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Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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07 Apr 2015, 05:21
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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13 Apr 2015, 06:21
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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Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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07 Apr 2015, 20:47
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. let the speed of a correct minute hand be S (1 revolution in 60mins) . Speed of watch1 = \(\frac{3}{4}*S\) Speed of watch2 relative to watch1 =\(\frac{5}{4} *\frac{3}{4} * S\) Speed of watch3 relative to watch2 =\(\frac{2}{3}*\frac{5}{4} *\frac{3}{4} * S\) = \(\frac{5}{8}* S\) Speed of Watch4 relative to watch3= \(\frac{5}{8}* \frac{4}{3}* S\) =\(\frac{5}{6} * S\) i.e. watch4 will lose 1/6 of 1 hour ,every hour. total loss in 12 hours = 1/6 * 12 hrs = 2hrs. So Watch4 will show 1000 PM . Answer A.




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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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07 Apr 2015, 14:09
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. Quite long way to solve but I can't find another: Watch1 from 12 noon till 12 midnight = 12 hours; delay 15 min 12 * 15 = 180 = 3 hours; 12 3 = 9 hours pass on Watch1 Watch2 +15 minutes on each hour from Watch1: 9 * 15 = 135 minutes: 9 hours + 135 minutes = 11 hours 15 minutes pass on Watch2 Watch 3 20 minutes on each hour from Watch2: 11 hours 15 minutes * 20 = 225 minutes; 11 hours 15 minutes  225 minutes = 7 hours 30 min pass on Watch3 Watch 4 +20 minutes on each hour from Watch3: 7 hours 30 minutes * 20 = 150 minutes; 7 hours 30 minutes + 150 minutes = 10 hours pass on Watch4 So on Watch4 from 12 noon till 12 midnight will pass 10 hours and answer is A
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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07 Apr 2015, 21:01
Harley1980 wrote: 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. Quite long way to solve but I can't find another: Watch1 from 12 noon till 12 midnight = 12 hours; delay 15 min 12 * 15 = 180 = 3 hours; 12 3 = 9 hours pass on Watch1 Watch2 +15 minutes on each hour from Watch1: 9 * 15 = 135 minutes: 9 hours + 135 minutes = 11 hours 15 minutes pass on Watch2 Watch 3 20 minutes on each hour from Watch2: 11 hours 15 minutes * 20 = 225 minutes; 11 hours 15 minutes  225 minutes = 7 hours 30 min pass on Watch3 Watch 4 +20 minutes on each hour from Watch3: 7 hours 30 minutes * 20 = 150 minutes; 7 hours 30 minutes + 150 minutes = 10 hours pass on Watch4 So on Watch4 from 12 noon till 12 midnight will pass 10 hours and answer is A Check: http://www.veritasprep.com/blog/2014/12 ... thegmat/Question 2 in this post discusses this question using the concept of relative speed.
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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07 Apr 2015, 21:40
VeritasPrepKarishma wrote: Harley1980 wrote: 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. Quite long way to solve but I can't find another: Watch1 from 12 noon till 12 midnight = 12 hours; delay 15 min 12 * 15 = 180 = 3 hours; 12 3 = 9 hours pass on Watch1 Watch2 +15 minutes on each hour from Watch1: 9 * 15 = 135 minutes: 9 hours + 135 minutes = 11 hours 15 minutes pass on Watch2 Watch 3 20 minutes on each hour from Watch2: 11 hours 15 minutes * 20 = 225 minutes; 11 hours 15 minutes  225 minutes = 7 hours 30 min pass on Watch3 Watch 4 +20 minutes on each hour from Watch3: 7 hours 30 minutes * 20 = 150 minutes; 7 hours 30 minutes + 150 minutes = 10 hours pass on Watch4 So on Watch4 from 12 noon till 12 midnight will pass 10 hours and answer is A Check: http://www.veritasprep.com/blog/2014/12 ... thegmat/Question 2 in this post discusses this question using the concept of relative speed. Hi karishma On gmat what is the level of this question? 600650 ?



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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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08 Apr 2015, 21:18
Lucky2783 wrote: On gmat what is the level of this question? 600650 ?
No. 650 to 700 (closer to 700).
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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09 Apr 2015, 07:09
W1(watch 1: loss by 15 min)=12(15*12)/60=9 PM W2(gain by 15 min with respect to W1) = 9+ (15*540/60)=9+ 135 min=11 hr +15 min W3 (Loss by 20 min WRT W2) = 11hr 15min  (20*135/60)=11hr 15min –(180min)= 7hr30min W4(Gain by 20 min WRTW3)= 7hr30min+(20*450/60)= 7hr30min+150min=10 hrs
Hence the time on W4 WRT to W1 will be 10 PM
Hence answer is A Thanks,



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Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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09 Apr 2015, 07:54
I must have missed something... w1 = 45/h (45m/h) w2 = 45+15/h (60m/h) w3 = 45+1520/h (40m/h) w4 = w2= 45+1520+20/h (60m/h) Now we ignore everything but w4. which has 60m/h. As such, it will move just like any ordinary clock. I thought 12:00 would be correct, but I must have misread something. edit: Found my mistake when reading through the replies. But I figure I'll leave this up here for anyone interested in one mistake that easily can be done.
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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09 Apr 2015, 19:49
MarkusKarl wrote: I must have missed something... w1 = 45/h (45m/h) w2 = 45+15/h (60m/h) w3 = 45+1520/h (40m/h) w4 = w2= 45+1520+20/h (60m/h) Now we ignore everything but w4. which has 60m/h. As such, it will move just like any ordinary clock. I thought 12:00 would be correct, but I must have misread something. edit: Found my mistake when reading through the replies. But I figure I'll leave this up here for anyone interested in one mistake that easily can be done. Whenever you add/subtract something, ensure that the units of the quantities are same and that the units make sense. What is m/h? You don't have any quantity with units Min/hour. That is time/time and leads to no units. You cannot depict speed using min/hour. Also, to that, you cannot add 15 mins.
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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09 Apr 2015, 21:17
VeritasPrepKarishma wrote: MarkusKarl wrote: I must have missed something... w1 = 45/h (45m/h) w2 = 45+15/h (60m/h) w3 = 45+1520/h (40m/h) w4 = w2= 45+1520+20/h (60m/h) Now we ignore everything but w4. which has 60m/h. As such, it will move just like any ordinary clock. I thought 12:00 would be correct, but I must have misread something. edit: Found my mistake when reading through the replies. But I figure I'll leave this up here for anyone interested in one mistake that easily can be done. Whenever you add/subtract something, ensure that the units of the quantities are same and that the units make sense. What is m/h? You don't have any quantity with units Min/hour. That is time/time and leads to no units. You cannot depict speed using min/hour. Also, to that, you cannot add 15 mins. Yeah, I figured that out. I thought he question meant that w1 in relation to w2 was 45:60, not 60:75. But thanks for your help!
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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Re: Mark owns four low quality watches. Watch1 loses 15 minutes every hour
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