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08 Oct 2018, 07:12
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:27) correct
29%
(02:29)
wrong
based on 792
sessions
History
Date
Time
Result
Not Attempted Yet
Two hoses (A and B) are filling a pool. Working alone at their individual constant rates Hose A could fill the pool in 6 hours, and Hose B could do it in 4 hours. If the two hoses work together until half the pool is full, and then Hose B finishes the pool alone, what fraction of the pool will have been filled by Hose B?
A. 1/2
B. 3/5
C. 3/4
D. 4/5
E. 7/8
A. 1/2
B. 3/5
C. 3/4
D. 4/5
E. 7/8
08 Oct 2018, 08:04
Kudos
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Bunuel
B completes half by himself and in other half he will do more as his speed is faster than A's.
So he would do atleast \(\frac{1}{2}\)+\(\frac{1}{2}\)*\(\frac{1}{2}\) so >1/2+1/4...>3/4..
So only D and E are left....
Now E is 7/8 so A does 1/8 work and B does 1/2-1/8=3/8.. so B does thrice the work, but the speed do not vary so much
Now D is 4/5 so A does 1/5 or 2/10 of work and B does 1/2-2/10=3/10.. so speeds 4:6 are giving similar work
D
Otherwise..
A and B together will do in..
1/4 +1/6=5/12
So work will be done in 12/5 and thus half work in 12/5*1/2=6/5
B in 6/5 days will do 6/(5*4)=3/10
Total work done is 1/2+3/10=(5+3)/10=4/5
D
03 May 2020, 01:23
Kudos
Bookmarks
Let's say work(W) = 1
Hence, R(A) = 1/6
and R(B) = 1/4
R(A+B) = 1/6 + 1/4 = 5/12
half pool was filled by both the hoses.
So, W = 1/2
t (time taken by both the hoses to fill half pool) = W/R = (1/2)/(5/12) = 6/5
Hence, in 6/5 hours, work done by hose B = R*T = (1/4)*(6/5) = 3/10
So, total work done by B = (3/10) + (1/2) = 4/5
Hence, R(A) = 1/6
and R(B) = 1/4
R(A+B) = 1/6 + 1/4 = 5/12
half pool was filled by both the hoses.
So, W = 1/2
t (time taken by both the hoses to fill half pool) = W/R = (1/2)/(5/12) = 6/5
Hence, in 6/5 hours, work done by hose B = R*T = (1/4)*(6/5) = 3/10
So, total work done by B = (3/10) + (1/2) = 4/5
