Question
Sum of present ages of P, Q and R is 79 years. Before 4 years, the age of P was three times the age of Q at that time and the age of R is (
- Y more than the present age of Q (Y < 5). Find the sum of present ages of P and R. (All ages are natural numbers)
Solution
Let the present ages of P, Q and R be (a), (b) and (c) years respectively. According to the question,
a + b + c = 79
Before 5 years,
a - 4 = 3(b - 4)
a - 4 = 3b - 12
or, a = 3b - 8 Also, c = b + Y (Y < 5) Substituting in the total age equation:
(3b - 8) + b + (b + Y) = 79
5b + Y - 8 = 79
5b + Y = 87
Since (Y < 5), taking (Y = 2):
5b = 85 Or, b = 17 Then, a = 3(17) – 8 = 43
c = 17 + 2 = 19 Required sum = a + c = 43 + 19 = 62
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