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A not-so-good clockmaker has four clocks on display in the w

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A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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10 Nov 2011, 16:15
9
42
00:00

Difficulty:

95% (hard)

Question Stats:

39% (03:18) correct 61% (02:49) wrong based on 329 sessions

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A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15). Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day)?

A. 5:00
B. 5:34
C. 5:42
D. 6:00
E. 6:24

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10 Nov 2011, 23:19
31
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23
enigma123 wrote:
A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15). Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day)?

A)5:00
B)5:34
C)5:42
D)6:00
E)6:24

Guys - again the official answer is not provided. But I have done the Maths to get D. Do you think D is the right answer? Can someone explain to me if you think D is not the right answer?

Here is my approach to such problems:

1. Loses 15 mins every hour (i.e. 60 mins) means the clock's speed is a fourth less than the normal speed. So the clock's speed is (3/4)th the normal speed. It covers only 45 mins in the time in which a correct clock covers 60 mins.
2. Gains 20 mins every hour means the clock's speed is a third more than normal speed. So the clock's speed is (4/3) times the normal speed. It covers 1 hr 20 mins in the time in which a correct clock covers 1 hr.

Let speed of a correct clock = s
Speed of clock 1 = (3/4)s
Speed of clock 2 = (3/4)s * (5/4) = (15/16)s
Speed of clock 3 = (15/16)s * (2/3) = (5/8)s
Speed of clock 4 = (5/8)s * (4/3) = (5/6)s

Speed of the fourth clock is (5/6)th the normal speed. If a correct covers 6 hrs, the fourth clock will cover 5 hrs. Hence, the time shown will be 5:00 pm
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10 Nov 2011, 22:09
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C1 loses 15 minutes every hour. So after 60 minutes have passed, C1 displays that 60-15 = 45 minutes have passed.

C2 gains 15 minutes for every 60 minutes displayed on C1. Thus, the time displayed on C2 is 75/60 = 5/4 the time displayed on C1. So after 60 minutes have passed, C2 displays the passing of (5/4 * 45) minutes.

C3 loses 20 minutes for every 60 minutes displayed on C2. Thus, the time displayed on C3 is 40/60 = 2/3 the time displayed on C2. So after 60 minutes have passed, C3 displays the passing of (2/3 * 5/4 * 45) minutes.

C4 gains 20 minutes for every 60 minutes displayed on C3. Thus, the time displayed on C4 is 80/60 = 4/3 the time displayed on clock 3. So after 60 minutes have passed, C4 displays the passing of 4/3 * 2/3 * 5/4 * 45 = 50 minutes.

C4 loses 10 minutes every hour.
In 6 hours, C4 will lose 6*10 = 60 minutes = 1 hour.
Since the correct time after 6 hours will be 6pm, C4 will show a time of 6-1 = 5pm.

General Discussion
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11 Nov 2011, 21:40
Good explanation.......
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12 Nov 2011, 13:48
1
Here is my take on this question. My answer is D.
I'll solve this with examples. Below is the movement of each cocks based on 12:00 noon as a reference.

Clock #1 moves from 12:00 to 1:00
Clock #2 moves from 12:00 to 1:15
Clock #3 moves from 12:00 to 12:55
Clock #4 moves from 12:00 to 1:15
Note - movement of Clock #2 and Clock #4 are same that makes life easier.

Time in each Clock at actual 6:00 pm that same day -
Clock #1 moves from 12:00 to 4:30
Clock #2 moves from 12:00 to 6:00
Clock #3 moves from 12:00 to 12:55 (Desb't matter)
Clock #4 moves from 12:00 to 6:00

Based on the calculation above. My answer is D.
Cheers!
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12 Nov 2011, 14:41
_________________
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12 Nov 2011, 18:01
My way is identical to Karishma's.
(1) 3/4 speed
(2) 4/3 speed

let s = speed of clock
1) 3/4s
2) (3/4)(5/4)s = 15/16s
3) (15/16)(2/3)s = 5/8s
4) (5/8)(4/3)s = 5/6s

so 5/6s * 6 = 5 o'clock
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12 Nov 2011, 18:02
enigma123 wrote:

@enigma123 - Thanks for responding. My assumption were incorrect and i'm convinced that answer should be A.
Cheers!
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Re: A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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16 Jun 2014, 19:07
cleverly written problem, I was also confused by the (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15), as this seems to indicate that Clock #2 properly offsets Clock #1's 15min/hr loss and displays the the real time

however, the 15 minutes clock 1 loses is relative to an hour of real time and not an hour of clock 1's distorted time. So when it reads 1:00 on clock 1, it is not 1:15 in real time, but rather 1:20 --> RT/C1 = (1m)/(3/4m) --> RT = C1(4/3)

So for D to be correct, it would have to read something like as Clock #1 moves from 12:00 to 12:45, Clock #2 moves from 12:00 to 1, etc etc
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Re: A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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24 Jul 2014, 03:36
Does such questions appear on actual Gmat test ?
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Re: A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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24 Jul 2014, 04:46
GmatDestroyer2013 wrote:
Does such questions appear on actual Gmat test ?

There is nothing out of the ordinary in this question. It is based on relative speed concepts so you should know how to do it.
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Re: A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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30 Nov 2014, 23:48
VeritasPrepKarishma wrote:
enigma123 wrote:
A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15). Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day)?

A)5:00
B)5:34
C)5:42
D)6:00
E)6:24

Guys - again the official answer is not provided. But I have done the Maths to get D. Do you think D is the right answer? Can someone explain to me if you think D is not the right answer?

Here is my approach to such problems:

1. Loses 15 mins every hour (i.e. 60 mins) means the clock's speed is a fourth less than the normal speed. So the clock's speed is (3/4)th the normal speed. It covers only 45 mins in the time in which a correct clock covers 60 mins.
2. Gains 20 mins every hour means the clock's speed is a third more than normal speed. So the clock's speed is (4/3) times the normal speed. It covers 1 hr 20 mins in the time in which a correct clock covers 1 hr.

Let speed of a correct clock = s
Speed of clock 1 = (3/4)s
Speed of clock 2 = (3/4)s * (5/4) = (15/16)s
Speed of clock 3 = (15/16)s * (2/3) = (5/8)s
Speed of clock 4 = (5/8)s * (4/3) = (5/6)s

Speed of the fourth clock is (5/6)th the normal speed. If a correct covers 6 hrs, the fourth clock will cover 5 hrs. Hence, the time shown will be 5:00 pm

Responding to a pm:

Quote:
how did you get (5/4) for the speed of clock 2?

I understand that clock 1 looses 15 min every hour hence 3/4s
Clock 2 gains 20 mins every hour . hence 4/3 s ?

We are given that "Clock #2 gains 15 minutes every hour relative to Clock #1"

This means that if the speed of clock 1 is s1, the speed of clock 2 is s1 + s1/4 (i.e. clock 2 covers 15 extra mins for every 60 mins covered by clock 1). This gives us that speed of clock 2 is 5s1/4.

Since the speed of clock 1 itself is 3s/4, the speed of clock 2 will be (5/4)*(3s/4) = 15s/16
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Re: A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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09 Nov 2015, 03:55
Hi all can ull help me in clarifying this doubt?
I think for clock 1: is losing 15 min means it will show 45 min instead of 1hr but for clock 2 which given in relation with with and mentioned in the question as well it is 15 min ahead of clock 1 and not the real time.
So inf the question mentioned as (clock #1 moves from 12:00 to 12:45,clock2# moves from 12:00 to 1:00) this example considers actual 1 hour

Questionz example (clock#1 moves from 12:00 to 1:00, clock2# from from 12:00 to 1.15) this shows that actual for clock 1 is 1.15 as it loses 15 min every hour and thus implies clock 2 is on time.

Same logic for clock 3 & 4

Hence practically ans marked as A could be a miss print, since we never consider the probability of a miss print, leaving clock 4 # with ans D in opinion.

Please clarify if i have missed any point?
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A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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09 Nov 2015, 20:54
60 minutes x [(3/4)(5/4)(2/3)(4/3)]=60(5/6)=50 minutes
clock 4 loses 10 minutes per hour➡1 hour per 6 hours
at actual 6:00pm clock 4 will show 5:00pm
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Re: A not-so-good clockmaker has four clocks on display in the w  [#permalink]

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18 Nov 2016, 00:21
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Actual time 06:00 pm

Clock 1 (losses 15 mins/hr relative to actual time) shows : 04:30 pm

Clock 2 (gains 15 mins/hr relative to clock 1) shows : 04:30 pm + (4x15 + (15/2) = 67.5 mins) = 05:37.5 pm

Clock 3 (losses 20 mins/hr relative to clock 2) shows : 05:37.5 pm - (5x20 + [(1/3)x37.5= 12.5] = 112.5 mins) = 03:45 pm

Clock 4 (gains 20 mins/hr relative to clock 3) shows : 03:45 + (3x20 + [(1/3)x45= 15] = 75 mins) = 05:00 pm

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Re: A not-so-good clockmaker has four clocks on display in the  [#permalink]

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21 Aug 2019, 00:07
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Re: A not-so-good clockmaker has four clocks on display in the   [#permalink] 21 Aug 2019, 00:07
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