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Bunuel
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Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2 07 May 2017, 13:49
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Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2500 holes an hour. If the two machines together must bore 45,000 holes, and if each machine can operate for at most 9 hours, what is the least amount of time, in hours, Drill Press N must operate?

A. 6
B. 7
C. 36/5
D. 8
E. 90/11
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C
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Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2 07 May 2017, 14:47
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if machine M bore 3000holes/hours * 9 hours= 27000holes
45000 -27000= 18000 holes left
18000/2500 and rearranging = 36000/5000 = 36/5 ---->[C]
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Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2 07 May 2017, 15:50
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Drill Press M: 3000 holes/hour
Drill Press N: 2500 holes/hour
To find the least amount of operation time of N, we must run the press M for the maximum duration possible which is 9 hours i.e.
3000*9= 27000 holes.
Now, holes left= 45000-27000 i.e. 18000 which need to be drilled by N.
So, time taken would be 18000/2500 = 36/5 hours which is option C.

Hope it helps :)
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Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2 07 May 2017, 22:50
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Drill Press M: 3000 holes/hour
Drill Press N: 2500 holes/hour
To find the least amount of operation time of N, we must run the press M for the maximum duration possible which is 9 hours i.e.
3000*9= 27000 holes.
Now, holes left= 45000-27000 i.e. 18000 which need to be drilled by N.
time taken =18000/2500 = 36/5 hours
ans C
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Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2 08 May 2017, 00:55
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If we have to minimize the work of N, then we need to maximize the work of M
So M working for all 9 hours will bore 3000 x 9 = 27000 holes
So pending holes = 45000 – 27000 = 18000

So N has to work for 18000 / 2500 = 36 / 5 hours
Option C
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Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2 01 Jul 2019, 17:27
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Bunuel
Drill Press M can bore 3000 holes an hour and Drill Press N can bore 2500 holes an hour. If the two machines together must bore 45,000 holes, and if each machine can operate for at most 9 hours, what is the least amount of time, in hours, Drill Press N must operate?

A. 6
B. 7
C. 36/5
D. 8
E. 90/11
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We are given that the rate of Drill Press M is 3000 holes/hour and the rate of Drill Press N is 2500 holes/hour. Since each machine can operate for at most 9 hours and we need to determine the minimum number of hours of operating time for N, we can let M work for 9 hours, and thus 3000 x 9 = 27,000 holes are completed, leaving 45,000 - 27,000 = 18,000 holes to be completed by N.

Since time = work/rate, N must operate for 18,000/2,500 = 180/25 = 36/5 hours. (Note the typo for choice C, which should be 36/5.)

Answer: C
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