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by Ozlemg » Sat Aug 20, 2011 1:58 am
Machince A and B are each used to manufacture 660 sprockets. It takes A 10 hours longer to produce 660 sprockets than machine B. B produces 10 percent more sprockets per hour than A. How many sprockets per hours does machine A produce?
A. 6
B. 6.6
C. 60
D. 100
E 110

OA A

sorry, typo! I editted the OA.
Last edited by Ozlemg on Sat Aug 20, 2011 2:52 am, edited 2 times in total.
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by akhilsuhag » Sat Aug 20, 2011 2:26 am
I think the OA should be[spoiler] A(6)[/spoiler]

Total Work: 660


Time: For A :- x+10 ; For B :- x ( Because: A takes 10 hrs longer to produce)
Rate: For A :- y ; For B :- 1.1y ( Because: B produces 10 percent more than A)

Now since the total work is the same for both we can equate:

y(x+10) = 1.1yx --> xy+10y = 1.1xy --> x=100.

Now y(x+10) = 660 --> 110y = 660 --> y=6.

Now Rate of A is 6 and that of B is 6.6 (Wither the question asks rate of B if the OA is B or the OA is incorrect.... IMO).
Ofcourse I am no expert, so can be wrong.
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by knight247 » Sat Aug 20, 2011 2:50 am
The answer is actually A

For Machine A
Work=660
Time=T1
Rate=R1=660/T1


For Machine B
Work=660
Time=T2
Rate=R2=660/T2


Given that It takes A 10 hours longer to produce 660 sprockets than machine B
T1-T2=10
aka

660/R1-660/R2=10......(1)

Also, it is given that B produces 10 percent more sprockets per hour than A. Meaning that B's rate of production/hr is 10% greater than A's rate of production/hr

R2=R1(1+10/100)

R2=1.1R1

Substituting this value in (1)
600/R1-600/1.1R1=10

Solving we get R1=6 and using this value we can find the value of R2 which is 6.6. Hence [spoiler]A[/spoiler]
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by GMATGuruNY » Sat Aug 20, 2011 3:16 am
Ozlemg wrote:Machince A and B are each used to manufacture 660 sprockets. It takes A 10 hours longer to produce 660 sprockets than machine B. B produces 10 percent more sprockets per hour than A. How many sprockets per hours does machine A produce?
A. 6
B. 6.6
C. 60
D. 100
E 110

OA B
We can plug in the answers, which represent the rate for A.

Since the difference between the time for A and the time for B is an integer (10 hours), the correct answer must yield an integer value when divided into 660.
Eliminate D.

B produces 10% more sprockets per hour than A.
Thus, when the correct answer is increased by 10% and divided into 660, the result must be an integer.
Eliminate B, since (1.1)*(6.6) = 7.26, which does not divide evenly into 660.
Eliminate E, since (1.1)*(110) = 121, which does not divide evenly into 660.

The correct answer must be C or A.

Answer choice C: 60 sprockets per hour for A.
Time for A = w/r = 660/60 = 11.
Since B produces 10% more sprockets per hour, rate for B = (1.1)*60 = 66.
Time for B = w/r = 660/66 = 10.
Difference in times = 11-10 = 1.
Eliminate C.

The correct answer is A.

Please note that we had to try only ONE answer choice to determine the correct answer -- a VERY efficient way to solve this problem.

For the skeptical, answer choice A: 6 sprockets per hour for A.
Time for A = w/r = 660/6 = 110.
Since B produces 10% more sprockets per hour, rate for B = (1.1)*6 = 6.6.
Time for B = w/r = 660/6.6 = 100.
Difference in times = 110-100 = 10.
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by saketk » Sat Aug 20, 2011 9:22 am
Hi -- You can quickly solve this question if you use the answer options.

Logically, you can directly eliminate D and E because the total time should be more that 10 in any case.
Also if you look at B, its in decimal.. not that easy to pick to solve the question.
Choose A, and Hey, this is the right answer. :)
Plugging 6 gives us 110 hrs and for B it gives 100 hrs. Suffice the condition in the stem.
Ozlemg wrote:Machince A and B are each used to manufacture 660 sprockets. It takes A 10 hours longer to produce 660 sprockets than machine B. B produces 10 percent more sprockets per hour than A. How many sprockets per hours does machine A produce?
A. 6
B. 6.6
C. 60
D. 100
E 110

OA A

sorry, typo! I editted the OA.
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