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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] New post 01 Nov 2005, 06:27
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Question Stats:

57% (06:28) correct 43% (02:15) wrong based on 64 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% 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: machine-a-and-machine-b-are-each-used-to-manufacture-98696.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Aug 2013, 03:16, edited 2 times in total.
Added the OA.
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 [#permalink] New post 03 Nov 2005, 20:06
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This is a very difficult rate problem, and the best approach is backsolving. Start with the middle choice (C). If machine A produces 60 sprockets per hour, then machine B produces 60 plus 10% of 60, or 66 sprockets per hour. According to the rate formula, the amount of work done, divided by the rate at which it is done, equals the time it takes to do the work. So it takes machine A 660/60, or 11 hours, while it takes machine B 660/66, or 10 hours. It doesn`t take machine A 10 hours longer than machine B at these rates, so choice (C) is out. At a slower rate it will take both machines longer, so try a smaller answer choice. Choice (A) is an interger, so it will be easier to work with than choice (B). If machine A produces 6 sprockets per hour, then machine B produces 6 plus 10% of 6, or 6.6 sprockets per hour. At these rates it takes machine A 660/6, or 110 hours, while it takes machine B 660/6.6, or 100 hours. Since in this case it does take machine A 10 hours longer than machine B, choice (A) is correct.
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Re: Machine A and machine B are each used to manufacture 660 [#permalink] New post 30 Aug 2013, 03:15
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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% 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

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: machine-a-and-machine-b-are-each-used-to-manufacture-98696.html
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 [#permalink] New post 01 Nov 2005, 06:41
B 6.6

Let A produces x sprockets per hour
therefore, B produces 1.1x sprockets (> 10%)

information given:

660/x - 660/1.1x = 10
solve for x = 6

B produces 6 * 1.1 = 6.6 sprockets
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Re: PS Rates Kap800 [#permalink] New post 01 Nov 2005, 06:53
time taken by B = t
time taken by A = t+10

qty produced by A = q
qty produced by B = 1.1 q

for B: t(1.1q) =660
qt=600

for A: (t+10)(q) = 660
qt+10q=660
600+10q=660
q=6
so A can produce 6/hour.

then B can produce = 6(1.1)=6.6/hour.
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 [#permalink] New post 03 Nov 2005, 19:31
first of all those who say its 6...read the question it is asking rate of machine B

(1)660=Ra(t+10)<--machine A

(2)660=1.1RaT <---Machine B

I am just going to right 1.1 as 110/100

660=Rat + 10Ra (1)

RaT=660*100/110=600 (2)

660=600+10Ra; 60=10Ra; Ra=6 if Ra=6 then Rb=1.1(6)=6.6

Kapish...I dont know what work formula you guys are using....dont just post an answer...show your steps...so we can follow what you are doing...
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 [#permalink] New post 03 Nov 2005, 19:46
In 1 hr, B produces -> 660/B sprockets
In 1 hr, A produces -> 660/(B+10) sprockets

Given,
1.1*(# produced by A) = # produced by B

1.1*660/(B+10) = 660/B -> 1.1B = B + 10
=> B = 100

So,
In 1 hr, B produces -> 660/100 = 6.6 sprockets
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 [#permalink] New post 03 Nov 2005, 20:25
[Edit] I'm editing this post out because I believe the tone of this post is entirely not appropriate. My apologies to all. :)
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Last edited by HongHu on 13 Nov 2005, 19:20, edited 1 time in total.
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 [#permalink] New post 03 Nov 2005, 20:55
HongHu wrote:
This is NOT a very difficult rate problem. And DON'T do back solving unless you are absolutely clueless. Backsolving needs time. If you can express what is given in the question using algebra expressions, and if you can solve an algebra equation (both are needed if you are going to a b-school). then you would have no problem solving this question.

(10+t)A=tB=660
B=1.1A
=>10A+tA=tA+0.1tA
10A=0.1tA
t=100
(10+100)A=660
A=6

How simple is that?


Kaplan claims that this is a difficult rate problem, and therefore Kaplan does not offer an algebraic formula/solution to this particular problem, advocating the "backsolving" approach instead; which is exactly why I was confused and posted it in this math forum.

BTW: thanks for your condescending words of encouragement Honghu.
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 [#permalink] New post 03 Nov 2005, 20:58
This was extremly difficult before -
1. The EDIT
2. "B is not the OA"

Ok after -
1. The EDIT
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 [#permalink] New post 03 Nov 2005, 20:59
:lol: Yes exactly. Don't believe in every words Kaplan says. In fact don't believe in every words anybody says. ;) Trust your own informed judgement is much better. :)
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 [#permalink] New post 03 Nov 2005, 21:27
I'm not trying to sound condescending but this work rate problem is as straight forward as they come. Don't try to backsolve every problem - I don't think that's a fundamentally sound approach - you can't backsolve most of the time unless you can visualize the algebraic setup. Use the RTW (Rate, Time, Work) chart to help you visualize the relationships.
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Last edited by Titleist on 03 Nov 2005, 21:44, edited 1 time in total.
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 [#permalink] New post 03 Nov 2005, 21:44
GMATT73 wrote:
BTW: thanks for your condescending words of encouragement Honghu.


Hmmm. Oops, I didn't meant to be condescending at all. :oops: I mean some kind of bashing of the authorities on the field should be allowed shouldn't it? :-D I apologize if I sounded that way to you. Sorry. :) I will try to remind myself about this next time. :)
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 [#permalink] New post 03 Nov 2005, 21:51
BTW: thanks for your condescending words of encouragement Honghu.[/quote]

I think she was refering to Kaplan's opinion of this particular question as "a difficult rate problem." I also beg to differ with Kaplan's opinion of this problem's level of difficulty. But there was no implication about your ability whatsoever. We're all here to help one another because we all have our strengths and weaknesses - except if you're Duttsit and you appear to have no glaring weaknesses. :lol:
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 [#permalink] New post 03 Nov 2005, 22:59
Two equation and Two unkowns

(a) TimeA = 10 + TimeB

(b) RateA = (1.1)RateB


Combine (a) and (b)

660/TimeB = (1.1)* 660/(10+TimeB)

TimeB = 100 hours

Therefore,

RateA = 660/110 = 6
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 [#permalink] New post 04 Nov 2005, 04:59
HongHu wrote:
This is NOT a very difficult rate problem. And DON'T do back solving unless you are absolutely clueless. Backsolving needs time. If you can express what is given in the question using algebra expressions, and if you can solve an algebra equation (both are needed if you are going to a b-school). then you would have no problem solving this question.

(10+t)A=tB=660
B=1.1A
=>10A+tA=tA+0.1tA
10A=0.1tA
t=100
(10+100)A=660
A=6

How simple is that?


Sis HongHu ...I know you're trying to analyze things but I think make a personal judgement on others' abilities is meaningless. We're all here to improve ourselves out of our weekness, we need encouragement from others rather than lead-to-nowhere judgements,right?! I'm a junior to you ...I'm not trying to be arrogant to a senior ... Just that I do hope you understand my point of view ....
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 [#permalink] New post 13 Nov 2005, 19:16
laxieqv wrote:
HongHu wrote:
This is NOT a very difficult rate problem. And DON'T do back solving unless you are absolutely clueless. Backsolving needs time. If you can express what is given in the question using algebra expressions, and if you can solve an algebra equation (both are needed if you are going to a b-school). then you would have no problem solving this question.

(10+t)A=tB=660
B=1.1A
=>10A+tA=tA+0.1tA
10A=0.1tA
t=100
(10+100)A=660
A=6

How simple is that?


Sis HongHu ...I know you're trying to analyze things but I think make a personal judgement on others' abilities is meaningless. We're all here to improve ourselves out of our weekness, we need encouragement from others rather than lead-to-nowhere judgements,right?! I'm a junior to you ...I'm not trying to be arrogant to a senior ... Just that I do hope you understand my point of view ....


Guys, I realize that the tone in that post of mine is not appropriate at all. Please believe me that I did not mean to be judgemental about anybody's capability at all. I was commenting about Kaplan, because I did not feel their recommendation in that case was appropriate, and it may very much mislead people. However, you are very right. The post didn't get that point accrossed. In fact it seemed to be discouraging to people. Now that I think about it, my latest posts do have the same problem. We are all learning, no matter how much we think we know. It is the helping from each other that makes us grow. Thanks for helping me with this problem. This is exactly what I needed. I promise I will try harder from now on to be more careful about my comments. We all need to learn to be respectful to our fellow posters. :)

(BTW sorry I didn't respond sooner. I've been a bit busy so didn't check the board for several days.)
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 [#permalink] New post 13 Nov 2005, 19:27
Here is what I did


Work = Rate * Time
660 = 1 * (t + 10) <-- For A
660 = 1.1 * t <-- For B

Equate A and B

we get t = 100; Put this value in (A)
A takes 110 hrs to produce 660 spokets so in one hour it produces 6.
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Re: Machine A and machine B are each used to manufacture 660 [#permalink] New post 29 Aug 2013, 16:13
ONE EQUATION solution
660/x-660/x+10
(whole divided by ) * 100=10%
660/x+10


solve for x . we get x=10
and substitute in 660/x+10 to get the answer
Re: Machine A and machine B are each used to manufacture 660   [#permalink] 29 Aug 2013, 16:13
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