WAEC - Mathematics (2000 - No. 9)
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
45 years
72 years
108 years
216 years
Explicación
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.