Each year, the age of the boy increases by 1. Each year, the sum of the ages of the two girls increases by 2 (as each girl gets older by one year, and there are two of them).
Let's say that the age of the boy today is equal to x, while the combined ages of the girls today is equal to y.
Then, next year the figures will be x + 1 and y + 2, respectively. The problem states that these two figures will be equal, which yields the following equation:
x + 1 = y + 2 which can be simplified to x = y + 1
(This is consistent with the fact that the sum of the ages of the two girls today is smaller than the age of the boy today.)
Three years from now, the combined age of the girls will be y + 3(2) = y + 6. Three years from now, the boy's age will be x + 3. Using the fact (from above) that x = y + 1, the boy's age three years from now can be written as x + 3 = (y + 1) + 3 = y + 4.
The problem asks for the difference between the age of the boy three years from today and the combined ages of the girls three years from today. This difference equals y + 4 – (y + 6) = –2.
The correct answer is D.
Plug in real numbers to see if this makes sense.
Let the girls be 4 and 6 in age. The sum of their ages today is 10. The boy's age today is then (10 + 1) = 11. Three years from today, the girls will be 7 and 9 respectively, so their combined age will be 16. Three years from today, the boy will be 14.
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Thanks & Regards,
Anaira Mitch