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# M18-12

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Math Expert
Joined: 02 Sep 2009
Posts: 55802

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16 Sep 2014, 01:03
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Difficulty:

35% (medium)

Question Stats:

67% (01:36) correct 33% (01:32) wrong based on 61 sessions

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It takes Computer A 36 hours to process data. If Computer A starts working at 7 pm on Monday and after half the data are processed Computer B, three times as fast as Computer A, joins in the task, when will the data processing be finished?

A. 4:30 pm Tuesday
B. 5:30 pm Tuesday
C. 7:00 pm Tuesday
D. 8:30 pm Tuesday
E. 10:30 pm Tuesday

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Math Expert
Joined: 02 Sep 2009
Posts: 55802

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16 Sep 2014, 01:03
Official Solution:

It takes Computer A 36 hours to process data. If Computer A starts working at 7 pm on Monday and after half the data are processed Computer B, three times as fast as Computer A, joins in the task, when will the data processing be finished?

A. 4:30 pm Tuesday
B. 5:30 pm Tuesday
C. 7:00 pm Tuesday
D. 8:30 pm Tuesday
E. 10:30 pm Tuesday

Computer A will process the first half of the data in 18 hours. Computer A and Computer B working together will process the remaining half four times quicker than will Computer A working alone. The entire task will require $$18 + \frac{18}{4} = 22.5$$ hours which is 1.5 hours short of the 24-hour cycle. If the processing commences at 7 pm on Monday, it will finish at 5:30 pm on Tuesday.

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Joined: 01 Nov 2013
Posts: 289
GMAT 1: 690 Q45 V39
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15 Dec 2014, 11:52
Bunuel wrote:
Official Solution:

It takes Computer A 36 hours to process data. If Computer A starts working at 7 pm on Monday and after half the data are processed Computer B, three times as fast as Computer A, joins in the task, when will the data processing be finished?

A. 4:30 pm Tuesday
B. 5:30 pm Tuesday
C. 7:00 pm Tuesday
D. 8:30 pm Tuesday
E. 10:30 pm Tuesday

Computer A will process the first half of the data in 18 hours. Computer A and Computer B working together will process the remaining half four times quicker than will Computer A working alone. The entire task will require $$18 + \frac{18}{4} = 22.5$$ hours which is 1.5 hours short of the 24-hour cycle. If the processing commences at 7 pm on Monday, it will finish at 5:30 pm on Tuesday.

How is B possible when it is clearly mentioned that B is 3 times faster;B should take 6 hours to complete the remaining job i.e. 18/3=6 hours and not 18/4=4.5 Hours........
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Math Expert
Joined: 02 Sep 2009
Posts: 55802

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16 Dec 2014, 04:09
1
samichange wrote:
Bunuel wrote:
Official Solution:

It takes Computer A 36 hours to process data. If Computer A starts working at 7 pm on Monday and after half the data are processed Computer B, three times as fast as Computer A, joins in the task, when will the data processing be finished?

A. 4:30 pm Tuesday
B. 5:30 pm Tuesday
C. 7:00 pm Tuesday
D. 8:30 pm Tuesday
E. 10:30 pm Tuesday

Computer A will process the first half of the data in 18 hours. Computer A and Computer B working together will process the remaining half four times quicker than will Computer A working alone. The entire task will require $$18 + \frac{18}{4} = 22.5$$ hours which is 1.5 hours short of the 24-hour cycle. If the processing commences at 7 pm on Monday, it will finish at 5:30 pm on Tuesday.

How is B possible when it is clearly mentioned that B is 3 times faster;B should take 6 hours to complete the remaining job i.e. 18/3=6 hours and not 18/4=4.5 Hours........

You should read a question more carefully.

We are told that after half the data are processed Computer B joins in the task. Hence after the data are processed A and B are working together on the remaining half. Since B is three times as fast as A, then A and B together are 4 times as fast as A.

Hope it's clear.
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Joined: 25 Jun 2013
Posts: 10

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09 Jun 2016, 14:53
Can someone show the math that proves that A+B = 4x as fast as A alone? I conceptually understand (I think) but it would be nice to see the numbers work out.
Math Expert
Joined: 02 Sep 2009
Posts: 55802

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09 Jun 2016, 22:35
glochou wrote:
Can someone show the math that proves that A+B = 4x as fast as A alone? I conceptually understand (I think) but it would be nice to see the numbers work out.

A does 1 unit of work in the same time as B does 3 units of work. Thus together they do 4 units of work in the same time interval.
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Joined: 06 May 2016
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22 Aug 2016, 09:28
1
A= 36hrs
B= three times more efficient than A = 36/3=12 Hrs

1/2 the work is already completed because A has done work for 18hrs

so equ will be

Rate * t= work
(1/36+1/12)t = 1/2

t=4.5 Hrs

18+4.5 =22.5 Hrs
Intern
Joined: 09 Jul 2016
Posts: 17
GMAT 1: 730 Q50 V39

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01 Jun 2017, 22:46
Question stem is quite confusing.
Intern
Joined: 19 Nov 2012
Posts: 28

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17 Jun 2018, 04:28
Hello Experts Bunuel and chetan2u

I typically get confused between the following kind of question prompts:

Computer B three times as fast as Computer A
The way I understand it is A takes 36 hours. B is 3 times faster so should take 1/3 A's time so B takes 12 hours.

Computer B three times faster than Computer A
How do you go about writing out this one?
Please provide an elaborate reply for understanding purposes rather than some really smart short cut trick or something.

Thanks!
Intern
Joined: 10 Aug 2018
Posts: 8
Location: India
Schools: Erasmus '20, NTU '20
GMAT 1: 570 Q43 V25
WE: Real Estate (Real Estate)

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30 Nov 2018, 00:29
Brilliant question. Didn't see the word 'joins.' So ended up with 7Pm Tuesday, which is wrong
Re: M18-12   [#permalink] 30 Nov 2018, 00:29
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# M18-12

Moderators: chetan2u, Bunuel