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A not-so-good clockmaker has four clocks on display in the w 10 Nov 2011, 16:15
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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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A
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A not-so-good clockmaker has four clocks on display in the w Updated on: 06 Sep 2023, 22:54
Originally posted by KarishmaB on 10 Nov 2011, 23:19.
Last edited by KarishmaB on 06 Sep 2023, 22:54, edited 1 time in total.
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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

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?
Show more

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

Check this post for more details: https://anaprep.com/arithmetic-gaining- ... on-clocks/
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A not-so-good clockmaker has four clocks on display in the w 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.

The correct answer is A.
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A not-so-good clockmaker has four clocks on display in the w 11 Nov 2011, 21:40
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Good explanation.......:)
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A not-so-good clockmaker has four clocks on display in the w 12 Nov 2011, 13:48
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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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A not-so-good clockmaker has four clocks on display in the w 12 Nov 2011, 14:41
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Thanks Capricorn369. Please see explanation from Karishma and Blink 005. Answer is A.
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A not-so-good clockmaker has four clocks on display in the w 12 Nov 2011, 18:01
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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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A not-so-good clockmaker has four clocks on display in the w 12 Nov 2011, 18:02
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enigma123
Thanks Capricorn369. Please see explanation from Karishma and Blink 005. Answer is A.
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@enigma123 - Thanks for responding. My assumption were incorrect and i'm convinced that answer should be A.
Cheers!
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A not-so-good clockmaker has four clocks on display in the w 16 Jun 2014, 19:07
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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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A not-so-good clockmaker has four clocks on display in the w 24 Jul 2014, 03:36
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Does such questions appear on actual Gmat test ?
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A not-so-good clockmaker has four clocks on display in the w 24 Jul 2014, 04:46
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Does such questions appear on actual Gmat test ?
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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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A not-so-good clockmaker has four clocks on display in the w 30 Nov 2014, 23:48
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VeritasPrepKarishma
enigma123
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
Show more

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 ?
Show more

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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A not-so-good clockmaker has four clocks on display in the w 09 Nov 2015, 03:55
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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 09 Nov 2015, 20:54
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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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A not-so-good clockmaker has four clocks on display in the w 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

Winkherehere :!:
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A not-so-good clockmaker has four clocks on display in the w 31 Jul 2020, 07:25
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enigma123
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
Show more
Solution:

When it is actually 6:00 pm, Clock #1 will display 4:30pm because this clock loses 15 minutes for each hour, and in six hours, it will lose 6 x 15 = 90 minutes. Notice that we find 4:30 pm if we go back 90 minutes = 1.5 hours from 6:00 pm.

When Clock #1 is displaying 4:30pm (i.e. when Clock #1 moved a total of 4.5 hours), Clock #2 will have gained 4.5 * 15 = 9/2 * 15 = 67.5 minutes relative to Clock #1. Thus, Clock #2 will display 4:30 + 67.5 minutes = 5:30 + 7.5 minutes = 5:37:30.

We see that Clock #2 moved a total of 5 hours + 37.5 minutes = 337.5 minutes. Since Clock #3 loses 20 minutes every hour relative to Clock #2, Clock #3 will have lost (337.5/60) * 20 = 337.5/3 = 112.5 minutes. Thus, Clock #3 will have moved 337.5 - 112.5 = 225 minutes = 3 hours 45 minutes. Thus, it will display 12 pm + 3 hours 45 minutes = 3:45 pm.

Finally, when Clock #3 moved 3 hours 45 minutes, Clock #4 will have gained 3 hr 45 min * 20 = 15/4 * 20 = 75 minutes relative to Clock #3. Thus, Clock #4 will display 3:45 pm + 75 minutes = 5:00 pm.

Answer: A
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A not-so-good clockmaker has four clocks on display in the w 01 Nov 2020, 03:16
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VeritasKarishma


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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shouldn't it be 1/3 instead of 2/3 VeritasKarishma ?
cause clock three losses 20 minutes out of 60 minutes
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A not-so-good clockmaker has four clocks on display in the w Updated on: 03 Nov 2020, 05:03
Originally posted by Fdambro294 on 01 Nov 2020, 09:53.
Last edited by Fdambro294 on 03 Nov 2020, 05:03, edited 1 time in total.
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Starting with Clock 1:
For every 1 hour that passes—————-> loses - 1/4 hour

For 6 hours that pass—————-> loses -(1/4) * 6 = -(3/2)hr

Actual hours that pass on Clock 1:
(6 hours) - (3/2 hr) = (12/2) - (3/2) = 9/2 hours pass on Clock 1


Clock 2:
For every 1 hour that passes on Clock 1 ———-> Gains +(1/4) hour

9/2 hours pass on Clock 1 ———> Gains (9/2) * (1/4) = +(9/8) hr

Hours that Pass on Clock 2 = (9/2) + (9/8) = (36/8) + (9/8) = 45/8 hours pass on Clock 2


Clock 3:
For every 1 hour that passes on Clock 2 ———-> Loses - (1/3) hour

45/8 hours pass on Clock 2 ———> Losses (45/8) * (1/3) = -(15/8) hr

Hours that Pass on Clock 3 = (45/8) - (15/8) = 30/8 hours pass on Clock 3



Clock 4:
For every 1 hour that passes on Clock 3 ———-> Gain +1/3 hour

30/8 hours pass on Clock 3 ——-> Gains (30/8) * (1/3) = +(10/8) hr

Hours that pass on Clock 4 = (30/8) + (10/8) = (40/8) = 5 hours pass



Clock 4 is set at 12 Noon

+ 5 hours pass
_____________________

Time shown on Clock 4 is

5:00 PM

-A-

Posted from my mobile device
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A not-so-good clockmaker has four clocks on display in the w 03 Nov 2020, 00:09
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hkkat
VeritasKarishma


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

shouldn't it be 1/3 instead of 2/3 VeritasKarishma ?
cause clock three losses 20 minutes out of 60 minutes
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hkkat: Clock 3 loses 20 min out of 60 means it covers only 40 mins when it should cover 60 mins. So its speed is only (2/3)rd of the speed of a correct clock.
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Schools: McCombs '26 (D) McCombs (WD) Tepper '26 (WL) Owen '26 (D) ISB '25 (D) Foster (D) Marshall (D) Goizueta '26 (D) Questrom '26 (WL) Mendoza '26 (D) Rice Jones '26 (WL) Kelley '26 (D) McDonough '26 (WL)
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WE:Analyst (Other)
Schools: McCombs '26 (D) McCombs (WD) Tepper '26 (WL) Owen '26 (D) ISB '25 (D) Foster (D) Marshall (D) Goizueta '26 (D) Questrom '26 (WL) Mendoza '26 (D) Rice Jones '26 (WL) Kelley '26 (D) McDonough '26 (WL)
GMAT 2: 650 Q46 V35
Posts: 107
Kudos: 18
A not-so-good clockmaker has four clocks on display in the w 28 Aug 2021, 08:58
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Is there any way to do this by listing out the times for every hour?

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