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Machine A and machine B are each used to manufacture 660

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Director
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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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New post 19 Jul 2016, 19:19
let t=B's time
r=A's rate
1=total work
equation1: 1/r=t+10
equation2: 1/1.1r=t
subtracting e2 from e1,
1/r-1/1.1r=10
r=.1/11
.1/11*660=6 sprockets per hour

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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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New post 10 Sep 2016, 01:53
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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

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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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New post 02 Feb 2017, 03:37
1) Rate of machine A: \((t+10)a=660; a=\frac{660}{t}+10\)
Rate of machine B: \(t*b=660; b=\frac{660}{t}\)
2) \(\frac{660}{t+10}=x\)
\(\frac{660}{t}=1.1x; x=\frac{660}{t/11/10}=\frac{660}{t}*\frac{10}{11}=\frac{600}{t}\)
3) \(\frac{660}{t+10}=\frac{600}{t}; 660t=600t+6000; 60t=6000; t=100\)
4) \(\frac{660}{100+10}=\frac{660}{110}=6\)

The correct answer is A.

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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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New post 08 Feb 2017, 10:21
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


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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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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New post 06 Jun 2017, 17:26
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

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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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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New post 25 Sep 2017, 06:57
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 \(a-10\) 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.



I don't understand why you multiple equation *1.1
if 1.1 is work, should not divide?

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Re: Machine A and machine B are each used to manufacture 660 [#permalink]

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New post 25 Sep 2017, 07:04
soodia 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 \(a-10\) 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.



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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Re: Machine A and machine B are each used to manufacture 660   [#permalink] 25 Sep 2017, 07:04

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