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Difficulty:
75%
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Question Stats:
63% (01:49) correct
37%
(01:58)
wrong
based on 216
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Two autonomous cleaning robots move along a long circular corridor. They start at the same time from points diametrically opposite each other on the corridor, one moving clockwise and the other moving counterclockwise. They continue without stopping at their own constant speeds. After how many seconds after they start do they meet for the second time?
(1) They meet for the first time 26 seconds after they start.
(2) One robot moves at a speed that is 3/4 the speed of the other.
I31-10
(1) They meet for the first time 26 seconds after they start.
(2) One robot moves at a speed that is 3/4 the speed of the other.
I31-10
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GMAT Club Official Explanation:
Two autonomous cleaning robots move along a long circular corridor. They start at the same time from points diametrically opposite each other on the corridor, one moving clockwise and the other moving counterclockwise. They continue without stopping at their own constant speeds. After how many seconds after they start do they meet for the second time?
The robots start from points directly opposite each other on a circular corridor and move in opposite directions at their own constant speeds. For the first time, they will meet each other after together covering half the corridor’s length. Following this first meeting, they continue to move and will meet each other again after together covering an additional full corridor length. Consequently, by the time of their second meeting, they would together have covered 0.5 + 1 = 1.5 corridor lengths in total.
(1) They meet for the first time 26 seconds after they start.
This says that their initial meeting takes place 26 seconds after they start. This implies that together it takes them 26 seconds to cover half the corridor length. Hence, to cover 1.5 corridor lengths, they would need 26*3 = 78 seconds. Sufficient.
(2) One robot moves at a speed that is 3/4 the speed of the other.
This is irrelevant because it fails to provide specific details about their individual speeds or the actual time each of them spent moving. Not sufficient.
Answer: A.
General Discussion
19 Dec 2025, 09:27
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Let the corridor circumference be c.
Let the speeds be v and u.
They start diametrically opp, so initial separation is c/2
I. First meeting at 26 sec
=> (v+u) *26 =c/2
=> c = 52(v+u)
Not sufficient
II. Speed ratio is 3:4
This give only relative speeds, no scale or anything
Not sufficient
Using I & II
With fix ratio and known c in terms of v+u, so time to cover full circle is uniquely determined. Hence the second meeting time is fixed.
Both statement together are sufficient, but neither alone is.
Answer: C
Let the speeds be v and u.
They start diametrically opp, so initial separation is c/2
I. First meeting at 26 sec
=> (v+u) *26 =c/2
=> c = 52(v+u)
Not sufficient
II. Speed ratio is 3:4
This give only relative speeds, no scale or anything
Not sufficient
Using I & II
With fix ratio and known c in terms of v+u, so time to cover full circle is uniquely determined. Hence the second meeting time is fixed.
Both statement together are sufficient, but neither alone is.
Answer: C
