The ages of three men are in the ratio 3:4:5. If the difference between the ages of the oldest and youngest is 18 years, find the sum of the ages of the three men

  • A 45 years
  • B 72 years
  • C 108 years
  • D 216 years
waec 2000mathematics

The correct answer is C. 108 years

Given that the ages are in the ratio 3: 4: 5.

Let the sum of their ages be t.

\(\therefore\) The youngest age = \(\frac{3}{12} t\)

The eldest age = \(\frac{5}{12} t\)

\(\implies \frac{5}{12} t - \frac{3}{12} t = 18\)

\(\frac{2t}{12} = 18 \implies t = \frac{18 \times 12}{2}\)

t = 108 years.

The sum of their ages = 108 years.

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