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Computer A requires 36 hours to process data. If Computer A starts

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mba9now
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Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   07 Aug 2008, 22:31
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Computer A requires 36 hours to process data. If Computer A starts working at 7 pm on Monday, and Computer B, which is three times as fast as Computer A, joins in after half the data is processed, when will the data processing be complete?

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


M18-12

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B
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PareshGmat
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   12 Nov 2014, 02:25
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Rate of Computer A = \(\frac{1}{36}\)

Rate of Computer B = \(\frac{1}{12}\)

\(\frac{1}{2}\) Job completed by Computer A alone; means time consumed by Computer A \(= \frac{1}{2} * 36 = 18\) Hrs

Combined rate of A & B \(= (\frac{1}{36} + \frac{1}{12}) = \frac{1}{9}\)

Time required to complete remaining \(\frac{1}{2}\) Job \(= \frac{1}{2} * \frac{9}{1} = 4.5 Hrs\)

Total Time required to complete job = 18 + 4.5 = 22.5 Hrs

24 Hrs is SAME TIME next day, so 22.5 Hrs would be

7PM Tuesday - 1.5 Hrs = 5:30 PM Tuesday

Answer = B
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paneer
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   07 Aug 2008, 22:40
mba9now wrote:
It would take Computer A 36 hours to process the data. If Computer A starts working at 7 pm on Monday and after half the data is processed Computer B, three times as fast as Computer A, is joined in the task, when will the data processing be finished?


4:30 pm Tuesday
5:30 pm Tuesday
7:00 pm Tuesday
8:30 pm Tuesday
10:30 pm Tuesday


comp A takes 36h and comp B takes 12h

comp A starts at 7pm on monday, processes half the data so it finishes at 7pm+(36/2)h which is 1pm on tuesday. Comp B takes up the next half so it finishes at 1pm+(12/2)h=7pm on Tuesday
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   07 Aug 2008, 22:54
The answer is B (5:30 PM Tuesday). Becasue it is said that Comp B joins A, never says only B did the work alone.
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   08 Aug 2008, 08:42
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mba9now wrote:
It would take Computer A 36 hours to process the data. If Computer A starts working at 7 pm on Monday and after half the data is processed Computer B, three times as fast as Computer A, is joined in the task, when will the data processing be finished?


4:30 pm Tuesday
5:30 pm Tuesday
7:00 pm Tuesday
8:30 pm Tuesday
10:30 pm Tuesday


computer A take 36 hr
computer B takes 12 hr

18 hrs (half job done by computer) + Half job dont by both
= 18 hrs + 18 / (3+1) = 22.5 hrs.

B is the answer
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   08 Aug 2008, 21:17
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mba9now wrote:
It would take Computer A 36 hours to process the data. If Computer A starts working at 7 pm on Monday and after half the data is processed Computer B, three times as fast as Computer A, is joined in the task, when will the data processing be finished?


4:30 pm Tuesday
5:30 pm Tuesday
7:00 pm Tuesday
8:30 pm Tuesday
10:30 pm Tuesday


work done by A in one hr = w/36
by B in 1 hr=w/12
by both in 1 hr=w/12 + w/36=w/9

hence half the work is already done by A in 18 hrs

half the work w/2 is to be done by both hence time talken is w/2/w/9=9/2=4.5 hrs
total time taken is 18+4.5 hrs=22.5 hrs
5.30pm tues is answer
IMO B
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   08 Aug 2008, 23:07
mba9now wrote:
It would take Computer A 36 hours to process the data. If Computer A starts working at 7 pm on Monday and after half the data is processed Computer B, three times as fast as Computer A, is joined in the task, when will the data processing be finished?


4:30 pm Tuesday
5:30 pm Tuesday
7:00 pm Tuesday
8:30 pm Tuesday
10:30 pm Tuesday


Total hrs = 18 + Hours required to do half job = 18 + (18*6)/24 = 22.5 hrs = 5:30pm Tue
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Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   10 Nov 2014, 02:38
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Computer A requires 36 hours to process data. If Computer A starts working at 7 pm on Monday, and Computer B, which is three times as fast as Computer A, joins in after half the data is processed, when will the data processing be complete?

A. 4:30 pm on Tuesday
B. 5:30 pm on Tuesday
C. 7:00 pm on Tuesday
D. 8:30 pm on Tuesday
E. 10:30 pm on 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 faster (since Computer B is three times as fast as Computer A, together they are four times as fast as Computer A alone). The entire task will require \(18 + \frac{18}{4} = 22.5\) hours, which is 1.5 hours short of the 24-hour cycle. Therefore, if the processing starts at 7 pm on Monday, it will complete at 5:30 pm on Tuesday.


Answer: B
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   20 Dec 2015, 11:46
A= 1/36
B= 1/12 ( Ratio is 3:1 )

A+B= 1/9

HALF WORK DONE BY A only
w=r*t
1/2 = 1/36*t
t= 18 hrs

Rest work done by A & B together

w=r*T
1/2 = 1/9*T
T= 4.5 hrs


Total Time= t+T

18+ 4.5 hrs = 22.5 hrs

Count from 7 pm Monday, answer is 5:30 pm Tuesday
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   20 Dec 2015, 22:18
Step wise:

A: 36 hours
B: 12 hours ( 3 times faster)

after half work done: 18 hours

W/2 left: Now W work can be done by both machines in 12*36/48 = 9 hours. So, W/2 cqan be done in 4.5 hours.

Total = 22.5 hours
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   20 Dec 2015, 22:20
x2suresh wrote:
mba9now wrote:
It would take Computer A 36 hours to process the data. If Computer A starts working at 7 pm on Monday and after half the data is processed Computer B, three times as fast as Computer A, is joined in the task, when will the data processing be finished?


4:30 pm Tuesday
5:30 pm Tuesday
7:00 pm Tuesday
8:30 pm Tuesday
10:30 pm Tuesday


computer A take 36 hr
computer B takes 12 hr

18 hrs (half job done by computer) + Half job dont by both
= 18 hrs + 18 / (3+1) = 22.5 hrs.

B is the answer


Hey, can you please explain how did you get 18/3+1 ?
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   12 May 2018, 03:22
mba9now wrote:
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

M18-12


"R" stands for rate

R(A) = 1/36

Time taken to complete half the job = 18 hours

R(B) = 1/12 (given)

Job left after 18 hrs = 1/2

(Combined rate)* T = 1/2

1/36 + 1/12 * T = 1/2

1/9 * T = 1/2

T = 4.5

Total time = 18 + 4.5 = 22.5

7 p.m. Monday + 22.5 hrs

5:30 pm Tuesday

(B)
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink] New post   19 Oct 2023, 12:23
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Re: Computer A requires 36 hours to process data. If Computer A starts [#permalink]
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