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Three copying machines A, B, and C, working together at their

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Three copying machines A, B, and C, working together at their  [#permalink]

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New post Updated on: 13 Aug 2018, 05:56
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Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #1

Three copying machines A, B, and C, working together at their respective constant rates, can do a copying work in 2 hours. B and C, working together at their respective constant rates, can do the same copying job in 4 hours. How many hours would it take A, working alone at its constant rate, to do the same job?

    A. 2 hours
    B. 3 hours
    C. 4 hours
    D. 5 hours
    E. 6 hours


To solve question 2: Question 2

To read the article: Solve Time and Work Problems Efficiently using Efficiency Method!

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Originally posted by EgmatQuantExpert on 23 May 2018, 04:12.
Last edited by EgmatQuantExpert on 13 Aug 2018, 05:56, edited 1 time in total.
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Re: Three copying machines A, B, and C, working together at their  [#permalink]

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New post Updated on: 26 May 2018, 13:11
2

Solution



Given:
In this question it is mentioned that
    • Three printing presses A, B, and C, working together at their respective constant rates, take 2 hours to do a certain printing job.
    • However, if only B and C are working, they can complete the same printing job in 4 hours.

To find:
    • The number of days A alone will take to complete the job

Approach and Working:
As the job remains same, we can assume the total job to be the 12 units (multiple of the LCM of 2 and 4)
    • A, B, and C together take 2 hours to complete 12 units of work
      • Therefore, in 1 hour all of them will do \(\frac{12}{2}\) = 6 units of work
    • B and C together take 4 hours to complete 12 units of work
      • Therefore, in 1 hour they will do \(\frac{12}{4}\) = 3 units of work
Now, in 1-hour time, A, B, and C together do 6 units, out of which B and C do 3 units work
    • Therefore, in 1 hour, A will do (6 – 3) = 3 units of work
So, working alone at this constant rate, A will take \(\frac{12}{3}\) days = 4 hours to complete the same job.

Hence, the correct answer is option C.

Answer: C

Important Observation


    • As mentioned in the article, when we assume the total job to be the LCM of the given number of hours, the calculation becomes much easy as we can avoid fractional calculations. However, in this case we have assumed the total job to be a multiple of the LCM, which shows not only the LCM but any multiple of LCM value can give you the same answer.

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Originally posted by EgmatQuantExpert on 26 May 2018, 11:35.
Last edited by EgmatQuantExpert on 26 May 2018, 13:11, edited 2 times in total.
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Re: Three copying machines A, B, and C, working together at their  [#permalink]

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New post 26 May 2018, 11:48
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The rate at which A,B & C work together is \(\frac{1}{2}\) [i.e. half of one work is done in one hour by A, B & C working together]
The rate at which B & C work together is \(\frac{1}{4}\) [i.e. one-fourth of one work is done in one hour by B & C working together]
The rate at which A works is \(\frac{1}{2}-\frac{1}{4} = \frac{1}{4}\) [i.e. one-fourth of one work is done in one hour by Machine A alone]
Therefore the time taken by A alone is 4 hours.
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Re: Three copying machines A, B, and C, working together at their  [#permalink]

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New post 23 May 2018, 09:03
EgmatQuantExpert wrote:
Three copying machines A, B, and C, working together at their respective constant rates, can do a copying work in 2 hours. B and C, working together at their respective constant rates, can do the same copying job in 4 hours. How many hours would it take A, working alone at its constant rate, to do the same job?

    A. 2 hours
    B. 3 hours
    C. 4 hours
    D. 5 hours
    E. 6 hours


Let the total work be 4 units...

So, Efficiency of A + B + C = 2 unit/hr.

Further Efficiency of B + C = 1 unit/hr

So, Efficiency of A = 1 unit/Hr

So, the time required by A to complete the work will be 4 Hours, answer must be (C)
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Re: Three copying machines A, B, and C, working together at their  [#permalink]

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New post 26 May 2018, 12:52
LCM Method:-

A, B & C working together take 2 hours
B & C working together take 4 hours
A alone ?

LCM of 2 & 4 is 4 therefore:-
A, B & C working together make 4/2, i.e. 2 units per hour
B & C working together make 4/4, i.e. 1 units per hour
There A alone will make 1 units per hour. To make 4 units, Machine A will take 4 hours at the rate of 1 units per hour
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Re: Three copying machines A, B, and C, working together at their  [#permalink]

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New post 03 Oct 2018, 00:24
Rate of work of A,B & C together=1/2 (work/hr)
Rate of work of B & C together=1/4 (work/hr)
Rate of work of A=1/2-1/4= 1/4 (work/hr)
So, time taken by A to finish job alone= 4 hrs

Ans-(C)
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Re: Three copying machines A, B, and C, working together at their  [#permalink]

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New post 13 Apr 2019, 19:15
EgmatQuantExpert wrote:
Solve Time and Work Problems Efficiently using Efficiency Method! - Exercise Question #1

Three copying machines A, B, and C, working together at their respective constant rates, can do a copying work in 2 hours. B and C, working together at their respective constant rates, can do the same copying job in 4 hours. How many hours would it take A, working alone at its constant rate, to do the same job?

    A. 2 hours
    B. 3 hours
    C. 4 hours
    D. 5 hours
    E. 6 hours




Let a, b, and c be the number of hours A, B, and C take to finish the job alone, respectively. Their respective rates are 1/a, 1/b, and 1/c. We have:

1/a + 1/b + 1/c = 1/2

and

1/b + 1/c = 1/4

Subtracting the second equation from the first, we have:

1/a = 1/4

a = 4

Answer: C
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Re: Three copying machines A, B, and C, working together at their   [#permalink] 13 Apr 2019, 19:15
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