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Water flows into a 25-liter bucket through a hose and out
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24 Nov 2017, 20:56
24 Nov 2017, 20:56
Expert Reply
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Question Stats:
80% (01:57) correct
20% (01:42) wrong based on 50 sessions
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Water flows into a 25-liter bucket through a hose and out through a hole in the bottom of the bucket. The rate of flow through the hose is 1 liter per minute. If the bucket is filled to capacity in 40 minutes, at what rate, in liters per minute, was water flowing out of the bucket through the hole?
A. \(\frac{3}{8}\)
B. \(\frac{3}{5}\)
C. \(\frac{5}{8}\)
D. \(\frac{8}{5}\)
E. \(\frac{13}{8}\)
A. \(\frac{3}{8}\)
B. \(\frac{3}{5}\)
C. \(\frac{5}{8}\)
D. \(\frac{8}{5}\)
E. \(\frac{13}{8}\)
Drill 2
Question: 7
Page: 525
Question: 7
Page: 525
Re: Water flows into a 25-liter bucket through a hose and out
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01 Dec 2017, 08:51
01 Dec 2017, 08:51
2
sandy wrote:
Water flows into a 25-liter bucket through a hose and out through a hole in the bottom of the bucket. The rate of flow through the hose is 1 liter per minute. If the bucket is filled to capacity in 40 minutes, at what rate, in liters per minute, was water flowing out of the bucket through the hole?
A. \(\frac{3}{8}\)
B. \(\frac{3}{5}\)
C. \(\frac{5}{8}\)
D. \(\frac{8}{5}\)
E. \(\frac{13}{8}\)
A. \(\frac{3}{8}\)
B. \(\frac{3}{5}\)
C. \(\frac{5}{8}\)
D. \(\frac{8}{5}\)
E. \(\frac{13}{8}\)
Drill 2
Question: 7
Page: 525
Question: 7
Page: 525
We can solve the problem using the equation
A = R x T
Where A= 25 L
R (Total) = R (inlet) - R (outlet) (Since we have rate of water entering the bucket and the rate of water flowing out the bucket)
R (inlet) = 1 Litre/minute
R (outlet) = Let us assume as r Litre/minute
T = 40 minute
Therefore we have
25 = (1 - r) X 40
or \(\frac{25}{40}\)= 1 - r
or r = 1 - \(\frac{5}{8}\)
or r =\(\frac{3}{8}\)
Hence the rate at which the the water was flowing out is = \(\frac{3}{8}\)
Re: Water flows into a 25-liter bucket through a hose and out
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29 Dec 2017, 13:59
29 Dec 2017, 13:59
2
Expert Reply
Explanation
Remember that amount = rate × time. So, 25 liters = rate × 40 minutes.
The rate was \(\frac{25}{40}=\frac{5}{8}\) liters/min.
The net rate at which the bucket is filling is the difference between the hose’s rate and the leaking rate. So, (1 liter/min) – (leaking rate) = \(\frac{5}{8}\).
Solve for the leaking rate to find the leaking rate is \(\frac{3}{8}\)liters/min; the answer is choice A.
Remember that amount = rate × time. So, 25 liters = rate × 40 minutes.
The rate was \(\frac{25}{40}=\frac{5}{8}\) liters/min.
The net rate at which the bucket is filling is the difference between the hose’s rate and the leaking rate. So, (1 liter/min) – (leaking rate) = \(\frac{5}{8}\).
Solve for the leaking rate to find the leaking rate is \(\frac{3}{8}\)liters/min; the answer is choice A.
