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

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A not-so-good clockmaker has four clocks on display in the [#permalink] New post 07 Jul 2003, 05:23
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

33% (00:00) correct 67% (00:35) wrong based on 15 sessions
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
[Reveal] Spoiler: OA

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 [#permalink] New post 07 Jul 2003, 06:52
D. is NOT correct. Hint: Say you buy stock for X amount. It then falls by 25%, then subsequently rises by 25%, do you get back to the same place?
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 [#permalink] New post 07 Jul 2003, 09:47
Hi,

I answer i got, is 324 minutes that is 5:24.

Now, i am doubtful about the options.

Can you through more light on that...
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 [#permalink] New post 07 Jul 2003, 12:33
The correct answer is (A). Clock #1 is 15 minutes slow. This means that after one actual hour, the clock shows that only 45 minutes have gone by.

Clock #2 is 15 minutes fast relative to Clock #1. That means that after one hour on Clock #1, Clock #2 moves ahead 60 + 15 or 75 minutes. This ALSO means that Clock #2 is running 75/60 = 5/4 times as fast as Clock #1. Therefore, after one actual hour of elapsed time, Clock #1 moves 45 minutes and Clock #2 moves 45 * 5/4 minutes.

Clock #3 is 20 minutes slow relative to clock one. That means that after one hour on Clock #2, Clock #3 moves ahead 60 - 20 or 40 minutes. This ALSO means that Clock #3 s running 40/60 = 2/3 times as fast as Clock #2. Hence, after one actual hour of elapsed time, Clock #1 moves 45 minutes, Clock #2 moves 45 * 5/4 minutes, and Clock #3 moves 45 * 5/4 * 2/3 minutes.

Clock #4 is 20 minutes fast relative to Clock #3. That means that after one hour on Clock #3, Clock #4 moves ahead 60 + 20 or 80 minutes. This ALSO means that Clock #4 is running 80/60 = 4/3 times as fast as Clock #3. Hence, after one actual hour of elapsed time, Clock #1 moves 45 minutes and Clock #2 moves 45 * 5/4 minutes, Clock #3 moves 45 * 5/4 * 2/3 minutes, and Clock #4 moves 45 * 5/4 * 2/3 * 4/3 = 50 minutes which is equivalent to saying that Clock #4 loses 10 minutes per actual hour.

At 6 p.m., six actual hours have gone by since all of the clocks were reset, hence Clock #4 loses 6 * 10 = 60 minutes, i.e., Clock #4 is an hour slow. Hence, the apparent time on Clock #4 is 5:00 p.m. and the correct answer is (A).
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 [#permalink] New post 08 Jul 2003, 04:40
Hey, i just learnt my mistake here...

I was calculating it as 125/100*80/100*120/100(time lost by clock 1).

The base chosen was wrong. I used 100 as base instead of 60. Ooops...!!!

Well thanks Akamaibrah.... for a lovely problem.
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A not-so-good clockmaker has four clocks on display in the [#permalink] New post 06 Jan 2004, 23:13
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


could you explain your method please?
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 [#permalink] New post 07 Jan 2004, 07:40
The answer is A.

Let me expound:

Afer six actual hours :

I clock: after six hours , the first clock displayed 4:30, because it lost 6 x 15 min = 90 min. Thus 6:00 - 1:30 =4:30

II clock gains 15 min per hour relative to Clock 1. It gained 15 x 4.5 = 67,5 minutes. => Clock 2 displayed 4:30:00 +1:07:30 = 5 hous 37 minutes and 30 seconds.
III clock: its loss relative to Clock 2 was 5 x 20 + (37,5 / 60) x 20= 100 +0,625 x 20=112,5 minutes. => 5:37:30 -1:52:30 = 3:45

Finally : Clock # 4 - gained 3 x 20 + 0,75 x 20 = 1 h 15 min => 3:45 +1:15 = 5:00 => A
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 [#permalink] New post 07 Jan 2004, 07:45
Even I found the answer to be A. Is this a GMAT problem ? I took more than five mins to solve this rather understand this.
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 [#permalink] New post 07 Jan 2004, 18:03
Geethu wrote:
Even I found the answer to be A. Is this a GMAT problem ? I took more than five mins to solve this rather understand this.



A is the correct answer.

this is a MGMAT problem
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 [#permalink] New post 26 Jan 2004, 13:17
6 * ( 1-1/4) (1+1/4)(1-1/3)(1+1/3)
1/4 hr is 15 minutes and 1/3 hr is 20 minutes

6 * ( 1-1/16) * ( 1-1/9 ) = 6 * 15/16 * 8/9 = 6 * 5/6 = 5
Sothe 4th clock should just display 5 hrs after 12
  [#permalink] 26 Jan 2004, 13:17
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