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26 Jun 2024, 01:30
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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:
35%
(medium)
Question Stats:
79% (01:41) correct
21%
(01:42)
wrong
based on 214
sessions
History
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A can complete a piece of work in x days and B can do the same piece of work in y days. If B starts working a day after A and then they work together, in how many days can they complete the work?
(1) x = 100
(2) x:y = 2:1
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) x = 100
(2) x:y = 2:1
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
26 Jun 2024, 03:51
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In one day B does: \(\frac{1}{y}\)
A and B working on the rest: \(\frac{1}{x}+\frac{1}{y} = \frac{\frac{y-1}{y}}{c}\)
The question asks for us to solve for \(c+1\)
(1) x = 100
Plugging this into the second equation gives: \(\frac{1}{100}+\frac{1}{y} = \frac{\frac{y-1}{y}}{c}\)
Without knowing \(y\) one cannot solve for \(c\), and therefore cannot solve for \(c+1\)
INSUFFICIENT
(2) x:y = 2:1
This tells us that \(x = 2y\). Plugging this into the second equation gives: \(\frac{1}{2y}+\frac{1}{y} = \frac{\frac{y-1}{y}}{c}\)
which simplifies to: \(\frac{3}{2y} = \frac{\frac{y-1}{y}}{c}\)
Without having a value for \(y\), it is impossible to solve for a numeric value for \(c\).
INSUFFICIENT
(1+2)
Combining the two, we get that \(x = 100\) and \(y = 50\).
Plugging this into the second equation: \(\frac{1}{100}+\frac{1}{50} = \frac{\frac{49}{50}}{c}\)
From here one can easily solve for \(c\), and \(c+1\)
SUFFICIENT
Math:
\(\frac{1}{100}+\frac{1}{50} = \frac{\frac{49}{50}}{c}\)
\(\frac{3}{100} = \frac{\frac{49}{50}}{c}\)
\(\frac{3c}{100} = \frac{49}{50}\)
\(3c = 98\)
\(c = 32.666666\) as we are dealing with days, and fraction must be rounded up = \(c = 33\)
Thus \(c+1 = 34\)
ANSWER C
A and B working on the rest: \(\frac{1}{x}+\frac{1}{y} = \frac{\frac{y-1}{y}}{c}\)
The question asks for us to solve for \(c+1\)
(1) x = 100
Plugging this into the second equation gives: \(\frac{1}{100}+\frac{1}{y} = \frac{\frac{y-1}{y}}{c}\)
Without knowing \(y\) one cannot solve for \(c\), and therefore cannot solve for \(c+1\)
INSUFFICIENT
(2) x:y = 2:1
This tells us that \(x = 2y\). Plugging this into the second equation gives: \(\frac{1}{2y}+\frac{1}{y} = \frac{\frac{y-1}{y}}{c}\)
which simplifies to: \(\frac{3}{2y} = \frac{\frac{y-1}{y}}{c}\)
Without having a value for \(y\), it is impossible to solve for a numeric value for \(c\).
INSUFFICIENT
(1+2)
Combining the two, we get that \(x = 100\) and \(y = 50\).
Plugging this into the second equation: \(\frac{1}{100}+\frac{1}{50} = \frac{\frac{49}{50}}{c}\)
From here one can easily solve for \(c\), and \(c+1\)
SUFFICIENT
Math:
\(\frac{1}{100}+\frac{1}{50} = \frac{\frac{49}{50}}{c}\)
\(\frac{3}{100} = \frac{\frac{49}{50}}{c}\)
\(\frac{3c}{100} = \frac{49}{50}\)
\(3c = 98\)
\(c = 32.666666\) as we are dealing with days, and fraction must be rounded up = \(c = 33\)
Thus \(c+1 = 34\)
ANSWER C
26 Jun 2024, 12:55
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