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

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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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.

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

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

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

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

put options

y = 4 --> x = 8
satisfies.
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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

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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Machine A 8 hr work = 8 * 9000 = 72000
remaining work = 100000-72000 = 28000

B takes 7000 in 1 hour = 28000/7000 = 4 hr

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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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

Bunuel ScottTargetTestPrep
If a and b are working together, wouldn't their collective rate be different? in this case-9k+7k=16k?

When they are working simultaneously, their combined rate is indeed 7,000 + 9,000 = 16,000 pencils per hour. However, it is clear from the question stem that the two machines are not necessarily working simultaneously all the time. In fact, in this question, the two machines may never work together simultaneously; machine A may run for a certain amount of time and then machine B may start running, or vice versa. That is why rather than calculating the combined rate, we should assume machine A operates for the maximum possible amount of time, which is 8 hours, and calculate the number of hours for machine B to finish the remaining number of pencils.
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