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Difficulty:
35%
(medium)
Question Stats:
81% (01:51) correct
19%
(02:08)
wrong
based on 121
sessions
History
Date
Time
Result
Not Attempted Yet
John usually walks to work at a certain speed, but today, due to walking at only 3/5 of his usual speed, he arrives 10 minutes late. How long would John have taken to reach his destination if he had walked at his usual speed?
A. 10 minutes
B. 15 minutes
C. 20 minutes
D. 25 minutes
E. 30 minutes
A. 10 minutes
B. 15 minutes
C. 20 minutes
D. 25 minutes
E. 30 minutes
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M47-31
14 Feb 2026, 10:15
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Bunuel
GMAT Club Official Solution:
Let John’s usual time be t minutes. Since time is inversely proportional to speed for the same distance, walking at 3/5 of his usual speed makes his time t/(3/5) = 5/3 * t.
5/3 * t = t + 10
2/3 * t = 10
t = 15
Answer: B.
General Discussion
06 Feb 2026, 10:41
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This is a classic speed-distance-time problem, and once you set it up correctly, the math is very clean.
Step 1: Define variables.
Let John's usual speed be s, and let the usual time to reach work be t. Then the distance to work is d = s * t.
Step 2: Set up the equation for today.
Today he walks at 3/5 of his usual speed, so his speed is (3/5)s. Since the distance doesn't change:
d = (3/5)s * t_new
Since d = s * t, we get:
s * t = (3/5)s * t_new
t_new = (5/3) * t
Step 3: Use the "10 minutes late" information.
The extra time is:
t_new - t = (5/3)t - t = (2/3)t = 10 minutes
Solving: t = 10 * (3/2) = 15 minutes.
Answer: B (15 minutes)
Common trap: Some students set up the ratio backwards and get 25 minutes. Remember — slower speed means more time, so t_new must be greater than t. The multiplier for time is the inverse of the speed fraction: speed is 3/5 of normal, so time is 5/3 of normal.
Takeaway: Whenever a speed-distance-time problem gives you a fraction of the usual speed and the resulting time difference, set up the equation (new time) - (old time) = difference. The distance cancels out, and you're left with a simple equation.
Step 1: Define variables.
Let John's usual speed be s, and let the usual time to reach work be t. Then the distance to work is d = s * t.
Step 2: Set up the equation for today.
Today he walks at 3/5 of his usual speed, so his speed is (3/5)s. Since the distance doesn't change:
d = (3/5)s * t_new
Since d = s * t, we get:
s * t = (3/5)s * t_new
t_new = (5/3) * t
Step 3: Use the "10 minutes late" information.
The extra time is:
t_new - t = (5/3)t - t = (2/3)t = 10 minutes
Solving: t = 10 * (3/2) = 15 minutes.
Answer: B (15 minutes)
Common trap: Some students set up the ratio backwards and get 25 minutes. Remember — slower speed means more time, so t_new must be greater than t. The multiplier for time is the inverse of the speed fraction: speed is 3/5 of normal, so time is 5/3 of normal.
Takeaway: Whenever a speed-distance-time problem gives you a fraction of the usual speed and the resulting time difference, set up the equation (new time) - (old time) = difference. The distance cancels out, and you're left with a simple equation.
