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?
- (a) 6
- (b) 7
- (c) 8
- (d) 9 ✓ UPSC's answer
Why the answer is (d)
• Let $f$ be the number of forward jumps and $b$ be the number of backward jumps.
• The total number of jumps is given as 31, so $f + b = 31$.
• The net displacement is 15 feet forward, where each forward jump adds 5 feet and each backward jump subtracts 2 feet, so $5f - 2b = 15$.
• Multiply the first equation by 2 to get $2f + 2b = 62$.
• Add this to the second equation: $(2f + 2b) + (5f - 2b) = 62 + 15$, which simplifies to $7f = 77$.
• Solving for $f$ gives $f = 11$. Substituting back, $b = 31 - 11 = 20$.
• The difference between the number of forward and backward jumps is $|f - b| = |11 - 20| = 9$.
Why the other options are wrong
- (a) 6
- Option (a) is incorrect because the calculated difference between forward (11) and backward (20) jumps is 9, not 6.
- (b) 7
- Option (b) is incorrect because the calculated difference between forward (11) and backward (20) jumps is 9, not 7.
- (c) 8
- Option (c) is incorrect because the calculated difference between forward (11) and backward (20) jumps is 9, not 8.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2026, held on 24 May 2026. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.