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Question 239
The average of four consecutive numbers A, B, C and D respectively is 49.5. What is the product of B and D?
Solution
Since the ages of A, B, C, D are consecutive
Let the ages of A, B, C, D be n, n+1,n+2,n+3
$$\frac{n+n+1+n+2+n+3}{4} = 49.5$$
4n+6 = 49.5*4 = 198
4n = 192
n = 48
Ages of A, B, C, D = 48, 49, 50 ,51
Product of ages of B and D = 49*51 = (50-1)(50+1) =$$50^2-1$$ = 2500-1 = 2449.
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