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26 Dec 2024, 09:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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70% (02:39) correct
30%
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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
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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26 Dec 2024, 10:00
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Bunuel
GMAT Club's Official Explanation:
From the stem:
- All three machines worked together for the first 10 hours.
- Machines B and C worked together for the next x hours.
- Only Machine C worked during the final 15 hours.
This gives the equation:
10 * (1/50 + 1/60 + 1/75) + x * (1/60 + 1/75) + 15 * (1/75) = 1
10 * (1/20) + x * (3/100) + (1/5) = 1
x = 10
10 * (1/20) + x * (3/100) + (1/5) = 1
x = 10
Thus, the total time required is (first 10 hours) + (next x = 10 hours) + (last 15 hours) = 35 hours.
Answer: C.
General Discussion
26 Dec 2024, 09:20
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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)
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)
