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Running at their respective constant rates, machine X takes

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Manager
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Running at their respective constant rates, machine X takes [#permalink] New post 29 Oct 2008, 23:54
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A. 4
B. 6
C. 8
D. 10
E. 12
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Re: Combine work Q [#permalink] New post 30 Oct 2008, 07:08
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E.

Let t be number of days.
Rate of Y = w/t
Rate of X = w/(t+2)

When they work together, the number of widgets produced is (combined rate)*time_taken
(X+Y)*3 = 5w/4

substituting X and Y with their rate
w/(t+2) + w/t = 5w/4

Simplifying above equation...
5(t^2) - 12t - 24 = 0

Solving above equation you will get
t = 4

So the time taken to produce 2w widgets by X is 12.

Note:
Two solutions of quadratic equation of form
a(x^2) + bx + c = 0, where a, b and c are constants,are

x = (-b + sqrt(b^2 -4ac))/2a
x = (-b - sqrt(b^2 - 4ac))/2a
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Re: Combine work Q [#permalink] New post 30 Oct 2008, 19:46
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OA is E

Tnx a lot :)
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Re: Combine work Q [#permalink] New post 04 Nov 2008, 08:50
@mbajingle
Thanks for the detailed solution - I always get confused by rate of work questions. Kudos.
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Re: Combine work Q [#permalink] New post 05 Nov 2008, 01:43
mbajingle wrote:
E.

Let t be number of days.
Rate of Y = w/t
Rate of X = w/(t+2)

When they work together, the number of widgets produced is (combined rate)*time_taken
(X+Y)*3 = 5w/4

substituting X and Y with their rate
w/(t+2) + w/t = 5w/4

Simplifying above equation...
5(t^2) - 12t - 24 = 0

Solving above equation you will get
t = 4

So the time taken to produce 2w widgets by X is 12.

Note:
Two solutions of quadratic equation of form
a(x^2) + bx + c = 0, where a, b and c are constants,are

x = (-b + sqrt(b^2 -4ac))/2a
x = (-b - sqrt(b^2 - 4ac))/2a


GOOD goin man........great explanation...thanks!
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Re: Combine work Q [#permalink] New post 05 Nov 2008, 02:03
My answer is B

@GODSPEED:

Check this

When they work together, the number of widgets produced is (combined rate)*time_taken
(X+Y)*3 = 5w/4

substituting X and Y with their rate
(w/(t+2) + w/t)*3 = 5w/4

and also, t is a number of days for Y
Manager
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Schools: MIT / INSEAD / IIM - ABC
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Re: Combine work Q [#permalink] New post 05 Nov 2008, 02:16
lylya4 wrote:
My answer is B

@GODSPEED:

Check this

When they work together, the number of widgets produced is (combined rate)*time_taken
(X+Y)*3 = 5w/4

substituting X and Y with their rate
(w/(t+2) + w/t)*3 = 5w/4

and also, t is a number of days for Y


Didn't get the point man???....you got the same equation as @mbajingle....i think he missed out on 3....rest remains the same

Anyways, thanks for this!
Manager
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Re: Combine work Q [#permalink] New post 05 Nov 2008, 02:20
GODSPEED wrote:
lylya4 wrote:
My answer is B

@GODSPEED:

Check this

When they work together, the number of widgets produced is (combined rate)*time_taken
(X+Y)*3 = 5w/4

substituting X and Y with their rate
(w/(t+2) + w/t)*3 = 5w/4

and also, t is a number of days for Y


Didn't get the point man???....you got the same equation as @mbajingle....i think he missed out on 3....rest remains the same

Anyways, thanks for this!


The equation is his, I just quoted to show he missed 3, that will give the incorrect calculation and the other thing is that t is for Y, so after find out t, he will need to add it by 2.
Re: Combine work Q   [#permalink] 05 Nov 2008, 02:20
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