Sonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?
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Correct Answer: Option n
Explanation:
Let Wale's present age = x.
Then Sonny's age = 2x
Four years ago, Wale = x - 4
Sonny = 4(x - 4) = 2x - 4
\(4x - 16 = 2x - 4 \implies 4x - 2x = -4 + 16\)
\(2x = 12 \implies x = 6\)
Sonny = 2x = 2(6) = 12 years
In z years' time, the sum of their ages = 66
Wale = 6 + z
Sonny = 12 + z
\(6 + z + 12 + z = 66 \implies 18 + 2z = 66\)
\(2z = 66 - 18 = 48 \implies z = 24\)
Therefore, in 24 years time, the sum of their ages will be 66.
Let Wale's present age = x.
Then Sonny's age = 2x
Four years ago, Wale = x - 4
Sonny = 4(x - 4) = 2x - 4
\(4x - 16 = 2x - 4 \implies 4x - 2x = -4 + 16\)
\(2x = 12 \implies x = 6\)
Sonny = 2x = 2(6) = 12 years
In z years' time, the sum of their ages = 66
Wale = 6 + z
Sonny = 12 + z
\(6 + z + 12 + z = 66 \implies 18 + 2z = 66\)
\(2z = 66 - 18 = 48 \implies z = 24\)
Therefore, in 24 years time, the sum of their ages will be 66.