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megafan
Joined: 28 May 2009
Last visit: 15 Oct 2017
Posts: 137
Given Kudos: 91
Location: United States
Concentration: Strategy, General Management
Schools: Booth PT '18 Stern PT Fall'18
GMAT Date: 03-22-2013
GPA: 3.57
WE:Information Technology (Consulting)
24 Mar 2013, 10:47
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
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Question Stats:
71% (01:48) correct
29%
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Pat’s watch gains an extra 10 seconds every 2 hours. Kim’s watch loses 5 seconds every 3 hours. If both watches are set to the correct time at 8 o’clock in the morning and run without interruption, after 72 hours, what will be the difference in time between Pat’s watch and Kim’s watch?
(A) 4 min
(B) 6 min
(C) 6 min 40 sec
(D) 7 min 30 sec
(E) 8 min
PS03258
(A) 4 min
(B) 6 min
(C) 6 min 40 sec
(D) 7 min 30 sec
(E) 8 min
PS03258
24 Mar 2013, 10:53
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In 72 hours Pat's watch will gain 360 seconds, since in 2 hours it gains 10 seconds, that means in 1 hour 5 seconds and in 72 hours 72*5 seconds, i.e. 360 seconds or 6 mins.
Kim's watch loses 5 seconds in 3 hours, in 72 hours it will loose 72/3 * 5 = 24*5 = 120 seconds = 2 mins.
Total difference 6 - (-2, it lost 2 mins ) = 6 + 2 = 8 mins.
Please give a kudo if you like my explanation.
Kim's watch loses 5 seconds in 3 hours, in 72 hours it will loose 72/3 * 5 = 24*5 = 120 seconds = 2 mins.
Total difference 6 - (-2, it lost 2 mins ) = 6 + 2 = 8 mins.
Please give a kudo if you like my explanation.
24 Mar 2013, 11:11
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First we have to bring everything to the same scale, to make life easier:
\(10 sec ->2h = 5 sec -> 1h\)
\(5 sec->3h = \frac{5}{3}sec->1h\)
Pat's watch will gain a total of \(5 sec * 72 h = 360 sec\)
Kim's watch will lose a total of\(5/3 sec * 72 h = 120 sec\)
<----------------|-------------------->
(-120sec)----- 0------------( +360 sec)
tot distance = \(120+360=480 sec = 8 min\)
E
\(10 sec ->2h = 5 sec -> 1h\)
\(5 sec->3h = \frac{5}{3}sec->1h\)
Pat's watch will gain a total of \(5 sec * 72 h = 360 sec\)
Kim's watch will lose a total of\(5/3 sec * 72 h = 120 sec\)
<----------------|-------------------->
(-120sec)----- 0------------( +360 sec)
tot distance = \(120+360=480 sec = 8 min\)
E
