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KM

· started a discussion

· 1 Months ago

capacity of B in two times that of A (Question: 177243.0 )

The solution tells the opposite of it.

Question:
A, B, C, D and E are five taps. The capacity of B is 2 times that of A, 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 first case A, C and E act as input pipes and B and D act as output pipes. In a second case, C, D, E act as input pipes and A and B act as output pipes. If A and B working together as input pipes can fill the tank in  4  hour,  then  what  is  the  difference  in  the  time required to fill the tank in the first and second case stated above ?
Options:
A) 4.44 hr
B) 3.33 hr
C) 2.22 hr
D) 1.11 hr
E) 2.67 hr
Solution:

Ans: (e) 

Let B take x hours to fill the tank

So according to question, time required by each

A = 2x

B = x

C = \(\cfrac{2x}{3}\)

D = \(\cfrac{2x}{4}\)

E = \(\cfrac{2x}{5}\)

A and B together fill the tank in 4 hours.

\(\cfrac{1}{2x}+\cfrac{1}{x} \ \ = \ \ \cfrac{1}{4}\)

x = 6

In first case, when A, C and E act as input pipes and B and D as output pipes.

Total time = \(\cfrac{1}{12}+\cfrac{1}{4}+\cfrac{5}{12}-\cfrac{1}{6}-\cfrac{1}{3}=\cfrac{1}{4}\)

\(\therefore\) 4 hours

In second case, when C, D and E act as input pipes and A and B as output pipes.

Total time = \(\cfrac{1}{4}+\cfrac{1}{3}+\cfrac{5}{12}-\cfrac{1}{12}-\cfrac{1}{6}=\cfrac{3}{4}\)

\(\therefore \cfrac{4}{3}\ \ hr\)

Difference = \(4-\cfrac{4}{3}\ \ = \ \ \cfrac{8}{3} \ \ 2.67\)

Pankaj Singh

· commented

· 1 Months ago

since capacity of B is twice so it takes half time.
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