MBA Exam

A, B, C, D and E are five taps. The capacity of B is 2 times that of A and the capacity of C is 3 times that of A. Capacities of D and E are 4 and 5 times that of A respectively. In the first case A, C and E act as input pipes and B and D act as output pipes. In the second case C, D and E act as input pipes and A and B act as output pipes. If A and D working together as input pipes can fill the tank in 4 hours, then what is the difference in time required to fill the tank in the first and second cases stated above?

1) 4.44 hrs
2) 3.33 hrs
3) 2.22 hrs
4) 1.11 hrs
5) 5.55 hrs

Read Solution (Total 1)

1) 4.44 Hrs
Given B=2A, C=3A, D=4A, E=5A
(A+D) 1Hr =1/4 =>A+4A 1Hr=1/4 => A 1Hr=1/20
So Capacity 1Hr [B=1/10], [C=3/20],[D=1/5],[E=1/4]
Ist Case time Required to fill=1/[(1/20)+(3/20)+(1/4)-(1/10)-(1/5)]=20/3 Hrs.
IInd Case time Required to fill=1/[(3/20)+(1/5)+(1/4)-(1/20)-(1/10)]=20/9 Hrs.
Difference in time in both cases=20/3 -20/9 =40/9 =4.44 Hrs.
11 years agoHelpfull: Yes(0) No(0)

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