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Machine job

Expert replies
by gmat009 » Thu Nov 13, 2008 7:14 pm
Machine A can complete a certain job in x hours. Machine B can complete the same job in y hours. If A and B work together at their respective rates to complete the job, which of the following represents the fraction of the job that B will not have to complete because of A's help?
a. x - y/x + y
b. x/y – x
c. x + y/xy
d. y/x – y
e. y/x + y
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Source: — Problem Solving |

by cramya » Thu Nov 13, 2008 7:37 pm
Is it E)? OA please?
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by cramya » Thu Nov 13, 2008 7:42 pm
Machine A can complete the job in x hours
B in y hours

Together they can complete it in xy/x+y hours

In x hours A can complete 1 job
In xy/x+y hours A completes 1/x * xy/x+y = y/x+y

So y/x+y of the job B does not have to do

E)

Other way of doing this would be to calculate fraction of job B completes which will be x/x+y and then subtract it from 1

i.e 1- x/x+y = y/x+y
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by gmat009 » Thu Nov 13, 2008 8:21 pm
cramya wrote:Machine A can complete the job in x hours
B in y hours

Together they can complete it in xy/x+y hours

In x hours A can complete 1 job
In xy/x+y hours A completes 1/x * xy/x+y = y/x+y

So y/x+y of the job B does not have to do

E)

Other way of doing this would be to calculate fraction of job B completes which will be x/x+y and then subtract it from 1

i.e 1- x/x+y = y/x+y
YES OA is E
This is exactly what I was looking for
Thanks
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by logitech » Thu Nov 13, 2008 8:36 pm
deleted
Last edited by logitech on Thu Nov 13, 2008 8:37 pm, edited 1 time in total.
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
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by logitech » Thu Nov 13, 2008 8:37 pm
cramya wrote: Together they can complete it in xy/x+y hours

In x hours A can complete 1 job
In xy/x+y hours A completes 1/x * xy/x+y = y/x+y
So y/x+y of the job B does not have to do
E)
Beautiful solution Cramya. 8)
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
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by raunekk » Fri Nov 14, 2008 12:41 am
Machine A can complete the job in x hours
B in y hours
Together they can complete it in xy/x+y hours

In x hours A can complete 1 job
In xy/x+y hours A completes 1/x * xy/x+y = y/x+y

So y/x+y of the job B does not have to do

E)

Other way of doing this would be to calculate fraction of job B completes which will be x/x+y and then subtract it from 1

i.e 1- x/x+y = y/x+y[/quote

sweet :) !!!!
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by canuckclint » Sat Nov 15, 2008 7:53 pm
I don't quite get it :(
I am trying to think of it in terms of work done:

Total work = 1 = 1/x + 1/y
A's work 1/x
B's work 1/y

Ans: total - A

1 - 1/x ??
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