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02 Dec 2020, 12:00
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Difficulty:
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There are four varieties of pipes Pipe A, Pipe B, Pipe C and Pipe D. Each pipe can be either an inlet pipe or an outlet pipe but cannot be both. there are 5 tanks of equal volume.
Tank P is connected to Pipe A and Pipe B
Tank Q is connected to Pipe A and Pipe C
Tank R is connected to Pipe A and Pipe D
Tank S is connected to Pipe B and Pipe C
Tank T is connected to Pipe C and Pipe D
Time taken for the first 3 tanks (P, Q and R) to get filled are in the ratio 1:2:4 and the time taken for the S and T tanks to be filled are in the ratio 7:10. Which of the following is (are) the outlet pipe?
A. Pipes C and D
B. Pipe D
C. Pipes A and B
D. Pipe A
E. Pipe A and C
Are You Up For the Challenge: 700 Level Questions
Tank P is connected to Pipe A and Pipe B
Tank Q is connected to Pipe A and Pipe C
Tank R is connected to Pipe A and Pipe D
Tank S is connected to Pipe B and Pipe C
Tank T is connected to Pipe C and Pipe D
Time taken for the first 3 tanks (P, Q and R) to get filled are in the ratio 1:2:4 and the time taken for the S and T tanks to be filled are in the ratio 7:10. Which of the following is (are) the outlet pipe?
A. Pipes C and D
B. Pipe D
C. Pipes A and B
D. Pipe A
E. Pipe A and C
Are You Up For the Challenge: 700 Level Questions
03 Dec 2020, 06:53
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Bunuel
There are 4 pipes and all are getting filled in different times.
Tank T is connected to Pipe c and Pipe d. If both c and d are outlet pipe as given in OPTION A, tank T will never get filled. So, discard option A.
Similarly, you can discard OPTIONS C and E.
Now, we are left with Pipe a or Pipe d. We cannot answer straightway as 1:2:4 and 7:10 are not related to each other, although the Probability of answering correctly has gone up to 50% in 30 seconds.
But let us find the answer.
Time taken for the first 3 tanks (P, Q and R) to get filled are in the ratio 1:2:4 or say x, 2x and 4x hrs.
The time taken for the S and T tanks to be filled are in the ratio 7:10 or say 7y and 10y.
Let us take two tanks each where all four pipes are in use-
1) Tank P is connected to Pipe a and Pipe b : time taken x hr, so one hour work of a and b = \(\frac{1}{x}\)
Tank T is connected to Pipe d and Pipe c : time taken 10y hr, so one hour work of c and d = \(\frac{1}{10y}\)
Combined 1 hour work of a, b, c and d => \(\frac{1}{x}+\frac{1}{10y}\)....(i)
2)Tank R is connected to Pipe a and Pipe d : time taken 4x hr, so one hour work of a and d = \(\frac{1}{4x}\)
Tank S is connected to Pipe b and Pipe c : time taken 7y hr, so one hour work of c and b = \(\frac{1}{7y}\)
Combined 1 hour work of a, b, c and d => \(\frac{1}{4x}+\frac{1}{7y}\)...(ii)
Both (i) and (ii) should be equal
\(\frac{1}{x}+\frac{1}{10y}\)= \(\frac{1}{4x}+\frac{1}{7y}\)
Multiply by LCM(x,4x,7y,10y) or 140xy and solve the equation to get x=13y.
Now we will get all times in terms of y......x:2x:4x=13y:26y:52y
All 5 tanks can be arranged now as 7y:10y:13y:26y:52y or S>T>P>Q>R
Or pipes(b+c)>pipes(c+d)>pipes(a+b)>pipes(a+c)>pipes(a+d)
We can easily see that b>c>d>a, and so a is the outlet pipe, as it leads to slower rate filling of all tanks.
D
General Discussion
09 Feb 2025, 11:48
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