Bunuel wrote:
Mark owns four low quality watches. Watch1 loses 15 minutes every hour. Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15). Watch3 loses 20 minutes every hour relative to watch2. Finally, watch4 gains 20 minutes every hour relative to watch3. If Mark resets all four watches to the correct time at 12 noon, what time will watch4 show at 12 midnight that day?
(A) 10:00
(B) 10:34
(C) 11:02
(D) 11:48
(E) 12:20
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:Before we look at the solution, think about the concept being tested here – clocks? Circular motion?
Neither!
Solution: Note that when giving data about watch1, you are told how it varies with the actual time. Data about all other watches tells us about the time they show relative to the incorrect watches. The concept being tested here is Relative Speed!
What do we mean by “gains 15 mins” or “loses 20 mins” etc? When a watch gains 15 mins every hour, it means that even though it should show that one hour has passed, it shows that 1 hr 15 mins have passed. So the watch runs faster than it should. Hence the speed of the watch is more than the speed of a correct watch. Now the question is how much more? The minute hand of the correct watch travels one full circle in one hour. The minute hand of the incorrect watch travels one full circle and then a quarter circle in one hour (to show that 1 hour 15 mins have passed even when only an hour has passed). So it is 5/4 times the speed of a correct watch. On the same lines, let’s analyze each watch.
Say the speed of a correct watch is s.
– “Watch1 loses 15 minutes every hour. “
Watch1 covers only three quarters of the circle in an hour.
Speed of watch1 = (3/4)*s
– “Watch2 gains 15 minutes every hour relative to watch1 (that is, as watch1 moves from 12:00 to 1:00, watch2 moves from 12:00 to 1:15).”
Now we have the speed of watch2 relative to speed of watch1. Speed of watch2 is (5/4) times the speed of watch1.
Speed of watch2 = (5/4)*(3/4)s = (15/16)*s
– “Watch3 loses 20 minutes every hour relative to watch2.”
Watch3 loses 20 mins every hour means its speed is (2/3)rd the speed of watch2
Speed of watch3 = (2/3)*(15/16)*s = (5/8)*s
– “Finally, watch4 gains 20 minutes every hour relative to watch3.”
Speed of watch4 = (4/3)*Speed of watch3 = (4/3)*(5/8)*s = (5/6)*s
So the speed of watch4 is (5/6)th the speed of a correct watch. So if a correct watch shows that 6 hours have passed, watch4 will show that 5 hours have passed. If a correct watch shows that 12 hours have passed, watch4 will show that 10 hours have passed. From 12 noon to 12 midnight, a correct watch would have covered 12 hours. Watch4 will cover 10 hours and will show the time as 10:00.
Answer (A)
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