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Exercise 2.1

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Solve the following equations.

 1.  x - 2 = 7 

Solution: 

x - 2 = 7 

x = 7 + 2 

x = 9 

 2. y + 3 = 10 

Solution:

y + 3 = 10 

y = 10 - 3 

y = 7 

 3. 6 = z + 2 

Solution: 

6 = z + 2 

z = 6 - 2 

z = 4 

Solution: 

 5. 6x = 12 

Solution: 

x = 12/6

x = 2 

Solution: 

t = 3 × 10 

t = 30 

 7. 2x + 3 = 18 

Solution: 

2x = 18 - 3 

2x = 15 

x = 15/2

x = 7.5 

Solution: 

 x = 1.6 × 1.5

x =  2.4

9. 7x - 9 = 16 

Solution: 

7x = 16 + 9

7x = 25

x = 25/7

x = 3.5

10. 14x - 8 = 13

Solution: 

14x = 13 + 8

14x = 21

x = 21/14 = 3/2 = 1.5

11. 17 + 6p = 9 

Solution: 

6p = 9 - 17

6p = -8

p = -8/6 = - 1.3

Solution: 

 

Exercise 2.2

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Q1.   If you subtract   from a number and multiply the result by  , you get . What is the number?

Solution:

Let the number be x 

so, the equation will be 

Q2.   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?

Solution:

Let the breadth of the swimming pool x.

So, the equation will be  

so, the length  of pool 2x + 2

= 2×25 + 2

= 50 + 2

= 52

and the breadth of pool x = 25

Q3. The base of an isosceles triangle   cm the perimeter of a triangle is 4  cm what is the length of either remaining equal sides?

Solution:

Let the length of one remaining equal sides x.So, equation wil  ​

Q4.   Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution:

Let the second number is x.

So, the equation will be  x + x + 15 = 95

So, the first number will be x + 15

= 40 + 15

= 55

And second number: x = 40

Q5.   Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Solution:

Let,

Both numbers will be 5x and 3x

So, the equation will be 5x -  3x = 18

⇒2x = 18

so, the first number will be 5x

= 5 × 9

= 45

and the second number will be =  3 × 9

=  27

Q6.   Three consecutive integers add up to 51. What are these integers?

Solution:

Let, all numbers x, x+1, x+2 respectively.

So, the equation will be x+x+1+x+2=51

⇒ 3x +3=51

⇒ 3x=51-3

So, the first number will be x = 16

, second number will be x + 1

= 16 + 1

= 17

and the third  number will be x + 2 =  18

Q7. The sum of three consecutive multiples of 8 is 888. Find the multiples?

Solution:

Let, three consecutive multiples of 8 is x, x+8, and x+16 respectively

So, the equations will be : x + x + 8 + x + 16 = 888

⇒ 3x + 24 = 888

 ⇒ 3x = 888 - 24

So, the first multiple x = 288

Second multiple x + 8 = 288+8 = 296

Third multiple x + 16 = 288 + 16 = 304

Q8.   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.

Solution:

Let, all numbers be x, x+1, x+2

So, the equation will be: 2(x) + 3(x+1) +4(x+2) =74

 ⇒ 2x+ 3x+3+ 4x+8 = 74

 ⇒ 9x+11=74

 ⇒ 9x=74-11

So, first number is x = 7

Second number is x + 1

= 7 + 1 = 8

Third number is x + 2

= 7 + 2

= 9

Q9.   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?

Solution:

Let, the present ages of Rahul and Haroon be 5x and 7x respectively

So, the equation will be 5x + 4 + 7x + 4 = 56

⇒ 12x + 8 = 56

⇒ 12x = 56 - 8

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So, the present age of Rahul: 5x = 5 × 4 = 20

The present age of Haroon: 7x = 7 × 4 = 28

Q10.   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.

Solution:

Let, the age of Aman’s son x

⇒5(x - 10) = 3x - 10

⇒ 5x – 50 = 3x - 10

⇒5x – 3x = -10 + 50

⇒2x = 40

The present age of Aman: 3x = 3 × 20 = 60

 The present age of his son: x = 20

 

  

Q1. Amna thinks of a number and subtract  5/2  from it. She multiplies result from 8 the result now obtained is 3 times the same number she thought of.  What is the number?

Solution:

Let, the number be x

So, A.T.Q the equation will be: (x - 5/2 ) 8 = 3x

⇒  8x - 40/2  = 3x

⇒ 8x – 3x = 20

⇒ 5x = 20

⇒ x = 20/5  = 4

So, the number is x = 4.

Q2. A positive number is 5 times another number. If 21 are added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution:

 Let, the first number is x

And the second number is 5x

2(x + 21) = 5x + 21

⇒ 2x + 42 = 5x + 21

⇒ 5x – 2x = 42 – 21

⇒ 3x = 21

⇒ x = 21/7 =7

So, the both numbers will be x = 7 × 1 = 7 and 5x = 5 × 7 = 35 respectively.

Q3.   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?

Solution:

Let, the tens digit of number is = x

       The once digit of number is = y

So, the sum of digits x + y = 9---------------- (1)

So, the number will be (10 × x) + y

                                           = 10x + y ----------------- (2)

When we under change the digits then the numbers will be: 10y + x ------------ (3)

By equation (1) and (2)

(10x + y) – (10y + x) = 27

⇒ 10x + y – 10y – x = 27

⇒ 9x – 9y = 27

⇒ 9(x – y) = 27

⇒ x – y = 27/3  = 3

So, x – y = 3 -------------- (4)

Adding equation (3) and (4)

   x + y = 9

+ x – y = 3   .

 2x = 12 

⇒  x = 12/2 = 6

Putting x = 6 in equation (1)

6 + y = 9

⇒ y = 9 – 6 = 3

At last the number = 10x + y = 10 × 6 + 3 = 63

Q4. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Solution:

Let, the first digit be x

   So, the second digit will be 3x

So, the number will be 10 × x + 3x = 13x

Interchanging the digits 10 × 3x + x = 31x

A.T.Q the equation will be 13x + 31x = 88

⇒ 44x = 88

⇒  x = 44/88  = 2

So, the original number is 13x = 13 × 2 = 26

Q5. 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 his mother’s present age. What are their present ages?

Solution:

Let the age of Shobo be x

So, the age of her mother is 6x

ATQ, the equation will be: 3(x + 5) = 6x

⇒ 3x + 15 = 6x

⇒ 6x – 3x = 15

⇒ 3x = 15 

⇒ x = 15/3  = 5

so the age of shobo x = 5

and the age of shobo mother 6x = 6 × 5 = 30

Q6.   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 100 per meter it will cost the village panchayat 75000 to fence the plot. What are the dimensions of the plot?

Solution:

Let, the length and the breadth of the plot is 11x and 4x respectively.

So, the perimeter of plot = 2(l+b)

= 2(11x+4x) = 22x + 8x= 30x

Perimeter of plot = the total cost of plot/per meter cost

So, the length of the plot 11x = 11×25 = 275m

       The breadth of the plot 4x = 4×25 = 100m

Q7. Hasan buys two kinds of cloths materials for school uniforms, shirt materials that cost him R.s 50 per meter and trouser material that costs him R.s 90 per meter. For every 3 meters of the shirt material he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ` 36,600. How much trouser material did he buy?

Solution:

 

Q9.   Grandfather is ten times older than his granddaughter.

He is also 54 years older than her. Find their present ages.

Solution:

Let, the age of granddaughter be x and the age of grandfather  10x

According to the question the equation will be:

⇒ 10x – x = 54

⇒  9x = 54 

so, the age of grandfather is 10x = 10 × 6 = 60

and the age of granddaughter x = 6

Q10.   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.

Solution:

Let, the age of Aman’s son xv and the age of Aman be 3x

According to the question the equation will be 

5(x-10) = 3x-10

⇒ 5x – 50 = 3x - 10

⇒ 5x – 3x = -10 + 50

⇒ 2x = 40

So, the age of Aman's son x = 20

and the age of Aman 3x = 3 × 20 = 60

 

 

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