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06 Aug 2019, 04:37
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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:
55%
(hard)
Question Stats:
63% (03:11) correct
38%
(03:32)
wrong
based on 8
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Either of the team of A and B or team of C and D can do the a work in 15 days. Also the same work can be done by the team of A and D in 10 days and by the team of B and C in 30 days. If B alone takes 40 days to complete the work, find the ratio of the number of days taken by the team of B and D and that taken by the team of A and C to complete the work.
1. 1 : 1
2. 3 : 5
3. 3 : 4
4. 1 : 2
5. Cannot be determined
1. 1 : 1
2. 3 : 5
3. 3 : 4
4. 1 : 2
5. Cannot be determined
06 Aug 2019, 05:59
Kudos
Bookmarks
\(\frac{1}{A}\)+\(\frac{1}{B}\)=\(\frac{1}{15}\)
\(\frac{1}{C}\)+\(\frac{1}{D}\)=\(\frac{1}{15}\)
--> \(\frac{1}{A}\)+\(\frac{1}{D}\)=\(\frac{1}{10}\)
--> \(\frac{1}{B}\)+\(\frac{1}{C}\)=\(\frac{1}{30}\)
\(\frac{1}{B}\)=\(\frac{1}{40}\)
--> (\(\frac{1}{B}\)+\(\frac{1}{D}\))=\(\frac{1}{x}\)
--> (\(\frac{1}{A}\)+\(\frac{1}{C}\))=\(\frac{1}{y}\)
-----------------
\(\frac{x}{y}\)=?
\(\frac{1}{C}\)=\(\frac{1}{30}\)-\(\frac{1}{40}\)=\(\frac{1}{120}\)
\(\frac{1}{D}\)=\(\frac{1}{15}\)-\(\frac{1}{120}\)=\(\frac{7}{120}\)
\(\frac{1}{A}\)=\(\frac{1}{15}\)-\(\frac{1}{40}\)=\(\frac{(8-3)}{120}\)=\(\frac{5}{120}\)
\(\frac{1}{B}\)+\(\frac{1}{D}\)=\(\frac{1}{40}\)+\(\frac{7}{120}\)=\(\frac{10}{120}\)=\(\frac{1}{12}\)=\(\frac{1}{x}\) --> x=12
\(\frac{1}{A}\)+\(\frac{1}{C}\)=\(\frac{5}{120}\)+\(\frac{1}{120}\)=\(\frac{6}{120}\)=\(\frac{1}{20}\)=\(\frac{1}{y}\)--> y=20
--> \(\frac{x}{y}\)=\(\frac{12}{20}\)=\(\frac{3}{5}\)
The answer choice is B
\(\frac{1}{C}\)+\(\frac{1}{D}\)=\(\frac{1}{15}\)
--> \(\frac{1}{A}\)+\(\frac{1}{D}\)=\(\frac{1}{10}\)
--> \(\frac{1}{B}\)+\(\frac{1}{C}\)=\(\frac{1}{30}\)
\(\frac{1}{B}\)=\(\frac{1}{40}\)
--> (\(\frac{1}{B}\)+\(\frac{1}{D}\))=\(\frac{1}{x}\)
--> (\(\frac{1}{A}\)+\(\frac{1}{C}\))=\(\frac{1}{y}\)
-----------------
\(\frac{x}{y}\)=?
\(\frac{1}{C}\)=\(\frac{1}{30}\)-\(\frac{1}{40}\)=\(\frac{1}{120}\)
\(\frac{1}{D}\)=\(\frac{1}{15}\)-\(\frac{1}{120}\)=\(\frac{7}{120}\)
\(\frac{1}{A}\)=\(\frac{1}{15}\)-\(\frac{1}{40}\)=\(\frac{(8-3)}{120}\)=\(\frac{5}{120}\)
\(\frac{1}{B}\)+\(\frac{1}{D}\)=\(\frac{1}{40}\)+\(\frac{7}{120}\)=\(\frac{10}{120}\)=\(\frac{1}{12}\)=\(\frac{1}{x}\) --> x=12
\(\frac{1}{A}\)+\(\frac{1}{C}\)=\(\frac{5}{120}\)+\(\frac{1}{120}\)=\(\frac{6}{120}\)=\(\frac{1}{20}\)=\(\frac{1}{y}\)--> y=20
--> \(\frac{x}{y}\)=\(\frac{12}{20}\)=\(\frac{3}{5}\)
The answer choice is B
