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03 Nov 2023, 08:07
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C
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A tank is connected to 10 pipes that are either inlet or outlet types. Each inlet pipe can fill an empty tank in 8 hours, and each outlet pipe can empty a full tank in 6 hours. If all pipes are open, the tank fills in 12 hours. How many of the pipes are inlets?
A. 4
B. 5
C. 6
D. 7
E. 8
A. 4
B. 5
C. 6
D. 7
E. 8
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23 Jan 2024, 09:39
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A tank is connected to 10 pipes that are either inlet or outlet types. Each inlet pipe can fill an empty tank in 8 hours, and each outlet pipe can empty a full tank in 6 hours. If all pipes are open, the tank fills in 12 hours. How many of the pipes are inlets?
A. 4
B. 5
C. 6
D. 7
E. 8
Let's assume there are \(x\) inlet and \(y\) outlet pipes.
Since each inlet pipe can fill an empty tank in 8 hours, then the rate of one inlet pipe is \(\frac{1}{8}\) tank/hour, and consequently, the rate of \(x\) inlet pipes is \(\frac{x}{8}\) tank/hour.
Similarly, since each outlet pipe can empty a full tank in 6 hours, then the rate of one outlet pipe is \(\frac{1}{6}\) tank/hour, and consequently, the rate of \(y\) outlet pipes is \(\frac{y}{6}\) tank/hour.
Next, since with all pipes open, the tank fills in 12 hours, we have \(\frac{x}{8} - \frac{y}{6} = \frac{1}{12}\).
We also know that there are a total of 10 pipes, so \(x + y = 10\). Substituting \(y = 10 - x\) in the above equation, we get:
Multiply by the LCM of 8, 6, and 12, which is 24, to eliminate the fractions:
Therefore, there are 6 inlet pipes.
Answer: C
A. 4
B. 5
C. 6
D. 7
E. 8
Let's assume there are \(x\) inlet and \(y\) outlet pipes.
Since each inlet pipe can fill an empty tank in 8 hours, then the rate of one inlet pipe is \(\frac{1}{8}\) tank/hour, and consequently, the rate of \(x\) inlet pipes is \(\frac{x}{8}\) tank/hour.
Similarly, since each outlet pipe can empty a full tank in 6 hours, then the rate of one outlet pipe is \(\frac{1}{6}\) tank/hour, and consequently, the rate of \(y\) outlet pipes is \(\frac{y}{6}\) tank/hour.
Next, since with all pipes open, the tank fills in 12 hours, we have \(\frac{x}{8} - \frac{y}{6} = \frac{1}{12}\).
We also know that there are a total of 10 pipes, so \(x + y = 10\). Substituting \(y = 10 - x\) in the above equation, we get:
- \(\frac{x}{8} - \frac{10-x}{6} = \frac{1}{12}\)
Multiply by the LCM of 8, 6, and 12, which is 24, to eliminate the fractions:
- \(3x - 4(10-x)=2\)
\(x = 6\)
Therefore, there are 6 inlet pipes.
Answer: C
General Discussion
03 Nov 2023, 10:47
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Bunuel
Let the number of inlets be a
Let the number of inlets be b
Given that
a + b = 10 ---- eq (1)
Now if we open all inlet and outlet pipes we fill the tank in 12 hours, hence
\(a/8 - b/6 = 1/12\)
=> \((6a - 8b)/48 = 1/12\)
=> \(6a - 8b = 4\)
=> \(3a - 4b = 2\) ---- eq (2)
Using eq (1) and eq(2) we get
\(3a - 4b = 2\)
\(a + b = 10\) or \(3a + 3b = 30\)
Combining we get
\(7b = 28\)
=> \(b = 4\)
Using these two equations we get b = 4
Hence number of inlets = 10 - 4 = 6
Option C
