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12 Feb 2025, 01:30
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D
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Difficulty:
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Question Stats:
84% (01:14) correct
16%
(01:43)
wrong
based on 79
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When working together at their constant rates, machines A and B can complete a certain task in 4 hours. How many hours would it take machine B to complete the task, when working alone at its constant rate?
(1) When working alone at its constant rate, machine A can finish the task in 6 hours.
(2) When working alone at their respective rates, it takes machine B 6 hours longer than machine A to finish the task
(1) When working alone at its constant rate, machine A can finish the task in 6 hours.
(2) When working alone at their respective rates, it takes machine B 6 hours longer than machine A to finish the task
12 Feb 2025, 01:59
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Time taken by both = AB/(A+B) = 4
If we know A, we can find B
Statement 1
A = 6
6B/(6+B) = 4
6B = 24+4B
B = 12
Sufficient
Statement 2
A = B-6
(B-6)B/(B +B -6) = 4
B^2 - 6B = 8B -24
B^2 -14B+24 = 0
B= 12 or 2
B>6 as A is positive
B = 12
Sufficient
Answer D
If we know A, we can find B
Statement 1
A = 6
6B/(6+B) = 4
6B = 24+4B
B = 12
Sufficient
Statement 2
A = B-6
(B-6)B/(B +B -6) = 4
B^2 - 6B = 8B -24
B^2 -14B+24 = 0
B= 12 or 2
B>6 as A is positive
B = 12
Sufficient
Answer D
Bunuel
12 Feb 2025, 20:35
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When working together at their constant rates, machines A and B can complete a certain task in 4 hours. How many hours would it take machine B to complete the task, when working alone at its constant rate?
Hence \(Wa+Wb = \frac{1}{4}\)
(1) When working alone at its constant rate, machine A can finish the task in 6 hours.
Given Wa = \(\frac{1}{6}\), Hence \(Wb = \frac{1}{4}-\frac{1}{6}\)
Hence Sufficient
(2) When working alone at their respective rates, it takes machine B 6 hours longer than machine A to finish the task
Let Wa = \(\frac{1}{x}\)
Hence \(Wb = \frac{1}{(x+6)}\)
\( Hence \frac{1}{x} + \frac{1}{(x+6)} = \frac{1}{4}\)
On solving we get \(x^2 - 2*x - 24 = 0\), solving we get x = 6
Hence \(Wb = \frac{1}{12}\)
Hence sufficient
Ans IMO D
Hence \(Wa+Wb = \frac{1}{4}\)
(1) When working alone at its constant rate, machine A can finish the task in 6 hours.
Given Wa = \(\frac{1}{6}\), Hence \(Wb = \frac{1}{4}-\frac{1}{6}\)
Hence Sufficient
(2) When working alone at their respective rates, it takes machine B 6 hours longer than machine A to finish the task
Let Wa = \(\frac{1}{x}\)
Hence \(Wb = \frac{1}{(x+6)}\)
\( Hence \frac{1}{x} + \frac{1}{(x+6)} = \frac{1}{4}\)
On solving we get \(x^2 - 2*x - 24 = 0\), solving we get x = 6
Hence \(Wb = \frac{1}{12}\)
Hence sufficient
Ans IMO D
