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?
Detailed Explanation:
Let f = forward jumps and b = backward jumps. Total jumps: f + b = 31. Net displacement: 5f - 2b = 15.
Substitute b = 31 - f into the second equation: 5f - 2(31 - f) = 15 → 5f - 62 + 2f = 15 → 7f = 77 → f = 11. Therefore, b = 20.
Therefore, the difference is |11 - 20| = 9, making Option 4 the correct answer.
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