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# Machine A produces pencils at a constant rate of 9,000 pencils per hou

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Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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Updated on: 06 Oct 2017, 10:01
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Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

Originally posted by nitinneha on 04 Apr 2007, 17:09.
Last edited by Mahmud6 on 06 Oct 2017, 10:01, edited 1 time in total.
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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27 Jul 2010, 14:16
4
3
Machine A produces pencils at a constant rate of 9,000 pencils per hour, and machine B produces pencils at a constant rate of 7,000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

a) 4
b) 4 and 2/3
c) 5 and 1/3
d) 6
e) 6 and 1/4

To minimize the time that machine B must operate we must maximize the time machine A can operate, so make it operate 8 hours. In 8 hours machine A will produce $$8*9,000 = 72,000$$ pencils, so $$100,000 - 72,000 = 28,000$$ pencils are left to produce, which can be produced by machine B in $$\frac{28,000}{7,000}=4$$ hours.

Answer: A.
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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04 Apr 2007, 23:11
1
2
Ax + Bx = 100,000
x <= 8

Since it's asking for the least amount of time for B, A must work to its maximum.

100,000-(9000*8) = 28,000
28,000/7000 = 4

A.

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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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05 Apr 2007, 22:04
Total target = 100, 000

Since machine A is faster, and max is 8 hours, we shall allow machine A to orun for 8 hours. It shall thus produce 9000 * 8 = 72, 000 pencils.

We are short by 28, 000, machine B can work for 28K/7K = 4 hours to do it.

A
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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05 Apr 2007, 23:19
1
9,000 x 8 hours + 7,000 x X hours = 100,000
72,000 + 7,000 x X hours = 100,000
X hours =28,000 / 7,000 = 4

A it is
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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06 Apr 2007, 00:51
nitinneha wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines togather must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?
A)4
B)4 2/3
C)5 1/3
D)6
E)6 1/4

B should operate for least time so A must run for 8 hours and produce 72K. So remaining 28K will be produced by B in 4 hours
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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10 Apr 2015, 07:26
nitinneha wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines to gather must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

9000x + 7000y = 100000
or 9x + 7y = 100

put options

y = 4 --> x = 8
satisfies.
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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10 Apr 2015, 08:38
nitinneha wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines to gather must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

we want machine B to work minimum hours , for that we have to make machine A work maximum possible hours i.e. 8.
in 8hrs machine A will produce 72 pencils . now remaining 28 pencils will be produced by machine B in 4hrs .
answer A.
(72 + 28 = 100 ) *1000
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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29 Jun 2016, 10:16
nitinneha wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines to gather must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

In order to have Machine B work the least, we need to make Machine A work the maximum number of hours = 8 hrs

Machine A will produce 8*9000= 72,000 pencils in 8 hrs.

Machine B will have to produce remaining 100,000-72,000= 28,000 pencils.

If Machine B produces 7000 pencils in one hour, it will produce 28,000 pencils in 28,000/7= 4 hrs.

A is the answer
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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27 Sep 2016, 01:48
Machine A 8 hr work = 8 * 9000 = 72000
remaining work = 100000-72000 = 28000

B takes 7000 in 1 hour = 28000/7000 = 4 hr
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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05 Jun 2017, 16:13
nitinneha wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines to gather must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

If machine B’s operational time is to be minimized, we must maximize the time for machine A to operate. Since each machine can operate for at most 8 hours, we can let machine A operate for 8 hours. Since the rate of machine A is 9000 pencils per hour, machine A produces 8 x 9000 = 72,000 pencils, and thus 28,000 pencils are left to be produced.

Thus, it will take machine B 28,000/7,000 = 4 hours to produce the remaining pencils.

Answer: A
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou  [#permalink]

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09 Apr 2019, 07:59
Top Contributor
nitinneha wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?

A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4

To MINIMIZE machine B's operating time, we must MAXIMIZE the time machine A's operating time.
So, let machine A operate for the full 8 hours.
In 8 hours, machine A produces 72,000 pencils [8 hours x 9,000 pencils/hour =72,000 pencils]

So, the number of pencils machine B must make = 100,000 - 72,000 = 28,000

Time = output/rate
So, operating time = 28,000/7000 = 4 hours

Answer: A

Cheers,
Brent
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Re: Machine A produces pencils at a constant rate of 9,000 pencils per hou   [#permalink] 09 Apr 2019, 07:59
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# Machine A produces pencils at a constant rate of 9,000 pencils per hou

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