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23 Dec 2019, 22:32
Kudos
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E
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:
73% (02:18) correct
27%
(02:32)
wrong
based on 267
sessions
History
Date
Time
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Not Attempted Yet
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.
24 Dec 2019, 00:07
Kudos
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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
A. 15 minutes, 20 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
24 Dec 2019, 00:27
Kudos
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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!
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!
