Ncert Solutions for Linear equations Class 8 CBSE Part 1

In this page we have NCERT book Solutions for Class 8th Maths:Linear equations for EXERCISE 4 . Hope you like them and do not forget to like , social share and comment at the end of the page.

Question 1- Amina thinks of a number and subtracts   5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Answer - Let the number be x.

According to the given question,

8(x-5/2) = 3x

8x − 20 = 3x

Transposing 3x to L.H.S and −20 to R.H.S, we obtain

8x − 3x = 20

5x = 20

Dividing both sides by 5, we obtain

x = 4

Hence, the number is 4.

Question 2- A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Answer - Let the numbers be x and 5x. As per question,

21 + 5x = 2(x + 21)

21 + 5= 2x + 42

Transposing 2x to L.H.S and 21 to R.H.S, we obtain

5− 2x = 42 − 21

3x = 21

Dividing both sides by 3, we obtain

= 7

5x = 5 × 7 = 35

Hence, the numbers are 7 and 35 respectively.

Question 3- Sum of the digits of a two digit number is 9. When we interchange the digits it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Answer - Let the digits at tens place and one’s place be x and 9 − x respectively.

Therefore, original number = 10x + (9 − x) = 9x + 9

On interchanging the digits, the digits at ones place and tens place will be x and 9 − x respectively.

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

= 90 − 10x + x

= 90 − 9x

As per question,

New number = Original number + 27

90 − 9x = 9x + 9 + 27

90 − 9x = 9x + 36

Transposing 9x to R.H.S and 36 to L.H.S, we obtain

90 − 36 = 18x

54 = 18x

Dividing both sides by 18, we obtain

3 = and 9 − x = 6

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

Therefore, the two-digit number is 36

Question 4- One of the two digits of a two digit number is three times the other digit. If you interchange the digit of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Answer - Let the digits at tens place and one’s place be x and 3x respectively.

Therefore, original number = 10x + 3x = 13x

On interchanging the digits, the digits at ones place and tens place will be x and 3x respectively.

Number after interchanging = 10 × 3x + x = 30x + x = 31x

According to the given question,

Original number + New number = 88

13x + 31x = 88

44x = 88

Dividing both sides by 44, we obtain

x = 2

Therefore, original number = 13= 13 × 2 = 26

By considering the tens place and ones place as 3x and x respectively, the two-digit number obtained is 62.

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

Question 5- Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of this mother’s present age. What are their present ages?

Answer - Let Shobo’s age be x years. Therefore, his mother’s age will be 6x years.

According to the given question,

(x+5)=6x/3

x + 5 = 2x

Transposing x to R.H.S, we obtain

5 = 2x − x

5 = x

6x = 6 × 5 = 30

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

Question 6- There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs 100 per metre it will cost the village panchayat Rs 75, 000 to fence the plot. What are the dimensions of the plot?

Answer - Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be 11x m and 4x m 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 × 30x = 75000

3000x = 75000

Dividing both sides by 3000

x = 25

Length = 11x m = (11 × 25) m = 275 m

Breadth = 4x m = (4 × 25) m = 100 m

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

Question 7- Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36660. How much trouser material did he buy?

Answer - Let 2x m of trouser material and 3x m of shirt material be bought by him.

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 × (2x) + 55 × (3x) = 36660

201.60+ 165x = 36660

366.60x = 36660

Dividing both sides by 366.60

x = 100

Trouser material = 2x m = (2 × 100) m = 200 m

Question  8  Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Answer - Let the number of deer be x.

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

8x –(4x+3x)=72

x=72

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

Question 9- A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages

Answer - Let the granddaughter’s age be x years. Therefore, grandfather’s age will be

10x years.

According to the question,

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

10x = x + 54

Transposing x to L.H.S, we obtain

10x − x = 54

9x = 54

x = 6

Granddaughter’s age = x years = 6 years

Grandfather’s age = 10x years = (10 × 6) years = 60 years

Question 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.

Answer  Let Aman’s son’s age be x years. Therefore, Aman’s age will be 3x years. Ten years ago, their age were (x − 10) years and (3x − 10) years respectively.

According to the question,

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

3x − 10 = 5(x − 10)

3x − 10 = 5x − 50

Transposing 3x to R.H.S and 50 to L.H.S, we obtain

50 − 10 = 5x − 3x

40 = 2x

Dividing both sides by 2, we obtain

20 = x

Aman’s son’s age = x years = 20 years

Aman’s age = 3x years = (3 × 20) years = 60 years