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Work rate problem, tricky one

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Work rate problem, tricky one

by razorback » Thu Nov 03, 2011 7:11 pm
From an MGMAT CAT

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?

A) (x-y) / (x+y)

B) x / (y-x)

C) (x+y) / xy

D) y / (x-y)

E) y / (x+y)

OA: E

The MGMAT explains the problem by using the VIC method (plugging in numbers) which seems ridiculously time consuming.

Can someone demonstrate the solution using standard x and y work equations? Thanks.
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by saketk » Thu Nov 03, 2011 7:57 pm
razorback wrote:From an MGMAT CAT

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?

A) (x-y) / (x+y)

B) x / (y-x)

C) (x+y) / xy

D) y / (x-y)

E) y / (x+y)

OA: E

The MGMAT explains the problem by using the VIC method (plugging in numbers) which seems ridiculously time consuming.

Can someone demonstrate the solution using standard x and y work equations? Thanks.


Why do you think that plugging in the numbers would take more time?

If you replace x by 2 and y by 3 and solve this question then it will not take more than 60 secs to solve this.

Method 1: -

X= 2
Y= 3

Total time taken when A and B work together -

(3*2)/(3+2) = 6/5 hrs

Time taken by B alone = 3 hrs. i.e. in 1 hr work done = 1/3

So, work done in 6/5 hrs = 2/5
Work not done by B = 1-2/5 = 3/5

Put X =2 and Y=3 in the option and see which expression gives you 3/5

Correct answer OPTION E
Last edited by saketk on Thu Nov 03, 2011 8:01 pm, edited 1 time in total.
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by saketk » Thu Nov 03, 2011 8:01 pm
Method 2: -

Time taken when A and B work together = xy/(x+y)

in 1 hr B does 1/y work

so, in xy/(x+y) hrs work done by B = 1/y* [xy/(x+y)] = x/(x+y)

So, work not done by B = 1- [x/(x+y)] = y/(x+y)

OPTION E
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by fcabanski » Thu Nov 03, 2011 8:02 pm
That's a simple problem, and plugging in numbers takes less time.
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by user123321 » Fri Nov 04, 2011 4:12 am
A does 1/x work in 1 hr
B does 1/y work in 1 hr
Both does 1/x+1/y work in 1 hr or they take xy/(x+y) hrs to do 1 work

the fraction of work B will not complete if they do together is ...(1/x+1/y-1/y) = 1/x in 1 hr
But in xy/(x+y) hrs it will be (1/x).(xy/(x+y)) = y/(x+y)

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by Anurag@Gurome » Fri Nov 04, 2011 6:02 am
razorback wrote:From an MGMAT CAT

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?

A) (x-y) / (x+y)
B) x / (y-x)
C) (x+y) / xy
D) y / (x-y)
E) y / (x+y)

OA: E

The MGMAT explains the problem by using the VIC method (plugging in numbers) which seems ridiculously time consuming.

Can someone demonstrate the solution using standard x and y work equations? Thanks.
1 hr work of machine A = 1/x
1 hr work of machine B = 1/y
1 hr work of A + B together = 1/x + 1/y = (x + y)/xy
So, A + B together together work for xy/(x + y) hours
Now,fraction of the job that B will not have to complete means fraction of job done by machine A = (1/x) * xy/(x + y) = y/(x + y)

The correct answer is E.
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