The present age of a father is twice as that of his son. 10 years ago, the age of father was thrice the age of his son. If the ratio of present age of son’s mother to father is 9 : 8, then find the difference between present age of mother and son (in years)?

The correct answer is 25.

Let us break down the solution step-by-step:

Step 1: Set up variables for the present ages of the father and the son.
Let the present age of the son be s years.
According to the problem, the present age of the father, f, is twice that of his son:
$f=2s$

Step 2: Use the information about their ages 10 years ago.
10 years ago:
The son's age was $s-10$ years.
The father's age was $f-10$ years.
At that time, the father's age was thrice (three times) the son's age:
$f-10=3(s-10)$

Step 3: Solve for the present ages of the father and the son.
Substitute $f=2s$ into the equation from Step 2:
$2s-10=3(s-10)$
Expand the right side:
$2s-10=3s-30$
Rearrange the terms to solve for s:
$30-10=3s-2s$
$s=20$
So, the present age of the son is 20 years.
Now, find the present age of the father:
$f=2s=2\times 20=40$ years.

Step 4: Find the present age of the mother.
Let the present age of the mother be m years.
The ratio of the present age of the mother to the father is given as 9 : 8:
$\frac{m}{f}$
Substitute the father's age ($f=40$) into the ratio:
$\frac{m}{40}=\frac{9}{8}$
Multiply both sides by 40:
$m=\frac{9}{8}\times 40$
$m=9\times 5=45$
So, the present age of the mother is 45 years.

Step 5: Calculate the difference between the present age of the mother and the son.
Difference = Mother's age - Son's age
$\text{Difference}=45-20=25$ years.