- What is algebraic expression
- |
- What is algebraic equation
- |
- What is Linear equation in one Variable
- |
- Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
- |
- Word Problem
- |
- Solving Equations having the Variable on both Sides
- |
- Reducing Equations to Simpler Form
- |
- Equations Reducible to the Linear Form

In this page we have *NCERT Solutions for Class 8 maths - Linear equation in One Variable Chapter 2 * for
Exercise 2.4 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

According to the given question,

8(x-5/2) = 3

8

Transposing 3

8

5

Dividing both sides by 5, we obtain

Hence, the number is 4.

21 + 5

21 + 5

Transposing 2

5

3

Dividing both sides by 3, we obtain

5

Hence, the numbers are 7 and 35 respectively.

Therefore, original number = 10

On interchanging the digits, the digits at ones place and tens place will be

Therefore, new number after interchanging the digits = 10(9 −

= 90 − 10

= 90 − 9

As per question,

New number = Original number + 27

90 − 9

90 − 9

Transposing 9

90 − 36 = 18

54 = 18

Dividing both sides by 18, we obtain

3 =

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

Therefore, the two-digit number is 36

Therefore, original number = 10

On interchanging the digits, the digits at ones place and tens place will be

Number after interchanging = 10 × 3

According to the given question,

Original number + New number = 88

13

44

Dividing both sides by 44, we obtain

Therefore, original number = 13

By considering the tens place and ones place as 3

Therefore, the two-digit number may be 26 or 62.

According to the given question,

(x+5)=6x/3

Transposing

5 = 2

5 =

6

Therefore, the present ages of Shobo and Shobo’s mother will be 5 years and 30 years respectively.

Perimeter of the plot = 2(Length + Breadth) It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.

So

100 × Perimeter = 75000

100 × 30

3000

Dividing both sides by 3000

Length = 11

Breadth = 4

Hence, the dimensions of the plot are 275 m and 100 m respectively.

Per metre selling price of trouser material =Original Price + Profit

=90 + 12% Profit

=90 + 12X90/100

=Rs 100.80

Per metre selling price of shirt material

=Original Price + Profit

=50 + 10% Profit

=50 + 10X50/100

=Rs 55

We know from question that total amount of selling = Rs 36660,So

100.80 × (2

201.60

366.60

Dividing both sides by 366.60

Trouser material = 2

Number of deer grazing in the field = x/2

So remaining would be x/2

Now Number of Deer playing near by= (3/4)(x/2)= 3x/8

Number of deer drinking water from the pond = 9

So

Multiplying both sides by 8, we obtain

Hence, the total number of deer in the herd is 72.

10

According to the question,

Grandfather’s age = Granddaughter’s age + 54 years

10

Transposing

10

9

Granddaughter’s age

Grandfather’s age = 10

Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

According to the question,

10 years ago, Aman’s age = 5 × Aman’s son’s age 10 years ago

3

3

Transposing 3

50 − 10 = 5

40 = 2

Dividing both sides by 2, we obtain

20 =

Aman’s son’s age =

Aman’s age = 3

Download NCERT Solutions Linear equation Exercise 2.4 as pdf

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