Question
A is four times as old as B. C was twice-as old as A six
years ago. In five years' time, A will be 41. Find the sum of the present ages of B and C.Solution
Let the age of A = A, B = B and C = C According to question, => A = 4B 6 years ago, => C - 6 = 2(A - 6) After 5 years, => A + 5 = 41 => A = 36 So, B = 36/4 = 9 => C = 2(A - 6) + 6 => C = 2(36 - 6) + 6 => C = 60 + 6 = 66 Sum of the present ages of B and C = 9 + 66 = 75 years
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