Pipe A and B running together can fill a cistern in 6 minutes. If Pipe

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Pipe A and B running together can fill a cistern in 6 minutes. If Pipe B takes 5 more minutes than Pipe A to fill the cistern, then time in which A and B can fill the cistern separately will be respectively?

A. 15 minutes, 20 Minutes.
B. 15 minutes, 10 Minutes.
C. 12 minutes, 7 minutes.
D. 25 minutes, 20 minutes.
E. 10 minutes, 15 minutes.

Leadership · Answer

NOTE:  then time in which A and B can fill the cistern separately will be respectively?  ... We know that A will fill faster and B will take 5 more minutes than A. So, we can also eliminate options B, C and D because they do not fall under the A and B respectively condition. The problem solving is between option A and E.

A. 15 minutes, 20 Minutes.
 B. 15 minutes, 10 Minutes.
C. 12 minutes, 7 minutes.
D. 25 minutes, 20 minutes .
E. 10 minutes, 15 minutes.

Let the time taken in minutes when A fills separately - 'T'
Then, the time taken in minutes when B fills separately - 'T+5'

Work formula for A and B -  \frac{Time.taken.by.A * Time.taken.by.B }{ Time.taken.by.A + Time.taken.by.B} = 6

\frac{T (T+5) }{ T + T+5} = 6 
 \frac{T^2 + 5T) }{ 2T +5} = 6 
 T^2 + 5T = 6 (2T +5) 
 T^2 + 5T = 12T +30 
 T^2 + 5T - 12T -30 = 0 
 T^2 - 7T -30 = 0 
 T^2 - 10T + 3T -30 = 0  ..factor 
 T(T-10) + 3(T -10) = 0 
 (T-10) (T+3) = 0 
 T = 10 or -3  (T cannot be negative) so, Work by A is 10 minutes and work by B is 15 minutes. 'E' is the winner

CrackverbalGMAT · Answer

Let the  capacity of the tank be x litres (which is the work to be done) . The question says that pipes A and B can fill the cistern in 6 minutes, when working together. Therefore, combined rate of A and B =  \frac{x}{6}  litres per minute.

Remember that  for questions on Time and Work ,  the fundamental equation is Work = Time * Rate . When you know two of the variables, it's not hard to find the third.

If A can fill the cistern in 'a' minutes, B can fill it in (a+5) minutes. So,

Rate of pipe A =  \frac{x}{a}  litres per minute and Rate of pipe B =  \frac{x}{(a+5)}  litres per minute. Substituting this in the equation representing the combined rate of A and B, we have,

\frac{x}{a}  +  \frac{x}{(a+5)}  =  \frac{x}{6} .

The LCM of the denominators is a(a+6). Simplifying the equation above, we have,

\frac{(2a + 5)}{ a(a+5)}  =  \frac{1}{6}  which yields the values of a as 10 and -3 respectively. Since 'a' represents the time taken for pipe A to do the work, we discard the negative value and keep a=10.

This means, pipe A can fill the cistern in 10 minutes. There is only one option like that and that is option E.
 The correct answer option is E .

Hope that helps!

madgmat2019 · Answer

Let A take x hours to fill.....then B takes x+5 hrs

so 1/x+1/(x+5) = 1/6.....from this equation we get x=10....so B=15,A=10

OA:E

exc4libur · Answer

rate=work/time
(rate of a + b)•time=work
 6=1
6(2t+5)=t(t+5)
t^2-7t-30=0
(t-10)(t+3)=0
t≥0:t=10
a's time=10, b's time=10+5=15

Ans (E)

Archit3110 · Answer

given
a+b/ab= 1/6
and b= 5+a
solve for a & b
we get 
a^2-7a-30=0
a= 10 and b = 15
IMO E ; 
10 minutes, 15 minutes

Pipe A and B running together can fill a cistern in 6 minutes. If Pipe B takes 5 more minutes than Pipe A to fill the cistern, then time in which A and B can fill the cistern separately will be respectively?

A. 15 minutes, 20 Minutes.
B. 15 minutes, 10 Minutes.
C. 12 minutes, 7 minutes.
D. 25 minutes, 20 minutes.
E. 10 minutes, 15 minutes.

eakabuah · Answer

Let time taken for pipe A to fill the cistern be x, then time taken for pipe B, y, is x+5. We know that both pipes working together can fill the tank in 6 minutes. Hence, 1/6 = 1/x +1/(x+5)
(2x+5)/(x^2+5x)=1/6
12x+30=x^2+5x
x^2-7x-30=0
(x+3)(x-10)=0
x=10
Hence y=10+5=15

The answer is E.

unraveled · Answer

Pipe A and B running together can fill a cistern in 6 minutes. If Pipe B takes 5 more minutes than Pipe A to fill the cistern, then time in which A and B can fill the cistern separately will be respectively?

A. 15 minutes, 20 Minutes.
B. 15 minutes, 10 Minutes.
C. 12 minutes, 7 minutes.
D. 25 minutes, 20 minutes.
E. 10 minutes, 15 minutes.

Here B = A +5
 \frac{1}{A} + \frac{1}{B} = \frac{1}{6} 
 \frac{A + B}{AB} = \frac{1}{6} 
6A + 6B = AB 
 A^2 - 7A - 30 = 0  (as B = A +5)
A = 10 and A = - 3

A = 10 B = 15

Answer E.

ikneet · Answer

A & B together 
6 min -> 1 unit
1 min ->1/6 unit..........(1)

A individual
X min -> 1 unit
1 min -> 1/x unit............(2)

B individual
X+5 min -> 1 unit
1 min -> 1/(x+5) unit........(3)

So, from (2) and (3) A&B in 1 minutes -> 1/x+1/(x+5) ..............(4)

from (1) and (4)
1/x+1/(x+5) = 1/6

which gives us the following equation

x^2-7x-30 = 0
which gives us x=10 & -3
only positive value possible so take x = 10 min.

A takes 10 and B takes 15 minutes separately. Hence (E).

gracie · Answer

let individual pipe rates equal 1/A and 1/B respectively
1/A and 1/B must sum to 1/6 combined rate
plugging in choices, only B and E work
B is not possible as A's time is given as<B's time
so E

bumpbot · Answer

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Pipe A and B running together can fill a cistern in 6 minutes. If Pipe

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