This page 10th cbse maths solution for exercise 3.3 part 5 is going to provide you solution for every problems that you find in the exercise no 3.3

(v) A fraction becomes 9/11,if 2 is added to both numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:

Let “x/y” be the required fraction

If 2 is added to both numerator and denominator the fraction becomes 9/11

(x + 2)/(y+2) = 9/11

11 (x + 2) = 9 (y + 2)

11 x + 22 = 9 y + 18

11 x – 9 y = 18 – 22

11 x – 9 y = -4 ------ (1)

3 is added to both the numerator and the denominator it becomes 5/6

(x + 3)/(y+3) = 5/6

6 (x + 3) = 5 (y + 3)

6 x + 18 = 5 y + 15

6 x – 5 y = 15 - 18

6 x – 5 y = - 3 ------ (2)

Substitute x = 7 in the first or second equation to get eh value of y

6 (7) – 5 y = -3

42 – 5 y = -3

-5y = -3 – 42

- 5 y = -45

y = 45/5

y = 9

Therefore the required fraction is 7/9

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution:

Let “x” be the age of Jacob

Let “y” be the age of his son

5 years hence their ages will be

x + 5 and y + 5

x + 5 = 3 (y + 5)

x + 5 = 3 y + 15

x – 3y = -5 + 15

x – 3 y = 10 ------(1)

5 years ago, their ages was

x – 5 and y – 5

x – 5 = 7 (y – 5)

x – 5 = 7 y – 35

x – 7 y = 5 - 35

x – 7 y = -30 ------(2)

y = 40/4

y = 10

Substitute y = 10 in the first or second equation to get the value of x

x – 7(10) = -30

x – 70 = -30

x = -30 + 70

x = 40

Therefore age of Jacob's son = 40

Age of Jacob = 10

**Exercise 3.3 Part 1****Exercise 3.3 Part 2****Exercise 3.3 Part 3****Exercise 3.3 Part 4****Exercise 3.3 Part 5****Go to exercise 3.4**

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