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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% (01:40) correct
21%
(01:42)
wrong
based on 213
sessions
History
Date
Time
Result
Not Attempted Yet
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
(1) x = 100
(2) x:y = 2:1
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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Explaination:
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26 May 2026, 02:31
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Equation as per Question Stem: (1/x) * 1 day + (1/x + 1/y)*(T-1)days = 1
Claim 1: x = 100; we dont know y so cant determine T.
Insuff -> A and D eliminated
Claim 2: x = 2k and y=k; yet again 2 variables so cant determne T.
Insuff -> B eliminated
C: mixing yes. k dissolves into integer and we can determine T. Suff.
Claim 1: x = 100; we dont know y so cant determine T.
Insuff -> A and D eliminated
Claim 2: x = 2k and y=k; yet again 2 variables so cant determne T.
Insuff -> B eliminated
C: mixing yes. k dissolves into integer and we can determine T. Suff.
