\left\{a+\frac{1}{a}-a+\frac{1}{a}\right\}^2\)

⇒ \((2 a)^2 \cdot\left(\frac{2}{a}\right)^2=4 a^2 \cdot \frac{4}{a^2}=16\)

∴ 16

Example 11. If x + \(\frac{1}{4}\) = 4, then show that, x² + 1/x² = 14

Solution:

L.H.S. = \(x^2+\frac{1}{x^2}\)

= \(\left(x+\frac{1}{x}\right)^2-2 \cdot x \cdot \frac{1}{x}\)

= (4)² – 2 = 16 – 2 = 14 = R.H.S. (Proved).

Class 7 Maths Exercise 5 Solved Examples

Example 12. If 6x² – 1 = 4x, then show that, 36x² +1/x² = 28.

Solution:

6x² -1 = 4x

⇒ \((6 x)^2-2.6 x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^2=(4)^2\)

⇒ \(36 x^2-12+\frac{1}{x^2}=16\)

⇒ \(36 x^2+\frac{1}{x^2}=16+12\)

∴ \(36 x^2+\frac{1}{x^2}=28\)  (Proved).

Example 13. If \(\frac{p}{6}\) + \(\frac{1}{p}\) = \(\frac{7}{6}\) then prove that \(p^2+36 / p^2=37\)

Solution:

⇒ \(\frac{p}{6}\) or, p + \(\frac{6}{p}\) = 7

[Multiplying both sides by 6]

Squaring both sides,

⇒ \((p)^2+2 \cdot p \cdot \frac{6}{p}+\left(\frac{6}{p}\right)^2=(7)^2\)

or, \(p^2+12+\frac{36}{p^2}=49\)

⇒ \(p^2+\frac{36}{p^2}=49-12\)

∴ \(p^2+\frac{36}{p^2}=37\) (Proved).

Example 14. If a + b = 9 and a – b = 5 then what is the value of a+b²/2ab?

Solution:

⇒\(\frac{a^2+b^2}{2 a b}=\frac{2\left(a^2+b^2\right)}{4 a b}\)

= \(\frac{(a+b)^2+(a-b)^2}{(a+b)^2-(a-b)^2}\)

= \(\frac{(9)^2+(5)^2}{(9)^2-(5)^2}=\frac{81+25}{81-25}=\frac{106}{56}=\frac{53}{28}=1 \frac{25}{28}\)

Example 15. If x² + y² = 4xy then prove that at⁴ + y⁴ = 14x²y².

Solution:

⇒ \(x^2+y^2=4 x y\)

or, \(\frac{x^2}{x y}+\frac{y^2}{x y}=4 \text { or, } \frac{x}{y}+\frac{y}{x}=4\)

Squaring both sides we get,

⇒ \(\frac{x^2}{y^2}+2 \cdot \frac{x}{y} \cdot \frac{y}{x}+\frac{y^2}{x^2}=16\)

or, \(\frac{x^2}{y^2}+2+\frac{y^2}{x^2}=16\)

or, \(\frac{x^2}{y^2}+\frac{y^2}{x^2}=14\)

or, \(\frac{x^4+y^4}{x^2 y^2}=14\)

or, \(x^4+y^4=14 x^2 y^2\)

Example 16. If 2a + \(\frac{1}{3a}\) = \(\frac{2}{3}\)

Solution:

2a+ \(\frac{1}{3a}\) = \(\frac{2}{3}\)

or, 3a + = \(\frac{1}{2a}\) = 1 [Multiplying both sides by \(\frac{3}{2}\)]

Squaring both sides,

⇒ \((3 a)^2+2 \cdot 3 a \cdot \frac{1}{2 a}+\left(\frac{1}{2 a}\right)^2=1\)

⇒ \(9 a^2+3+\frac{1}{4 a^2}=1 \text { or, } 9 a^2+\frac{1}{4 a^2}=1-3\)

∴ \(9 a^2+\frac{1}{4 a^2}=-2\)   (Proved).

Example 17. If a² + b² = 5ab then show that, a²/b² + b²/a²= 23.

Solution:

Given:

⇒ \(a^2+b^2=5 a b\)

or, \(\frac{a^2+b^2}{a b} \text { or, } \frac{a}{b}+\frac{b}{a}=5\)

Squaring both sides,

⇒ \(\left(\frac{a}{b}\right)^2+2 \cdot \frac{a}{b} \cdot \frac{b}{a}+\left(\frac{b}{a}\right)^2=(5)^2\)

or, \(\frac{a^2}{b^2}+2+\frac{b^2}{a^2}=25\)

or, \(\frac{a^2}{b^2}+\frac{b^2}{a^2}=25-2\)

or, \(\frac{a^2}{b^2}+\frac{b^2}{a^2}=23\) (Proved).

Example 18. Find the continued product: \((a+b)(a-b)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\).

Solution:

The given expression = \((a+b)(a-b)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\)

= \(\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\)

= \(\left(a^4-b^4\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\)

= \(\left(a^8-b^8\right)\left(a^8+b^8\right)\)

∴ \(a^{16}-b^{16}\)

Example 19. Multiply with the help of the formula : \(\left(a^2+a+1\right)\left(a^2-a+1\right)\left(a^4-a^2+1\right)\).

Solution:

Given:

⇒ \(\left(a^2+a+1\right)\left(a^2-a+1\right)\left(a^4-a^2+1\right)\)

⇒ \(\left(a^2+a+1\right)\left(a^2-a+1\right)\left(a^4-a^2+1\right)\)

= \(\left\{\left(a^2+1\right)+a\right\}\left\{\left(a^2+1\right)-a\right\}\left\{\left(a^4-a^2+1\right)\right\}\)

= \(\left\{\left(a^2+1\right)^2-(a)^2\right\}\left(a^4-a^2+1\right)\)

= \(\left(a^4+2 a^2+1-a^2\right)\left(a^4-a^2+1\right)=\left(a^4+a^2+1\right)\left(a^4-a^2+1\right)\)

= \(\left\{\left(a^4+1\right)+a^2\right\}\left\{\left(a^4+1\right)-a^2\right\}\)

= \(\left(a^4+1\right)^2\left(a^2\right)^2\)

= \(a^8+2 a^4+1-a^4=a^8+a^4+1\)

∴ \(a^8+a^4+1\).

Example 20. If x + y = 5 and x – y = 1, then find the value of \(8 x y\left(x^2+y^2\right)\)

Solution:

Given:

x + y = 5 and x – y = 1

⇒ \(8 x y\left(x^2+y^2\right)\)

= \(4 x y \cdot 2\left(x^2+y^2\right)\)4xy.2(x²+y²)

= \(\left\{(x+y)^2-(x-y)^2\right\}\left\{(x+y)^2+(x-y)^2\right\}\)

= \(\left\{(5)^2-(1)^2\right\}\left\{(5)^2+(1)^2\right\}\)

= (25 – 1) (25 + 1) = 24 x 26 = 624

∴ 624.

∴ \(8 x y\left(x^2+y^2\right)\) = 624

Example 21. If x= \(\frac{a}{b}\) – \(\frac{b}{a}\) and y= \(\frac{a}{b}\) – \(\frac{b}{a}\) then show that, \(x^4+y^4-2 x^2 y^2=16\).

Solution:

= \(\left(x^2-y^2\right)^2=\{(x+y)(x-y)\}^2\)

= \(\left\{\left(\frac{a}{b}+\frac{b}{a}+\frac{a}{b}-\frac{b}{a}\right)\left(\frac{a}{b}+\frac{b}{a}-\frac{a}{b}+\frac{b}{a}\right)\right\}^2\)

= \(\left\{\left(2 \cdot \frac{a}{b}\right)\left(2 \cdot \frac{b}{a}\right)\right\}^2=\left(2 \cdot \frac{a}{b} \cdot 2 \cdot \frac{b}{a} \cdot\right)^2\)

= \((4)^2\) = 16 = R.H.S.

Example 22. If m – \(\frac{1}{m-3}\) = 5, then what is the value of (m – 3)² + 1/(m-3)²?

Solution:

m – \(\frac{1}{m-3}\) = 5

⇒ \((m-3)^2+\left(\frac{1}{m-3}\right)^2-2 \cdot(m-3) \frac{1}{(m-3)}=4\)

or, \((m-3)^2+\frac{1}{(m-3)^2}-2=4\)

or, \((m-3)^2+\frac{1}{(m-3)^2}=6\)

∴ (m – 3)² + 1/(m-3)² = \((m-3)^2+\frac{1}{(m-3)^2}=6\)

Algebra Chapter 5 formulae And Their Applications Exercise 5 Some Examples On a² – b² = (a + b)(a – b).

Example 1. Find the product of (2a – 5b) and (2a + 5b).

Solution:

The required product = (2a – 5b) (2a + 5b)

= (2a)² – (5b)²

= 4a² – 25b² 

(2a – 5b) and (2a + 5b) = 4a² – 25b² 

Example 2.  Find the product of (1 + 5pq) and (1 – 5pq).

Solution:

The required product = (1 + 5pq) (1 – 5pq)

= (1)²-(5pq)² =1

∴ 1 – 25p²q².

(1 + 5pq) X (1 – 5pq) = 1 – 25p²q².

Example 3. Find the value of 255² – 254².

Solution:

255² – 254² = (255 + 254)(255 – 254)

= 509 x 1 = 509

∴ 509.

255² – 254² = 509.

Example 4. Find the continued product of \((a+b),(a-b),\left(a^2+b^2\right),\left(a^4+b^4\right)\).

Solution:

The required product = \((a+b)(a-b)\left(a^2+b^2\right)\left(a^4+b^4\right)\)

= \(\left(a-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\)= \(\left\{\left(a^2\right)^2-\left(b^2\right)^2\right\}\left(a^4+b^4\right)\)

= \(\left(a^4-b^4\right)\left(a^4+b^4\right)\)

= \(\left(a^4-b^4\right)\left(a^4+b^4\right)\)

∴ \(a^8-b^8\).

∴ \((a+b),(a-b),\left(a^2+b^2\right),\left(a^4+b^4\right)\) = \(a^8-b^8\).

Math Solution Of Class 7 Wbbse

Example 5. Find the product of (x -y + a) and (x +y-z).

Solution:

The required product = (x-y + z) (x+y-z)

= {x- (y – z)} {x + (y – z)}

= \((x)^2-(y-z)^2=x^2-\left(y^2-2 y z+z^2\right)\)

= \(x^2-y^2-z^2+2 y z\)

∴ \(x^2-y^2-z^2+2 y z\).

⇒ (x -y + a) a X (x +y-z) = \(x^2-y^2-z^2+2 y z\).

Example 6. Find the product of (x + 2y + 3z) and (x + 2y – 3z).

Solution:

The required product = (x + 2y + 3 z) (x + 2y – 3 z)

= {(x + 2y) + 3z} {(x + 2y) – 3z}

= \((x+2 y)^2-(3 z)^2=(x)^2+2 \cdot x \cdot 2 y+(2 y)^2-(3 z)^2\)

= \(x^2+4 x y+4 y^2-9 z^2\)

= \(x^2+4 y^2-9 z^2+4 x y\)

∴ \(x^2+4 y^2-9 z^2+4 x y\).

⇒ (x + 2y + 3z) X (x + 2y – 3z) = \(x^2+4 y^2-9 z^2+4 x y\).

Step-by-Step Solutions for Class 7 Algebra Problems

Example 7. Multiply with the help of the formula 1001 x 999.

Solution:

1001 x 999 = (1000 + 1) X (1000 – 1)

= (1000)² – (1)² = 1000000 – 1 = 999999

∴ 999999.

1001 x 999 = 999999.

Math Solution Of Class 7 Wbbse

Example 8. Simplify with the help of the formula : \((2 x+3 y+4 z)^2-(3 x-2 y+4 z)^2\).

Solution:

⇒ \((2 x+3 y+4 z)^2-(3 x-2 y+4 z)^2\)

=\({(2 x+3 y+4 z)+(3 x-2 y+4 z)} x{(2 x+3 y+4 z)-(3 x-2 y+4 z)}\)

= (2x + 3y + 4z + 3x – 2y + 4z) x (2x + 3y + 4z – 3x + 2y – 4z)

= (5x + y + 8z) (- x + 5y)

= \(-5 x^2-x y-8 z x+25 x y+5 y^2+40 y z\)

= \(-5 x^2+5 y^2+24 x y+40 y z-8 z x\)

∴ \(-5 x^2+5 y^2+24 x y+40 y z-8 z x\).

∴ \((2 x+3 y+4 z)^2-(3 x-2 y+4 z)^2\) = \(-5 x^2+5 y^2+24 x y+40 y z-8 z x\).

Example 9. Express as the product of two expressions  \(a^2-4 a b+4 b^2-4\)

Solution:

⇒ \(a^2-4 a b+4 b^2-4\)a² – 4ab + 4b² – 4

= \((a)^2-2 \cdot a \cdot 2 b+(2 b)^2-4\)

= \((a-2 b)^2-(2)^2=(a-2 b+2)(a-2 b-2)\).

∴ \(a^2-4 a b+4 b^2-4\) = \((a-2 b)^2-(2)^2=(a-2 b+2)(a-2 b-2)\).

Algebra Chapter 3 Laws Exercise 3 Solved Example Problems

Algebra Introduction

In this chapter, we shall discuss some phenomena known to you in a different manner.

Actually, out of the four basic operations namely addition, subtraction, multiplication, and division only addition and multiplication (and in some special cases division also) abide by some laws.

WBBSE Class 7 Algebra Laws Solutions

Read and Learn More WBBSE Solutions For Class 7 Maths

These laws are:

1. Commutative law
2. Associative law and
3. Distributive law.

These laws play an important role in higher mathematics, especially in classical and modern algebra and also in the theory of sets. Let us have a look at these laws, starting with very simple examples.

Wbbse Class 7 Maths Solutions

Commutative law 

You know that, 7 + 8 = 15

Again, 8 + 7 = 15

Therefore, 7 + 8 = 8 + 7.

So, it may be said that, if a and 6 be any two integers then, a+b = b+a.

In the above equation, we see that on the left-hand side a is in the first position and b is after a while on the right-hand side b is in the first position and a is after b.

This law, where a and b interchange their positions is known as commutative law.

Also, you know that, 7 x 4 = 28 and also, 4 x 7 = 28

Therefore, 7 x 4 = 4 x 7.

So, it may be said that, if a and b are two integers then, axb= bxa.

Thus, the commutative law is also applicable in the case of multiplication.

Illustration: 1. Suppose, you have 6 mangoes and your brother has 4 mangoes. Thus the total number of mangoes can be obtained by adding your mangoes with your brother’s share i.e., 4 + 6 = 10.

Again we can find the total number of mangoes by adding your brother’s mangoes with your mangoes i.e., 6 + 4 = 10. Therefore, 4 + 6 = 6 + 4.

Hence, we can say that the addition of 6 with 4 and the addition of 4 with 6 give the same result. Thus if 4 and 6 change their places, their sum remains unaltered.

2. Suppose, in a garden, there are 4 rows of trees and in each row, there are 3 trees.

Then the total number of trees in the garden = 3 x 4 = 12.

Also in another garden, there are 3 rows of trees and in each row, there are 4 trees.

Then the total number of trees in the garden is 4 x 3 = 12. Thus we see that, 3×4 = 4×3.

Associative law

The sum 4 + 5 + 7 may be written in different ways.

If, at first 5 is added with 4 and then 7 is added with this result of addition then it may be written as (4 + 5) + 7 = 9 + 7=16; again, if at first 7 is added with 5 and then the result of this addition is added with 4 then it may be written as 4 + (5 + 7) = 4 + 12 = 16.

So, it is clear that (4 + 5) + 7 = 4 + (5 + 7).

Therefore, in general, if a, 6 and c be three integers then, (a + 6) + c = a + (6 + c).

This is known as the associative law of addition.

The associative law is also applicable in the case of multiplication.

For example, we see that (2 x 3) x 4 = 6 x 4 = 24 and 2 x (3 x 4) = 2×12 = 24.

Therefore, in general, if a, b and c be three integers then it may be written in general that, (a x b) x c = a x (b x c).

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

Illustration: 1. Suppose, you have 4 pens. Your brother has 3 pens and your sister has 2 pens.

At first, adding your and your brother’s pens the total number of pens becomes 4 + 3 = 7.

Now adding this number the number of your sister’s pens the total number of pens becomes (4 + 3) + 2 = 7 + 2 = 9.

Again, adding your brother’s and sister’s pens, the total number of pens = 3+2 = 5.

Now adding with your pens, the sum of your brother’s and sister’s pens the total number of pens becomes 4+(3+2) = 4+5 – 9.

Thus counting the total number of pens in both ways we get, (4 + 3) + 2 = 4 + (3 + 2).

2. Suppose, a boy can do 5 sums per day and we are to find the number of sums that can be done by the boy in 4 weeks.

Since in 1 day the boy can do 5 sums therefore in 1 week (i.e., 7 days) the boy can do 5 x 7 = 35 sums.

Thus in 4 weeks the boy can do (5 x 7) x 4 = 35 x 4 = 140 sums.

Also we may think the issue from another point of view. The total number of days in 4 weeks = 7 x 4 = 28.

Since in 1 day the boy can do 5 sums, in 7 x 4 days the boy can do 5 x (7 x 4) = 5 x 28 = 140 sums.

Thus the total number of sums done by the boy is same in both cases.

Math Solution Of Class 7 Wbbse

Distributive law

You know that, 15 x 6 = 90

Again, since 6 = 4 + 2

∴ 15 x 6 = 15 x (4 + 2) = 15 x 4 + 15 x 2 = 60 + 30 = 90.

So, it may be said in general that, if a, b, and c are three integers then,

a x (b+c) = a x b + a x c and a x (b-c) = a x b -a x c.

So, in the case of multiplication, the distributive law is always applicable.

The distributive law of multiplication may also have the following forms:

(a+b) x c=a x c+b x c and (a-b) x c=a x c- b x c.

Let us now see that, how the distributive law may be applied in the case of division.

You know that, 35 ÷ 5 = 7 and 35 = 25 + 10

Now, (25+10) ÷ 5 = 25÷5 + 10 ÷ 5 = 5 + 2 = 7.

So, it may be said in general that, if a, b, and c are three integers then,

(a+b) ÷ c= a ÷ c + b ÷ c and (a-b) ÷ c= a ÷ c- b ÷c.

This distributive law of division is only applicable to the right-hand side.

If a, b, and c be three integers then it may be shown that,

c ÷ (a + b) ≠ c ÷ a + c ÷ b and c ÷ (a-b) ≠ c ÷a – c ÷ b.

Therefore, the distributive law of division is not applicable in this case.

Illustration: Suppose you can do 12 sums daily and your brother can do 8 sums daily.

In 10 days what will be the total number of sums done by you? We may solve this problem in two ways.

In one day the total sum done by you and your brother is (12 + 8) = 20.

Hence, the number of sums done in 10 days = {(12 + 8) x 10} = 20 x 10 = 200.

Again in 10 days, you can do 12 x 10 – 120 sums and in 10 days your brother can do 8 x 10 = 80 sums.

Hence, the total number of sums done by you and your brother in 10 days = 12 x 10 + 8 x 10 = 120 + 80 = 200.

Thus, we can say that, (12 + 8) x 10 = 12 x 10 + 8 x 10.

Use of brackets: In algebra, brackets are used in the same manner as in arithmetic. In general, four types of brackets are used.

They are:

1. Line bracket —
2. First bracket ( )
3. Second bracket { }
4. Third bracket [ ]

If in a question all the brackets are present then first of all, the operation under the line bracket is performed. After that, the operation under first bracket, second bracket, and third bracket are done in series after finding values inside the corresponding brackets.

Algebra Chapter 3 Laws Exercise 3 Some Problems With Commutative, Associative, And Distributive Laws And The Use Of Brackets

Example 1. Prove on the number axis that 4 + 3 = 3 + 4.

Solution:

Moving 4 units to the right of 0

we get to point 4. After this, moving 3 more units in the same direction we get point 7.

Thus, 4 + 3 = 7.

Again, at first moving 3 units to the right of 0 we get point 3. After this, moving 4 more units in the same direction we get to point 7. Thus, 3 + 4 = 7.

Hence, 4 + 3 = 3 + 4 (proved).

Example 2. Prove that : 4 x 5 = 5 x 4.

Solution:

4 x 5 = 5 + 5 + 5 + 5 = 20

5 x 4=4+4+4+4+4=20

Hence, 4 x 5 = 5 x 4 (proved).

Solved Problems for Class 7 Algebra Chapter 3

Example 3. Rewrite 2 + 3 + 5 in two different ways such that a single bracket is used in each case.

Solution: 2 + 3 + 5 = (2 + 3) + 5 and 2 + 3 + 5 = 2 + (3 + 5).

Example 4. Rewrite 5 x 6 x 7 in two different ways such that a single bracket is used in each case.

Solution:  5 x 6 x 7 = (5 x 6) x7 and 5 x 6 x 7 = 5 x (6 x 7).

Example 5.  In each of the following cases, determine which law of which operation has been used.

1. a + b = b + a.
2. a x b – b x a,
3. (a + y) + z – x + (y + z).
4. m x (n x p) = (m x n) x p.
5. a x (6 + c) = a x b + a x c.
6. (a + b) ÷ c = a ÷ c + b ÷ c.

Solution:

1. Commutative law of addition.
2. Commutative law of multiplication.
3. Associative law of addition.
4. Associative law of multiplication.
5. Distributive law of multiplication,
6. Distributive law of division.

Example 6. Add the sum of 3 and 7 with 5 and find the result.

Solution:

Given:

5 + (3 + 7) = 5 + 10 = 15

∴ 15.

The result is 15.

Example 7. From 50, subtract the sum of 25 and 15 and find the result.

Solution:

Given:

From 50, subtract the sum of 25 and 15

50 – (25 + 15) = 50 – 40 = 10

∴ 10.

The result is 10.

Class 7 Maths Laws Exercise Solutions

Example 8. Using the result (a + b). c = a.c + b.c prove that,

1. (a – b).c=a.c -b.c.
2. a. (b-c) = a.b- a.c.

Solution:

Given:

Using the result (a + b). c = a.c + b.c

1. (a – b).c + b.c = x.c + b.c [Assuming x = a – b]= (x + b).c [Using the given relation]

= {(a – b) + b}.c [Putting the value of x] = [a + {(-b + b)}].c [By associative property]

= [a + o].c [According to the definition of zero]

= a.c + o.c [Using the given relation]

= a.c + o = a.c [According to the definition of zero]

Now, a.c – b.c = {(a – b).c + b.c} – b.c [Putting the value of a.c]

= (a – b).c + {b.c – b.c} [By associative property]

= (a – b).c + o = (a – b).c

Hence, it is proved that (a – b).c = a.c – b.c.

2. a.(b – c)= (b – c).a [By commutative property]

= b.a – c.a [From the proof of 1st part]= a.b – a.c [By commutative property]

Hence, it is proved that a.(b – c) = a.b – a.c.

Example 9. Prove that,

1. (a + b + c).x = a.x+ x + c.x.
2. (a + b + c) = x.a+ x.b + x.c.

Step-by-Step WBBSE Class 7 Algebra

Example 10. Prove that, (a + b) ÷ x = (a ÷ x) + (b ÷ x).

Example 11. Simplify xyz (x + y + z).

Solution:

The given expression = xyz (x + y + z)

= xyz X x + xyz X y + xyz X z

∴ x²yz + xy²z + xyz²

∴ xyz (x + y + z) = x²yz + xy²z + xyz²

Example 12. Simplify : a(a² + 5a – 6).

Solution:

The given expression

= a(a² + 5a – 6)

= a x a² + a x 5a – a x 6 = a³ + 5a² – 6a

∴ a³ + 5a² – 6a.

a(a² + 5a – 6) = a³ + 5a² – 6a.

Class 7 Maths Algebra Exercise Solutions

Example 13.  Simplify: \(\frac{x}{x-y}\)+\(\frac{y}{y-x}\)

Solution:

Example 14. Simplify 
Solution:

Example 15. Simplify
Solution :

Example 16. Simplify: 

Solution:

Arithmetic Chapter  Square Measure Exercise 6 Solved Example Problems

Plane

If we keep the face of a solid on a flat white paper with which it coincides and indicate its boundary with a pencil, we obtain a plane.

Keeping a rectangular parallelopiped on a white paper, and moving a pencil through its edge we obtain a rectangle. If a cube is taken and the same process is applied we obtain a square.

In other words, the portion of a plane enclosed by three or more straight lines or by one or more curves is called a plane.

Read and Learn More WBBSE Solutions For Class 7 Maths

Area

The measure of the portion of a plane enclosed by a plane figure is called its area. The plane area enclosed by the straight lines is called a polygon.

The measure of the portion enclosed by the sides of a polygon is called its area.

Perimeter

The sum of the lengths of the sides of a polygon is called its perimeter.

Rectangle

A quadrilateral whose opposite sides are equal and the angles are right angles is called a rectangle.

 

 

Square

The quadrilateral whose four sides are equal and the angles are right angles is called a square.

 

 

Some useful measurements

1. Area of rectangle Length x Breadth.
2. Perimeter of rectangle = 2 x (Length + Breadth).
3. Area of square = (Side)².
4. Perimeter of square = 4 x Side.

Areas in connection with a room: A rectangular or square-shaped room has the following parts

1. One floor and one roof
2. Two walls along the length of the room (equal in size & shape and placed opposite to each other)
3. Two walls along the breadth of the room (equal in size & shape and placed opposite to each other)
4. There may be one or more doors and windows in the four walls.

The following images gives a clear idea about the above parts

 

 

Useful measurements in connection with a room:

1. Area of the roof or floor of the room = Length x breadth
2. Area of two walls along the length of the room = 2 x length x height
3. Area of two walls along the breadth of the room = 2 x breadth x height
4. Area of the four walls of the room =2 (length + breadth)x height = Perimeter of room (or floor) x height
5. Number of floor tiles = \(\frac{Area of floor}{Area of one tile}\)
6. One has to consider the total area of the four walls and that of the roof to either paint or whitewash or plaster the room.
7. If the room contains windows and/or door(s), then their total area has to be subtracted from the total area of roof and four walls to get the effective total area.

Height and Area of a triangle

The perpendicular drawn from any vertex on its opposite side is called the height of the triangle. In the image a perpendicular AD is drawn from the vertex A on its opposite side BC in the triangle ABC.

 

 

Hence AD is the height of the triangle. Here BC is the base of the triangle. If AC is taken as the base of the triangle, then the perpendicular drawn on AC from the opposite vertex B shall be taken as the height of the triangle.

Similarly, if AB is taken as the base of the triangle, then the perpendicular drawn on AB from the opposite vertex C shall have to be taken as the height of the triangle ABC.

The following images show the heights of an equilateral triangle, an isosceles triangle and a right-angled triangle.

 

 

It is not mandatory that the height of a triangle should lie within its area enclosed by the three sides.

 

 

The adjacent figure shows that if one of the adjacent sides of the obtuse angle of an obtuse-angled triangle is taken as the base of the triangle, then the height of the triangle shall lie outside the area of the triangle bounded by its three sides.

The images shows that AD lies outside the bounded area of the triangle ABC when BC is taken as the base.

Some useful measurements in connection with a triangle:

1. Perimeter of a triangle = Sum of lengths of three sides = AB + BC + CA
2. Half or semiperimeter of a triangle = \(\frac{AB + BC + CA}{2}\)
3. Area of triangle = \(\frac{1}{2}\) X base x height = \(\frac{1}{2}\) X BC X AD
4. Area of a right-angled triangle = \(\frac{1}{2}\) X BC X AB

Unit of area

While writing the unit of area we have to write the word, ‘square’.

For example square metre, square centimetre etc.

Some points to note

1. To indicate the area of rectangle having length 10 cm and breadth 6 cm we write 10 cm x 6 cm, and read it as 10 cm by 6 cm.

2. Metre x Metre = Square metre.

10 square metres + 5 metres = 2 metres.
10 sq metres + 8 sq metres = 18 sq metres.
10 sq metres-8 sq metres = 2 sq metres.
25 sq metres + 5 sq metres = 5.

3. 12 inches = 1 foot, 144 sq inches = 1 sq foot.

3 feet = 1 yard, 9 sq feet = 1 sq yard.

4840 sq yards = 1 acre.

640 acres = 1 sq mile.

4. 10 mm = 1 cm, 100 sq mm = 1 sq cm

10 cm = 1 dm, 100 sq cm = 1 sq dm.

10 dm = 1 m, 100 sq dm 1 sq m.

10 m = 1 Dm, 100 sq m = 1 sq Dm.

10 Dm 1 Hm,  100 sq Dm = 1 sq Hm.

10 Hm = 1 Km, 100 sq Hm = 1 sq km.

1 are = 1 sq Dm = 100 sq m.

100 centiares = 1 are.

100 are 1 hectare.

100 hectares = 1 sq Kilometres.

5. To calculate area, both length and breadth must be reduced to same unit. The area is expressed in terms of sq unit.

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

Example 1.  The length of a rectangle is 15 m 25 cm and its breadth is 10 m 2 cm. Find its area.

Solution:

Given:

The length of a rectangle is 15 m 25 cm and its breadth is 10 m 2 cm.

Here, length = 15 m 25 cm = 1525 cm breadth = 10 m 2 cm = 1002 cm

∴ Area of the rectangle = length x breadth = 1525 x 1002 sq cm

= 1528050 sq cm

= 152 sq m 80 sq dm 50 sq cm

∴ 152 sq m 80 sq dm 50 sq cm.

Area = 152 sq m 80 sq dm 50 sq cm.

Example 2.  The page of an exercise book is 15 cm long and 12 cm broad. A margin of width 2 cm is drawn on all sides and something was written on the remaining portion. What is the area of the written portion? What is the area of the portion where nothing was written?

Solution:

Given:

The page of an exercise book is 15 cm long and 12 cm broad. A margin of width 2 cm is drawn on all sides and something was written on the remaining portion.

Area of the page = 15 cm x 12 cm = 180 sq cm.

The portion on which something was written, was of length (15-2x 2) cm 11 cm and breadth (12-2 × 2) cm = 8 cm

∴ Area = 11 cm x 8 cm = 88 sq cm.

 

 

The portion on which nothing was written was of area (180 -88) sq cm = 92 sq cm.

∴ 88 sq cm, 92 sq cm.

= (4 × 100 + 50) sq cm = 450 sq cm. its breadth = 10 cm

The area of the portion where nothing was written = (4 × 100 + 50) sq cm = 450 sq cm. its breadth = 10 cm

WBBSE Class 7 Arithmetic Square Measure Examples

Example 3. The area of a rectangle is 4 sq dm 50 sq cm Find the area of the path. and its breadth 10 cm. Find its length.

Solution:

Given:

The area of a rectangle is 4 sq dm 50 sq cm

Area of the rectangle = 4 sq dm 50 sq cm

= (40 x 100 + 50) sq cm

= 450 sq cm.

its breadth = 10 cm

∴ Its length = \(\frac{Area}{breadth}\) = \(\frac{450 sq cm}{10 cm}\)

∴ 45 cm.

Length =  45 cm.

Example 4. The length and breadth of a rectangular plot are 36 m and 24 m. There is a path 2 m wide all around outside of the plot.

1. Find the length and breadth of the plot of land including the path.
2. Find the area of the plot of land excluding the path.
3. What is the area of the path?

Solution:

1. Length of the rectangular plot including the path (36+2×2) m = 40 m

Breadth of the rectangular plot including path = (24+2×2)m = 28 m

∴ 40 m, 28 cm.

2. Excluding the path, area of the rectangular plot = 36 x 24 sq m = 864 sq m

∴ 864 sq m.

3. Area of the path = (40 x 28-36 x 24) sq m = (1120-864) sq m = 256 sq m.

Example 5. How many tiles, each 5 dm by 2 dm will be required to cover the floor of a room 15 m by 12 m?

Solution:

The area of the floor = 15 x 12 sq m = 180 sq m = 18000 sq dm.

The area of each tile = 5 dm x 2 dm

∴ The number of tiles required = \(\frac{18000}{10}\) = 1800

∴ 1800

The number of tiles required = 1800.

Example 6. There is a square plot of land of side 20 m. A path 1 m wide runs all around outside of it.

Solution:

Given:

There is a square plot of land of side 20 m.

Including the path, length of the square plot of land (20+2 x 1) m = 22 m.

Including the path, area of the square plot of land = 22 x 22 sq m = 484 sq m.

Excluding the path, area of the square plot of land = 20 x 20 sq m = 400 sq m.

∴ Area of the path = (484-400) sq m = 84 sqm.

∴ 84 sq m.

Area of the path 84 sq m.

Solved Problems for Class 7 Square Measure

Example 7. Find the perimeter of a rectangle whose length is 15 m 5 dm and breadth is 8 m 7 dm.

Solution:

The perimeter of the rectangle = 2 x (length + breadth)

= 2x (155+ 87) dm

= 2x (242) m = 484 dm = 48 m 4 dm

∴ 48 m 4 dm.

The perimeter of a rectangle 48 m 4 dm.

Example 8. The area of a square plot of land is 6400 sq m. Find the cost of fencing all around the land at 3.50 per metre.

Solution:

Given:

The area of a square plot of land is 6400 sq m.

Area of the square plot of land = 6400 sq m

length of side of the square plot of land = √6400 m = 80 m

The perimeter of the square plot of land = 4 x 80m = 320 m

Cost of fencing 1 m = ₹ 3.50

Cost of fencing 320 m = ₹ 3.50 x 320 = ₹ 1120

Example 9. One plot is 4 sq metres and another plot is 4 metres square. Which one is of the greater area? 

Solution:

Given:

One plot is 4 sq metres and another plot is 4 metres square.

By 4 sq metres is meant a plot whose area rectangle. is 4 sq metres.

[It may be a square, the length of whose side is 2 m and hence its area = 2 m x 2 m = 4 sq m

or, it may be a rectangle whose length is 4 m and breadth is 1 m and hence its area = 4

or, it may be triangular or circular or it may have the shape of any plane whose area is equal to 4 sq m.

On the other hand, by the sentence ‘a plot is 4 metres square’ we mean only a square the length of whose side is 4 metres i.e., its area (4)2 sq m = 16 sq m.

∴ The plot ‘4 metres square’ is of the greater area.

Example 10. The length of a rectangular plot of land is twice its breadth and its area is 578 sq m. Find the length, breadth and perimeter of the land.

Solution:

Given:

The length of a rectangular plot of land is twice its breadth and its area is 578 sq m.

Let, the breadth of the rectangular plot of land = x m then its length = 2x m.

∴ 2x x x = 578

or, 2x² = 578 or, x² = 289

or, x= √28917

∴ Breadth 17 m, Length = 2 x 17 m = 34 m Perimeter = 2 × (34 +17) m = 102 m

∴ Length is 34 m, breadth is 17 m and perimeter is 102 m.

Example 11. How much fencing will be required to fence a square field of area 625 sq metres?

Solution:

The area of the square field = 625 sq metres

∴ The length of its side = √625 m = 25 m .

∴ Its perimeter = 4 x 25 m = 100 metres.

Since the total length of fencing is equal to the 4 x 80 m = 320 m perimeter, therefore the required length of fencing 100 metres.

Cost of fencing 1 m = 3.50 Cost of fencing 320 m = 3.50 x 320=1120

Class 7 Maths Exercise 6 Solutions on Square Measure

Example 12. The length of a rectangle is 1 \(\frac{1}{2}\) times its breadth. If the area of a square be 100 sq m and its perimeter be equal to the perimeter of the rectangle then find the length of the rectangle.

Solution:

Area of the square = 100 sq m

Length of side of the square = √100 m = 10m

∴ Perimeter of the square= 4 x 10 m = 40 m

Let, the breadth of the rectangle = xm, 3х then length of the rectangle = \(\frac{3x}{2}\) m

∴ 2 X (x + \(\frac{3x}{2}\)) = 40

or, \(\frac{5x}{2}\)=20

or, x =\(\frac{2 X 20}{5}\) = 8

∴ Length of the rectangle = \(\frac{3 X 8}{2}\)

∴ 12 m.

Length of the rectangle =12 m.

Example 13. The length of a room is 8 m, its breadth is 6 m and its height is 5 m. What will be the cost of polishing the floor of the room at 70 per sq m?

Solution:

Given:

The length of a room is 8 m, its breadth is 6 m and its height is 5 m.

Area of the floor of the room = 8 x 6 sq m = 48 sq m

Cost in 1 sq m = ₹ 70

Cost in 48 sq m = ₹ 70 x 48 = ₹ 3360

∴ ₹ 3360

Cost in 48 sq m = ₹ 3360

Example 14. A rectangular plot of length 200 m and breadth 120 m has been divided in 4 equal parts by two paths 10 m wide parallel to length and breadth. What is the area of that path?

Solution:

Given:

A rectangular plot of length 200 m and breadth 120 m has been divided in 4 equal parts by two paths 10 m wide parallel to length and breadth.

Area of the path parallel to length = 200 x 10 sq m = 2000 sq m

Area of the path parallel to breadth = 120 x 10 sq m = 1200 sq m

Area of the portion where the two paths intersect 10 x 10 sq m = 100 sq m

∴ The required area of the path =(2000+ 1200-100) sq m = 3100 sq m

∴ 3100 sq m

The required area of the path = 3100 sq m

Example 15. If both length and breadth of a rectangle are doubled then how many times will be its area?

Solution:

Let, length of the rectangle = x m and its breadth = y m

∴ Area of the rectangle = x y sq m

Changed length = 2x m and breadth = 2ym

∴ Changed area = 2x × 2y sq m = 4 xy sq m

∴ Area of the rectangle will be 4 times

∴ 4 times.

Example 16. The length of a rectangular stage of a theatre is twice its breadth. It costs ₹ 6048 to cover the whole stage with tarpaulin. If one sq m of tarpaulin costs ₹21, find the length and breadth of the stage.

Solution:

Given:

The length of a rectangular stage of a theatre is twice its breadth. It costs ₹ 6048 to cover the whole stage with tarpaulin. If one sq m of tarpaulin costs ₹21

Since the total expenditure is ₹ 6048 to cover the stage at the rate ₹21 per square metre, therefore, the area of the stage = (6048 ÷ 21) sq m = 288 sqm

Let, the breadth of the stage =  x m then its length  = 2x m.

∴ 2x X x = 288  or, 2x² = 288

or, x² = \(\frac{288}{2}\) = 144

or, x=√144 = 12

∴ Breadth = 120m and length = 2 x 12 m = 24 m

∴ Length is 24 m and breadth is 12 m

Example 17. A garden is 30 m long and 20 m broad. All round outside of it, there is a path 3 m broad. What is the area of the path?

Solution:

Given:

A garden is 30 m long and 20 m broad. All round outside of it, there is a path 3 m broad

Length of the garden including the path = (30+3×2) m = (30 + 6) m = 36 m

Breadth of the garden including the path = (20 + 3 x 2) m = (20 + 6) m = 26 m

 

 

Area of the garden including the path = 36 m x 26 m = 936 sq m

Area of the garden excluding the path = 30 m x 20 m = 600 sq m.

Area of the path = (936- 600) sq m = 336 sq m

∴ 336 sq m

Area of the path = 336 sq m

Example 18. The length and breadth of a rectangular garden are 60 m and 40 m respectively. Two 5 m wide path is divided Find the cost of paving the path at ₹80 per rectangular garden into four equal parts. sq. metre. Find also the area of each part of the land.

Solution:

Given:

The length and breadth of a rectangular garden are 60 m and 40 m respectively. Two 5 m wide path is divided Find the cost of paving the path at ₹80 per rectangular garden into four equal parts. sq. metre.

Area of the path parallel to the length = 60 x 5 sq m = 300 sq m.

Area of the path parallel to the breadth = 40 x 5 sq m = 200 sq m.

Area of the portion where two paths intersect each other = 5 x 5 sq m = 25 sq m.

Area of the path = (300 + 200- 25) sq m = 475 sq m

cost of paving 1 sq m path = ₹ 80

cost of paving 475 sq m path = ₹ 80 x 475 = ₹ 38000

 

 

length of one part of the garden =\(\frac{60 -5}{2}\) m = \(\frac{55}{2}\)

breadth of one part of the garden = \(\frac{40 -5}{2}\) m = \(\frac{35}{2}\)

area of one part of the garden = \(\frac{55}{2}\) x \(\frac{35}{2}\) sq m

= \(\frac{1925}{4}\) sq m = 481.25 sq m

∴ ₹ 38000, 481 .25 sq m.

West Bengal Board Class 7 Square Measure Assistance

Example 19. A garden is 25 metres square. All around inside of it, there is a path 2\(\frac{1}{2}\) m broad. What is the area of the path?

Solution:

Given:

A garden is 25 metres square. All around inside of it, there is a path 2\(\frac{1}{2}\) m broad.

Area of the garden = 25 x 25 sq m = 625 sq m

Length of the side of the garden excluding the path

= (25 – \(\frac{5}{2}\) x 2) m = 25 -5 m = 20 m

∴ Area of the garden excluding the path = 20 m x 20 m = 400 sq m.

∴ Area of the path = (625-400) sq m = 225 sq m.

∴ 225 sq m.

Area of the path = 225 sq m.

Example 20. The path that leads to a man’s house is 2 m wide. The path divides the garden in front of the house in two equal squares. The path is constructed for 8000 at the rate of 500 per sq m. Calculate the area of each square portion of the garden. The house is built in rectangular land. If the length of the rectangular land is 4 m, calculate the area occupied by the house.

Solution:

Given:

The path that leads to a man’s house is 2 m wide. The path divides the garden in front of the house in two equal squares. The path is constructed for 8000 at the rate of 500 per sq m. Calculate the area of each square portion of the garden.

Since the total expenditure of constructing the path is ₹ 8000 at the rate of ₹ 500 per sq m therefore, area of the path

\(\frac{8000}{500}\) sq m = 16 sq m

Since the breadth of the path = 2 m

therefore its length = \(\frac{16}{2}\) m = 8 m

∴ Area of each square portion of the garden = 8 x 8 sq m = 64 sq m

Length of the portion occupied by the house = (8 + 2 + 8)m = 18m and its breadth = 4 m.

Therefore, its area = 18 x 4 sq m = 72 sq m

∴ 64 sq m, 72 sq m

Area= 64 sq m, 72 sq m

Example 21. The length, breadth and height of a room are 4 m, 3 m and 5 m respectively. What will be the cost of cementing the floor of that room at? ₹15 per square metre? What will be the cost of white washing the four walls at ? ₹5 per square metre?

Solution:

Given:

The length, breadth and height of a room are 4 m, 3 m and 5 m respectively.

Area of the floor of the room = 4 x 3 sq m = 12 sq m.

cost of cementing 1 sq m =  ₹ 15

cost of cementing 12 sq m =  ₹ 15 x 12 = ₹ 180

Again, the area of the four walls of the room = 2 x (4+3) x 5 sq m

= 2x7x5 sq m = 70 sq m

cost of whitewashing 1 sq m = ₹ 5

cost of whitewashing 70 sq m = ₹ 5 x 70 = ₹ 350

∴ Cost of cementing the floor =₹ 180.

Cost of whitewashing the four walls = ₹ 350.

WBBSE Class 7 Chapter 6 Square Measure Guide

Example 22.  The cost of cultivating a 30 m long piece of land is ₹ 150. Had the breadth of the land been 5 m less, the cost would have been ₹ 120. Calculate the breadth of the land.

Solution:

Given:

The cost of cultivating a 30 m long piece of land is ₹ 150. Had the breadth of the land been 5 m less, the cost would have been ₹ 120.

Let the breadth of the land = xm

The area of the land = 30x sq m

Had the breadth of the land been 5 m less the breadth would have been (x-5) m.

The area would have been 30 (x-5) sq m

According to the question,

\(\frac{30x}{30(x -5)}\) = \(\frac{x}{x – 5}\) = \(\frac{5}{4}\)

or, 5x-25= 4x or, 5x-4x= 25 or, x = 25

∴ The breadth of the land is 25 m.

Example 23. The area of four walls of a room is 154 sq m If its length and breadth be 6 m and 5 m respectively then what will be its height?

Solution:

Given:

The area of four walls of a room is 154 sq m If its length and breadth be 6 m and 5 m respectively

2 × (length + breadth) x height = area of four walls.

or, 2 x (6+ 5) m x height = 154 sq m or, 22 m x height = 154 sq m

or, height = \(\frac{154}{22}\) m = 7 m

∴ 7 m.

Height = 7 m.

Example 24. The length and breadth of a rectangular hall are 30 m and 18 m respectively. Find how many square tiles of side 3 dcm will be required for flooring the hall.

Solution:

Given:

The length and breadth of a rectangular hall are 30 m and 18 m respectively.

Area of the floor of the hall = 30 x 18 sq m = 540 sq m.

Area of one tile = 3 dcm x 3 dcm = .3 mx .3 m = .09 sq m

∴ Number of tiles = \(\frac{540}{.09}\) = 6000

∴ 6000 tiles will be required.

Example 25. Find the greatest measure of each stone of the same square shape to cover a rectangular courtyard 36 m long and 28 m broad. Find the number of such stones.

Solution:

The H.C.F. of 36 m and 28 m is 4 m.

∴ Each stone of the greatest area will be 4 metres square.

∴ Area of each stone = (4)² sq m = (4×4) sq m

∴ The required number of stones = \(\frac{36 X 28}{4 X 4}\) =63

∴ Each stone is 4 metres square and a number of stones is 63.

Example 26. Mahim’s family has an 18 m x 14 m rectangular plot of land. There is a square portion of the garden of side 3.4 m within the rectangular plot of land. Calculate the area of the plot of land without the garden. Find how many square tiles of side 2 dcm will be required to cover the empty portion of land.

Solution:

Given:

Mahim’s family has an 18 m x 14 m rectangular plot of land. There is a square portion of the garden of side 3.4 m within the rectangular plot of land.

Area of a rectangular plot of land = (18 x 14) sq m = 252 sq m.

Area of the square portion of the garden of side 3.4 m = (3.4 x 3.4) sq m = 11.56 sq m.

Area of the empty portion of land excluding the garden (252-11.56) sq m = 240.44 sq m.

Area of one tile = 2 dcm x 2 dcm = .2 m x .2 m = .04 sq m

∴ Number of tiles = \(\frac{240.44}{.0.4}\) = 6011

∴ Area of empty plot of land = 240.44 sq m, 6011 tiles will be required.

Understanding Square Measure for Class 7 Students

Example 27. Find the perimeter of a square field having twice the area of a rectangular field of length 25 m and breadth 8 m.

Solution: Area of the rectangular field = 25 x 8 sq m = 200 sq m

∴ Area of the square field = 2 x 200 sq m = 400 sq m

∴ Each side of the square field = √400 m = 20 m

∴ Perimeter of the square field = 4 × 20 m = 80 m

∴ 80 metres.

Perimeter of the square field = 80 metres.

Example 28. The length, breadth and height of a room of a school are 8 m, 6 m and 5 m respectively.

1. Find the cost of cementing the floor at ₹ 75 per square metre.
2. Find the cost of whitewashing the ceiling at ₹ 52 per square metre.
3. The room has 2 doors each 1.5 m wide and 1.8 m high and has 2 windows each 1.2 m wide and 1.4 m high. Calculate the cost of painting doors and windows at ₹ 260 per square metre.
4. Calculate the total cost of plastering the four walls without doors and windows at ₹ 95 per square metre and painting the walls at ₹ 40 per square metre.

Solution:

1. Area of the floor = (8 x 6) sq m = 48 sq m.

The cost of cementing 48 sq m at the rate ₹ 75 per square metre ₹ 75 x 48= ₹3600

∴ ₹ 3600.

2. Area of the ceiling area of the floor = 48 sq m.

Cost of whitewashing 48 sq m at the rate ₹ 52 per square metre = ₹ 52 x 48= ₹ 2496

∴ ₹ 2496.

3. Total area of two doors each 1.5 m wide and 1.8 m high = 2x (1.5 x 1.8) sq m = 5.4 sq m.

Total area of two windows each 1.2 m wide and 1.4 m high = 2 × (1.2 x 1.4) sq m = 3.36 sq m.

Total area of the doors and windows = (5.4+ 3.36) sq m = 8.76 sq m.

Cost of painting doors and windows at the rate ₹ 260 per square metre ₹ (260 x 8.76)= ₹ 2277.60

∴ ₹ 2277.60.

4.  Total area of the four walls = 2 × (length +breadth) x height = 2 x (6+8) x sq m = 140 sq m

area of the four walls without doors and windows (140-8.76) sq m = 131.24 sq m.

cost of plastering and painting per square metre (95+40) = ₹ 135

Total cost in 131.24 sq m at the rate of  ₹135 per square metre ₹ (135 x 131.24)= ₹ 17717.40

∴ ₹ 17717.40.

Example 29. What will be the cost of paper 3 m wide, at 25 paise per metre for covering the four walls of a room 15 m long, 12 m broad and 10 m high?

Solution:

The area of the paper =the area of the four walls

= 2 x (length + breadth) x height

= 2 x (15 m + 12 m) x 10 m

= 2 x 27 x 10 sq m

The length of the paper used = \(\frac{2 X 27 X 10}{3}\) m = 180 m

The required cost = 180 x 25 paise = 4500 paise = ₹ 45

∴ ₹45.

The required cost ₹45.

Class 7 Maths Exercise 6 Solved Examples

Example 30. A club room is square in the shape of side 15 m long and 5 m high. There are 4 doors in the club room each 1.5 m wide and 2 m high. Calculate the cost of oil painting its four walls without doors at 350 per square metre.

Solution:

Given:

A club room is square in the shape of side 15 m long and 5 m high. There are 4 doors in the club room each 1.5 m wide and 2 m high

Area of the 4 walls of the club room =2x (15+15) x 5 sq m = 300 sq m.

Area of 4 doors each 1.5 m wide and 2 m high = 4 x (1.5 × 2) sq m = 12 sq m.

Area of 4 walls without doors (300-12) sqm=288 sqm

Cost of oil painting 4 walls without doors at the rate ₹350 per sq m = ₹350 x 288 = ₹100800.

∴ ₹ 100800.

Cost of oil painting 4 walls without doors at the rate ₹350 per sq m = ₹ 100800.

Example 31. The length, breadth and height of a room are equal. If the area of four walls be 100 sq m then find its length, breadth and height.

Solution:

Given:

The length, breadth and height of a room are equal. If the area of four walls be 100 sq m

Let, the length of the room = its breadth = its height = a metres.

∴ 2x (a + a) x a = 100

or, 2 x 2a x a = 100

or, 4a² = 100

or, a² = 100 = 25

or, a = √25 = 5

∴ Each of length, breadth and height is equal to 5 m.

Example 32. What will be the cost of preparing a road of length 12 km and breadth 4 m at ₹ 3.50 per square metre?

Solution:

Area of the road= 12000×4 sq m. = 48000 sq m.

∴ Cost of preparing the road

= ₹ 48000 x 3.50 = ₹ 168000

∴ ₹ 168000

Cost of preparing the road is ₹ 168000

Example 33. ₹57800 is required to fence a square park with iron railing at 50 per metre. Find the length of one side of the park.

Solution:

Since, to fence the square park with iron railing at ₹ 50 per metre, ₹ 57800 is needed therefore, the perimeter of the square park

= \(\frac{57800}{50}\) metres = 1156 metres

Hence, the length of its one side = \(\frac{1156}{4}\) metres = 289 metres.

∴ 289 metres.

The length of its one side 289 metres.

Example 34. The length of a rectangular field is three times its breadth and its area is 192 sq metres. How much greater will be the area of the square field than that of the rectangular field if the perimeter of the square field is equal to that of the rectangular field?

Solution:

Given:

The length of a rectangular field is three times its breadth and its area is 192 sq metres.

Area of the rectangle = 192 sq m

∴ length x breadth = 192 sq m.

or, 3 x breadth x breadth = 192 sq m

or, (breadth)² = \(\frac{192}{3}\) sq m = 64 sq m

or, breadth = 8 m

∴ Length = 3 x 8 m = 24 m

Hence, the perimeter of the rectangular field = 2 (24 + 8) m = 64 m

Therefore, the perimeter of the square field is also 64m

∴ Length of each side of the square field = \(\frac{64}{3}\) = 16 m

∴ Area of the square field = (16)² sq m = 256 sq m

∴ Area of the square field is greater than that of the rectangular field by (256- 192) sq m = 64 sq m

∴ Greater by 64 sq m

Step-by-Step Solutions for Class 7 Square Measure Problems

Example 35. The length, breadth and height of a room are 6 m, 5 m and 3.5 m respectively. There are two doors each measuring 1.75 m by 1.2 m and two windows each measuring 1.4 m by 1 .2 m. What will be the cost of cementing the floor at? ₹65 per sq m? What will be the cost of whitewashing the four walls? ₹15 per sqm leaving the doors and the windows?

Solution:

Given:

The length, breadth and height of a room are 6 m, 5 m and 3.5 m respectively. There are two doors each measuring 1.75 m by 1.2 m and two windows each measuring 1.4 m by 1 .2 m.

Area of the floor of the room = 6 x 5 sq m = 30 sq m.

Total cost of cementing the floor at the rate of  ₹ 65 per sq m = ₹ 65 x 30 = ₹ 1950.

The area of the four walls of the room = 2 x (6 + 5) x 3.5 sq m = 77 sq m

Total area of two doors and two windows = (2 x 1.2 x 1.75 + 2 x 1.4 x 1.2) sqm

= (4.2 + 3.36) sq m = 7.56 sq m

∴ Area of the four walls excepting the doors and the windows = (77-7.56) sq m = 69.44 sq m.

Total cost of whitewashing at the rate of  ₹ 15 per square metre = ₹ 69.44 × 15 = ₹ 1041.60.

∴ ₹ 1950, 1041.60.

Example 36. There is a path 1 metre broad all round outside of a square garden of side 20 m. Find the cost of paving the path at ₹ 35.50 per sq m.

Solution:

Given:

There is a path 1 metre broad all round outside of a square garden of side 20 m.

Length of the square garden including the path = (20 + 2) m = 22 m.

∴ Area of the square garden including the path = (22)² sq m = 484 sq m.

area of the square garden excluding the path = (20)² sq m = 400 sq m.

∴ Area of the path =(484-400) sq m = 84 sq m.

The total cost of paving the path at ₹ 35.50 per square metre = ₹ (35.50 x 84) = ₹ 2982.

∴ The required cost is  ₹ 2982.

Example 37. Boundary walls are made along the bank of a pond to surround it. The length of the pond is 6 m more than its breadth. Now widening the edge of this pond by 8 m, a railing of length 192 m was made towards the waterside. What is the outside length and breadth of the boundary walls along the bank? What is the perimeter of the pond along the outside boundary?

Solution:

Given:

Boundary walls are made along the bank of a pond to surround it. The length of the pond is 6 m more than its breadth. Now widening the edge of this pond by 8 m, a railing of length 192 m was made towards the waterside.

Considering the water side 2 x (length + breadth) = 192 m

or, length + breadth = 96 m

or, breadth + 6 m + breadth = 96 m

 

 

or, 2 x breadth = 90 m or, breadth =\(\frac{90}{2}\) 45 m

Hence, length (45+6) m = 51 m.

Now, the outside length of the boundary wall along the bank = (51 +16) m = 67 m and the outside breadth = (45+16) m = 61 m.

Perimeter of the pond along the outside boundary=2x (67 +61) m = 256 m.

Example 38. Find the height of a triangle if its area and length of the base be x sq. units and y units respectively.

Solution:

Area of triangle =\(\frac{1}{2}\) x base x height

or, x = \(\frac{1}{2}\)  x y x height

or, height = \(\frac{2x}{y}\)

∴ \(\frac{2x}{y}\) unit.

Example 39. The perimeter of a triangle is (3a + b)cm. If the length of its two sides are (2a-b)cm and b cm, then find the length of the third side.

Solution:

Given:

The perimeter of a triangle is (3a + b)cm. If the length of its two sides are (2a-b)cm and b cm

Sum of lengths of two sides = (2a – b) + b = 2a cm.

∴ Length of third side = (3a + b) – 2a = (a + b) cm.

∴ (a + b) cm.

Length of third side = (a + b) cm.

Example 40. If the adjacent sides of the right angle in a right-angled triangle measure 4 cm and 10 cm in length, then find the area of the triangle.

Solution:

If one side adjacent to the right angle be considered as the base, then the other side becomes the height or altitude of the right angles triangle.

∴ Area of the triangle

= \(\frac{1}{2}\) x 4 x 10 sq.cm = 20 cm

Area of the triangle = 20 cm.

Example 41. The length of the base of a triangular-shaped land is thrice its height. A sum of ₹4,50,000 has been expended to cultivate the land at a rate of ₹30 per sq.m. Calculate the length of height of the land.

Solution:

Given:

The length of the base of a triangular-shaped land is thrice its height. A sum of ₹4,50,000 has been expended to cultivate the land at a rate of ₹30 per sq.m.

Area of land = sq.m 15000 sq.m.

Let the height of the triangular-shaped land be x m.

∴ Length of its base = 3x m.

∴ Area of triangle = \(\frac{1}{2}\) × 3x X x = 3x²/2 sq.m

Therefore, 3x²/2= 15000

or, x² = 3 x 15000 = 10,000

or, x=√10,000 = 100

∴ Height of land, x = 100m and length of the base, 3x=300m.

∴ 300m and 100m.

Height of land = 300m and 100m.

Example 42. The ratio of the lengths of the two sides = (2a – b) + b = 2a cm. adjacent to the right angle in a right-angled triangle is 3 4. If the area of the triangle is 1014 sq. cm, then find out the length of these sides.

Solution:

Given: 

The ratio of the lengths of the two sides = (2a – b) + b = 2a cm. adjacent to the right angle in a right-angled triangle is 3 4. If the area of the triangle is 1014 sq. cm

Let the lengths of the two sides be 3x and 4x.

∴ Area of triangle =\(\frac{1}{2}\) x 3x x 4x=\(\frac{12}{2}\) x² = 6x²

Therefore, 6x² =1014

or, x² = \(\frac{1014}{6}\) = 169

or, x =√169 = 13

∴ Length of two sides, 3×13 = 39 cm and 4×13 = 52 cm.

∴ 39 cm and 52 cm.

Length of two sides = 39 cm and 52 cm.

Arithmetic Chapter 5 Time And Distance Exercise 5 Solved Example Problems

Introduction:

The problem of time and distance arises from travelling. In our daily life, we have to travel from one place to the other. Travelling may be of different types. Some people walk on foot and some travel by cycle.

Other than these there are several vehicles by which we may go from one place to the other. There are buses, trams, trains, boats, ships, aeroplanes, etc., for our travelling.

In every case of travelling, to cover some distance, some time elapse. Thus to work out problems on time and distance we shall have to concentrate on two main things-

1. The distance travelled
2. The time required to travel that distance.

Read and Learn More WBBSE Solutions For Class 7 Maths

Uniform and non-uniform speed

When a body covers a particular distance in a particular time, we say that the body is travelling with uniform speed; otherwise, the speed of the body is said to be non-uniform.

In the case of uniform speed, if a car travels 40 km in one hour, it will travel 80 km in two hours and 120 km in three hours. This indicates that the car is travelling at a uniform speed of 40 km per hour.

But if a car travels 40 km in the first hour, 45 km in the second hour, and 35 km in the third hour then say that the speed of the car is non-uniform.

In this chapter, we shall assume that the objects are moving with uniform speed. We shall use ratio and proportion and unitary methods to solve the problems on uniform speed.

Relation between time, distance and speed

The distance traversed by any vehicle or person in unit time is called speed. By the term ‘unit time’ we mean ‘per hour’ or ‘per minute’ or ‘per second’.

If a train travels 50 kilometres per hour then its speed is written as, “50 km per hour” or, “50 km/hour”.

We get the following relations among time, distance and speed.

1. Distance travelled = Speed x Time. Distance Travelled

2. Speed= \(\frac{Distance travelled}{Time}\)

3. Time= \(\frac{Distance travelled}{Speed}\)

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

Two important rules

1. Since a tree or a telegraph post does not possess any length, they can be represented either by a dot or a point. Hence, when a train crosses a tree (or a telegraph post), it travels a distance equal to its own length.

2. When a train crosses a platform (or a bridge), it travels a distance equal to the sum of the lengths of the train and the platform (or the bridge).

 

 

Relative speed

Let, Ram and Shyam, start walking from the same place in the same direction. Let the speed of Ram be 4 km/hour and that of Shyam be 3 km/hour. Then in 1 hour, Ram travels (4-3) km or, 1 km more than Shyam. So it is clear that

When two men walk in the same direction the difference of their speeds is their relative speed.

 

 

Again, let Jadu and Madhu start walking from the same place in the opposite direction.

Let the speed of Jadu be 3 km/hour and that of Madhu be 2 km/hour. Then in 1 hour the distance between Jadu and Madhu increases by (3+2) km or, 5 km. So it is clear that

When two men walk in the opposite direction the sum of their speeds is their relative speed.

 

 

Thus from the idea of relative speed, we which the high-speed train crosses the slow can say that:

1. Between two trains moving towards the same direction in parallel tracks, the time in which the high-speed train crosses the slow-speed train is, speed train is,

 

 

2. When two trains are moving towards opposite directions in parallel tracks, the time in which the trains would cross each other is,

 

 

Average speed

When the speed of a body is not uniform throughout the journey, the average speed of the same is considered.

Boat and Stream

When a boat moves in a river then we have to consider two things:
1. Speed of the boat.
2. Speed of the river current.

The actual speed of a boat: By actual speed of a boat we mean the distance covered by a boat in unit time in still water (i.e., when there is no current) under the influence of oars.

 

 

Speed of a boat under the current of a river: When a boat moves under the current of a river, there may arise two cases.

If the boat moves along the current (downstream) the effective speed is greater than the actual speed of the boat and if the boat moves against the current (upstream) the effective speed is less than the actual speed of the boat.

 

 

When the boat rows along the current, then the joint effective speed = actual speed of the boat + speed of the river current.

When the boat rows against the current, then the joint effective speed = actual speed of the boat-speed of the river current.

 

 

Arithmetic Chapter 5 Time And Distance Exercise 5 Some Problems On Time And Distance

Example 1. A train travels 24 km in 4 hours. What distance will it travel in 10 minutes?

Solution:

Given:

A train travels 24 km in 4 hours.

Time
4 x 60 minutes
10 minutes

Distance
24 km
x km (say)

Since distance is directly proportional to time,

∴\(\frac{40 X 60}{10}\) = \(\frac{24}{x}\)

or, x = \(\frac{24 X 10}{4 X 60}\) = 1

∴ The required distance = 1 km.

Alternative method:

In 4 x 60 minutes, the train travels 24 km

In 1 minute the train travels \(\frac{24}{4 X 60}\) km

In 10 minutes the train travels \(\frac{24 X 10}{4 X 60}\) = 1 km

WBBSE Class 7 Time and Distance Solutions

Example 2. On Monday, Ram was cycling at a speed of 13 km/hour. He covered a certain distance in 2 hours. On Tuesday, cycling for the same time at a speed of 11 km/hour he covered a certain distance. On which day, Ram travelled how much more distance in 2 hours?

Solution:

Given:

On Monday, Ram was cycling at a speed of 13 km/hour. He covered a certain distance in 2 hours. On Tuesday, cycling for the same time at a speed of 11 km/hour he covered a certain distance.

On Monday In 1 hour he travelled 13 km In 2 hours he travelled 13 x 2 = 26 km

On Tuesday – In 1 hour he travelled 11 km In 2 hours he travelled 11 x 2 km = 22 km

On Monday he travelled (26-22) km or 4 km more distance.

∴ On Monday, he travelled 4 km more distance.

Example 3. A man can travel 5 metres in 6 seconds by a cycle. Find his speed in km/hour.

Solution:

Given:

A man can travel 5 metres in 6 seconds by a cycle.

Time
6 seconds
60 x 60 seconds

Distance
5 metres
x metres (say)

Since distance is directly proportional to time,

∴ \(\frac{6}{60 X 60}\)

or, 6x= 5 x 60 x 60

or, x= \(\frac{5 X 60 X 60}{6}\) = 3000

∴ In 1 hour the main travel of 3000 metres = 3km

∴ The speed of the man is 3km/hour.

Alternative method:

In 6 seconds the man travels 5 m

In 1 second the man travels \(\frac{5}{6}\) m

In 60 x 60 seconds, the man travels \(\frac{5}{6}\) x 60 x 60

= 3000 metres = 3km

Speed in km/hour = 3km

WBBSE Class 7 Arithmetic Time and Distance Examples

Example 4. On Saturday, Jadu went to the market cycling at a speed of 12 km/hour. But on Sunday he went to market cycling at a speed of 15 km/hour. If the distance between the house and the market is 2 km, on which day, Jadu take how much less time to reach the market?

Solution:

Given:

On Saturday, Jadu went to the market cycling at a speed of 12 km/hour. But on Sunday he went to market cycling at a speed of 15 km/hour. If the distance between the house and the market is 2 km,

On Saturday, he travelled 12 km in 60 minutes

On Saturday, he travelled 1 km in \(\frac{60}{12}\) x 2 = 10 minutes

On Sunday, he travelled 15 km in 60 minutes

On Sunday, he travelled 1 km in \(\frac{60}{15}\) minutes

On Sunday, he travelled 2 km in \(\frac{60}{15}\) x 2 minutes = 8 minutes

On Sunday he took (108) min or 2 minutes to reach the market.

∴ On Sunday, he took 2 min less time.

Example 5. Find the time taken to travel 600 km by train whose speed is 25 km/hour.

Solution

Distance
25 km
600 km

Time
1 hour
x hours (say)

Since distance is proportional to time,

∴ \(\frac{25}{600}\) = \(\frac{1}{x}\)

or, 25x= 600 or, x = \(\frac{600}{25}\) = 24

∴ The required time is 24 hours.

Alternative method:

The train travels 25 km in 1 hour

The train travels 1 km in \(\frac{1}{25}\) hour

The train travels 600 km in \(\frac{600}{25}\) hour = 24 hours.

Example 6. A man covered a distance of 12 km in 40 minutes by bus. Find the speed of the bus.

Solution:

Given:

A man covered a distance of 12 km in 40 minutes by bus.

That man, in 40 minutes travelled 12 km

That man, in 1 minute travelled \(\frac{12}{40}\) km

That man, in 60 minutes travelled \(\frac{12}{40}\) x 60 = 18 km

∴ The speed of the bus is 18 km/hour.

Solved Problems for Class 7 Time and Distance

Example 7.  A train went from Kolkata to Burdwan at a speed of 45 km/hour and then returned from Burdwan to Kolkata at a speed of 36 km/hour. Find the average speed of the train.

Solution:

Given:

A train went from Kolkata to Burdwan at a speed of 45 km/hour and then returned from Burdwan to Kolkata at a speed of 36 km/hour.

While going from Kolkata to Burdwan

The train travelled 45 km in 1 hour

The train travelled 1 km in \(\frac{1}{45}\) hour

While returning from Burdwan to Kolkata

The train travelled 36 km in 1 hour

∴ The train travelled 1 km in \(\frac{1}{36}\) hour

∴ To travel 2 km the train takes

(\(\frac{1}{45}\) + \(\frac{1}{36}\)) hr = \(\frac{4 + 5}{180}\) hr = \(\frac{1}{20}\) hour.

In \(\frac{1}{20}\) hour the train travels 2 km

In 1 hour the train travels 2 x 20 km = 40 km

∴ The average speed of the train is 40 km/hour.

Alternative method:

Let the distance between Kolkata and Burdwan be 180 km. (L.C.M. of 45 and 36 is 180).

∴ To travel from Kolkata to Burdwan the train \(\frac{180}{45}\) hours = 4 hours.

To return from Burdwan to Kolkata the train takes \(\frac{180}{36}\) hours = 5 hours.

∴ In (4+5) hours or 9 hours, the train travels (180+180) km = 360 km.

∴ Average speed of the train is \(\frac{1}{36}\)km/hour = 40 km/hour.

Example 8. A train of length 100 metres passed a tree moving at a speed of 60 km/hour. How much time did it take?

Solution:

Given:

A train of length 100 metres passed a tree moving at a speed of 60 km/hour.

When a train crosses a tree it travels its own length

100 metres = \(\frac{100}{1000}\) km = \(\frac{1}{10}\) km.

The train travels 60 km in 60 x 60 sec

The train travels 1 km in \(\frac{60 X 60}{60}\) sec

The train travels \(\frac{1}{10}\) km in \(\frac{1}{10}\) x 60 sec = 6 sec

Class 7 Maths Exercise 5 Solutions on Time and Distance

Example 9. A train of length 210 metres crosses a telegraph post in 7 seconds. Express the speed of the train in the “km/hour” unit.

Solution:

Given:

A train of length 210 metres crosses a telegraph post in 7 seconds.

When a train crosses a telegraph post it travels a distance equal to its own length.

Time
7 seconds
60 x 60 seconds

Distance
210 metres
x metres (say)

Since distance is directly proportional to time,

∴ \(\frac{7}{60 X 60}\) = \(\frac{210}{x}\)

or, 7x= 210 × 60 × 60

∴ in 1 hour the train travels 108000 metres = 108 km.

∴ The speed of the train is 108 km/hour.

Alternative method:

In 7 seconds the train crosses 210 m

In 1 second the train crosses \(\frac{210}{7}\) m

In 60 x 60 second the train crosses \(\frac{210}{7}\) x 60 x 60 m

= 108000 metres = 108 km.

Example 10. Moving with a uniform speed a taxi takes 6 hours 12 min to cover a distance of 217 km. How much time will it take to cover a distance of 273 km?

Solution:

Given:

Moving with a uniform speed a taxi takes 6 hours 12 min to cover a distance of 217 km.

6 hour 12 min (6 x 60+12) min = 372 min

To travel 217 km time required = 372 min

To travel 1 km time required = \(\frac{372}{217}\) min

To travel 273 km time required = \(\frac{372}{217}\) x 273

= 468 mon = 7 hours 48 min

∴ 7 hour 48 min.

7 hour 48 min time will it take to cover a distance of 273 km

Example 11. A train crosses a platform in 30 seconds travelling at a speed of 60 km/hour. If the length of the train be 200 metres, then find the length of the platform.

Solution:

Given:

A train crosses a platform in 30 seconds travelling at a speed of 60 km/hour. If the length of the train be 200 metres

When a train crosses a platform, it travels a distance equal to the sum of the lengths of the train and the platform.

Time
60 x 60 seconds
30 seconds

Distance
60 km
x km (say)

Since the distance is directly proportional to time,

∴  \(\frac{60 X 60}{30}\) = \(\frac{60}{x}\)

or, x X 60 x 60 = 30 × 60

or, x = \(\frac{30 X 60}{60 X 60}\) = \(\frac{1}{2}\)

∴ Distance travelled by train in 30 seconds

= \(\frac{1}{2}\) km = \(\frac{1}{2}\) x 1000 metres = 500 metres.

Now, the length of the train + length of the platform = 500 metres

∴ length of the platform = (500 – 200) metres = 300 metres.

Alternative method:

The speed of the train is 60 km/hour

∴ In 60 x 60 seconds the travel 60 x 1000 m

In 1 second the travel \(\frac{60 X 1000}{60 X 60}\) m

In 30 seconds the travel \(\frac{60 X 1000 X 30}{60 X 60}\) m= 500 metres

∴ length of the platform = (500-200) metres = 300 metres

West Bengal Board Class 7 Time and Distance Assistance

Example 12. Ram covered a distance of 100 km in 2 hours and 5 min on his motorbike. Shyam covered the same distance on his cycle in 6 hours and 40 min. Find the ratio of the speeds of the motorbike and cycle.

Solution:

Given:

Ram covered a distance of 100 km in 2 hours and 5 min on his motorbike. Shyam covered the same distance on his cycle in 6 hours and 40 min.

2 hour 5 min = (2 x 60+ 5) min = 125 min.

In 125 min Ram travels 100 km

In 1 min Ram travels \(\frac{100}{125}\) = \(\frac{4}{5}\) km

6 hour 40 min = (6 x 60+40) min = 400 min

In 400 min Shyam travels 100 km

In 1 min Shyam travels \(\frac{100}{400}\) km = \(\frac{1}{4}\) km

Ratio of the speeds of the motorbike and cycle

= \(\frac{4}{5}\) ÷ \(\frac{1}{14}\)

= \(\frac{4}{5}\) X \(\frac{1}{14}\)

= \(\frac{16}{5}\) = 16: 5

The ratio of the speeds of the motorbike and cycle = 16: 5

Example 13. A train of length 160 metres crosses a platform of length 320 metres in 36 seconds. Find the speed of the train.

Solution:

Given:

A train of length 160 metres crosses a platform of length 320 metres in 36 seconds.

In order to cross the platform, the train has to travel (160+320) metres = 480 metres.

Time
36 seconds
60 X 60 seconds

Distance
480 metres
x metres(say)

Since the distance is directly proportional to time,

∴ \(\frac{36}{60 X 60}\) = \(\frac{480}{x}\)

or, 36x = 480 X 60X 60

or, x = \(\frac{480 X 60 X 60}{36}\) = 48000

∴ The speed of the train is 48000 metres/hour = 48 km/hour

Alternative method:

∴ In 36 seconds the train travels 480 m

In 1 seconds the train travels \(\frac{480}{36}\) m

In 60 X 60 seconds the train travels \(\frac{480 X 60 X 60}{36}\)

= 48000 metres = 48 km.

Example 14. Moving with a uniform speed a goods train covers a distance of 49.5 km in 2 hours 45 min. How long will it take to reach a station situated at a distance of 58.5 km?

Solution:

Given:

Moving with a uniform speed a goods train covers a distance of 49.5 km in 2 hours 45 min.

2 hr 45 min = 2 \(\frac{45}{60}\) hr = 2 \(\frac{3}{4}\) hour = \(\frac{11}{4}\) hr.

To travel 49.5 km time taken = \(\frac{11}{4}\) hour

To travel 1 km time taken = \(\frac{11}{4 X 49.5}\) hour

To travel 58.5 km time taken = \(\frac{11 X 58.5}{4 X 49.5}\) hour

= \(\frac{11 X 585}{4 X 495}\) hour = \(\frac{13}{4}\) hour

= 3 \(\frac{1}{4}\)

∴ 3 hours 15 min

3 hours 15 min will it take to reach a station situated at a distance of 58.5 km

WBBSE Class 7 Chapter 5 Time and Distance Guide

Example 15. Two trains are travelling towards each other on parallel lines. Their speeds are 40 km/hour and 30 km/hour respectively. They reached their destinations after 4 and 5 hours respectively from the time they met each other. Find the distance between the destinations of the two trains. (The lengths of the trains may be neglected in comparison with the distance travelled by them.)

Solution:

Given:

Two trains are travelling towards each other on parallel lines. Their speeds are 40 km/hour and 30 km/hour respectively. They reached their destinations after 4 and 5 hours respectively from the time they met each other.

Let the destination of the first train be point B and that of the second train be A.

Let O be the point at which they meet.

In 4 hours the first train travels the distance OB, with a speed of 40 km/hour.

∴ OB = 40 x 4 km = 160 km.

In 5 hours the second train travels the distance OA, with a speed of 30 km/hour.

∴ OA 30 x 5 km = 150 km.

Hence, AB = OB + OA = (160 + 150) km

∴ 310 km

The distance between the destinations of the two trains is 310 km.

Example 16. Your father went to Bandel on his motorcycle, worked there for an hour and returned home in 3 hours and 30 minutes. If the uniform speed of the motorcycle is 40 km/hour find the distance of Bandel from your home.

Solution:

Given:

Your father went to Bandel on his motorcycle, worked there for an hour and returned home in 3 hours and 30 minutes. If the uniform speed of the motorcycle is 40 km/hour

Since your father has worked for 1 hour, therefore the time is taken to Bandel and return 2 hours 30 min.

Therefore, the time taken only to go to Bandel = 2 hours 30 min ÷ 2 = 1 hour 15 min = 75 min.

At 60 min he travels 40 km

In 1 min he travels \(\frac{40}{60}\) km

In 75 min he travels \(\frac{40}{60}\) X 75 km = 50 km

∴ The required distance = 50 km

Example 17. Two trains start at the same time towards each other from two stations 120 km apart. The speeds of the trains are 50 km/hour and 40 km/hour respectively. Find the distance between them after 1 hour and 15 minutes. Find also the distance between them after 2 hours. (Here assume that the two trains are travelling on parallel lines.)

Solution:

Given:

Two trains start at the same time towards each other from two stations 120 km apart. The speeds of the trains are 50 km/hour and 40 km/hour respectively. Find the distance between them after 1 hour and 15 minutes.

In 1 hour the distance between the two trains decreases by (50+40) km = 90 km.

Now, 1 hr 15 m = 1 \(\frac{15}{60}\) hrs = 1 \(\frac{1}{4}\)hrs = \(\frac{5}{4}\)hrs.

In 1 hour distance between the two trains decreases by 90 km

∴ In \(\frac{5}{4}\) hours distance between the two trains decreases by 90 x \(\frac{5}{4}\) km = 112.5 km.

∴ The distance between them after 1 hour 15 minutes will be (120-112.5) km = 7.5 km.

Again, in 2 hours the two trains advance towards each other by 90 x 2 km = 180 km.

∴ The distance between them after 2 hours will be (180120) km = 60 km.

∴ 7.5 km, 60 km.

The distance between them after 2 hours is 7.5 km, 60 km.

Understanding Time and Distance for Class 7 Students

Example 18. A 70-metre-long train runs at a speed of 75 km/hour. How long will it take to cross a platform of length 105 metres?

Solution:

Given: 

A 70-metre-long train runs at a speed of 75 km/hour.

To cross a platform of length 105 metres a train of length 70 metres will travel (70+105) metre = 175 metres.

The train travels 75 x 1000 m in 60 x 60 sec

The train travels 1 m in \(\frac{60 X 60}{75 X 1000}\) sec

The train travels 175 m in \(\frac{60 X 60 X 175}{75 X 1000}\) sec

= \(\frac{42}{5}\) sec = 8 \(\frac{8}{5}\) sec.

∴ 8 \(\frac{8}{5}\) sec.

8 \(\frac{8}{5}\) sec. it will take to cross a platform of length 105 metres.

Example 19. A passenger train and a goods train start from the same station on parallel lines at the same time and in the same direction. The speed of the passenger train is 50 km/hour. After 30 minutes it is found that the passenger train is 5 km ahead of the goods trains. Find the speed of the goods trains.

Solution:

Given:

A passenger train and a goods train start from the same station on parallel lines at the same time and in the same direction. The speed of the passenger train is 50 km/hour. After 30 minutes it is found that the passenger train is 5 km ahead of the goods trains

In 60 minutes the passenger train goes 50 km

In 1 minute the passenger train goes \(\frac{50}{60}\) km

In 30 minutes the passenger train goes \(\frac{50}{60}\) x 30 km = 25 km

By this time, the goods train goes (25-5) km = 20 km

In 30 minutes the goods train goes 20 km.

In 1 minute the passenger train goes \(\frac{20}{30}\) km

In 60 minutes the passenger train goes \(\frac{20}{30}\) x 60 km = 40 km

∴ Speed of the goods train is 40 km/hour.

Example 20. A 90-metre-long train takes 25 sec to cross a pillar. Calculate the speed of the train in km/ hour.

Solution:

Given:

A 90-metre-long train takes 25 sec to cross a pillar.

When a train crosses a pillar it crosses a distance equal to it own length.

In 25 sec the train crosses \(\frac{90}{1000}\) km

In 1 sec the train crosses \(\frac{90}{25 X 1000}\) km

In 60 X 60 sec the train crosses \(\frac{90 X 60 X 60}{25 X 1000}\) km

= \(\frac{324}{25}\) km = 12.96 km

∴ 12.96 km/hour.

The speed of the train in km/ hour is 12.96 km/hour.

Example 21. Two trains of lengths 200 metres and 240 metres approach each other in two parallel lines running side by side with speeds of 42.5 km/hour and 36.7 km/hour respectively. What time will they take to cross each other after they meet?

Solution:

Given:

Two trains of lengths 200 metres and 240 metres approach each other in two parallel lines running side by side with speeds of 42.5 km/hour and 36.7 km/hour respectively.

In 1 hour the distance between the two trains decreases by (42.5+ 36.7) km = 79.2 km.

The two trains will have to travel a total distance of (200+ 240) metres = 440 metres in order to cross each other.

79.2 X 1000 metres distance decreases in 60 x 60 seconds

1-metre distance decreases in \(\frac{60 X 60}{792 X 1000}\) sec

440-metre distance decreases in \(\frac{60 X 60 X 400}{79.2 X 1000}\) seconds 20 seconds.

∴ 20 seconds.

20 seconds time will they take to cross each other after they meet.

Example 22. To cross a bridge of length 250 metres a train of length 150 metres takes 30 sec. How long will this train take to cross a platform 130 metres long?

Solution:

Given:

To cross a bridge of length 250 metres a train of length 150 metres takes 30 sec.

When a train of length 150 metres crosses a bridge of length 250 metres then it travels (150+250) metre = 400 metres.

When a train of length 150 metres crosses a platform of length 130 metres then it travels (150+130) metres 280 metres.

The train travels 400 metres in 30 sec

The train travels 1 metre in \(\frac{30}{400}\) sec

The train travels 280 metre in \(\frac{30 X 280}{400}\) sec = 210 sec

Class 7 Maths Exercise 5 Solved Examples

Example 23. A bus starts from Kolkata at 7-30 A.M. and reaches Digha at 12 noon. If the speed of the bus be 45 km/hour, what is the distance between Kolkata and Digha?

Solution:

Given:

A bus starts from Kolkata at 7-30 A.M. and reaches Digha at 12 noon. If the speed of the bus is 45 km/hour

Total time from 7-30 A.M. to 12 noon

= 4 hours 30 minutes = 4 hours = hours.

Now, in 1 hour the bus goes 45 km

Now, in \(\frac{9}{2}\) hours the bus goes 45 Χ \(\frac{9}{2}\) km = 202.5 km

Example 24. A train takes 4 sec to cross a telegraph post and 20 sec to cross a 264-metre-long bridge. Find the length of the train and its speed.

Solution:

Given:

A train takes 4 sec to cross a telegraph post and 20 sec to cross a 264-metre-long bridge.

Let the length of the train be x metre. Then the train travels x metre in 4 sec and (x+264) metre in 20 sec.

∴ \(\frac{4}{20}\) = \(\frac{x}{x + 264}\)

or, 5x=x+264 or, 5x-x= 264

or, 4x = 264 or, x = 264 = 66

∴ The length of the train is 66 metres.

In 4 sec the train travels \(\frac{66}{1000}\) km

In 1 sec the train travels \(\frac{66}{4 X 1000}\) km

In 60 X 60 sec the train travels \(\frac{66 X 60 X 60}{4 X 1000}\) km = 59.4 km

∴ The length of the train is 66 metres and its speed is 59.4 km/hour.

Example 25. An up train and a down train started their journey at the same time after meeting at a station. After 24 minutes, the two trains reached two opposite stations, the distance between them being 30 km. If the speed of the up train is 40 km/hour, find the speed of the down train.

Solution:

Given:

An up train and a down train started their journey at the same time after meeting at a station. After 24 minutes, the two trains reached two opposite stations, the distance between them being 30 km. If the speed of the up train is 40 km/hour

In 60 minutes the up train travels 40 km

In 1 minute the up train travels \(\frac{40}{60}\) km

In 24 minutes the up train travels \(\frac{40}{60}\) X 24 km = 16 km.

Therefore, the down train travels (30-16) km = 14 km.

In 24 minutes the down train travels 14 km

In 1 minute the down train travels \(\frac{14}{24}\) km

In 60 minutes the down train travels \(\frac{14}{24}\) X 60 km = 35 km

∴ The speed of the down train is 35 km/ hour.

Example 26. A 100-metre-long train, at a speed of 48 km/ hour crosses a tunnel in 21 sec. Calculate the length of the tunnel.

Solution:

Given:

A 100-metre-long train, at a speed of 48 km/ hour crosses a tunnel in 21 sec

In 60 x 60 sec, the train travels 48000 metre

In 1 sec the train travels \(\frac{48000}{60 X 60}\) metre

In 21 sec the train travels \(\frac{48000 X 21}{60 X 30}\) = 280 metres.

Since the length of the train is 100 metres, therefore, the length of the tunnel = (280-100) metres = 180 metres.

Example 27. A 250-metre-long goods train is travelling at a speed of 33 km/hour. A 200 metre-long mail train approaches it from the back side in a parallel line with a speed of 60 km/hour. How long will it take to cross the goods train after catching it?

Solution:

Given:

A 250-metre-long goods train is travelling at a speed of 33 km/hour. A 200 metre-long mail train approaches it from the back side in a parallel line with a speed of 60 km/hour.

In 1 hour the distance between the two trains (60-33) km = 27 km.

In order to overtake the goods train, the mail train will have to travel (250 + 200) metres = 450 metres.

Now, the distance is 27 x 1000 m in 60 min.

the distance is 1 m in \(\frac{60}{27 X 1000}\) min.

the distance is 450 m in \(\frac{60 X 450}{27 X 1000}\) = 1 minute.

Example 28. A train takes 10 sec to cross a man standing on a platform 150 metres long and crosses the platform in 22 sec. Calculate the length and the speed of the train.

Solution:

Given:

A train takes 10 sec to cross a man standing on a platform 150 metres long and crosses the platform in 22 sec.

Let, the length of the train be x metre. In 10 sec the train crosses x metre and in 22 sec the train crosses (x + 150) metre.

∴ \(\frac{10}{22}\) = \(\frac{x}{x + 150}\)

or, 11x=5x+750

or, 11x-5x=750

or, 6x = 750

or, x = \(\frac{750}{6}\) = 125

The length of the train is 125 metres

In 10 sec the train crosses 125 metre

In 1 sec the train crosses \(\frac{125}{1000 X 10}\) km

In 60 x 60 sec the train crosses \(\frac{125 X 60 X 60}{1000 X 10}\) km = 45 km

∴ The length of the train is 125 metres and its speed is 45 km/hour.

Example 29. A train crosses two bridges of lengths 210 metres and 122 metres in 25 seconds and 17 seconds respectively. Find the length and the speed of the train.

Solution:

Given:

A train crosses two bridges of lengths 210 metres and 122 metres in 25 seconds and 17 seconds respectively.

Let, the length of the train be x

Then in 25 seconds, the train travels (x+210) metres and in 17 seconds it travels (x + 122) metres.

Hence, \(\frac{25}{17}\) = \(\frac{x + 210}{x + 122}\)

or, 25x+3050 = 17x+3570

or, 25x-17x=3570-3050

or, 8x = 520

or, x = \(\frac{520}{8}\) = 65

∴ Length of the train = 65 metres.

Now, in 25 seconds the train travels (65+ 210) metres 275 metres.

In 25 seconds it travels 275 metres

In 1 second it travels \(\frac{275}{25}\) metres

In 60 X 60 seconds it travels \(\frac{275 X 60 X 60}{25}\) metres

= 39600 metres = 39.6 km

∴ The speed of the train is 39.6 km/hour

∴ The length of the train = 65 metres and its speed is 39.6 km/hour

Step-by-Step Solutions for Class 7 Time Problems

Example 30. The distance between A and B along a river is 96 km. The speed of the river current is 4 km/ hour from A towards B. A boat takes 8 hours to travel from A to B. What time will it take to return from B to A?

Solution:

Given:

The distance between A and B along a river is 96 km. The speed of the river current is 4 km/ hour from A towards B. A boat takes 8 hours to travel from A to B.

For the journey from A to B,

In 8 hours the boat travels 96 km.

In 1 hours the boat travels \(\frac{96}{8}\) km = 12 km

∴ The actual speed of the boat = (12-4) km/ hour = 8 km/hour.

Therefore, the speed of the boat against the current = (8-4) km/hour = 4 km/hour.

So, for the journey from B to A,

To travel 4 km the boat takes 1 hour

To travel 1 km the boat takes \(\frac{1}{4}\) hour

To travel 96 km the boat takes \(\frac{96}{4}\) hours = 24 hours

Example 31. A boat can travel 8 km/hour in still water. But upstream it takes 4 times of that time to cover the same distance. What time will the boat take to cover 56 km downstream?

Solution:

Given:

A boat can travel 8 km/hour in still water. But upstream it takes 4 times that time to cover the same distance.

Against the Current-

In 4 hours the boat goes 8 km.

In 1 hour the boat goes \(\frac{8}{4}\) = 2 km

Also, in still water, the boat travels 8 km/hour.

Hence, the speed of the river current is (8-2) km/hour 6 km/hour.

Therefore, downstream, in 1 hour the boat travels (8+6) km = 14 km.

So, along the current, the boat covers-

14 km in 1 hour

1 km in \(\frac{1}{14}\) hour

56 km in \(\frac{56}{14}\) hours = 4 hours.

Example 32. P and Q are two places on the bank of a river. The speed of the river current is from P towards Q. At the same time, a boat starts from P towards Q with an actual speed of 8 km/ hour and another boat starts from Q towards P with an actual speed of 7 km/hour. If the distance between P and Q is 150 km, when will the two boats meet?

Solution:

Given:

P and Q are two places on the bank of a river. The speed of the river current is from P towards Q. At the same time, a boat starts from P towards Q with an actual speed of 8 km/ hour and another boat starts from Q towards P with an actual speed of 7 km/hour. If the distance between P and Q is 150 km

Let the speed of the river current be x km/hour.

In 1 hour the first boat covers (8 + x) km and in 1 hour the second boat covers (7-x) km

In 1 hour the distance between the two boats decreases by (8+x+7-x) km = 15 km

The distance decreases by 15 km in 1 hour

1 km in \(\frac{1}{15}\) hour

150 km  in \(\frac{150}{15}\) hours = 10 hours

∴ The two boats will meet after 10 hours.

Example 33. Speeds of a boat along and against the current are 12 km/hour and 8 km/hour respectively. Find the actual speed of the boat and the speed of the current.

Solution:

Given:

Speeds of a boat along and against the current are 12 km/hour and 8 km/hour respectively.

Actual speed of the boat = \(\frac{20}{2}\) km/hr = 10 km/hr

∴ Speed of the current = (120- 10) km/hr = 2 km/hr.

∴ Actual speed of the boat = 10 km/hr.

Speed of the current = 2 km/hr

Example 34. The speed of a boat in still water is 10 km/ hour. The boat can cover 90 km in 6 hours downstream. What time will the boat take to cover the same distance upstream?

Solution:

Given:

The speed of a boat in still water is 10 km/ hour. The boat can cover 90 km in 6 hours downstream.

Downstream, in 6 hours the boat covers 90 km

Downstream, in 1 hour the boat covers \(\frac{90}{6}\) km = 15 km

Therefore, the speed of the current = (15- 10) km/hour = 5 km/hour.

Hence, against the current in 1 hour, the boat covers (10-5) km = 5 km.

Therefore, upstream the boat covers—

5 km in 1 hour

1 km in \(\frac{1}{5}\) hour

90 km in \(\frac{90}{5}\) hours = 18 hours.

∴ 18 hours.

18 hours Time will the boat take to cover the same distance upstream

Arithmetic Chapter 4 Square Root Of Fraction Exercise 4 Solved Example Problems

 Introduction

You have learnt the method of finding the square root of the perfect square integers in the previous class. Here, you will be able to learn the method of finding the square root of fractions and decimals.

As in our practical life we have to deal not only with integers but also with fractions and decimals so it is necessary to know the method of finding the square root of fractions and decimals as well.

The square roots of fractions and decimals are calculated in a similar way as that of integers but here the method is slightly complicated.

Read and Learn More WBBSE Solutions For Class 7 Maths

Square root of decimal fraction

If a decimal fraction is multiplied by itself then the product is called the square of the decimal fraction and the original decimal fraction is called the square root of the product.

Example:

Square of 0.2 0.2 0.2 = 0.04; the therefore square root of 0.04 = √0.04= 0.2.

Square of 0.7 0.7 x 0.7 = 0.49; therefore square root of 0.49 = √0.49 = 0.7.

Similarly, (0.12)²= 0.12 x 0.12 = 0.0144, hence √0.0144 = 0.12

(0.24)² = 0.24 x 0.24 = 0.0576, hence √0.0576 = 0.24

Rule:

1. Mark pair of digits, for an integral part, starting from the digit in the units’ place (i.e., from the digit immediately before the decimal point), towards the left in a similar way as you have done in the case of integers.

Then, similarly, mark the digits immediately after the decimal point in pairs towards the right.

2. If the decimal fraction do not contain any integral part then perform marking the pairs after the decimal point.

3. If the given decimal fraction has an odd number of decimal places, put a zero at the end to make the number of digits after decimal point even.

4. The decimal point should be put in the square root immediately when the pair of digits from the decimal part is brought down at the end of an integral part.

WBBSE Class 7 Arithmetic Square Root Examples

Example: Find the square root of: 150.0625

∴ The required square root is 12.25.

Example: (0.1)² = 0.01; Here, 0.01 < 0.1

And (1.5)² = 2.25; here 2.25 > 1.5

The square root of a vulgar fraction

If a vulgar fraction is multiplied by itself then the product is called the square of that vulgar fraction and the original vulgar fraction is called the square root of the product.

Example: Square of \(\frac{1}{2}\) = \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\)

therefore square root of \(\frac{1}{2}\) = = \(\frac{1}{2}\)

Arithmetic Chapter 4 Square Root Of Fraction Exercise 4 Some Problems On Square Root Of Decimal Fractions

Example 1. The area of a square is 32.49 sq cm. What is the length of one side of the square?

Solution:

Given:

The area of a square is 32.49 sq cm.

The length of the side of the square = √32.49 cm = 5.7 cm.

∴ 5.7 cm

The length of one side of the square 5.7 cm.

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

 Example 2. What least number must be subtracted from 0.0582 so that the result of subtraction be a perfect square decimal number?

Solution:

Given:

0.0582

Hence, it is found that the number 0.0582 exceeds the square of 0.24 by 0.0006.

Hence, at least 0.0006 should be subtracted from 0.0582 so that the result of subtraction will be a perfect square decimal number.

∴ At least 0.0006 should be subtracted.

Example 3. Find the length of one side of a square whose area is equal to the sum of the areas of two rectangles whose areas are 2.1214 sq m and 2.9411 sq m.

Solution:

Given:

2.1214 sq m And 2.9411 sq m

Total area of the two rectangles = (2.1214+2.9411) sq m = 5.0625 sq m

∴ Area of the square = 5.0625 sq m

∴ Length of the side of the square = √5.0625 m = 2.25 m

∴ 2.25 m

Length of the side of the square 2.25 m

Example 5. What must be added with 0.28 so that the square root of 1?

Solution:

Given:

0.28

Since, square root of the sum is 1 therefore, sum is also 1.

Number to be added = (1-0.28) = 0.72 = 0.72

∴ 0.72

Solved Problems for Class 7 Square Roots of Fractions

Example 6. Find the square root of the product of 0.032 and 0.2.

Solution:

Given:

0.032 And 0.2

The required square root

= √0.032×0.2= √0.0064=0.08

= 0.08

The square root of the product 0.08

Example 7. Find the value of √240.25 + √2.4025 + √0.024025

Solution:

Given:

√240.25 + √2.4025 + √0.024025

∴ √240.25 = 15.5

Similarly, √240.25 = 1.55 and,

√0.024025 = 0.155

∴ The required sum = 15.5 + 1.55 + 0.155 = 17.205

∴ 17.205

The value of √240.25 + √2.4025 + √0.024025 = 17.205

Class 7 Maths Exercise 4 Solutions on Square Roots

Example 8. Which number should be subtracted from 48.03 so that the square root of the difference is 5.7?

Solution:

Since, the square root of the difference is 5.7 therefore, the difference = (5.7)² = 32.49.

Hence, the number to the subtracted = 48.03 – 32.49 = 15.54

Example 9. Find the decimal number which when multiplied by itself gives 1.1025 as the product.

Solution: The required number will be the square root of 1.1025 = √1.1025 = 1.05

∴ 1.05

Example 10. Find the square root of 2 upto 3 places of decimals.

Solution:

Given:

Here, although 2 is an integer it is not a perfect square number.

Hence, the square root of 2 must be a decimal fraction.

Also, we may consider 2 as 2.000000…

∴ The required square root = 1.414.

Example 11. Which decimal number is to be added with 0.75 so that square root of the sum will be 2?

Solution:

Since, the square root of the sum will be 2 therefore the sum will be 4.

∴ Number to be added = 4-0.75 = 3.25

∴ 3.25.

Example 12. Find the square root of 5.842 up to 3 places of decimals.

Solution:

∴ The required square root is 2.417.

Example 13. Find the approximate value of √15 upto 2 decimal places. Find how greater or less than 15 is the square of this approximate value.

Solution:

Therefore, the approximate value of upto 2 decimals places is 3.87.

Now, square of 3.87 = (3.87)² = 14.9769.

Therefore, the square is less than 15 by (15-14.9769)=0.0231.

∴  3.87, 0.0231 less.

Example 14. Find the square root of 52.983841 correct upto two places of decimals

Solution:

Hence, the square root of 52.983841 correct up to two places of decimals is 7.28

Example 15. Find the value of √0.5 correct up to three decimal places.

Solution:

∴ The value of √0.5 correct up to three decimal places is 0.707.

Example 16. Which decimal fraction multiplied by itself will give a product 7.5625?

Solution: The required decimal fraction will be the square root of 7.5625

∴ 2.75

West Bengal Board Class 7 Square Root Assistance

Example 17. Find the least number which when added to 31.69 will make it a perfect square having 4 places after the decimal.

Solution:

Example 18. Find the difference of the lengths of the sides The area of the new square of the squares whose areas are 1.4641 sq m and 1.0609 sq m.

Solution:

The length of side of the first square. Each side of the new square = √14641 m = 1.21 m.

The length of side of the second square = √1.0609 m = 1.03 m

Example 19. Find the length of the side of the square whose area is equal to the sum of the areas of two rectangles having areas 3.24 sq m and 2.52 sq m.

Solution:

The sum of the areas of two rectangles (3.24 + 2.52) sq m = 5.76 sq m.

∴ Length of the side of the square = √5.76 m = 2.4 m

∴ The length of the side of the square is 2.4 m.

Example 20. The sides of two squares are 16.5 and 22 metres respectively. Find the side of a square whose area is equal to the sum of the areas of the two squares.

Solution:
Area of the first square = (16.5)² sq m = 272.25 sq m

Area of the second square = (22)² sq m = 484 sq m

The area of the new sqaure = (272.25 + 484)  sq m = 756.25 sq m

∴ Each side of the new square = √756.25 m = 27.5 m

∴ 27.5 meters.

Example 21. A man spent ₹ 506.25 and each day he spent as many rupees as the number of days in which the money was spent. How much did he spend each day?

Solution:

Since the number of days and the number of rupees spent each day are equal, therefore the product of two equal numbers = 506.25

∴ Each number = √506.25 = 22.5

Example 22. A ladder 36 metres long, stands against a wall so that its bottom is 21.6 metres away from the wall. How high is its top on the wall?

Solution:

Let Bc = 36 metres be the ladder and AB = 21.6 metres.

Let AC = h metres.

Now h²+ (21.6)² = (36)²

or, h = (36)² – (21.6)²

= 1296 – 466.56 = 829.44

or, h =√829.44 = 28.8

 

Arithmetic Chapter 4 Square Root Of Fraction Exercise 4 Some Problems On Square Root Of Vulgar Fractions

Example 1. Find the square root of \(\frac{324}{625}\)

Solution:

Square root of

∴ \(\frac{18}{25}\)

WBBSE Class 7 Chapter 4 Square Roots Guide

Example 2. The area of a square is \(\frac{1089}{825}\) sq. cm.What is 825 the length of one side of the square?

Solution:

Length of the side of the square

∴ Length of one side of the square is \(\frac{33}{25}\) cm.

Example 3. A vulgar fraction multiplied by itself gives the product \(\frac{841}{2025}\) Find the fraction.

Solution:

Here the required vulgar fraction is the Square root of \(\frac{841}{2025}\)

∴ \(\frac{29}{45}\)

Example 4. Find the square root of 6 \(\frac{433}{676}\)

Solution:

 

∴ The required square root = 2 \(\frac{15}{26}\)

Example 5. Extract the square root of \(\frac{2}{5}\) to two places of decimals.

Solution:

∴ 0.63

Understanding Square Roots of Fractions for Class 7

Example 6. By what should the square root of \(\frac{121}{169}\) be multiplied so that the product will be 1?

Solution: Square root of

Now, \(\frac{11}{13}\) should be multiplied by \(\frac{13}{11}\) so that the product will be 1.

∴ To be multiplied by \(\frac{13}{11}\)

Example 7. The product of two numbers is \(\frac{75}{121}\) and one of them is threice the other. Find the  numbers

Solution:

Greater number x smaller number = \(\frac{75}{121}\)

or, 3 x smaller x smaller number = \(\frac{75}{121}\)

or,(smaller number)² = \(\frac{75}{3 x 121}\) = \(\frac{25}{121}\)

or, smaller number =

∴ greater number = 2 x \(\frac{7}{8}\) = \(\frac{7}{4}\)

∴ The two numbers are \(\frac{7}{4}\) and \(\frac{7}{8}\)

Example 9. Find the smallest whole number by which \(\frac{64}{125}\) should be multiplied so that the product will be a fraction which is a perfect square.

Solution:

\(\frac{64}{125}\) = \(\frac{8 x 8}{5 x 5 x 5}\)

Clearly, it should be multiplied at least by 5 so that the product will be a fraction which is a perfect square.

∴ 5

Example 10. Find a fraction, which when multiplied by itself gives 6 \(\frac{145}{256}\)

Solution:

6 \(\frac{145}{256}\) = \(\frac{1681}{256}\)

The required fraction will be the sqaure root of \(\frac{1681}{256}\)

 

∴ The required fraction = \(\frac{41}{61}\)

Class 7 Maths Exercise 4 Solved Examples

Example 11. Find the square root of

Solution:

= \(\frac{9}{25}\) + \(\frac{16}{25}\) = \(\frac{9 + 16}{25}\) = \(\frac{25}{25}\) = 1

the square root of 1 is 1

Question 12. By which fraction should \(\frac{49}{91}\) be multiplied so that the square root of the product is 1?

Solution:

Square root of the product will be 1. Therefore, product will also be 1.

\(\frac{49}{91}\) should be multiplied by \(\frac{91}{49}\) sp, that the product is 1.

∴To be multiplied by \(\frac{91}{49}\)

Example 13.  =what?

Solution:

∴ 2

Example 14. By which fraction should \(\frac{35}{42}\) be multiplied so that the square root of the product is 2?

Solution:

\(\frac{35}{42}\) = \(\frac{5}{6}\)

If the square root of the product is 2 then the product is 4.

∴ The required number = 4 ÷ \(\frac{5}{6}\)

= 4 x \(\frac{5}{6}\) = \(\frac{24}{5}\) = 4 \(\frac{4}{5}\)

∴ To multiplied by 4 \(\frac{4}{5}\)

Step-by-Step Solutions for Class 7 Square Root Problems

Example 15. By what number should the square root of \(\frac{625}{144}\) be multiplied nso that the product will be 1?

Solution:

Square root of

Now, the number by which \(\frac{25}{12}\) should be multiplied so that the product will be 1 is \(\frac{12}{25}\)

∴ \(\frac{12}{25}\)

Example 16. Find the least positive integer by which \(\frac{9}{50}\) should be multiplied so that the product will be a perfect square.

Answer:

\(\frac{9}{50}\) = \(\frac{3 X 3}{5 X 5 X 2}\)

Therefore, it should be multiplied at least by 2, so that the product will be a perfect sqaure.

∴ To be multiplied by 2.

Example 17. One number is thrice the other, If the product of the two numbers be 15\(\frac{3}{16}\) then what are the numbers?

Solution:

The greater number x smaller number = 15 \(\frac{3}{16}\)

or, 3x smaller number x smaller number= \(\frac{243}{16}\)

or, (smaller number)² = \(\frac{243}{16 X 3}\) = \(\frac{81}{16}\)

or, smaller number =

∴ greater number = 3 x \(\frac{9}{4}\) = \(\frac{27}{4}\) = 6 \(\frac{3}{4}\)

∴ The numbers are 2 \(\frac{1}{4}\) and 6 \(\frac{3}{}\)

Example 18. The product of two positive numbers is \(\frac{14}{15}\) and their quotient is \(\frac{35}{24}\). Find the numbers.

Solution:

Let the two numbers be x and y.

∴ The two numbers are \(\frac{7}{6}\) and \(\frac{4}{5}\)

Example 19. Out of the three numbers, the product of the first and the second number is 3, that of the second and the third number is 8 \(\frac{2}{5}\) and that of the third and the first number is 4 \(\frac{3}{8}\). Find the three numbers.

Solution:

First number x second number=3…………..(1)

Second number x third number = \(\frac{42}{5}\) …..(2)

Third number x first number = \(\frac{35}{8}\) ……(3)

By (1) x (2) x (3) we get,

(First number x second numnber x third number)²

= 3 x \(\frac{42}{5}\) x \(\frac{35}{8}\) = \(\frac{21 X 21}{4}\)

∴ First number x second number x third number =   = \(\frac{21}{2}\)

By (4) ÷ (1) we get, the third number

= \(\frac{21}{2}\) x \(\frac{1}{3}\) = \(\frac{7}{2}\) = 3 \(\frac{1}{2}\)

By (4) ÷ (2) we get, first number

= \(\frac{21}{2}\) x \(\frac{5}{42}\) = \(\frac{5}{4}\) = 1 \(\frac{1}{4}\)

By (4) ÷ (3) we get, second number

= \(\frac{21}{2}\) x \(\frac{8}{35}\) = \(\frac{12}{5}\) = 2 \(\frac{2}{5}\)

∴ The numbers are 1 \(\frac{1}{4}\), 2 \(\frac{2}{5}\), 3 \(\frac{1}{2}\)

Example 20. Find by what magnitude is (√25+ √81) more than (√16 + √36).

Solution:

√16+ √36=4+6=10

√25+√815+9=14

Now, 14-10=4

∴  Is more by 4.

Example 21. The area of a square is 4225 sq metres. If it is divided into 9 equal parts then each part becomes a square. What is the length of each side of such a square?

Solution:

If a square of area 4225 sq metres is divided into 9 equal parts then area of each part becomes \(\frac{4225}{9}\) sq metres.

If these parts are squares, then length of each side of such a square= meters =

= \(\frac{65}{3}\) metres = 21 \(\frac{2}{3}\)

∴ 21 \(\frac{2}{3}\)

Example 22. Find the value of

Solution:

= \(\frac{1}{2}\) + \(\frac{1}{3}\) – \(\frac{1}{4}\) – \(\frac{1}{5}\)

⇒ 2 x 3 x 10 = 60

= (\(\frac{1}{2}\) + \(\frac{1}{3}\)) – (\(\frac{1}{4}\) + \(\frac{1}{5}\))

= \(\frac{5}{6}\) – \(\frac{9}{20}\) = \(\frac{50- 27}{60}\) = \(\frac{23}{60}\)

∴ \(\frac{23}{60}\)

Example 23. Area of the square ‘A’ is \(\frac{225}{289}\)th of the area of the square ‘B’. What fraction of the side of ‘B’ is the side of ‘A’?

Solution:

Area of the square ‘A’ = Area of the square ‘B’ x \(\frac{225}{289}\)

∴ Length of the side of ‘A’ = Length of the side of ‘B’ x

∴ Length of the side of ‘A’ = Length of the side of ‘B’ x \(\frac{15}{17}\)

Hence, length of the side of the square ‘A’ is \(\frac{15}{17}\) th part of the length of the side of the square ‘B’.

∴ \(\frac{15}{17}\)
Example 24. Arrange the following in the descending order of their magnitude:

Solution:

∴ In descending order of magnitude the numbers are:

Example 25. The areas of two rectangles are \(\frac{24}{25}\) sq metres and \(\frac{12}{25}\) sq metres respectively, Find the length of each side of a square whose area is equal to the sum of the areas of these two rectangles.

Solution:

Given:

The areas of two rectangles are \(\frac{24}{25}\) sq metres and \(\frac{12}{25}\) sq metres respectively

The sum of the areas of the two rectangles = (\(\frac{24}{25}\) + \(\frac{12}{25}\)) sq metres = \(\frac{36}{25}\) sq meters.

Hence, the area of the square = \(\frac{35}{25}\) sq metres.

Hence, the length of each side of the square

= metres

= \(\frac{6}{5}\) metres = 1 \(\frac{1}{5}\) metres.

∴ 1 \(\frac{1}{5}\) metres.

Example 26. 308 square stones are required to pave a square of area 1925  square metres. Find the length of each side of such a stone.

Solution:

Given:

308 square stones are required to pave a square of area 1925  square metres.

By 308 stones 1925 sq metres is paved

By 1 stones \(\frac{1925}{308}\) sq metres is paved

= \(\frac{25}{4}\)  sq metres is paved

Therefore, area ofeach stone = \(\frac{25}{4}\) sq metres.

∴ Length of each side of a stone = metres =\(\frac{5}{2}\) metres = 2 \(\frac{1}{2}\) metres.

∴ 2 \(\frac{1}{2}\) metres.

Example 27. The sides of two squares are \(\frac{12}{5}\) metres. metres and \(\frac{9}{5}\) metres.respectively. Find the side of a a square whose area is equal to the sum of the areas of the two squares.

Solution:

Given:

The sides of two squares are \(\frac{12}{5}\) metres. metres and \(\frac{9}{5}\) metres.respectively.

The area of the first square = (\(\frac{1}{2}\))² sq metres

= \(\frac{144}{25}\) = sq metres

∴ The area of the second square = (\(\frac{9}{5}\))²

= \(\frac{81}{25}\) sq metres.

∴ Area of the new square

=(\(\frac{144}{25}\) + \(\frac{81}{25}\))

= \(\frac{225}{25}\) sq metres = 9 sq metres.

∴ Each side of the new square=√9 metres = 3 metres

∴ 3 metres.

The side of a a square whose area is equal to the sum of the areas of the two squares is 3 metres.

Arithmetic Chapter 3 Approximate Value Exercise 3 Solved Example Problems

Approximate Value

In our daily life, in many cases, it is not possible to calculate the actual value of a quantity or the actual price or weight of a body accurately.

Therefore, in such cases for practical purpose, a value is taken nearest to the actual value, actual price, or actual weight.

It is called the approximate value of that quantity or the approximate price or weight of that body.

Suppose, ₹262.50 is to be divided among 50 boys. Then each boy will get ₹262.50 +50 =₹5.25.

Now observe that, if each boy is given ₹5 then each boy gets 25 paise less than what he would have obtained.

Again, if each boy is given ₹6 then each boy gets 75 paise more than his due.

Therefore, neither ₹5 nor ₹6 is due for that boy. Hence, this less or more is to be termed an error.

Read and Learn More WBBSE Solutions For Class 7 Maths

Since there is an error in both cases, therefore, the less is the amount of error more is the degree of purity.

In the above case, if ₹5 is given then an error is 25 paise and if ₹6 is given the error is 75 paise. Hence, if each is given 5 then he will be given the nearest money due for him.

Again, let ₹287.50 is to be divided among 50 boys equally. Then each boy will get ₹287.50÷ 50= ₹5.75.

Therefore, in this case, if each boy is given 6 then he will be given the nearest money due for him.

WBBSE Class 7 Approximate Value Notes

Rule: In order to calculate the approximate value instead of the actual value of any quantity, if the digit after the decimal be less than 5 (₹ 5 and paise 25 = ₹ 5.25) then it is to be ignored and if it be 5 or greater than 5 (₹ 5 and paise 75 =₹ 5.75) then 1 is to be added with the digit in the units’ place.

Approximate value of the integer

If the value of the number 5832 is written as 5000 then 832 less than the actual value is written. Again if 6000 is written instead of 5832 then, (60005832) or 168 more than the actual value is written.

Now observe that between 6000 and 5000 the former is nearer to the actual value of 5832.

Therefore, the value of 5832 in the approximate thousands of places is 6000. Similarly, the value of 5832 in the approximate hundreds’ place is 5800, and in the approximate tens’ place is 5830.

Approximate value of decimal fraction

In order to find the approximate value of any decimal fraction up to a definite decimal place, write the digits up to that decimal place and reject the next remaining digits.

But if the first digit from the left-hand side of the rejected digits be 5 or more than 5 then 1 is to be added with the last digit of that particular place.

For example, the approximate value of 3.65038 up to integer = 4. (Since, the number 6 at the left end of the rejected portion after 3 is greater than 5 therefore, 4 is the result after adding 1 with 3).

Its approximate value up to 1 decimal place =3.7 (Here, since the first digit of the rejected portion is 5 therefore, 1 is added with the previous digit.

Its approximate value up to 2 decimal places = 3.65.
(Here, since the first digit of the rejected portion is 0 therefore, nothing has been added to the previous digit.)

Its approximate value up to 3 decimal places = 3.650.

Its approximate value up to 4 decimal places = 3.6504.

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

Examples:

1. Find the correct value of 0.655172 up to 4 decimal places.

Solution: 0.6552 (first digit of the rejected portion is 7)

2. Express 3628753 in approximate hundred.

Solution: 3628800

Class 7 Maths Exercise 3 Solved Examples

3. Find the correct values of 1.218713 upto 3 and 4 decimal places and express the approximate value of their difference corrected to 2 decimal places.

Solution:

The subtracted value, when corrected upto 2 decimal places, is 0.00

Significant digit

The numbers are formed by the digits 1 to 9. They are called significant digits. If between two significant digits, there is/are one or more than one zeroes (0) then those zeroes are also taken as significant digits.

If in any decimal fraction, there is a decimal point in the beginning and there are some zeroes after the decimal point then digits other than zeroes are called significant digits. The last zero or zeroes of an integer or decimal fraction are sometimes significant digits and sometimes they are not so.

Example:

1. The approximate value of 30.23046 up to 4 significant digits 30.23. But its approximate value corrects up to 4 decimal places = 30.2305.

2. In the case of the number 0.12065 the value is correct up to 4 decimal places and also the approximate value up to 4 significant digits are both equal to 0.1207.

3. In the case of the number 0.00147, the value corrects up to one significant digit = 0.001 and the value corrects up to two significant digits= 0.0015.

4. The approximate value of 0.000240079 up to five decimal places= is 0.00024. But its approximate value up to 5 significant digits= 0.00024008.

5. In the case of the number 0.44598 the correct value up to 4 decimal places and the approximate value up to 4 significant digits are both equal to 0.4460 (Here 0 is the significant digit; here since the rejected digit is 8, therefore, adding 1 with the last digit 9 of 0.4459 we obtained 0.4460.)

6. The approximate value of 163289 up to 4 significant digits is 163300. (Here, the two zeroes are not significant digits).

7. In case of 705769, the approximate value up to a thousand and the approximate value up to a hundred are 706000 and 705800 respectively. (In both cases the last zeroes are not significant digits).

Error

If a value other than the actual value is taken then there occurs an error. In practical case this error should be as small as possible. Let us roughly take the weight of 100 kg of coal.

If the weight is taken more accurately then it may be found that the weight of coal is 99 \(\frac{1}{2}\) kg. In this \(\frac{1}{2}\) kg of coal, the loss is not significant.

But if it is gold in and that would not be negligible. Therefore, in lieu of coal then there would be a great loss if a goldsmith is compelled to measure gold more accurately.

Arithmetic Chapter 3 Approximate Value Exercise 3 Absolute Error, Relative Error, And Percentage Error

The difference between the actual value and the approximate value is called the absolute error of the approximate value.

That means, absolute error = Actual value ∼ approximate value.

The ratio of absolute error and the actual value is called the relative error.

That means, relative error = \(\frac{Absolute error}{Actual value}\).

The percentage error of the approximate value is that percentage of the actual value which the absolute error is.

That means, percentage error = \(\frac{Absolute error}{Actual value} \)

Example 1. Express 4325 kg in approximate hundred kg and find an absolute error, relative error, and percentage error.

Solution:

Given:

4325 kg

4325 kg = 4300 kg (in approximately 100 kg)

∴ absolute error = 4325 kg-4300 kg = 25 kg 25 kg

relative error = \(\frac{25 \mathrm{~kg}}{4325 \mathrm{~kg}}=\frac{1}{173}\)

and percentage error = relative error x 100 = \(\frac{1}{173}\) x100= 0.58 (correct up to 2 decimal places).

WBBSE Class 7 Arithmetic Approximate Values Examples

Example 2. Find the approximate value up to first two significant digits and the absolute error of 3.068.

Solution:

Given:

3.068

The approximate value of 3.068. The required sum = 13.67489.
up to first two significant digits = 3.1. absolute error = 3.068-3.1 = 0.032.

Example 3. A man measured a path of length 12 \(\frac{7}{200}\) km and told its length as 12.01 km. At this find his absolute error, relative error, and percentage error.

Solution:

Given:

A man measured a path of length 12 \(\frac{7}{200}\) km and told its length as 12.01 km.

12 \(\frac{7}{200}\) = 12.035

Absolute error = 12.035 km -12.01 km = 0.025 km

Relative error = \(\frac{0.025 \mathrm{~km}}{12035 \mathrm{~km}}=\frac{5}{2407}\)

and percentage error = \(\frac{5}{2407}\) x 100 = 0.21%

(correct up to 2 decimal places).

Arithmetic Chapter 3 Approximate Value Exercise 3 To Find The Addition And Subtraction In Approximate Value Of Decimal

Rule: When the result of addition and subtraction are to be known approximately up to a definite decimal place, it is necessary to know the next digit.

Therefore, we have to find the result of addition and subtraction up to two more digits than the digits up to which approximation is required to be done. Then the answer is to be given in approximate value as per requirement.

Example 1. What is the sum of 4.3074, 0.0028391, and 9.364 approximate up to five decimal places?

Solution:

Given:

4.3074, 0.0028391, And 9.364

Here, to get the approximate 5 decimal place the 6th decimal place must be known correctly.

So, writing up to one more decimal place i.e., 7th decimal place the sum is to be determined.

∴ The required sum = 13.67489

Solved Problems for Class 7 Approximate Values

Example 2. Find the difference of 30.406 and 23.139 approximately up to 4 decimal places.

Solution:

Given:

30.406 And 23.139

∴ The required difference of 7.2670

Example 3. Find the sum of 7.302, 0.0865, 32.87, and 3.02 approximately up to 3 significant digits.

Solution:

Given:

7.302, 0.0865, 32.87, And 3.02

∴The required sum = 43.3.

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2

Solved Example Problems

Integers are all natural numbers including Zero (0) i.e., all whole numbers with a positive or negative sign.

For example 0, 3, (-3), 5, (-8), etc., are all integers.

Addition and Subtraction of the integers

You have already learned the number line and the representation of the numbers in the number line to attain the value of an expression in class 6.

Let us find the value of the expression 0+(+2)+( + 5)+(- 10)+(-6).

WBBSE Class 7 Math Solutions

Read and Learn More WBBSE Solutions For Class 7 Maths

 

 

The value of the expression 0 + 2 + 5-10-6 = (-9).

Step 1 >	(+ 0) + (+ 2) = + 2.
Step 2 >	(+ 2 ) + (+ 5) = + 7	(adding both integers with the same signs gives the result with that sign).
Step 3 >	(+ 7) + (- 10) = 7- 10 =-3	(By adding the integers with different signs the result will be the difference between the numbers with the sign of the greater number.)
Step 4 >	(-3) + (-6) =  (-9)	(adding both integers with the same signs gives the result with that sign).

 

Let us find the value of {( + 3) – (- 3)} – ( – 8) from the number line.

{(+ 3) – (- 3)} – (- 8) = {(+ 3) + ( 3)} + ( + 8)= ▭

 

 

So, we get {( + 3) – (- 3)} – (- 8) = + 14.

Commutative Property

Class 7 Algebra Integer Problems

We know 3 + 5 = 5 + 3 = 8. This infers that whole numbers can be added in any order.

So, the addition of whole numbers is commutative.

But let us see what happens in the case of integers

We know 4 + (- 6)= – 2 and (-6) + 4 = -2.

So, 4 + (-6) = (-6) + 4.

So, it is known that addition is commutative for integers, a + b = b + a

But subtraction is not commutative for whole numbers as well as integers.

Let us consider the integers 6 and ( – 3 ).

Example 1. 6-(- 3) = 6 + 3 = 9.

Example 2. – 3 – 6 = – 9.

So, we can conclude that subtraction is not commutative for integers.

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

Example 1. By selling each kg of mango, a fruit seller gains ₹ 5 per kg but loses ₹4 per kg of lichi. If he sells 10 kg mango and 14 kg of lichi, find his overall profit or loss.

Solution:

Given:

By selling each kg of mango, a fruit seller gains ₹ 5 per kg but loses ₹4 per kg of lichi. If he sells 10 kg mango and 14 kg of lichi,

On selling 1 kg mango, he earns ₹ 5

So, profit in 10 kg of mango = ₹ 5 x 10 = ₹50

Again, on selling 1 kg of lichi, the fruit seller losses ₹ 4, in other words, he earns – ₹ 4.

So his earning on selling 14 kg lichi = – ₹ 4 x 14 = – ₹ 56

So, his overall profit = ₹ 50 + ₹ (- 56) = – ₹ 6, that means he losses ₹ 6 after the sale.

Example 2. In an examination, 6 marks are given for correct answers, and – 3 marks are given for incorrect answers. If Rabin makes 7 correct answers and 5 incorrect answers, then what is his score?

Solution :

Given:

In an examination, 6 marks are given for correct answers, and – 3 marks are given for incorrect answers. If Rabin makes 7 correct answers and 5 incorrect answers

For 7 correct answers Rabin gets = 7 x 6 = 42 marks

For 5 incorrect answers, Rabin gets = 5 x (- 3) = – 15 marks.

So, his total marks = 42 + (-15) = 27 marks.

His total Score  = 27 marks.

Step-by-Step Solutions for Class 7 Integers

Example 3. In a quiz competition, Manika’s scores in five successive rounds were 25, – 2, – 10,15 and 10 with negative markings for wrong answers. Find her total score at the end of 5th round.

Solution:

Given:

In a quiz competition, Manika’s scores in five successive rounds were 25, – 2, – 10,15 and 10 with negative markings for wrong answers.

Her total score = 25 + (- 2) + ( – 10) + 15 + 10

= 25-2-10+15+10

= 50 – 12 = 38.

Her total score at the end of 5th round = 38

 

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Associative Properties

⇔ Consider the integers + 3,-5, and – 7.

Example 1. Let us find out the value of the expression by number line {(+ 3) + (- 5)} + (- 7).

Solution:

 

Example 2. If the expression be (+ 3)+{(-5)+(-7)} then find the value.

Solution:

 

We get, {( + 3) + (- 5)} + ( – 7) = ( + 3) + {(- 5) + (- 7).

We find that addition is associative for integers.

In general, for any integers a, b, and c, we can say, (a + b) + c = a + (b + c)

 

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Multiplication Of Integers

We know that the multiplication of whole numbers is repeated addition.

Example: 4 x 2 = 2 + 24-2 + 2 = 8.

Example 1. Let us find the value of 3 x 2 (both the integers are +ve).

Solution:

 

We get from number line 3×2 = ( + 2) + ( + 2) + ( + 2) = + 6.

Example 2. Now, find the value of 2 x 3.

Solution:

 

We get from number line =2×3=(+3)+(+3)=+6

∴ We find that in both cases the value is same.

So, 3 x 2 = 2 x 3 or a x b = b x a

WBBSE Class 7 Maths Algebra Guide

Example 3. Now, let us find out the product value of 3 x (- 4) (One integer is +ve and the other is -ve).

Solution:

 

We get from number line 3x(-4)-(-4)+(- 4)+(-4)-12-(3×4)

Again, in case of 2x(- 4) = (-41) + (- 4) = (- 8) = (- 2 ) x 4

= (- 2) + (- 2)+ (-2 ) + (-2)

So, a x (-6) = (-a) x b = – (ab)

We thus find that while multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus (-) sign before the product. We thus get a negative integer.

So, we learn that when signs of integers are same, the answer is a +ve number.

( + )x( + )or(-)x(-) are always +ve.

When the signs of the integers are different, the answer is always – ve (+) x (-) or (-) x ( + ) are always – ve.

WBBSE Class 7 Integer Exercises

Multiplication of negative integers:

Now let us see the product of two negative integers

(-3) x (-2) = + 6

(-4) x (-5)= + 20

(- 4) x (- 3) = + 12.

So, after observing these products we understand that there is a rule prevailing over states that the product of two negative integers is always a positive integer.

So, (-a) x (-b) = a x b

Now, let us see the product of three or more negative integers:

1. (-5) x (-3) x (-2) = [-5 x -3] x (-2) = 15 x (-2) =-30.

2.( -3)x( -5)x( -2)x(-4) = [ -3 x -5] x [-2 x -4] = (+15) x ( + 8) = +120.

From the above results we observe that:

1. Product of two negative integers is a positive integer.
2. Product of three negative integers is a negative integer.
3. Product of four negative integers is a positive integer.

So, we can conclude that if the number of negative integers in a product is even then the product is always a positive integer. If the number of negative integers in a product is odd, then the product is always a negative integer.

Integer Operations for Class 7 Students

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Commutative Property Of Multiplication

Now we observe the following results

Pattern -1	Pattern – 2	Conclusion
3x(-4) =-12, (-15) x 10 =-150, (-4)x(-5) = 20	(-4)x3= -12, 10 x (-15)=-150, (-5)x(-4)=20	3x(-4) = (-4 )x3, (-15)x10 = 10 x (-15), (-4)x(-5) = (-5)x(-4)

 

All the cases shown above confirm that multiplication is commutative in the case of integers.

So, a x b = b x a (where a & b are integers)

Distributive property of multiplication: We know that, 15 x (8 + 2) =(15 x 8) + (15 x 2) = 150

Now, let us see the cases for negative integers also: (- 3) x (2 + 5) = – 3 x 7 = – 21.

Also, [(-3)x 2] + [(-3) x5] =-6+-15=-21.

So, we can conclude that

a x (b + c) = a x b + a x c

We know, 5(9 – 3) = 5 x 6 = 30

Again, ( 5 x 9) – (5 x 3) = 45 – 15 = 30.

So, we can conclude that if a, b, c are there integers then we can write, a x ( b – c ) = a x b – a x c

Similarly,

(-5) x 3 = (- 15) > 1. (-15) + (- 5) = 3,2. (-15) + 3 = (-5)

(-3) x (-5) = 15 > 1. 15 ÷ (- 3) = (- 5), 2. 15 + (- 5) = (- 3)

Class 7 Maths Integer Addition and Subtraction

Rule for division:

1. When we divide a +ve integer by a -ve integer or a -ve integer by a + ve integer, division is done considering whole numbers but putting a minus sign ( – ) before the quotient. Hence, the result is negative.

\(\frac{-15}{3}\) = – \(\frac{15}{3}\) = -5

So, a ÷ (-b) = (-a) ÷ b = – \(\frac{a}{b}\)

2. When we divide a negative integer by a negative integer, we divide the integers considering whole numbers but putting a ( + ) positive sign before the quotient.

\(\frac{-15}{5}\) = + \(\frac{15}{5}\) = +3

So, (- a) + (- b) = a ÷ b = \(\frac{a}{b}\)

3. Now let us take two integers like 24, ( – 3). If we divide one by other, then let us see the result.

24 ÷ ( – 3) = \(\frac{24}{-3}\) = – \(\frac{24}{3}\) = -8 ….(1)

Again, (-3) + 24= \(\frac{-3}{24}\) = – \(\frac{1}{8}\) …..(2)

We find (1) & (2) are not same.

So, we can write, a ÷ b ≠ b ÷ a

4. Let us take three integers like – 30, – 5, and + 2. We now set the number in two ways.

 

Set 1	Set 2
(-30) ÷ {(-5) + 2} = (-30) ÷ (-3) = 10	(-30) ÷ (-5)+ (-30) ÷ 2 = 6 +(-15) = 6 – 15 = – 9

 

∴ (-30) {(-5) + 2} ≠ (-30) ÷ (-5) + (-30) ÷ 2

So, we can write, a ÷ ( b + c) ≠ a ÷ b + a ÷ c

5. We now take three integers – 30, + 2, and 7. Let us set the numbers in two ways.

Set 1	Set 2
{(- 30) + 2 } ÷ 7 = – 28 ÷ 7 = – 4	{(-30) + 2}÷ 7 = \(\frac{-30}{7}\) + \(\frac{2}{7}\) = \(\frac{-30+2}{7}\) = \(\frac{-28}{7}\) = -4

 

So, we can write, (b + c) ÷ a = b ÷ a + c ÷ 4

 

Algebra Chapter 2 Addition Subtraction Multiplication And Division Of Integers Exercise 2 Examples For Multiplication And Division

Example 1. Find the value of the following products:

1. (- 18) x (- 9) x 9
2. (- 20) x (- 2) x { – 5) x 6
3. (- 1) x (- 5) x ( – 4) x (- 6)

Solution:

1. (- 18) x (- 9) x 9 = [(- 18) x (-9)] x 9 = 162 x 9 = 1458.
2. (- 20) x(-2)x(-5)x6 = (- 20 x – 2) x (- 5) x 6 = 40 x (- 5) x 6 = – 200 x 6 . = -1200.
3. (- 1) x (_ 5) x (- 4) x (- 6) = [(- 1) x (- 5)] x [(- 4) x (- 6)] = 5×24 = 120.

Solved Examples for Class 7 Algebra

Example 2.  In a test containing 12 questions, 4 marks are given for every correct answer, and ( – 2) marks are given for every incorrect answer. Raju attempted all questions but only 8 of his answers were correct. Find his total score.

Solution:

Given:

In a test containing 12 questions, 4 marks are given for every correct answer, and ( – 2) marks are given for every incorrect answer. Raju attempted all questions but only 8 of his answers were correct.

Marks are given for each correct answer = 4

So, marks for 8 correct answers = 8 x 4 = 32

Marks given for one incorrect answer = – 2

So, marks for (12 – 8) = 4 incorrect answers = (-2) x 4 = -8.

Therefore, Raju’s total score = 32 + (- 8) = 32-8 =24.

Class 7 Maths Exercise 2 Solutions

Example 3. A shopkeeper earns a profit of ₹2 by sealing one pen and incurs a loss of ₹ 1 per pencil. In a particular month, he incurs a loss of ₹ 15, During this period, he sold 45 pens. How many pencils did he sell in this period?

Solution:

Given:

A shopkeeper earns a profit of ₹2 by sealing one pen and incurs a loss of ₹ 1 per pencil. In a particular month, he incurs a loss of ₹ 15, During this period, he sold 45 pens.

Profit earned by selling one pen = ₹ 2.

Profit earned by selling 45 pens = ₹2 x 45 = ₹ 90.

Loss incurred =₹ 15, which is denoted by – ₹ 15.

We know profit earned + loss incurred = Total loss

Therefore, Loss incurred =  Total loss – Profit earned = -₹ 15 – ₹90 = – ₹ 105.

Loss incurred by selling one pencil = ₹ 1.

So, number of pencils sold = – ₹ 105 ÷ – ₹1 = 105.

Number of pencils sold in this period = 105.

 

Algebra Chapter 1 Some Problems On Symbols

Symbols Introduction:

In the previous class, you have known the contribution of famous mathematicians in the study of algebra.

Here we shall not repeat it. In this chapter, our aim is to discuss briefly the algebraic system of numbers, directed numbers, the use of algebraic symbols and four basic operations which you have learnt in the previous class.

Wbbse Class 7 Maths Solutions

Algebraic system of numbers and directed numbers

Read and Learn More WBBSE Solutions For Class 7 Maths

In arithmetic, we form all the numbers with the help of the ten digits 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0. But we cannot write “any number” by these digits. Also, the numbers considered in arithmetic are positive numbers.

In arithmetic + sign denotes addition and – sign denotes subtraction. In algebra, these two signs are used in much wider sense.

In arithmetic when we subtract 5 from 10 then we say 10 – 5 = 5.

But when we are asked to subtract 15 from 10 then we say that 10 – 15 cannot be calculated as a greater number cannot be subtracted from a smaller number.

Thus, 10-15 becomes meaningless. But in algebra 10 – 15 = – 5. The consideration of negative numbers in algebra helps us to overcome many difficulties.

Also to express some quantities of same kind but of opposite nature, we use positive and negative numbers.

Thus to express profit and loss, increase and decrease rise and fall, and income and expenditure we take one as + quantity and the other as -quantity.

In arithmetic, there is no place of negative numbers. The numbers +1, +2, +3…. and -1, -2, -3…. considered in algebra are called directed numbers since the idea of direction is associated with these numbers.

Here, the signs + and – are called the signs of affection.

WBBSE Class 7 Algebra Symbols Notes 

Use of algebraic symbols

To express ‘any number’ we use the letters of the alphabet as symbols. Thus we use the letters x, y, z etc., as symbols to represent numbers.

When such alphabetic symbols are used instead of numbers then those symbols express numbers— although their values are not fixed.

For example, x + 5 represents a number but its value is not fixed as the value of x is not fixed.

We may express all the 90 numbers of two digits with the help of a single symbol 10b + a in which a represents the digit in the units’ place and b represents the digit in the tens’ place.

In arithmetic, we find the relation, Dividend = Divisor x Quotient + Remainder. This relation cannot be expressed, in general, by the ten digits of arithmetic.

But if we assume that, if the number a is divided by b then the quotient is c and the remainder is d, then we may write, a = b c + d in this relation a, b, c, d may represent any number.

WBBSE Class 7 Geography Notes	WBBSE Solutions For Class 7 History
WBBSE Solutions For Class 7 Geography	WBBSE Class 7 History Multiple Choice Questions
WBBSE Class 7 Geography Multiple Choice Questions	WBBSE Solutions For Class 7 Maths

 

Four basic operations

Addition, Subtraction, Multiplication and Division are the four basic operations.

We use these operations in algebra in the same way as we use them in arithmetic.

For example:

1. Sum of a and b is represented by a + b.
2. Result of subtracting b from a is represented by a – b.
3. Product of a and b is represented by a x b.
4. Result of dividing a by b is represented by \(\frac{a}{b}\).

Opposite numbers: If x is a natural number then + x and – x are opposite to each other.

Example: -7 is the opposite number of +7.

Absolute value: The value of a positive or negative number without its sign is called its absolute value. For example, the absolute value of both -10 and +10 is 10.

Addition:

1. To add two positive numbers, the absolute values of the numbers are added and the sign of the sum is +.

For Example:

(+6) + (+4) = + 10

(+15) +(+6) = + 21.

2. To add two negative numbers, the absolute values of the numbers are added and the sign of the sum is -.

For Example:

(-7) + (- 5) = -12

(-12) + (- 8) = -20.

3. To add one positive and one negative number the smaller absolute value is subtracted from the greater absolute value and the sign of the result is that of the greater number.

For Example:

(+8) + (-3) = +5

(-12) + (+4) – -8.

Subtraction:

Important Definitions Related to Algebraic Symbols

Subtracting the integral number y from the integral number x means adding the opposite of the integral number y to the integral number x. For example, to subtract -5 from +7 we have to add +5 with +7.

So, (+7) – (-5) = (+7) + (+5) = (+12).

Similarly, (+8) – (+3) = (+8) + (-3) = (+5).

Multiplication:

In order to find the product of two numbers we have to multiply their absolute values and the sign of the product will be “+’ if the given two numbers are of the same sign and the sign of the product will be —’ if the given two numbers are of the opposite sign.

Example: (+5)x(+4) = + 20
(- 4) x (+ 7) = – 28
(+ 3) x (- 8) = – 24
(-7) x (- 2) = + 14

Division:

In order to find the quotient of two numbers we have to divide their absolute values and the sign of the quotient will be + if the dividend and the divisor are of the same sign and the sign of the quotient will be – if they are of opposite sign.

Example:

(+30) ÷ (+ 5) = (+ 6)
(+ 28) ÷ (- 4) = (- 7)
(- 42) ÷ (+ 6) = (- 7)
(- 40) ÷ (- 5) = (+ 8)

Algebra Chapter 1 Revision Of Previous Lessons Exercise 1 Some Problems On Symbols

Example 1. Write in words:

1. x + y
2. \(\frac{x}{3}\)
3. 2x + 3
4. x ≥ 14
5.  x ≤ 8.

Solution:

1. The sum of x and y.
2. One-third of x.
3. 3 more than twice x.
4. The value of x is greater than or equal to 14.
5. The value of x is less than or equal to 8.

Example 2. Write with the help of a sign 

1. 7 less than p
2. 20 times q
3. 3 less than one-fourth of x
4. The value of x is not greater than 25
5. The value of y is not less than 30.

Solution:

1.  p – 7
2. 20 q
3. \(\frac{x}{4}\) – 3
4. x 25
5. y 30.

Conceptual Questions on Operations with Algebraic Symbols

Example 3.  If a = 2, b = 5 and c = 8 then write 582 by the alphabetic symbol.

Solution : 582 = 5 x 100 + 8 x 10 + 2 = b x 100 + c x 10 + a = 100b + 10c + a

∴ 1006 + 10c + a.

Example 4. Ram had ₹ x, and Ram’s father gave him ₹y more. How many rupees has Ram at present?

Solution: ∴(x + y).

Example 5. The product of two numbers is 30. If one of them be x then find the other.

Solution: \(\frac{30}{x}\)

Example 6. 20 years ago, a man’s age was (x- 10) years. What will be his age after 20 years?

Solution: 20 years ago, the man’s age was (x -10) years.

So at present his age = (x – 10 + 20) years = (x + 10) years.

∴ After 20 years his age will be (x + 10 + 20) years = (x + 30) years

∴ (x + 30) years.

Real-Life Scenarios Involving Algebraic Problem Solving

Example 7. Ram is x years older than Shyam, and y years younger than Jadu. If Jadu is 2 years old then find the ages of Ram and Shyam.

Class Vii Math Solution Wbbse
Solution: Jadu is z years old.

Ram is y years younger than Jadu.

So Ram’s age = (z – y) years.

Again, Shyam is x years younger than Ram.

So, Shyam’s age = (z-y-x) years

∴ Ram’s age is (z-y) years and Shyam’s age is (z – y – x) years.

Example 8. Hari’s age will be b years, after a year. What was his age c years ago?

Solution:

Given:

Hari’s age will be b years, after a year.

At present, Hari’s age is (b – a) years, c years ago his age was (b – a-c) years

∴ (b- a-c) years.

Example 9. The number of members of a club is x. If each member contributes? ₹y, what will be the total contribution? If z number of footballs are bought by that money, find the price of each football.

Solution:

Given:

The number of members of a club is x. If each member contributes? ₹y,

1 member contributes ₹y

x members contribute ₹ xy

The price of z footballs is ₹ xy

The price of 1 football is ₹ \(\frac{xy}{z}\)

∴ ₹ \(\frac{xy}{z}\), ₹ xy

Class Vii Math Solution Wbbse

Example 10. The product of two numbers is p; if their H. C. F. be m then what will be their L. C. M.?

Solution:

Given:

The product of two numbers is p; if their H. C. F. be m

L. C. M. X H. C. F. = Product of the numbers

or, L. C. M. X m =p

or, L, C. M. = \(\frac{p}{m}\)

∴ \(\frac{p}{m}\)

Concept of Index

In case of the number xⁿ, we say that x is the main number and n is power or index.

Three basic rules of the index are:

1. x^m X xⁿ= x^m+n,

2. x^m ÷ xⁿ= x^m-n,

3. (x^m)ⁿ = x^mn.

Examples of Real-Life Applications of Algebraic Symbols

Example: x^mn x x⁴ = x⁸⁺⁴ = x¹²,

x⁸ ÷ x⁴ = x^8-4 =  x⁴ and (x⁸)⁴ = X^{8 x 4} = x³².

Also, x° = 1, because x° = x^n-n = xⁿ ÷ xⁿ= 1.

Any integer can also be expanded is terms of the index as follows:

915 = 9 x 100 + 10 x 1 + 5 = 9 x 10² + 10¹+ 5

Also, 81 = 3 x 3 x 3 x 3 = 3⁴

Here, 3 is the main number and 4 is the index.

Some problems on directed numbers:

Example 1. Arrange the following numbers in the ascending order of their values: 5,-3,-7, 0, 4,-13.

Solution: In the ascending order of their values the numbers are: -13, -7, -3, 0, 4, 5.

Example 2. Arrange the following numbers in the descending order of their values: -7,- 15,-8, 5, 0,-20, 25, 15.

Solution: In the descending order of their values, the numbers are: 25, 15, 5, 0, – 7, – 8, – 15, – 20.

Example 3. Write the opposite of the following expressions.

1. Expenditure of ₹ 30.
2. – 3 km towards north.
3. A capital of ₹ 50.
4. Rise of temperature by 42°C.
5. Descending – 20 metres.
6. Fall of temperature by – 20°C.

Solution:

1.  An income of ₹ 30.
2. 3 km towards the north.
3. A loan of ₹ 50.
4. Fall of temperature by 42°C.
5. Descending 20 metres.
6. Fall of temperature by 20°C.

Example 4. What do the following expressions mean?

1. A loan of  ₹ -10.
2. An income of ₹ -40.
3. Ascending -70 metres.
4. A profit of  ₹-100.
5. Fall of temperature by -5°C.
6. -15 km towards north.

Solution:

1. A capital of  ₹ 10.
2. An expenditure of ₹ 40.
3. Descending 70 metres.
4. A loss of  ₹ 100.
5. Rise of temperature by 5°C.
6. 15 km towards south.

Example 5. If +x represents an income of  ₹ 20. then which will represent an expenditure of  ₹ 100?

Solution:

Given:

+x represents an income of  ₹ 20.

An income of ₹ 20 is represented by + x

An income of ₹ 1 is represented by \(\frac{+x}{20}\)

An income of ₹ 100 is represented by \(\frac{+x}{20}\) X 100 = + 5x

∴ An expenditure of  ₹ 100 will be represented by – 5x.

Short Answers on Algebraic Expressions and Symbols

Example 6. If x = a profit of ₹ 7 and y = a loss of ₹ 15 then find what amount of profit or loss will be represented by 12x + 3y?

Solution:

Given:

x = a profit of ₹ 7 and y = a loss of ₹ 15

12x = 12 X a profit of ₹ 7 = a profit of ₹ 84

3y = 3 X a loss of ₹ 15 = a loss of ₹ 45 .

∴ 12x+3y = a profit of ₹ (84 – 45) = a profit of ₹ 39

Solution: A profit of ₹ 39.

Example 7.  If an expenditure of ₹ 50 is denoted by -5 \(\frac{1}{2}\) then what will be denoted by + 33?

Solution: – \(\frac{11}{2}\) denotes an expenditure of ₹ 50

-1 denotes an expenditure of ₹ 50 X \(\frac{2}{11}\)

∴ -33 denotes an expenditure of ₹ 50 X \(\frac{2}{11}\) X 33

+ 33 denotes an income of  ₹ 300

∴ An income of ₹ 300.

Common Questions on Mathematical Symbols in Algebra

Example 8. If a man earns ₹ 10 daily and his total expenses in 7 days be ₹ 60, what will be his total savings after 7 days?

Solution:

Given:

A man earns ₹ 10 daily and his total expenses in 7 days be ₹ 60,

In 1 day the man earns ₹ 10

In 7 days the man earns ₹ 70

Also in 7 days his total expenses is ₹ 60

∴ his savings after 7 days = ₹ (70 – 60) = ₹ 10

∴ ₹ 10.

Total savings after 7 days ₹ 10.

Example 9. Find the product :

1. x⁷ X x³
2. (- x⁵) X x³
3. x⁴  X (- x⁵)
4. (-x⁴) X (- x⁷)
5. x^m X xⁿ

Solution:

1. x⁷ X x³ = x^{7 + 3} = x¹⁰.
2. x⁴  X (- x⁵) = -x^{5 + 3} = -x⁸.
3. x⁴  X (- x⁵) = -x^{4 + 5} = -x⁹.
4. (-x⁴) X (- x⁷) = x⁴⁺⁷=x¹¹
5. x^m X xⁿ = x^m+n.

Example 10. Find the quotient :

1. x⁸ ÷ x³
2. (- x¹¹) ÷ x³
3. x¹² ÷ (- x⁴)
4.  (-x⁵)÷ (-x²)
5. x^m ÷ xⁿ.

Solution:

1. x⁸ ÷ x³ = x^{8 – 3} = x⁵.
2. (- x¹¹) + x³=-x^11-3 = -x⁸.
3. x¹² + (- x⁴) = -x^{12- 4} = – x⁸.
4. (-x⁵)+ (-x²) = x^5-2 – x³.
5. x^m + xⁿ= x^m-n.

Example 11. If a =-2,b = 4 and c = – 5 then find the value of a³ + b³ + c³.

Solution:

Given:

a =-2,b = 4 and c = – 5

a³ + b³ + c³ = (- 2)³ + (4)³ + (- 5)³

= (- 2) x (- 2) x (- 2) + 4 x 4 x 4 + (- 5) x (- 5) x (- 5)

= – 8 + 64 – 125 = 64 – (8 + 125)

= 64- 133

= -69

∴ -69.

a³ + b³ + c³ = -69

Example 12. If a = 2, b = – 3, c = 4 find the value of a²b³c⁴.

Solution :

Given:

a = 2, b = – 3, c = 4

a²b³c⁴ = (2)² x (- 3)³ x (4)⁴

= 2x2x(-3)x(-3)x(-3)x 4 x 4 x 4 x 4 = 4 x (- 27) x 256

= – 27648

∴ -27648.

a²b³c⁴= -27648.