One year ago, a father was four times as old as his son. After six years his age exceeds twice his son’s age by 9 years. The ratio of their present age is
9 : 2
11 : 3
12 : 5
13 : 4
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2017
– One year ago, the son’s age was $S-1$ and the father’s age was $F-1$.
– According to the first condition: $F-1 = 4(S-1)$. This simplifies to $F-1 = 4S – 4$, so $F = 4S – 3$ (Equation 1).
– After six years, the son’s age will be $S+6$ and the father’s age will be $F+6$.
– According to the second condition: The father’s age ($F+6$) exceeds twice his son’s age ($2(S+6)$) by 9 years. So, $F+6 = 2(S+6) + 9$.
– This simplifies to $F+6 = 2S + 12 + 9$, which is $F+6 = 2S + 21$. So, $F = 2S + 15$ (Equation 2).
– Now we have a system of two linear equations for $F$ and $S$:
1) $F = 4S – 3$
2) $F = 2S + 15$
– Equating the expressions for $F$: $4S – 3 = 2S + 15$.
– Subtract $2S$ from both sides: $2S – 3 = 15$.
– Add 3 to both sides: $2S = 18$.
– Divide by 2: $S = 9$. The son’s present age is 9 years.
– Substitute $S=9$ into Equation 1 (or Equation 2) to find $F$:
$F = 4(9) – 3 = 36 – 3 = 33$. The father’s present age is 33 years.
– The ratio of their present age (Father : Son) is $F : S = 33 : 9$.
– This ratio can be simplified by dividing both numbers by their greatest common divisor, which is 3.
$33 \div 3 = 11$
$9 \div 3 = 3$
– The simplified ratio is 11 : 3.
Present ages: Father = 33, Son = 9.
One year ago: Father = 32, Son = 8. Is 32 four times 8? Yes, $32 = 4 \times 8$. (Condition 1 satisfied).
After six years: Father = $33+6 = 39$, Son = $9+6 = 15$. Is 39 equal to twice the son’s age plus 9? $2 \times 15 + 9 = 30 + 9 = 39$. Yes. (Condition 2 satisfied).
The calculated ages satisfy both conditions.