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12 Days of Christmas GMAT Competition - Day 9: In a factory, Machine A

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IssacChan

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27 Dec 2024, 07:57

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Bunuel

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

In a factory, Machine A, working alone, can complete a task in 50 hours, Machine B, working alone, can complete the same task in 60 hours, and Machine C, working alone, can complete it in 75 hours. All three machines started working together on the task. However, Machine A stopped working after 10 hours, and Machine B stopped 15 hours before the task was completed. How many total hours were required to finish the task?

A. 25
B. 30
C. 35
D. 40
E. 45

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Let x be the number of hour C required to completed

(10/50+x/60+(x+15)/75) = 1
(60+5x+4x+60)/300 = 1
120+9x = 300
9x = 180
x = 20

The total hours required to finish the task
= 10+x+15
= 10+20+15
= 45
Therefore the answer is E

nikiki

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27 Dec 2024, 08:06

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Let t be the total time to finish the task

work done by A = 10/50 = 1/5
work done by B = (t-15)/60
work done by C = t/75

total work done is 1
add all the work done by A,B,C and equate it to 1
simplifying we get 9t - 15 = 300
t = 35hours

Option C

Bunuel

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maddscientistt

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27 Dec 2024, 08:30

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Rate of A = 1/50
Rate of B = 1/60
Rate of C = 1/75

work done in first 10 hours
10(1/50+1/60+1/75) = 0.5 of the task

work remaining after 10 hours = 0.5

work done after A stops
in each hour B and C complete = 1/60+1/75= 3/100 of task
in t-25 hours, work done by B and C = 3(t-25)/100

total work =1
0.5 + 3(t-25)/100

===> t = 40hours

Answer = D

Bunuel

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BatrickPatemann

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3 machines working together for 10 hours = 1/2 of the task done.
(6/300 + 5/300 + 4/300)*10 = 1/2

Now if T is total time worked to complete task, then B and C work together for t - 15 hours and the remaining last 15 hours must be completed by C alone.

Which translates into B&C + C = 1/2 of the job.

(t-15)9/300 + 15*(1/75) = 1/2

solving for T = 25 hours Answer (A)

Nikhil17bhatt

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IMO C

To determine the total hours required to finish the task, we need to calculate the contributions of each machine and the total work done.

Calculate the work rates of each machine:
- Machine A's rate: 1/50 of the task per hour.
- Machine B's rate: 1/60 of the task per hour.
- Machine C's rate: 1/75 of the task per hour.
Calculate the work done by each machine:
- Machine A worked for 10 hours.
- Machine B worked until 15 hours before the task was completed.
- Machine C worked the entire time.

Let TT be the total time required to complete the task.

Work done by Machine A:
Work done by A=10×1/50=10/50=1/5
Work done by Machine B: Machine B stopped 15 hours before the task was completed, so it worked for T−15hours.
Work done by B=(T−15)×1/60
Work done by Machine C: Machine C worked the entire time T.
Work done by C=T×1/75
Total work done: The sum of the work done by all machines should equal 1 (the entire task).
1/5+(T−15)×1/60+T×1/75=1
Combine and solve the equation:
1/5+T−15/60+T/75
To combine the fractions, find a common denominator. The least common multiple of 5, 60, and 75 is 300.
Convert each term to have a denominator of 300:
15=60300,160=5300,175=430051=30060,601=3005,751=3004
T=35

Therefore, the total hours required to finish the task is 35.

mpp01

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Total hours is T

B+C work for T-15 hours
C works alone for 15 hours
A+B+C work for 10 hours now

Combining rates we get that all three combined accomplish half the work so B and C must finish it.

B+C*(t-15) + C*15 = 1/2 of the work and inserting here the combined rates we get T = 25 (a)

t-15 = 10, which is the number of hours worked by B and C alone
and the first 10 hours is done by the 3 of them.

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I kept doing the math on my whiteboard repeatedly not realizing your math is wrong.

siddhantvarma

Given:
Machine A = 50 hours (Rate = 1/50)
Machine B = 60 (Rate = 1/60)
Machine C = 75 hours (Rate = 1/75)

All three machines started working together on the task. However, Machine A stopped working after 10 hours.
In 10 hours, all the three machines together completed = (1/50+1/60+1/75)*10 work which simplifies to 1/5
So in the first 10 hours, all three machines together finished 1/5th of the work

Now 4/5th of work is remaining
Machine B stopped 15 hours before the task was completed => Machine C worked alone for 15 hours
In 15 hours, Machine C would complete : (1/75)*15 => 1/5th of the work => 4/5-1/5 = 3/5
Now 3/5th of the work was completed by both B and C before B stopped working.

3/5 = (1/60+1/75)t => t = 20
Thus Machine B and C worked together for 20 hours.

Adding them all up, we get 10 + 15 + 20 => 45
Thus total 45 hours were required. (E)