Question

Sum of the present ages of a father and his son is 48 years. If the product of their ages 5 years back is 165, what is the present age of the father?

• A. 38 years
• B. 36 years
• C. 31 years
• D. 28 years

Hint:

Let the father's age be \(f\) and the son's age be \(48-f\), then form a quadratic from the product of their ages 5 years back.

Answer 1

The Correct Option is A

Step 1: Set up variables for the present ages.
Let the father's present age be \(f\) years and the son's present age be \(s\) years.
We are told
\[ f+s = 48 \]
so \(s = 48-f\).

Step 2: Write the second condition using ages 5 years back.
5 years back, the father's age was \(f-5\) and the son's age was \(s-5\). Their product was 165:
\[ (f-5)(s-5) = 165 \]

Step 3: Substitute \(s = 48-f\) and expand.
\[ (f-5)(48-f-5) = 165 \]
\[ (f-5)(43-f) = 165 \]
\[ 43f - f^2 - 215 + 5f = 165 \]
\[ -f^2 + 48f - 215 = 165 \]
\[ f^2 - 48f + 380 = 0 \]

Step 4: Solve the quadratic equation.
Using the quadratic formula with \(a=1\), \(b=-48\), \(c=380\):
\[ f = \frac{48 \pm \sqrt{48^2 - 4(1)(380)}}{2} = \frac{48 \pm \sqrt{2304-1520}}{2} = \frac{48 \pm \sqrt{784}}{2} = \frac{48 \pm 28}{2} \]
This gives two values, \(f = 38\) or \(f = 10\).

Step 5: Choose the sensible value and check.
A father must be older than his son, and 10 is too young to be a father here, so we reject \(f=10\). With \(f=38\), the son's age is \(48-38=10\). Check: 5 years back the ages were 33 and 5, and \(33 \times 5 = 165\), which matches.
This also rules out options (b), (c) and (d), since none of 36, 31 or 28 satisfy the quadratic.

Final Answer:
The father's present age is 38 years.
\[ \boxed{38 \text{ years}} \]