Question

a's present age is to b's present age is 7:9.twelve years ago their ages were in the ratio 3:5.when would the ratio of the ages be 6:7

Answer 1

Let the present age of a = x years
and let the present age of b = y years
The ratio of the ages of a and b = 7:9
Therefore
x:y = 7:9
x/y = 7/9
9x = 7y
9x - 7y = 0 (Equation 1)

Twelve years ago
age of a = x - 12 years
age of b = y - 12 years
Ratio of ages of a and b 12 years ago = 3:5
Therefore
x - 12 : y - 12 = 3:5
(x - 12)/(y - 12) = 3/5
5x - 60 = 3y - 36
5x - 3y = 24 (Equation 2)

Taking equation 1 and 2 as simultaneous equation
5(9x - 7y = 0)
9(5x - 3y = 24)

45x - 35y = 0
45x - 27y = 216

Subtracting 45x - 27y = 216 from 45x - 35y = 0
45x - 35y = 0
- 45x + 27y = - 216

- 8y = - 216
y = - 216/- 8
y = 27

The present age of y = 27 years
9x - 7(27) = 0
9x = 189
x = 21 years

The present age of x = 21 years


Given that the ratio of ages of and b after sometime = 6:7
Let that after z years, the ratio of ages of a and b = 6:7
Therefore
(21 + z) : (27 + z) = 6:7
(21 + z)/(27 + z) = 6/7
147 + 7z = 162 + 6z
z = 15 years

Therefore after 15 years, the ratio of age of a and b = 6:7

Hope this helps you.

Answer 2

Answer:

the above answer is correct