A and B have travel-rate ratio A:B = 2:3. A takes 10 minutes more than B to cover the same distance. If A had walked at double speed, how long would A take?

Introduction / Context:
Rate–time inverse proportionality says time is proportional to 1/speed for fixed distance. With a speed ratio and time difference, we can find the individual times and then compute the effect of doubling A's speed.

Given Data / Assumptions:

• Speed ratio A:B = 2:3.
• A takes 10 minutes more than B for the same distance.
• Constant distance; uniform speeds.

Concept / Approach:
If speeds are 2k and 3k, times are proportional to 1/2k and 1/3k. Therefore t_A : t_B = 3 : 2. The difference t_A − t_B equals 10 minutes.

Step-by-Step Solution:
Let t_B = 20 min and t_A = 30 min (maintains 3:2 with 10 min difference).Doubling A's speed halves A's time → 30/2 = 15 min.

Verification / Alternative check:
Let distance be d. With speeds 2k and 3k, t_B = d/(3k) = 20 and t_A = d/(2k) = 30 are consistent for some d,k; the ratio holds and the difference is 10.

Why Other Options Are Wrong:
20, 25, 30 min do not reflect the halving correctly from A's 30-min original time.

Common Pitfalls:
Assuming time ratio equals speed ratio rather than the inverse.

Final Answer:
15 min