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08 May 2020, 08:35
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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:
55%
(hard)
Question Stats:
69% (02:46) correct
31%
(03:05)
wrong
based on 212
sessions
History
Date
Time
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Not Attempted Yet
Raman can do a piece of work in half the time taken by Kapil. Sunil can do the same work in one-third of the time taken by Raman. All three of them work on it for 30 days after which Kapil leaves. Sunil and Raman complete the remaining work in 18 more days. How many days would it take for Raman alone to complete the total work?
(A) 52
(B) 138
(C) 207
(D) 414
(E) 415
Are You Up For the Challenge: 700 Level Questions
(A) 52
(B) 138
(C) 207
(D) 414
(E) 415
Are You Up For the Challenge: 700 Level Questions
08 May 2020, 22:50
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Let efficiency of Raman, Kapil and Sunil are R,K and S respectively.
R:K=2:1
S:R=3:1
S:R:K = 6:2:1
which means S can do 6 units, K can do 2 units and S can do 1 unit in the same time(1 day).
Total work = [30(6+2+1)]+[18(6+2)] units
Time taken by Raman = \(\frac{[30(6+2+1)]+[18(6+2)]}{2 }\)
\(=15*9 +9*8= (15+8)*9 = 230-23= 207 days\)
R:K=2:1
S:R=3:1
S:R:K = 6:2:1
which means S can do 6 units, K can do 2 units and S can do 1 unit in the same time(1 day).
Total work = [30(6+2+1)]+[18(6+2)] units
Time taken by Raman = \(\frac{[30(6+2+1)]+[18(6+2)]}{2 }\)
\(=15*9 +9*8= (15+8)*9 = 230-23= 207 days\)
Bunuel
General Discussion
11 May 2020, 06:01
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Let the times of each of the individuals be tk,tr and ts
tr = tk/2, ts =tk/6
So in 1 day 9/tk of work is done
In 30 days, the amount of work done is 270/tk. So remaining work would be (270-tk)/tk
tr and ts together do 8/tk of work.
Which means (270-tk)/8
Plugging in options only 414 works. So tr = tk/2 = 207
PS: I must admit nick1816 's solution is more efficient
tr = tk/2, ts =tk/6
So in 1 day 9/tk of work is done
In 30 days, the amount of work done is 270/tk. So remaining work would be (270-tk)/tk
tr and ts together do 8/tk of work.
Which means (270-tk)/8
Plugging in options only 414 works. So tr = tk/2 = 207
PS: I must admit nick1816 's solution is more efficient
