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03 Oct 2024, 01:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
56% (01:57) correct
44%
(02:07)
wrong
based on 236
sessions
History
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Time
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Pump A and B together put water in the empty tank of which volume is x. If machine A put 1-gallon of water more than machine B for every minute, how many hours A and B work together?
(1) Machine A put 100 gallon more than machine B
(2) Two machines put total 800 gallon of water
(1) Machine A put 100 gallon more than machine B
(2) Two machines put total 800 gallon of water
This is a Data Sufficiency Butler Question
Check the links to other Butler Projects:
Check the links to other Butler Projects:
- Data Sufficiency Butler
- Problem Solving Butler
- Sentence Correction Butler
- Reading Comprehension Butler
- Integrated Reasoning Butler
05 Oct 2024, 03:17
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Quote:
Let \(R_A\) = rate of A and \(R_B\) = rate of B
We know that Work = rate*time
We can say: (\(R_A\) + \(R_B\))*t = x
We need to find time t.
Quote:
(\(R_A\))*1 = g +1
(\(R_B\))*1 = g
We can subtract the above equations to get: (\(R_A\)) - (\(R_B\)) = 1
Now let's look at the statements:
(1) Machine A put 100 gallon more than machine B
For the same time 't' working alone, Machine A puts 100 gallon more than machine B.
Let's write it as:
\(R_A\)*t = G + 100
\(R_B\)*t = G
We can subtract the above equations to get: (\(R_A\) - \(R_B\))*t = 100 => t = (100)/(\(R_A\) - \(R_B\))
We can substitute \(R_A\) - \(R_B\) = 1 to get t = 100 minutes.
Therefore, Statement (1) is sufficient.
Let's check statement (2) now.
(2) Two machines put total 800 gallon of water
This gives us the value of x = 800.
(\(R_A\) + \(R_B\))*t = x => (\(R_A\) + \(R_B\))*t = 800
We still don't know the sum of the rates of A and B so we can't calculate the time.
Therefore, Statement (2) is not sufficient.
Hence answer is (A).
28 Oct 2025, 13:48
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