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12 Jan 2026, 11:17
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At 9:00 AM, a man and his dog leave their home and walk along the same path toward a park. The man walks at a constant speed, while the dog runs at a faster constant speed. The dog reaches the park at 9:30 AM, spends 20 minutes playing with other dogs, and then runs back toward the man at its usual constant rate. They meet at 10:00 AM, and from there, both walk together at the man’s speed to the park. When will the man reach the park?
A. 10:10 AM
B. 10:15 AM
C. 10:20 AM
D. 10:30 AM
E. 10:40 AM
A. 10:10 AM
B. 10:15 AM
C. 10:20 AM
D. 10:30 AM
E. 10:40 AM
12 Jan 2026, 11:17
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Official Solution:
At 9:00 AM, a man and his dog leave their home and walk along the same path toward a park. The man walks at a constant speed, while the dog runs at a faster constant speed. The dog reaches the park at 9:30 AM, spends 20 minutes playing with other dogs, and then runs back toward the man at its usual constant rate. They meet at 10:00 AM, and from there, both walk together at the man’s speed to the park. When will the man reach the park?
A. 10:10 AM
B. 10:15 AM
C. 10:20 AM
D. 10:30 AM
E. 10:40 AM
When the man and the dog meet at 10:00 AM, the man has been walking for 60 minutes, while the dog has been running from 9:50 AM to 10:00 AM for 10 minutes. The distance from home to the park is the total distance covered by the man in 60 minutes and the dog in 10 minutes.
We know the dog covered the entire distance to the park in 30 minutes, so in 10 minutes, the dog would have covered one-third of this distance. This means that in 60 minutes, the man covered two-thirds of the distance to the park.
If the man covers two-thirds of the distance in 60 minutes, it would take him 90 minutes to cover the entire distance. Therefore, the man will reach the park at 9:00 AM plus 90 minutes, which means he will arrive at 10:30 AM.
Answer: D
At 9:00 AM, a man and his dog leave their home and walk along the same path toward a park. The man walks at a constant speed, while the dog runs at a faster constant speed. The dog reaches the park at 9:30 AM, spends 20 minutes playing with other dogs, and then runs back toward the man at its usual constant rate. They meet at 10:00 AM, and from there, both walk together at the man’s speed to the park. When will the man reach the park?
A. 10:10 AM
B. 10:15 AM
C. 10:20 AM
D. 10:30 AM
E. 10:40 AM
When the man and the dog meet at 10:00 AM, the man has been walking for 60 minutes, while the dog has been running from 9:50 AM to 10:00 AM for 10 minutes. The distance from home to the park is the total distance covered by the man in 60 minutes and the dog in 10 minutes.
We know the dog covered the entire distance to the park in 30 minutes, so in 10 minutes, the dog would have covered one-third of this distance. This means that in 60 minutes, the man covered two-thirds of the distance to the park.
If the man covers two-thirds of the distance in 60 minutes, it would take him 90 minutes to cover the entire distance. Therefore, the man will reach the park at 9:00 AM plus 90 minutes, which means he will arrive at 10:30 AM.
Answer: D
04 Mar 2026, 03:28
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Woah great question. Definitely worth 800+ purely because of the visualisation required to get to an elegant solution.
I took lot of time and did the algebraic method and landed at 30mins from 10 or 10:30.
We get multiple relations with D for both dog and the man
We eventually landed up with a speed relation that dog's speed is 3x that of the man's.
On resubstituting we land up with 1.5-1 from 10am or 0.5hrs or 30 mins.
I took lot of time and did the algebraic method and landed at 30mins from 10 or 10:30.
We get multiple relations with D for both dog and the man
We eventually landed up with a speed relation that dog's speed is 3x that of the man's.
On resubstituting we land up with 1.5-1 from 10am or 0.5hrs or 30 mins.
Bunuel
