**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?**

**Answer-**

Let the number be *x*.

According to the given question,

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?**

**Answer-**

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?**

**Answer-**

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?**

**Answer-**

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

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

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

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

According to the given question,

Original number + new number = 88

13*x *+ 31*x *= 88

44*x *= 88

Dividing both sides by 44, we obtain

*x *= 2

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

By considering the tens place and ones place as 3*x *and *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?**

**Answer-**

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

According to the given question,

*X *+ 5 = 2*x*

Transposing *x *to R.H.S, we obtain

5= 2*x-x*

5=*x*

6*x*= 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?**

**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 11*x *m and *4x*m respectively.

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

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

100 x 30*x* = 75,000

3000*x* = 75,000

Dividing both sides by3000, we obtain

*X* =25

Length = 11*x* m = (11×25)m = 275m

Breadth = 4*x* 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?**

**Answer-**

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

Per metre selling price of the trouser material =

Per metre selling price of trouser of shirt material =

Given that, total amount of selling= Rs 36660

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

201.60*x* + 165*x* = 36660

366.60*x* = 36660

Dividing both the sides by 366.60, we obtain

*X *= 100

Trouser material = 2*x*m = (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.**

**Answer-**

Let the number of deer be *x.*

Number of deer grazing in the field =

Number of deer playing nearby =

=

Number of deer drinking water from the pond = 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.**

**Answer-**

Let the daughter’s age be *x* years. Therefore grandfather’s age will be 10*x* years.

According to the question,

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

10*x* = *x*+54

Transposing *x* to L.H.S, we obtain

= 10*x*–*x *= 54

9*x *= 54

*X=*6

Granddaughter’s age = *x* years = 6 years

Grandfather’s age = 10*x *years = (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**

**Answer-**

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

According to the question,

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

3*x *– 10 = 5(*x*-10)

3*x*-10 = 5*x*-50

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

50 – 10 = 5*x*-3*x*

40= 2*x*

Dividing both sides by 2, we obtain

20= *x*

Arman’s son’s age = *x *years = 20 years

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