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M18-12

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M18-12  [#permalink]

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New post 16 Sep 2014, 00:03
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A
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Difficulty:

  45% (medium)

Question Stats:

65% (01:24) correct 35% (01:31) wrong based on 54 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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Re M18-12  [#permalink]

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New post 16 Sep 2014, 00:03
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.

Answer: B
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Re: M18-12  [#permalink]

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New post 15 Dec 2014, 10:52
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.

Answer: B


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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Re: M18-12  [#permalink]

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New post 16 Dec 2014, 03:09
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.

Answer: B


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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Re: M18-12  [#permalink]

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New post 09 Jun 2016, 13:53
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.
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Re: M18-12  [#permalink]

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New post 09 Jun 2016, 21:35
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M18-12  [#permalink]

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New post 22 Aug 2016, 08:28
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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
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Re M18-12  [#permalink]

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New post 01 Jun 2017, 21:46
Question stem is quite confusing.
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Re: M18-12  [#permalink]

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New post 17 Jun 2018, 03:28
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!
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Re: M18-12  [#permalink]

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New post 29 Nov 2018, 23:29
Brilliant question. Didn't see the word 'joins.' So ended up with 7Pm Tuesday, which is wrong :?
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Re: M18-12 &nbs [#permalink] 29 Nov 2018, 23:29
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