Question
Present average age of 4 friends: βAβ, βBβ,
βCβ and βDβ is 33 years. If their ages are in the arithmetic progression in the same order and βDβ is 4 years older than βBβ, then find the sum of present ages of βAβ and βCβ.Solution
Common difference between ages = (4/2) = 2 years. So, let present ages of βAβ, βBβ, βCβ and βDβ be βxβ, (x + 2) years, (x + 4) years and (x + 6) years, respectively According to question: x + x + 2 + x + 4 + x + 6 = 132 Or, 4x = 120 Or, x = 30 years (Age of A) Present age of βCβ = x + 4 = 30 + 4 = 34 years Required sum = 34 + 30 = 64 years
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