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03 Mar 2026, 02:25
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:56) correct
36%
(02:35)
wrong
based on 44
sessions
History
Date
Time
Result
Not Attempted Yet
At 12:00 noon yesterday, Car B passed by a monument at a constant speed of x + y miles per hour. Car A had passed the same monument m minutes earlier, traveling in the same direction as B but at a constant speed of x miles per hour. Did Car B catch up to Car A before 1:00 pm yesterday?
(1) y/(mx) = 1/45
(2) (my)/x = 5
(1) y/(mx) = 1/45
(2) (my)/x = 5
03 Mar 2026, 03:04
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kevincan
At 12:00 noon, car B crosses a monument at a speed of (x+y) miles per 60 minutes.
m minutes earlier, car A passes the monument (same direction) x miles per 60 minutes.
So, in m minutes, the distance travelled by car A is (x miles / 60 minutes) * m minutes = xm/ 60
Relative speed = (x+y) - x = y miles per 60 minutes.
Time to catch in minutes = (xm/60) ➗ (y)
= (mx/60y)
We need to find: Did Car B catch up to Car A before 1:00 pm yesterday?
Statement 1:
y/(mx) = 1/45
(mx/y) = 45
substituting, we get time of catch in minutes = (45/60) = 12:45 minutes. Which is less than 1 pm.
Hence, Sufficient
Statement 2:
(my)/x = 5
substituting, we get time of catch in minutes = m^2 / 300. m can take any values.
Not sufficient.
Option A
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