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A
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Difficulty:
35%
(medium)
Question Stats:
71% (02:12) correct
29%
(02:15)
wrong
based on 238
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A person X working alone can complete a work in 10 days. A person Y completes the same amount of work in 20 days and a person Z when working alone can complete the same amount of work in 30 days. How many more days are required if all the people started working together but X and Y leave the work after 2 days ?
a) 19
b) 20
c) 21
d) 22
e) 25
a) 19
b) 20
c) 21
d) 22
e) 25
18 Jun 2015, 21:36
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anurag356
ALTERNATE
Consider the Total work as Number of Units which is a common multiple of 10, 20 and 30 [to get away with the difficult calculation of Fractions]
Let, Total Work = 60 Units
X Finishes 60 Unit work in 10 days
i.e. Work done by X in a day = 60/10 = 6 Units
Y Finishes 60 Unit work in 20 days
i.e. Work done by Y in a day = 60/20 = 3 Units
Z Finishes 60 Unit work in 30 days
i.e. Work done by Z in a day = 60/30 = 2 Units
X and Y leave after two days while all start i.e. X, Y and Z work together for 2 days
Work done by X, Y and z in 2 days = 2*(6+3+2) = 22 Units
Work Remaining (to be finished by Z alone) = 60 - 22 = 38 Units
Time take by Z to finish 38 Units alone = 38/2 = 19 Days more
Answer: option
General Discussion
18 Jun 2015, 20:23
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anurag356
I am assuming that the question is "...X and Y leave the work after 2 days?"
Rate of work of A = 1/10
Rate of work of B = 1/20
Rate of work of C = 1/30
Combined rate of work of all 3 = 1/10 + 1/20 + 1/30 = 11/60
In 2 days, the work they complete is Rate*Time = 11/60*2 = 22/60 of the work
We have (1 - 22/60 ) = 38/60 work leftover.
Time taken by C alone to complete the remaining work = Work/Rate = (38/60)/(1/30) = 19 days
Answer (A)
