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# Machine A and machine B are each used to manufacture 660

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Machine A and machine B are each used to manufacture 660 [#permalink]

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07 Aug 2010, 05:37
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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
[Reveal] Spoiler: OA

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Re: Kaplan "800" rate: Machines [#permalink]

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07 Aug 2010, 05:45
[Reveal] Spoiler:
Time (B): 660/x
Time (A): [660/(x+10)]
660/x = [660/(x+10)] *110/100

660/x = (66*11 )/(x+10)

660 (x+10) = 66*11*x

660x +6600 = 66*11*x
x= 100

plug in back to time (A)
660/100+10 => 660/110 = 6

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Re: Kaplan "800" rate: Machines [#permalink]

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07 Aug 2010, 06:04
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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 $$rate_A*1.1=rate_B$$ --> $$\frac{660}{a}*1.1=\frac{660}{a-10}$$ --> $$a=110$$ --> $$rate_A=\frac{job \ done}{time}=\frac{660}{a}=6$$.

Hope it's clear.
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Re: Kaplan "800" rate: Machines [#permalink]

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13 Aug 2010, 07:21
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?

Thanks again!

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Re: Kaplan "800" rate: Machines [#permalink]

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13 Aug 2010, 07:28
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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?

Thanks again!

Sure:

$$\frac{660}{a}*1.1=\frac{660}{a-10}$$ --> reduce by 660 --> $$\frac{1.1}{a}=\frac{1}{a-10}$$ --> cross multiply --> $$1.1a-11=a$$ --> $$0.1a=11$$ -- > $$a=110$$.

Hope it's clear.
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Re: Kaplan "800" rate: Machines [#permalink]

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14 Aug 2010, 13:57
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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.
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Re: Kaplan "800" rate: Machines [#permalink]

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26 Nov 2011, 13:07
B takes x hours
A takes x + 10 hours
rate of A = 660/x+10
rate of B = 660/x

thus, 660/x = (660/x+10)*1.10
x = 100
so B = 100
A = 110 sprockets per hour
660/110 = 6
Ans. 6
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Last edited by Baten80 on 17 Dec 2011, 15:16, edited 1 time in total.

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Re: Kaplan "800" rate: Machines [#permalink]

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27 Nov 2011, 00:03
Baten80 wrote:
B takes x hours
A takes x + 10 hours
rate of A = 660/x+10
rate of B = 660/x

thus, 660/x = (660/x+10)*1.10
x = 100
so B = 100
A = 110 sprockets per hour
ans E.

110 is the time taken to produce 660 units. 660/110 =6 is the answer

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Re: Kaplan "800" rate: Machines [#permalink]

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27 Nov 2011, 05:26
A(t+10)=660
Bt=660
B=1.1A(10% faster)
1.1At=At+10A
.1t=10
t=100
A's speed per hour=6 sprockets

+1 for A

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Re: Kaplan "800" rate: Machines [#permalink]

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03 Dec 2011, 12:44
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Fro me MGMAT RTW matrix helped. From the attached matrix, we can solve for Tb.
Therefore, Ta = 100+10=110
Rare of Machine A= 660/110=6
Attachments

RTW.gif [ 1.89 KiB | Viewed 9695 times ]

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Re: Kaplan "800" rate: Machines [#permalink]

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17 Dec 2011, 15:25
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
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03 Feb 2013, 04:58
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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.

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03 Feb 2013, 23:06
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Considering the amount of work(660 sprockets) to be constant, we know that Work = rate x time

Thus,$$R*(t+10) = 1.1R*t$$

or 0.1t =10

or t = 100. Thus, A takes 110 hours for 660 sprockets. Thus in one hour, it can make $$\frac{660}{110} = 6$$sprockets.

A.
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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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04 Feb 2013, 10:41
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 8835 times ]

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Re: Kaplan "800" rate: Machines [#permalink]

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31 Jul 2013, 16:14
Bunuel 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...

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 $$rate_A*1.1=rate_B$$ --> $$\frac{660}{a}*1.1=\frac{660}{a-10}$$ --> $$a=110$$ --> $$rate_A=\frac{job \ done}{time}=\frac{660}{a}=6$$.

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,

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Re: Kaplan "800" rate: Machines [#permalink]

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31 Jul 2013, 22:47
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jjack0310 wrote:
Bunuel 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...

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 $$rate_A*1.1=rate_B$$ --> $$\frac{660}{a}*1.1=\frac{660}{a-10}$$ --> $$a=110$$ --> $$rate_A=\frac{job \ done}{time}=\frac{660}{a}=6$$.

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.
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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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01 Aug 2013, 01:26
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

7. From (5) and (6) x=6.
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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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24 Dec 2013, 12:46
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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

Hope it helps
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Machine A and machine B are each used to manufacture 660 [#permalink]

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22 Aug 2015, 13:49
Options C, D and E don't work because the numbers don't add up. It's just a matter of choosing between A and B

(E) 660/110 = 6 for A and -4 for B (X)

(D) 660/100 = 6.6 for A and -3.4 for B (X)

(C) 660/60 = 11 for A and 1 for B. But 1 is not 10% faster than A (X)

(B) 660/6.6 = 100 for and 90 for B. But 90 is not 10% faster than A - think 90*1.1=99. Almost there. (X)

Finally for option A

660/6 = 110 for A and 100 for B. 100*1.1 = 110. Correct.

The whole thing took less than a minute. As soon as you realise that C,D,E are too large it becomes a question of discarding B to get A. Similarly, at first sight 6.6 looks like a "clean number" in that it is obviously 10% greater than 6.

Hope it helps!

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Machine A and machine B are each used to manufacture 660 [#permalink]

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19 Jul 2016, 05:38
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

Quick way to do this one..we will use the concept of percentages..

B produces 10% more sockets per hour..thus if per hour A produces x..B produces 1.1x

The ratio of time of the two will be inverse of the ratio of efficiencies
or
1.1 : 1

The difference in these times has been given as 10 hours..
1.1y - y = 10

y = 100 hours

A's time = 110 hours
B's time = 100 hours

A's production per hour?...660/110 = 6 sprockets(A)..

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Machine A and machine B are each used to manufacture 660   [#permalink] 19 Jul 2016, 05:38

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