## 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**

- If same number is added to both sides of an equation, its sides remain equal.
- If same number is subtracted from both sides of an equation, its sides remain equal.
- If both sides of an equation is multiplied by the same number, its sides remain equal.
- 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.

**Math Solution Of Class 7 Wbbse**

**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**

- At first, each side has to be simplified.
- 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.
- Now each side has to be simplified again.
- 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.

**Class Vii Math Solution Wbbse**

**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.

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

Solution: \(\frac{x}{a}\) + b = \(\frac{x}{b}\)+ a .

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

Solution:

**Given:**

∴ -3.

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

Solution:

**Given:**

\(\frac{1}{2}\) (x +1) + (x + 2) + \(\frac{1}{4}\) (x + 3) = 16.

∴ x= 13.

**Wbbse Class 7 Maths 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:

**Given:**

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

∴ x = -7

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

Solution: \(\frac{1}{(x+1)(2)}\) + \(\frac{1}{(x+3)(3)}\) = 4

∴x= 3.

**Wbbse Class 7 Maths Solutions**

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

Solution:

**Given:**

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 \(\frac{x}{2}\) + \(\frac{1}{3}\) = \(\frac{x}{3}\) + \(\frac{1}{2}\)**

Solution:

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

∴ x = 1

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

Solution:

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

Solution:

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 = at^{2} and y = 2at, find the relation between x and y.

Solution:

**Given:**

x = at^{2} and y = 2at

x = at^{2} or, at^{2} = x or, t^{2} =\(\frac{x}{a}\)…… (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 Maths Solutions**

**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.

**Math Solution Of Class 7 Wbbse**

**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 bank. He received some five rupee notes and some ten rupee notes. If he received altogether 300 notes

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

∴ According to the question,

∴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

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.

**Math Solution Of Class 7 Wbbse**

**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 + \(\frac{x}{2}\))

Hence, \(\frac{x}{2}\) + (y + \(\frac{x}{2}\)).5 = 210

** ∴ **Number of 1 rupee coins is 20 and that of 5 rupee coins is 30.

** ****Math Solution Of Class 7 Wbbse**

**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