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28 Apr 2022, 01:30
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E
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Difficulty:
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Question Stats:
70% (02:56) correct
30%
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wrong
based on 244
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A takes 5 more days than B to do a certain job and 9 days more than C; A and B together can do the same job in the same time as C. How many days would A take to do it?
A. 10
B. 12
C. 18
D. 16
E. 15
Are You Up For the Challenge: 700 Level Questions
A. 10
B. 12
C. 18
D. 16
E. 15
Are You Up For the Challenge: 700 Level Questions
28 Apr 2022, 04:18
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\(a = b + 5\)
\(b = a - 5\) (1)
\(a = c + 9\)
\(c = a - 9\) (2)
Plugging (1) & (2) into \(\frac{1}{a} + \frac{1}{b} = \frac{1}{c}\) gives:
\(\frac{1}{a} + \frac{1}{a-5} = \frac{1}{a-9}\)
\(\frac{a-5+a}{a*(a-5)} = \frac{1}{a-9}\)
\(\frac{2a-5}{a^2-5a} = \frac{1}{a-9}\)
\(2a^2 - 18a - 5a + 45 = a^2 - 5a\)
\(2a^2 - 18a + 45 = 0\)
\((a - 15)(a - 3)\)
\(a = 15\) or \(a = 3\)
Answer E
\(b = a - 5\) (1)
\(a = c + 9\)
\(c = a - 9\) (2)
Plugging (1) & (2) into \(\frac{1}{a} + \frac{1}{b} = \frac{1}{c}\) gives:
\(\frac{1}{a} + \frac{1}{a-5} = \frac{1}{a-9}\)
\(\frac{a-5+a}{a*(a-5)} = \frac{1}{a-9}\)
\(\frac{2a-5}{a^2-5a} = \frac{1}{a-9}\)
\(2a^2 - 18a - 5a + 45 = a^2 - 5a\)
\(2a^2 - 18a + 45 = 0\)
\((a - 15)(a - 3)\)
\(a = 15\) or \(a = 3\)
Answer E
General Discussion
28 Apr 2022, 12:15
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Bunuel, I think the answer options are numbers and hence need to be in ascending or descending order, as per the PS questions on the GMAT.
Let the total work = 1 unit.
Let time taken by B to complete the work be b days; then, time taken by A will be b + 5 days.
Let time taken by C to complete the work be c days; then, time taken by A will be c + 9 days.
Therefore, b + 5 = c + 9 OR c = b – 4.
A and B can together do the job in the same time as C. Therefore, combined rate of A and B is equal to rate of C.
\(\frac{1 }{ a}\) + \(\frac{1 }{ b}\) = \(\frac{1 }{ c}\)
Substituting the values of a and c, we have,
\(\frac{1 }{ (b + 5) }\)+\( \frac{1 }{ b}\) =\( \frac{1 }{ (b – 4)} \)
Taking the LCM on the LHS and simplifying, we have,
\(\frac{b + b + 5 }{ b (b + 5)}\) = \(\frac{1 }{ (b – 4)}\)
Cross multiplying and simplifying, we have,
2\(b^2\) – 3b – 20 = \(b^2\) + 5b
Simplifying further, we have,
\(b^2\) – 8b – 20 = 0.
Factorising, (b-10) (b+2) = 0.
Therefore, b = 10 since b = -2 is not a valid proposition for the time taken by B.
If b = 10, a = b + 5 = 15.
A would take 15 days to complete the job.
The correct answer option is E.
Let the total work = 1 unit.
Let time taken by B to complete the work be b days; then, time taken by A will be b + 5 days.
Let time taken by C to complete the work be c days; then, time taken by A will be c + 9 days.
Therefore, b + 5 = c + 9 OR c = b – 4.
A and B can together do the job in the same time as C. Therefore, combined rate of A and B is equal to rate of C.
\(\frac{1 }{ a}\) + \(\frac{1 }{ b}\) = \(\frac{1 }{ c}\)
Substituting the values of a and c, we have,
\(\frac{1 }{ (b + 5) }\)+\( \frac{1 }{ b}\) =\( \frac{1 }{ (b – 4)} \)
Taking the LCM on the LHS and simplifying, we have,
\(\frac{b + b + 5 }{ b (b + 5)}\) = \(\frac{1 }{ (b – 4)}\)
Cross multiplying and simplifying, we have,
2\(b^2\) – 3b – 20 = \(b^2\) + 5b
Simplifying further, we have,
\(b^2\) – 8b – 20 = 0.
Factorising, (b-10) (b+2) = 0.
Therefore, b = 10 since b = -2 is not a valid proposition for the time taken by B.
If b = 10, a = b + 5 = 15.
A would take 15 days to complete the job.
The correct answer option is E.
