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Interesting Work Problem

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Interesting Work Problem

by GHong14 » Sat Nov 27, 2010 4:31 pm
I was just taking the Practice GMAT Test for the 2nd time and came across an interesting work problem that I was not sure if I solved correctly. It went something like:

There were three machines one can finish the job in 2 hours, one is 3 hours and one in 4 hours. If the job was finished in 1.5 hours. How many of the fastest machines was used to finish the job. THIS IS NOT THE EXACT WORDING.

Has anyone seen this problem before?!?!?!? If so what is the proper setup or approach to solve something like this?
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by goyalsau » Sat Nov 27, 2010 5:12 pm
GHong14 wrote:I was just taking the Practice GMAT Test for the 2nd time and came across an interesting work problem that I was not sure if I solved correctly. It went something like:

There were three machines one can finish the job in 2 hours, one is 3 hours and one in 4 hours. If the job was finished in 1.5 hours. How many of the fastest machines was used to finish the job. THIS IS NOT THE EXACT WORDING.

Has anyone seen this problem before?!?!?!? If so what is the proper setup or approach to solve something like this?
Machine 1, will do 1/2 work in one hour,
Machine 2 , will do 1/3 work in one hour
Machine 3, will do 1/4 in one work,

Now for the work to be completed in 1.5 hours 2/3 for the work should be completed in 1 hour.

And i am not getting that combination IN any Form ,

Machine 1 and Machine 2 Together,
1/2 + 1/3 = 5/6

Machine 2 and Machine 3 Together,
1/3 + 1/4 = 7/12

Machine 1 and Machine 3 Together,
1/2 + 1/4 = 3/4

In any Case we are not getting 2/3 so i don't understand How the work can be done in 1.5 hours,
Saurabh Goyal
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by diebeatsthegmat » Sat Nov 27, 2010 5:35 pm
GHong14 wrote:I was just taking the Practice GMAT Test for the 2nd time and came across an interesting work problem that I was not sure if I solved correctly. It went something like:

There were three machines one can finish the job in 2 hours, one is 3 hours and one in 4 hours. If the job was finished in 1.5 hours. How many of the fastest machines was used to finish the job. THIS IS NOT THE EXACT WORDING.

Has anyone seen this problem before?!?!?!? If so what is the proper setup or approach to solve something like this?
i find the answer is 1 ( my answer)
whats the answer you get??? and whats the explanation?
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by rishab1988 » Sun Nov 28, 2010 1:28 am
No one will ever get answer to this question because,according to me,the poster has posted a DS question.No matter what you do you can't solve it.

Here is my reasoning:

Let a be the no of machines that can complete the job in 2 hrs.
Let Ra denote rate at which each machine works

Then a *Ra*2= 1 [1 job]

Ra = 1/2a

Similarly Let b and c denote the number of machines that complete work in 3 hrs and 4 hrs.[You can't assume that the no of machines are same because the question doesn't say SO!]

Rb = 1/3b

Rc = 1/4c

Combined rate = Ra+Rb+Rc
= (6bc+4ac+3ab)/12abc

Time taken = 3/2 hrs

(6bc+4ac+3ab)/12abc *3/2 = 1

6bc+4ac+3ab = 8abc
6bc = 8abc-4ac-3ab
6bc = a [ 8bc-4c-3b]

a [no of fastest machines] = 6bc/(8bc-4c-3b)

See you need to know the no of machines b and c.

Even if you assume that the number of each machine is same as the others,the question doesn't make any sense.

b=c=a

a = 6a^2/(8a^2-7a)

8a^3 - 7a^2 -6a^2=0
8a^3-13a^2=0

a^2 [8a-13]=0

Since a cannot be 0 [ because then no work would be done]

8a-13=0
a=13/8

But number of machines CANNOT be a FRACTION.
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by goyalsau » Sun Nov 28, 2010 1:41 am
rishab1988 wrote:No one will ever get answer to this question because,according to me,the poster has posted a DS question.No matter what you do you can't solve it.

Here is my reasoning:

Let a be the no of machines that can complete the job in 2 hrs.
Let Ra denote rate at which each machine works

Then a *Ra*2= 1 [1 job]

Ra = 1/2a

Similarly Let b and c denote the number of machines that complete work in 3 hrs and 4 hrs.[You can't assume that the no of machines are same because the question doesn't say SO!]

Rb = 1/3b

Rc = 1/4c

Combined rate = Ra+Rb+Rc
= (6bc+4ac+3ab)/12abc

Time taken = 3/2 hrs

(6bc+4ac+3ab)/12abc *3/2 = 1

6bc+4ac+3ab = 8abc
6bc = 8abc-4ac-3ab
6bc = a [ 8bc-4c-3b]

a [no of fastest machines] = 6bc/(8bc-4c-3b)

See you need to know the no of machines b and c.

Even if you assume that the number of each machine is same as the others,the question doesn't make any sense.

b=c=a

a = 6a^2/(8a^2-7a)

8a^3 - 7a^2 -6a^2=0
8a^3-13a^2=0

a^2 [8a-13]=0

Since a cannot be 0 [ because then no work would be done]

8a-13=0
a=13/8

But number of machines CANNOT be a FRACTION.
I completely Agree with you Rishab , People post wrong question and expect Proper or correct solutions.............
hahahaha...........................................
Saurabh Goyal
[email protected]
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