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Machine A and machine B are each used to manufacture 660 [#permalink]
07 Aug 2010, 04:37

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Difficulty:

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Question Stats:

59% (02:36) correct
41% (01:42) wrong based on 170 sessions

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 produces?

Re: Kaplan "800" rate: Machines [#permalink]
07 Aug 2010, 05:04

3

This post received KUDOS

Expert's post

zisis wrote:

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 produces?

6 6.6 60 100 110

book give a backsolving solution which I am not a big fan of...........please explain method...

Let time needed for machine A to produce 660 sprockets be a hours, then the rate of machine A would be rate_A=\frac{job \ done}{time}=\frac{660}{a} sprockets per hour;

As "it takes machine A 10 hours longer to produce 660 sprockets than machine B" then time needed for machine B to produce 660 sprockets be a-10 hours and the rate of machine B would be rate_B=\frac{job \ done}{time}=\frac{660}{a-10} sprockets per hour;

As "machine B produces 10 percent more sprockets per hour than machine A" then rate_A*1.1=rate_B --> \frac{660}{a}*1.1=\frac{660}{a-10} --> a=110 --> rate_A=\frac{job \ done}{time}=\frac{660}{a}=6.

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 produces?

6 6.6 60 100 110

book give a backsolving solution which I am not a big fan of...........please explain method...

Let time needed for machine A to produce 660 sprockets be a hours, then the rate of machine A would be rate_A=\frac{job \ done}{time}=\frac{660}{a} sprockets per hour;

As "it takes machine A 10 hours longer to produce 660 sprockets than machine B" then time needed for machine B to produce 660 sprockets be a-10 hours and the rate of machine B would be rate_B=\frac{job \ done}{time}=\frac{660}{a-10} sprockets per hour;

As "machine B produces 10 percent more sprockets per hour than machine A" then rate_A*1.1=rate_B --> \frac{660}{a}*1.1=\frac{660}{a-10} --> a=110 --> rate_A=\frac{job \ done}{time}=\frac{660}{a}=6.

Answer: A.

Hope it's clear.

I did a similar approach, but what I did different was I said that Rate A = 660/(x+10) and Rate B = 1.1(660/x) This is basically saying B takes x hours and A takes x+10 hours. Why is this wrong? Because I don't get the same answer.

Thanks,

The highlighted part is not correct. Rate B = \frac{660}{x} and as Machine B makes more sprockets than Machine A, thus, by the given condition, Rate B = 1.1*Rate A.

Thus, \frac{660}{x} = 1.1*\frac{660}{(x+10)} = x+10 = 1.1x = x = 100. Thus, Per hour, Machine A would produce = \frac{660}{(100+10)} = \frac{660}{110)} = 6. _________________

Re: Kaplan "800" rate: Machines [#permalink]
13 Aug 2010, 06:28

1

This post received KUDOS

Expert's post

mrwuzzman wrote:

Thanks very much for the solution and explanation, Bunuel. One quick clarification though. In the explanation you make the jump from (660/a)*1.1 = 660/(a-10) to a = 110. Can you give a quick explanation for how you made that jump?

Suppose rate of B is b and rate of A is a. Suppose B takes x hours to produce 660 sprockets, so 660/b = x ( b = number of sprockets produced by B in one hour ) So A takes x + 10 hours to produce 660 sprockets or 660/a = x + 10. Now it is given that B produces 10% more sprockets than A in 1 hour, hence b = 110% of a or b = 1.1a 660/b = x and 660/a = x + 10 or 660/a - 10 = x From above, 660/b = 660/a - 10 ( since both of them equals x ) Since b = 1.1a 660/1.1a = 660/a - 10 Solving above equation will give us a = 6 sprockets/hour or we can say that A produces 6 sprockets per hour. Hence answer is option A

Re: Kaplan "800" rate: Machines [#permalink]
13 Aug 2010, 06:21

Thanks very much for the solution and explanation, Bunuel. One quick clarification though. In the explanation you make the jump from (660/a)*1.1 = 660/(a-10) to a = 110. Can you give a quick explanation for how you made that jump?

Re: Kaplan "800" rate: Machines [#permalink]
14 Aug 2010, 12:57

Expert's post

zisis wrote:

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 produces?

6 6.6 60 100 110

book give a backsolving solution which I am not a big fan of...........please explain method...

Hi zisis,

Sorry you're not a fan of Kaplan's backsolving methods, but in this case it can be really helpful.

Here, a little estimation goes a long way. We know that A works 10 hours longer than B does, so if A is making 100 or 110 sprockets per hour, it would be making 1000+ sprockets--impossible! Even 60/hour is clearly too high

Given that, the correct answer has to be either A or B. So, we start where it's easiest--the whole number. If A makes 6 sprockets/hour, then A will take 110 hours to produce 660 sprockets. Meanwhile, if A makes 6 sprockets per hour and B makes 10% more, B must make 6.6 sprockets/hour. B would therefore take 100 hours to make 660 sprockets.

The question stem tells us that A should work 10 more hours than B. When we plug 6 back into the question, A does work 10 more hours than B--that confirms that A is the correct answer, with a minimum of crunchy math. _________________

Re: Kaplan "800" rate: Machines [#permalink]
14 Aug 2010, 14:16

KapTeacherEli wrote:

zisis wrote:

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 produces?

6 6.6 60 100 110

book give a backsolving solution which I am not a big fan of...........please explain method...

Hi zisis,

Sorry you're not a fan of Kaplan's backsolving methods, but in this case it can be really helpful.

Here, a little estimation goes a long way. We know that A works 10 hours longer than B does, so if A is making 100 or 110 sprockets per hour, it would be making 1000+ sprockets--impossible! Even 60/hour is clearly too high

Given that, the correct answer has to be either A or B. So, we start where it's easiest--the whole number. If A makes 6 sprockets/hour, then A will take 110 hours to produce 660 sprockets. Meanwhile, if A makes 6 sprockets per hour and B makes 10% more, B must make 6.6 sprockets/hour. B would therefore take 100 hours to make 660 sprockets.

The question stem tells us that A should work 10 more hours than B. When we plug 6 back into the question, A does work 10 more hours than B--that confirms that A is the correct answer, with a minimum of crunchy math.

no offence......sometimes when i am stuck I will use backsolving but I try not to rely on it....

Re: Kaplan "800" rate: Machines [#permalink]
17 Dec 2011, 14:25

zisis wrote:

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 produces?

6 6.6 60 100 110

book give a backsolving solution which I am not a big fan of...........please explain method...

If i form the following equation from the condition is it wrong? 660/x - 660/x+10 =10/100 _________________

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

Source says - this is a very difficult rate problem hence choosing 700+. Please solve without back solving.

Re: Machine A and machine B are each used to manufacture 660 [#permalink]
04 Feb 2013, 09:41

Let 't' be the time.

Look at the attached RTW chart.

It is given Machine B produces 10% more per hour than machine, so the equation becomes-

\frac{660}{t} = \frac{660}{t+10}+\frac{66}{t+10}

This gives....> 66t=6600 Therefore t=100

Substituting t in rate of a... \frac{66}{t+10}gives the rate as 6.

Attachments

1.jpg [ 12.63 KiB | Viewed 2100 times ]

_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

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 produces?

6 6.6 60 100 110

book give a backsolving solution which I am not a big fan of...........please explain method...

Let time needed for machine A to produce 660 sprockets be a hours, then the rate of machine A would be rate_A=\frac{job \ done}{time}=\frac{660}{a} sprockets per hour;

As "it takes machine A 10 hours longer to produce 660 sprockets than machine B" then time needed for machine B to produce 660 sprockets be a-10 hours and the rate of machine B would be rate_B=\frac{job \ done}{time}=\frac{660}{a-10} sprockets per hour;

As "machine B produces 10 percent more sprockets per hour than machine A" then rate_A*1.1=rate_B --> \frac{660}{a}*1.1=\frac{660}{a-10} --> a=110 --> rate_A=\frac{job \ done}{time}=\frac{660}{a}=6.

Answer: A.

Hope it's clear.

I did a similar approach, but what I did different was I said that Rate A = 660/(x+10) and Rate B = 1.1(660/x) This is basically saying B takes x hours and A takes x+10 hours.

Why is this wrong? Because I don't get the same answer.

Re: Machine A and machine B are each used to manufacture 660 [#permalink]
01 Aug 2013, 00:26

1. Let the number of sprockets produced by machine A in 1 hour be x 2. Number of sprockets produced by machine B in 1 hour is 1.1x

3. Let machine A take y hours to produce 660 sprockets. In 1 hour it produces 660/y sprockets 4. Machine B takes y-10 hours to produce 660 sprockets. In 1 hour it produces 660/y-10 sprockets

5. Equating (1) and (3) -> xy=660 6. Equating (2) and (4) -> 1.1xy-11x=660

Re: Machine A and machine B are each used to manufacture 660 [#permalink]
24 Dec 2013, 11:46

zisis wrote:

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 produces?

A. 6 B. 6.6 C. 60 D. 100 E. 110

Let me chip in on this one

So we get that B manufactures the 660 sprockets in 10 hours less which indeed are 10%. Therefore total hours it takes is 100 Then A must take 10 hrs more hence 110 hours

Now, Total Work/Rate = 660/110 = 6 sprockets per hour

Answer is A Hope it helps Cheers! J

gmatclubot

Re: Machine A and machine B are each used to manufacture 660
[#permalink]
24 Dec 2013, 11:46