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Learn Exercise 2.2 with Free Lessons & Tips

Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Let the numbers are x and y.

From the given data we can write two equations with these variables x and y.

The sum of two numbers is 95. that is


One exceeds the other by 15. that is 


By giving (2) in (1)







So the numbers are 55 and 40




Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

In question, it is given that numbers are in the ratio 5:3. So that we can assume the numbers as 5k and 3k.

where k is a numerical constant. 

Now the ratio becomes 5k:3k=5:3

They are differed by 18, that means






first number=5k=5*9=45

second number=3k=3*9=27

Hence the numbers are 45 and 27



The organizers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3000. Find the number of winners, if the total number of participants is 63.

From the given question, we can obtain the following data.

prize of winner=100 Rupees

prize of a participant who doesn't win=25 Rupees

Total prize money=3000

total number of participant=63

Let number of winners=x

number of participant who doesn't win=y

total number of participant =x+y=63............(1)

total prize money=x*100+y*25=3000............(2)

from (1)



substitute (3) into (2)






ie number of winners=19


If you subtract from a number and multiply the result by, you get. What is the number?

Assume the number as x.

following equation is obtained from the given data



x=1/2 + 1/4=3/4



The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

Let the breadth be x m. The length will be (2x + 2) m.

Perimeter of swimming pool = 2(l + b) = 154 m

2(2x + 2 + x) = 154

2(3x + 2) = 154

Dividing both sides by 2, we obtain

3x + 2 = 77

On transposing 2 to R.H.S, we obtain

3x = 77 − 2

3x = 75

On dividing both sides by 3, we obtain

x = 25

2x + 2 = 2 × 25 + 2 = 52

Hence, the breadth and length of the pool are 25 m and 52 m respectively.


The base of an isosceles triangle is cm. The perimeter of the triangle is cm. What is the length of either of the remaining equal sides?

Isosceles triangles are triangle with 2 sides are equal. non equal side is called base

length of equal side=x

perimeter=2x+b=2x+4/3=4 2/15=(4*15+2)/15=62/15


2x= 62/15-4/3=(62-4*5)/15




Three consecutive integers add up to 51. What are these integers?

let the first integer be x.

then three consecutive integers are x,x+1,x+2





so the integers are 16,17,18



The sum of three consecutive multiples of 8 is 888. Find the multiples.

let the first multiple be x

next multiple of 8=x+8

next to next multiple of 8=x+16





hence multiples are 288,296and 304


Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Let three consecutive integers be x, x + 1, x + 2. According to the question,

2x + 3(x + 1) + 4(x + 2) = 74

2x + 3x + 3 + 4x + 8 = 74

9x + 11 = 74

On transposing 11 to R.H.S, we obtain

9x = 74 − 11

9x = 63

On dividing both sides by 9, we obtain

x = 7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

Hence, the numbers are 7, 8, and 9.


The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Let common ratio between Rahul’s age and Haroon’s age be x.

Therefore, age of Rahul and Haroon will be 5x years and 7x years respectively. After 4 years, the age of Rahul and Haroon will be (5x + 4) years and (7x + 4) years respectively.

According to the given question, after 4 years, the sum of the ages of Rahul and Haroon is 56 years.

∴ (5x + 4 + 7x + 4) = 56

12x + 8 = 56

On transposing 8 to R.H.S, we obtain

12x = 56 − 8

12x = 48

On dividing both sides by 12, we obtain

x = 4

Rahul’s age = 5x years = (5 × 4) years = 20 years

Haroon’s age = 7x years = (7 × 4) years = 28 years


The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Let the common ratio between the number of boys and numbers of girls be x.

Number of boys = 7x

Number of girls = 5x

According to the given question,

Number of boys = Number of girls + 8

∴ 7x = 5x + 8

On transposing 5x to L.H.S, we obtain

7x − 5x = 8

2x = 8

On dividing both sides by 2, we obtain

x = 4

Number of boys = 7x = 7 × 4 = 28

Number of girls = 5x = 5 × 4 = 20

Hence, total class strength = 28 + 20 = 48 students


Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Let Baichung's, his father's, his grandfather's ages are x,y, and z respectively.



substitute (2) in (1)



replace the values of y and z with (1) and (3) in (4)








so the ages are 17,46 and 72 respectively



Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Let Ravi’s present age be x years.

Fifteen years later, Ravi’s age = 4 × His present age

x + 15 = 4x

On transposing x to R.H.S, we obtain

15 = 4xx

15 = 3x

On dividing both sides by 3, we obtain

5 = x

Hence, Ravi’s present age = 5 years


A rational number is such that when you multiply it by and add to the product, you get. What is the number?

Let the number be x.

According to the given question,

On transposing to R.H.S, we obtain

On multiplying both sides by, we obtain

Hence, the rational number is.


Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4, 00,000. How many notes of each denomination does she have?

since the ratio of number of notes is 2:3:4

we can assume like this,

number of 100 notes=2k

number of 50 notes=3k

number of 10 notes= 5k

where k is an integer constant.

total cash=2k*100+3k*50+5k*10=400000

200k + 150k + 50k= 400000



number of 100 notes=2k=2*1000=2000

number of 50 notes=3k=3*1000=3000

number of 10 notes=5k=5*1000=5000


I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

number of rs 5 coins=x

number of rupees 3 coins=3x

since total number of coins is 160,

number of rs 1 coins=160-3x-x=160-4x

total cash in rupees=x*5+3x*2+(160-4x)*1=300




number of 5 rs coins=20

number of 2  rs coins=3x=3*20=60

number of 1 rs coins=160-4x=160-4*20=160-80=80


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