WBBSE Solutions For Class 7 Maths Algebra Chapter 1 Some Problems On Symbols

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.

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Algebraic system of numbers and directed numbers

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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.

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.

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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:

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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.

Example 3.  If a = 2, b = 5 and c = 8 then write 582 by the alphabetic symbol.

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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.

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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.

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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

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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.

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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.

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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.

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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.