Learn Exercise 2.2 with Free Lessons & Tips

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

x+y=95..............(1)

One exceeds the other by 15. that is 

x=y+15.............(2)

By giving (2) in (1)

x+y=95

y+15+y=95

2y=95-15=80

y=80/2=40

from(2)

x=y+15=40+15=55

So the numbers are 55 and 40

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

5k=3k+18

5k-3k=18

2k=18

k=18/2

k=9

first number=5k=5*9=45

second number=3k=3*9=27

Hence the numbers are 45 and 27

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)

x+y=63

y=63-x.........(3)

substitute (3) into (2)

x*100+y*25=3000

100x+25(63-x)=3000

100x-25x+25*63=3000

75x=3000-25*63=3000-1575=1425

x=1425/75=19

ie number of winners=19

Assume the number as x.

following equation is obtained from the given data

(x-)=

x-=2/8=1/4

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

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.

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+4/3=62/15

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

=(62-20)/15=42/15=14/5

x=7/5=1.4

let the first integer be x.

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

x+(x+1)+(x+2)=51

3x+3=51

3x=51-3=48

x=48/3=16

so the integers are 16,17,18

let the first multiple be x

next multiple of 8=x+8

next to next multiple of 8=x+16

x+x+8+x+16=888

3x+24=888

3x=888-24=864

x=864/3=288

hence multiples are 288,296and 304

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.

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

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

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

z=y+26............(1)

y=x+29..................(2)

substitute (2) in (1)

z=y+26=x+29+26=x+55.............(3)

x+y+z=135............(4)

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

x+y+z=135

x+x+29+x+55=135

3x+84=135

3x=51

x=51/3=17

y=x+29=17+29=46

z=x+55=72

so the ages are 17,46 and 72 respectively

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 = 4x − x

15 = 3x

On dividing both sides by 3, we obtain

5 = x

Hence, Ravi’s present age = 5 years

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.

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

400k=400000

k=400000/400=1000

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

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

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