# Solution of Question4and5

In this page solution of question4and5 we are going to see solutions for the questions 4th,5th and 6th.

Question 4:

The sum of two numbers is 135 and they are in the ratio 7 : 8 . Find those numbers.

Solution:

Let "7 x" be the first number

Let "8x" be the second number

Sum of two numbers = 135

7x + 8 x = 135

15 x = 135

x = 135/15

x = 9

So the first number 7 x = 7 (9) = 63

and the second number 8 x = 8 (9) = 72

Therefore the correct option is (C) 63 and 72

Question 5:

The difference of two numbers is 72 and the quotient obtained by dividing the one by other is 3. Find the numbers.

Solution:

Let "x" be the first number

Let "y" be the second number

The difference between two numbers = 72

So   x - y = 72

x = 72 + y -- (1)

If we divide first number by the second number we will get 3 as quotient.

So x/y = 3  --- (2)

Applying the value of x in the second equation

(72 + y)/y = 3

72 + y = 3 y

3y - y = 72

2y = 72

y = 36

Substituting the value of y in the first equation,we will get

x = 72 + 36

x = 108

Therefore the two numbers are 36 and 108.

Therefore the correct option is (B) 36 and 108.

Question 6:

In a certain fraction, the denominator is 4 less than the numerator. If the number 3 is added to both the numerator and denominator, the resulting fraction is equal to  9/7, find the original fraction.

Solution:

Let "x" be the number which is in numerator

Let "x - 4" be the number in the denominator (Because the denominator is 4 less than the numerator)

(x + 3)/(x - 4 + 3) = 9/7

(x + 3)/(x - 1) = 9/7

7 (x + 3) = 9 (x - 1)

7 x + 21 = 9 x - 9

7 x - 9x = -9 - 21

- 2x = - 30

x = -30/(-2)

x = 15

So the required fraction = 15/(15-4)

= 15/11

Therefore the correct option is (D) 15/11.

These are the solution of question4and5.

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