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# Cassandra sets her watch to the correct time at noon. At the actual ti

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Math Expert
Joined: 02 Sep 2009
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Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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20 Mar 2019, 22:59
00:00

Difficulty:

95% (hard)

Question Stats:

18% (02:57) correct 82% (02:42) wrong based on 62 sessions

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Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads 10:00 PM?

(A) 10:22 PM and 24 seconds
(B) 10:24 PM
(C) 10:25 PM
(D) 10:27 PM
(E) 10:30 PM

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Re: Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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20 Mar 2019, 23:39
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1
Wen clock reads 57.6 minutes,actual time is 60 minutes.
So,when clock reads 10 hours,i.e. 600 minutes, actual time is 60*600/57.6= 625 minutes= 10 hours 25 minutes.
Hence,C

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Re: Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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21 Mar 2019, 00:22
2
Bunuel wrote:
Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads 10:00 PM?

(A) 10:22 PM and 24 seconds
(B) 10:24 PM
(C) 10:25 PM
(D) 10:27 PM
(E) 10:30 PM

total time in watch by 1 PM = 12 :57: 36 sec so the watch has done = 57*60 + 36 = 3456 seconds
watch is lagging every hour by 3600-3456 = 144 seconds
so in 10 hrs watch will lag by 144* 10 = 1440 seconds

now by 10 PM total seconds moved = 10*60*60 = 36000 seconds
so the actual time in watch must be by 36000/1440 = 25 ahead ; i.e 10:25
IMO C
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Re: Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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21 Mar 2019, 01:55
1
Bunuel

should the seconds by which her watch is slow be divided by 60 so as to get the correct answer?
Her watch is slow by a total of 1440 seconds in 10 hours so that means if her watch shows 10 PM then the actual time has to be ahead by 1440 seconds right?

Pls correct me if im wrong
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Re: Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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22 Mar 2019, 04:08
Exactly my thoughts, if by every hour the watch lags by 144 seconds, at the end of 10 hours shouldn't it have lagged by 1440 seconds which equals 24 minutes?
Thus the time shown should be 10:24 p.m?
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Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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22 Mar 2019, 10:03
2
abdul96 wrote:
Exactly my thoughts, if by every hour the watch lags by 144 seconds, at the end of 10 hours shouldn't it have lagged by 1440 seconds which equals 24 minutes?
Thus the time shown should be 10:24 p.m?

The time of 10 pm is shown in the watch lagging time,not the watch running correctly.
For every 57 minutes 36 sec,i.e. 57.6 minutes on the incorrect watch,the correct time is 1 hour,i.e. 60 minutes.
So,for 10 hours or 600 minutes on incorrect watch,correct time is 60*600/57.6 minutes which is equal to 625 minutes or 10 hours 25 minutes

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Re: Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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23 Mar 2019, 11:13
Bunuel wrote:
Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads 10:00 PM?

(A) 10:22 PM and 24 seconds
(B) 10:24 PM
(C) 10:25 PM
(D) 10:27 PM
(E) 10:30 PM

let t=actual time
t/10*3600 sec=10/10*3456 sec
t=10.41 hours=10:25 PM
C
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Re: Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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23 Mar 2019, 19:50
the clock loses 2 minutes and 24 seconds every 60 minutes.

2.4/60=24/600=1/25. The clock loses 1 minute every 25 minutes. So when the correct clock goes 25 minutes the bad clock goes 24 minutes. We can think of it this way. The real clock GAINS a minute every time the bad clock goes 24 minutes.
Since the bad clock went for 600 minutes (10:00 pm is 10 hours after noon) we can create this proportion 1/24=x/600.
Solving for x we obtain 25 minutes. It must have gained 25 minutes. So 10:00+25 is 10:25.
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Re: Cassandra sets her watch to the correct time at noon. At the actual ti  [#permalink]

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24 Mar 2019, 18:34
1
1
Bunuel wrote:
Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads 10:00 PM?

(A) 10:22 PM and 24 seconds
(B) 10:24 PM
(C) 10:25 PM
(D) 10:27 PM
(E) 10:30 PM

We see that Cassandra’s watch loses 2 minutes and 24 seconds each hour or 144 seconds per hour. So if the correct time is 10 PM, her watch will lose 1440 seconds or 1440/60 = 24 minutes and the time on her watch would read 9:36 PM. Therefore, when her watch reads 10:00 PM (“24 minutes” later), the correct time should be slightly over 10:24 PM. Of the answer choices, we can eliminate A and B. Let’s guess that correct answer should be 10:25 PM, that is, for the correct time to elapse 25 minutes, the time on her watch only elapses 24 minutes. In other words, her watch loses 1 minute for every 25 minutes. Let’s verify that is indeed the case. We know that her watch losses 144 seconds every hour (i.e., every 60 minutes). So we can set up the proportion:

144 sec/60 min = x sec/25 min

60x = 144(25)

60x = 3600

x = 60

We see that her watch indeed losses 60 seconds (i.e., 1 minute) for every 25 minutes. So when her watch reads 10:00 PM, the correct time should be 10:25 PM.

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Re: Cassandra sets her watch to the correct time at noon. At the actual ti   [#permalink] 24 Mar 2019, 18:34
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