A takes 4 hours more than B to walk a fixed distance D. If A doubles her speed, she can complete D in 2 hours less than B. How much time (in hours) does B require to walk D at his usual speed?

Introduction / Context:This is a time comparison with speed changes. If A normally takes 4 hours longer than B, but halving A’s time (by doubling speed) makes A finish 2 hours faster than B, you can form equations in terms of the unknown times.

Given Data / Assumptions:

• Let B’s time = t hours.
• A’s time = t + 4 hours.
• If A doubles speed, her time halves: (t + 4)/2 hours, which is 2 hours less than B’s time t.

Concept / Approach:Equation: (t + 4)/2 = t − 2. Solve for t to get B’s time directly without needing distances or speeds explicitly.

Step-by-Step Solution:

Verification / Alternative check:A’s usual time = 12 h. If A doubles speed, A’s time becomes 6 h, which is indeed 2 hours less than B’s 8 h.

Why Other Options Are Wrong:4, 6, 7, 9 hours do not satisfy (t + 4)/2 = t − 2.

Common Pitfalls:Forgetting that doubling speed halves time; setting up the inequality rather than the strict equation.

Final Answer:8 h