Machines X and Y run at different constant rates, and

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Machines X and Y run at different constant rates, and [#permalink]

09 Jul 2008, 06:38

Machines X and Y run at different constant rates, and machine X can complete a certain job in 9 hours. Machine X worked on the job alone for the first 3 hours and the two machines, working together, then completed the job in 4 more hours. How many hours would it have taken machine Y, working alone, to complete the entire job?

(A) 18 (B)13+1/2 (C)7+1/5 (D)4+1/2 (E)3+2/3

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Re: work question [#permalink]

09 Jul 2008, 07:05

X can finish 1/9 of the job in 1 hr.
X works for 3 hrs and finishes 3/9 or 1/3 of the job.

2/3 of the job remains.

Y can finish 1/Y of the job in 1 hr
X and Y finish the remaining 2/3 of the job in 4 hrs.
3/2(1/9 + 1/Y) = 1/4
1/9 + 1/Y = 2/3*1/4
1/9 + 1/Y = 1/6
1/Y = 1/18
This means Y needs 18 hrs to finish the job alone.

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Re: work question [#permalink]

09 Jul 2008, 08:20

The answer is (A) that is 18hrs

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Re: work question [#permalink]

09 Jul 2008, 08:21

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Here is a non-standard way to look at the work problem:
assign numbers to the "job."

Take this particular question as an example.
Machines X can complete a certain job in 9 hours.
Let the job be: in this case, making 9 pieces of candies
So, Machine X's work rate = 1 candy per hour.

Then the question stated:
Machine X worked on the job alone for the first 3 hours
So, 6 more pieces of candies to go :-D

Next: The two machines, working together, then completed the job in 4 more hours
We know that Machine X can produce 4 pieces of candies in 4 hours
So, Machine Y produces 2 pieces of candies in 4 hours, which is 0.5 piece per hour.

Thus, it will take 18 hours to produce 9 pieces of candies for machine Y alone.

This is what I came up with on the fly during the actual GMAT on a work problem that kinda confused me when I tried with the standard work function way... wondering why I didn't think of this way earlier.....
Nice and easy, with no fraction to confuse you... not to mention that candies are sure yummy :-D

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Re: work question [#permalink]

09 Jul 2008, 14:58

stingraybullray wrote:

could you explain why you flip the 2/3 to get 3/2 ?

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Re: work question [#permalink]

09 Jul 2008, 19:56

I found it easier to set it up like this:

4(\frac{1}{9} + \frac{1}{x}) =\frac{6}{9}

4 because we're told the two machines work together for 4 hours. 6/9 because that's the same as 2/3 the job remaining as indicated in the stem.
multiply both sides by 1/4 or divide by 4, same operation, different look.
\frac{1}{9} + \frac{1}{x} =\frac{6}{9}*\frac{1}{4}
\frac{1}{9} + \frac{1}{x} =\frac{3}{18}
Change 1/9 to 2/18 so i can subtract it from both sides.
\frac{2}{18} + \frac{1}{x} =\frac{3}{18}
\frac{1}{x} =\frac{1}{18}
It's easy to see here that x = 18 and with work problems, the denominator is the number of units it takes that worker/machine to complete the portion of the work required.
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Re: work question [#permalink]

10 Jul 2008, 19:13

pmenon wrote:

could you explain why you flip the 2/3 to get 3/2 ?

I used the rate formula RT = W

R = 1/9 + 1/Y; W = 2/3; T = 4

(1/9 + 1/Y)4 = 2/3
3/2(1/9 + 1/Y) = 1/4 -- this is how I got it.

Re: work question [#permalink] 10 Jul 2008, 19:13

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Machines X and Y run at different constant rates, and

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Machines X and Y run at different constant rates, and

Machines X and Y run at different constant rates, and

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