Machine A and machine B are each used to manufacture 660

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 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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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

B takes \(x\) hours

So, A takes \(x + 10\) hours

Rate of \(A = \frac{660}{x+10}\)

Rate of \(B = \frac{660}{x}\)

Thus, \(\frac{660}{x} = \frac{660}{x+10}*1.10\)

\(x = 100\)

so \(B = 100\) Time taken by B alone

\(A = 110\) Time taken by A alone

Production of A per hour \(=\frac{660}{110} = 6\)

The answer is \(A\)
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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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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.

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

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.

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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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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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

Rate A : Rate B = 10: 11 (because machine B produces 10% more sprockets per hour. So if machine A produces 10 sprockets per hr, machine B produces 11 sprockets per hour)

Time A : Time B = 11 : 10 (Ratio of time taken by machine A and B is inverse of rate)

Now, given that it takes machine A 10 hours longer to produce 660 sprockets than machine B. So actual difference between the time taken is 10 hrs for 660 sprockets. Since difference on the ratio scale is 1, multiplier is 10.
Machine A takes 110 hrs to produce 660 sprockets and machine B takes 100 hrs to produce 600 sprockets.

So machine A makes 660/110 = 6 sprockets per hour.
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B produces 660 sprockets in t hours
A produces 660/1,1 = 600 sprockets in t hours since B produces 10% more sprockets than A per hour
Thus, there are 60 sprockets left for A to finish in 10 hours
=> Rate machine A = 60/10=6 sprockets/h

We are given that machine A and machine B are each used to manufacture 660 sprockets; thus the work of each machine is 660. We also are given that it takes machine A 10 hours longer to produce 660 sprockets than it takes machine B, and that machine B produces 10% more sprockets per hour than machine A. If we let the rate of machine A = r, then the rate of machine B = 1.1r.

Since time = work/rate:

The time of machine A = 660/r and the time of machine B = 660/(1.1r) = 600/r

Since machine A takes 10 hours longer to produce 660 sprockets than does machine B, we can create the following equation and determine r:

660/r = 600/r + 10

60/r = 10

60 = 10r

6 = r

Thus, machine A produces 6 sprockets per hour.

Answer: A
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1. Both the pieces of information talk about productivity. So we can equate them,
2. (x-10)/x = 1/1.1 . x=110 and A takes 110 hrs for producing 660 sprockets which is 6 sprockets per hour.
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I don't understand why you multiple equation *1.1
if 1.1 is work, should not divide?

We are told that Machine B produces 10 percent more sprockets per hour than machine A, so the rate of Machine B is 10 percent more than that of machine A: \(rate_A*1.1=rate_B\).

17. Work/Rate Problems

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For more check:
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