A toy T jumps forward or backward. In each forward jump, it moves 5' forward whereas in each backward jump, it moves 2' backward. If in 31 jumps, T moves exactly 15' forward, then what is the difference of the number of forward and backward jumps?

Correct Answer: (D) 9

Let $f$ be the number of forward jumps and $b$ be the number of backward jumps. The total number of jumps is $f + b = 31$, and the net forward displacement is $5f - 2b = 15$$\text{ feet}$. Substituting $b = 31 - f$ into the distance equation gives $5f - 2(31 - f) = 15$$\implies$ 7f - 62 = 15 $\implies$ 7f = 77 $\implies$ f = 11. Consequently, $b = 31 - 11 = 20$. The difference between the number of backward and forward jumps is $|20 - 11| = 9$.