# Ncert Solutions For Class 8 Maths Ex 2.2

## Ncert Solutions For Class 8 Maths Chapter 2 Ex 2.2

Question -1

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

Let the number be x.

According to the given question,

8(x52)=3x$8\begin{pmatrix} x-\frac{5}{2} \end{pmatrix}=3x$

8x-20=3x

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

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, the one of the new numbers become twice the other new number. What are the numbers?

Let the number be x and 5x. according to the question,

21+5x=2(x+21)

21+5x=2x+42

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

5x-2x=42-21

3x=21

Dividing both the sides by 3, we obtain

X=7

5x= 5×7= 35

Hence, the number are 7 and 35 respectively.

Question-3

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

Let the digits are tens place and ones place be x and 9-x respectively.

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

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

Therefore new number afterinterchanging the digits = 10(9-x)+x

= 90-10x+x

= 90-9x

According to the given 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 = x and 9-x=6

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

Therefore the two digits number is 9x+9 = 9 x 3+9 =36

Question-4

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

Let the digits at tens place and once place be x and 3x respectively.

Therfore the original number = 10x+3x = 13x

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

Number after interchanging  = 10 X 3x + x=31x

According to the given question,

Original number + new number = 88

13+ 31= 88

44= 88

Dividing both sides by 44, we obtain

= 2

therefore the original number = 13x = 13 X 2= 26

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

Let shobo’s age be years. Therefore, his mother’s age will be 6years.

According to the given question,

After5years,Shobosage=shobosmotherspresentage3$After\: 5\: years,\: Shobo’s\: age\: =\: \frac{shobo’s\: mother’s\: present\: age }{3}$ x+5=6x3$x + 5= \frac{6x}{3}$

+ 5 = 2x

Transposing to R.H.S, we obtain

5= 2x-x

5=x

6x= 6 X 5= 30

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

Question-6

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

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 11m and 4xm respectively.

Perimeter of the plot = 2(length + breadth) = [2(11x + 4x)]m = 30m

It is given that the cost of fencing the plot at the rate Rs 100 per metre is Rs 75,000.

100 x 30x = 75,000

3000x = 75,000

Dividing both sides by3000, we obtain

X =25

Length = 11x m = (11×25)m = 275m

Breadth = 4x m = (4×25)m = 100m

Hence, the dimension of the plot are 275m 100m respectively.

Question-7

For school uniform Hasan buys two kinds of cloth material, the shirt material cost him Rs50 per metre and trouser material that cost him Rs 90 per metre. For every 2 metre of the trouser material he buys 3 metres of the shirt material. He sells the material at 12% and 10% profit respectively. His total sales are Rs 36,660. How much trouser material did he buy?

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

Per metre selling price of the trouser material = Rs90+90×12100=Rs100.80$Rs  90+\frac{90\times 12}{100}= Rs 100.80$

Per metre selling price of trouser of shirt material = Rs50+50×10100=Rs55$Rs 50+\frac{50\times 10}{100} = Rs 55$

Given that, total amount of selling= Rs 36660

100.80 x (2x) + 55 x (3x) = Rs36660

201.60x + 165x = 36660

366.60x = 36660

Dividing both the sides by 366.60, we obtain

= 100

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

Question-8

Half of the herd of deer are gazing in the field and three fourth of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Let the number of deer be x.

Number of deer grazing in the field = x2$\frac{x}{2}$

Number of deer playing nearby = 34×numberofremainingdeer$\frac{3}{4}\times number\: of \: remaining \: deer$

= 34×(xx2)=34×x2=3x8$\frac{3}{4}\times \begin{pmatrix} x-\frac{x}{2} \end{pmatrix}=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}$

Number of deer drinking water from the pond = 9

= xx2+3x8=9$x- \frac{x}{2}+\frac{3x}{8} =9$

= x4x+3x8=9$x- \frac{4x+3x}{8} =9$

= x7x8=9$x-\frac{7x}{8}=9$

= x8=9$\frac{x}{8}= 9$

Multiplying both the sides by 8, we obtain

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 year older than her.  Find their present age.

Let the daughter’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= 54

9= 54

X=6

Granddaughter’s age = x years = 6 years

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

Question-10

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

Let Arman’s son’s age be x years. Therefore, Arman’s age will be 3x years. Ten years ago, their age was (x – 10)years and (3x-10)years respectively.

According to the question,

10 years ago, Arman’s age = 5 x Arman’s son’s age 10 years ago

3– 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

Arman’s son’s age = years = 20 years

Arman’s age = 3x years= (3×20)years = 6  years.