Question 71

In a family, the father's age is 5 times the age of one of his sons. Mother's age is 4 times the age of their other
son. Difference in the parents' ages is half the sum of the sons' ages. Then:

Solution

Let , Child 1's age is X and Child 2's age is Y.

And let us assume Father's age is F and Mother's age is M.

From the statements,

F=5X.....(i) and M= 4Y...(ii)

Again, F + M =5X + 4Y...(iii)

Again, F - M = (X + Y)/2...(iv)

Adding (iii) and (iv) and then replacing F of(i),

we get 9X/4=9Y/4

Hence it is proved that , X = Y. Hence twins.

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In a family, the father's age is 5 times the age of one of his sons. Mother's age is 4 times the age of their otherson. Difference in the parents' ages is half the sum of the sons' ages. Then: