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03 Feb 2021, 06:30
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B
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:
60% (03:20) correct
40%
(04:01)
wrong
based on 10
sessions
History
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Time
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Not Attempted Yet
A contractor employed three workers A, B, and C to complete a work. The workers can complete the entire work in 6 days, 8 days and 10 days alone, respectively. For the first three days A, B and C work individually in this order. After that thus work in pairs such that first A and B work together, followed by B and C, and then A and C and so on. Find the total time required to complete the work.
A. 5 3/8
B. 5 11/24
C. 5 7/11
D. 5 3/4
E. 5 5/6
A. 5 3/8
B. 5 11/24
C. 5 7/11
D. 5 3/4
E. 5 5/6
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03 Feb 2021, 10:56
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Continuing from the diagram,
\(Total \ Work =120\)
Given, For the first three days A, B and C work individually in this order. Therefore 3 days work \(=(A+B+C)=(20+15+12) = 47\) (Whereas efficiency of A=20, B=15, C=12)
Remaining Work after first 3 days\(= 120-47=73\)
The efficiency of \((A+B) = 20+15=35\), remaining work after \(4^{th}\) day \(=73-35=38\)
The efficiency of \((B+C) = 15+12 =27\), remaining work after \(5^{th}\) day \(= 38-27=11\)
The efficiency of \((C+A)=12+20=32\), remaining work \(=11\) which is completed by \(C+A\) in \(\frac{11}{32}\)
Total days \(= 3+1+1+\frac{11}{32}=5\frac{11}{32}\)
Dear Bunuel
I think the answer should be \(5\frac{11}{32} \ not \ 5\frac{11}{24}\)
Only 11 part of the work is remaining, even if we try the traditional method \(\frac{1}{6}+\frac{1}{10}=\frac{16}{60}=\frac{32}{120}\). Therefore, in conclusion \(\frac{24}{120} \ i.e. \ 24 \) can't be a choice.
So the answer should be B with \(5\frac{11}{32}\)
\(Total \ Work =120\)
Given, For the first three days A, B and C work individually in this order. Therefore 3 days work \(=(A+B+C)=(20+15+12) = 47\) (Whereas efficiency of A=20, B=15, C=12)
Remaining Work after first 3 days\(= 120-47=73\)
The efficiency of \((A+B) = 20+15=35\), remaining work after \(4^{th}\) day \(=73-35=38\)
The efficiency of \((B+C) = 15+12 =27\), remaining work after \(5^{th}\) day \(= 38-27=11\)
The efficiency of \((C+A)=12+20=32\), remaining work \(=11\) which is completed by \(C+A\) in \(\frac{11}{32}\)
Total days \(= 3+1+1+\frac{11}{32}=5\frac{11}{32}\)
Dear Bunuel
I think the answer should be \(5\frac{11}{32} \ not \ 5\frac{11}{24}\)
Only 11 part of the work is remaining, even if we try the traditional method \(\frac{1}{6}+\frac{1}{10}=\frac{16}{60}=\frac{32}{120}\). Therefore, in conclusion \(\frac{24}{120} \ i.e. \ 24 \) can't be a choice.
So the answer should be B with \(5\frac{11}{32}\)
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11 Feb 2021, 07:41
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meanup
Dear Bunuel,
After working out I too have tried at \(5\frac{11}{32}\) as the total time required.
If option B is the correct ans. then it should be changed to \(5\frac{11}{32}\).
Hope you will look into this.
Thank you.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
