Question

The present ages of A, B and C are in proportions 4:5:9. Nine years ago, sum of their ages was 45 years. Find their present ages in years.

• A. 15, 20, 35
• B. 20, 24, 36
• C. 16, 20, 36
• D. 20, 25, 45

Hint:

To solve this instantly, add up the ratio parts: $4 + 5 + 9 = 18$ parts.
If the sum of their ages 9 years ago was 45, their present sum must be $45 + (3 \times 9) = 72$ years.
So, 18 parts = 72 $\implies$ 1 part = 4.
Thus, the ages are $4(4) = 16$, $5(4) = 20$, and $9(4) = 36$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Problems on ages are solved by setting up a linear equation using ratio multipliers.
Detailed Explanation:
Let the present ages of A, B, and C be $4x$, $5x$, and $9x$ years respectively.
Nine years ago, their respective ages were:
- A: $4x - 9$
- B: $5x - 9$
- C: $9x - 9$
According to the problem, the sum of their ages 9 years ago was 45: \[ (4x - 9) + (5x - 9) + (9x - 9) = 45 \] Simplify the equation: \[ 18x - 27 = 45 \] \[ 18x = 72 \] \[ x = 4 \] Now, calculate their present ages: - Age of A = $4 \times 4 = 16$ years.
- Age of B = $5 \times 4 = 20$ years.
- Age of C = $9 \times 4 = 36$ years.

Step 2: Final Answer:
Their present ages are 16, 20, and 36 years, matching Option (C).