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Machine A and machine B are each used to manufacture 660 [#permalink]
 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]
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]
07 Aug 2010, 06:04
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. Answer: A. Hope it's clear.
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Re: Kaplan "800" rate: Machines [#permalink]
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]
13 Aug 2010, 07:28
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Re: Kaplan "800" rate: Machines [#permalink]
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+ 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]
14 Aug 2010, 15:16
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....
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Re: Kaplan "800" rate: Machines [#permalink]
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]
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]
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]
03 Dec 2011, 12:44
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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RTW.gif [ 1.89 KiB | Viewed 1421 times ]
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Re: Kaplan "800" rate: Machines [#permalink]
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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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.
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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
Please give a kudo if you like my explanation.
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The question is fairly easy. Considering the amount of work(660 sprockets) to be constant, we know that Work = rate x time
Thus, Rx(t+10) = 1.1Rxt
or 0.1t =10
or t = 100. Thus, A takes 110 hours for 660 sprockets. Thus in one hour, it can make 660/110 = 6 sprockets.
A.
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Re: Machine A and machine B are each used to manufacture 660 [#permalink]
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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1.jpg [ 12.63 KiB | Viewed 560 times ]
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Re: Machine A and machine B are each used to manufacture 660
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04 Feb 2013, 10:41
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