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Kap Advanced 800: Word Problems: # 26 pg 306

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Kap Advanced 800: Word Problems: # 26 pg 306 [#permalink] New post 29 Aug 2006, 08:19
I understand how to solve it by Backsolving. Is there a right straightforward way that doesn’t need to backsolve?

Machine A and Machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine A. How many sprockets per hour does machine A produce?
a. 6
b. 6.6
c. 60
d. 100
e. 110
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 [#permalink] New post 29 Aug 2006, 08:49
Let B= number of hours it takes B to produce 660 sprockets
Let B+10- number of hours it takes A to produce 660 sprockets

660/B=1.1(660/(B+10))
B=100 So number of sprockets produce by A per hour is 6
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 [#permalink] New post 29 Aug 2006, 11:02
apollo168 wrote:
Let B= number of hours it takes B to produce 660 sprockets
Let B+10- number of hours it takes A to produce 660 sprockets

660/B=1.1(660/(B+10))
B=100 So number of sprockets produce by A per hour is 6


Is this showing the sprockets per hour for B? = 660/B

How is 660/B EQUAL to 1.1(660/(B+10)) it didn't say both produce the same # of sprockets per hour?
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 [#permalink] New post 29 Aug 2006, 12:58
Because Machine B produces 10 percent more sprockets per hour than machine A thats what gives us the formula.

Let me go over it again for clarity:
Assume that B is the number of hours it takes B to make 660 sprockets.
It then takes A 10+B hrs to do the same.

Therefore
B sprockets per hour = 660/B
A sprockets per hour = 660/(B+10)

And B sprockets per hr = 1.1 (A sprockets per hr)

Leading to the formula.

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 [#permalink] New post 29 Aug 2006, 13:01
mikki0000 wrote:
Because Machine B produces 10 percent more sprockets per hour than machine A thats what gives us the formula.

Let me go over it again for clarity:
Assume that B is the number of hours it takes B to make 660 sprockets.
It then takes A 10+B hrs to do the same.

Therefore
B sprockets per hour = 660/B
A sprockets per hour = 660/(B+10)

And B sprockets per hr = 1.1 (A sprockets per hr)

Leading to the formula.

MG


Thanks so much. Funnily so I tried the same thing yesterday and got it wrong. Did it now and got it right.
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 [#permalink] New post 19 May 2007, 23:00
mikki0000 wrote:
Because Machine B produces 10 percent more sprockets per hour than machine A thats what gives us the formula.

Let me go over it again for clarity:
Assume that B is the number of hours it takes B to make 660 sprockets.
It then takes A 10+B hrs to do the same.

Therefore
B sprockets per hour = 660/B
A sprockets per hour = 660/(B+10)

And B sprockets per hr = 1.1 (A sprockets per hr)

Leading to the formula.

MG

thanks. this really helped.
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Re: Kap Advanced 800: Word Problems: # 26 pg 306 [#permalink] New post 27 May 2011, 16:42
Thank you for posting. I was going nuts trying to figure out how to set this problem up. Lame for the book to ignore the actual method for solving.
I'm not looking for a cheat here!
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Re: Kap Advanced 800: Word Problems: # 26 pg 306 [#permalink] New post 03 Jun 2011, 08:46
saumster wrote:
I understand how to solve it by Backsolving. Is there a right straightforward way that doesn’t need to backsolve?

Machine A and Machine B are each used to manufacture 660 sprockets. It takes machine A 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine A. How many sprockets per hour does machine A produce?
a. 6
b. 6.6
c. 60
d. 100
e. 110



here is one way to solve this problem

Suppose

Work rate of machine a is y spockets per hour( produces y spockets per hour)
then
work rate of Machine B is 1.1y spockets/ hr

then no. of hours req. to produce 660 spockets by machine
then time taken by machine B = 660/1.1y

then time taken by machine A to produce 660 spockets= 660/1.1y + 10

660/y=660/1.1y + 10

660/y* (1-10/11)=10

660/y* 1/11=10

=> y= 6

so machine A produces 6 spockets per hour
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Re: Kap Advanced 800: Word Problems: # 26 pg 306 [#permalink] New post 03 Jun 2011, 10:57
Pff I solved for t = 100 but fell for one of the most famous GMAT traps.. NOT PROPERLY ANSWERING THE QUESTION and ended up with 6.6 :@
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Re: Kap Advanced 800: Word Problems: # 26 pg 306 [#permalink] New post 29 Jun 2011, 12:16
Back-solving on these types of questions are very effective as well. Just look at the numbers and try and figure out what makes sense.


Of course we can always solve it straight up:

X -- time taken by B --- so per hour # produced --- (660/X)

X + 10 -- time taken by A --- so per hour # produced --- 660 / (x+10)

We are told B produces 10% more per hour:

660/x = (110/100)*(660/ (x+10))

660x + 6600 = 726x

66x = 6600

X = 100 *Remember this is the time taken by B and not the final answer. A trap choice D is staring at us!

Time taken by A = 110; Per hour # produced = 660/110 = 6;

Choice A!
Re: Kap Advanced 800: Word Problems: # 26 pg 306   [#permalink] 29 Jun 2011, 12:16
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Kap Advanced 800: Word Problems: # 26 pg 306

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