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VP  P
Joined: 07 Dec 2014
Posts: 1235
Re: Machine A and machine B are each used to manufacture 660  [#permalink]

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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
Intern  S
Joined: 14 May 2014
Posts: 16
Re: Machine A and machine B are each used to manufacture 660  [#permalink]

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1
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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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) $$\frac{660}{100+10}=\frac{660}{110}=6$$

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

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.

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

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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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 $$rate_A*1.1=rate_B$$ --> $$\frac{660}{a}*1.1=\frac{660}{a-10}$$ --> $$a=110$$ --> $$rate_A=\frac{job \ done}{time}=\frac{660}{a}=6$$.

Hope it's clear.

I don't understand why you multiple equation *1.1
if 1.1 is work, should not divide?
Math Expert V
Joined: 02 Sep 2009
Posts: 61283
Re: Machine A and machine B are each used to manufacture 660  [#permalink]

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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 $$rate_A*1.1=rate_B$$ --> $$\frac{660}{a}*1.1=\frac{660}{a-10}$$ --> $$a=110$$ --> $$rate_A=\frac{job \ done}{time}=\frac{660}{a}=6$$.

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

For more check:
ALL YOU NEED FOR QUANT ! ! !

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

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Machine A makes 660 units in X hours and it's rate is 660/X;
Machine B makes 660 units in X - 10 hours, as it's faster, and it's rate is 660/(X-10);

Since "machine B produces 10 percent more sprockets per hour than machine A", then:
1.1(660/X) = 660/(X-10) => X = 110 Hours; AND 660/110= 6 unit per hour. Re: Machine A and machine B are each used to manufacture 660   [#permalink] 28 Jan 2020, 01:07

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