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

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07 Jul 2003, 06:23

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00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

34% (02:58) correct
66% (02:03) wrong based on 254 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)?

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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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

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07 Jul 2003, 07:52

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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AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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

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07 Jul 2003, 13:33

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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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Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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

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07 Jan 2004, 00:13

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)?

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

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07 Jan 2004, 08:40

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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