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
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Correct Answer: Option C
Explanation:
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.
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.