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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
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Re: Kaplan "800" rate: Machines [#permalink]
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07 Aug 2010, 05:45
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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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 \(a10\) hours and the rate of machine B would be \(rate_B=\frac{job \ done}{time}=\frac{660}{a10}\) sprockets per hour; As "machine B produces 10 percent more sprockets per hour than machine A" then \(rate_A*1.1=rate_B\) > \(\frac{660}{a}*1.1=\frac{660}{a10}\) > \(a=110\) > \(rate_A=\frac{job \ done}{time}=\frac{660}{a}=6\). Answer: A. 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/(a10) 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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Re: Kaplan "800" rate: Machines [#permalink]
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14 Aug 2010, 13:57
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+ sprocketsimpossible! 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 easiestthe 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 Bthat 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
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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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
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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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Re: Rate problem. [#permalink]
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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. Hence answer is option A
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Re: Rate problem. [#permalink]
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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.
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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 \(a10\) hours and the rate of machine B would be \(rate_B=\frac{job \ done}{time}=\frac{660}{a10}\) sprockets per hour; As "machine B produces 10 percent more sprockets per hour than machine A" then \(rate_A*1.1=rate_B\) > \(\frac{660}{a}*1.1=\frac{660}{a10}\) > \(a=110\) > \(rate_A=\frac{job \ done}{time}=\frac{660}{a}=6\). 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,



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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 \(a10\) hours and the rate of machine B would be \(rate_B=\frac{job \ done}{time}=\frac{660}{a10}\) sprockets per hour; As "machine B produces 10 percent more sprockets per hour than machine A" then \(rate_A*1.1=rate_B\) > \(\frac{660}{a}*1.1=\frac{660}{a10}\) > \(a=110\) > \(rate_A=\frac{job \ done}{time}=\frac{660}{a}=6\). 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.
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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 y10 hours to produce 660 sprockets. In 1 hour it produces 660/y10 sprockets 5. Equating (1) and (3) > xy=660 6. Equating (2) and (4) > 1.1xy11x=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 Answer is A Hope it helps Cheers! J



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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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