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…

The correct option is D - 9.

Solution

Let the number of forward jumps be f and the number of backward jumps be b.
We are given two conditions:
1. Total number of jumps is 31:
$f + b = 31$f+b=31 --- (Equation 1)
2. Total distance moved is 15' forward (taking forward as positive and backward as negative):
$5f - 2b = 15$5f−2b=15 --- (Equation 2)

To solve for f and b, we can use the substitution or elimination method.
From Equation 1, we get: $b = 31 - f$b=31−f
Substitute this into Equation 2:
$5f - 2(31 - f) = 15$5f−2(31−f)=15
$5f - 62 + 2f = 15$5f−62+2f=15
$7f = 15 + 62$7f=15+62
$7f = 77$7f=77
$f = 11$f=11

Now, find b using Equation 1:
$11 + b = 31$11+b=31
$b = 20$b=20

The question asks for the difference between the number of forward and backward jumps:
Difference = $|f - b| = |11 - 20| = 9$∣f−b∣=∣11−20∣=9.

Why the other options are incorrect

• Option (a) - 6: This would imply a different distribution of jumps (e.g., 12.5 and 18.5), which is impossible as the number of jumps must be an integer, and it does not satisfy the distance equation.
• Option (b) - 7: If the difference were 7, the jumps would be 12 and 19. Calculating distance: $(12 \times 5) - (19 \times 2) = 60 - 38 = 22$(12×5)−(19×2)=60−38=22', which does not match the required 15'.
• Option (c) - 8: A difference of 8 with a total of 31 jumps is mathematically impossible for integers (31 is odd, so the difference between the two parts must also be odd).

Key Concept

Linear equations in two variables based on total count and weighted sum constraints.