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Joined: 20 Feb 2017
Posts: 2506
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In a certain bathtub, both the cold-water and the hot-water fixtures
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20 Nov 2021, 01:05
20 Nov 2021, 01:05
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Question Stats:
33% (02:30) correct
66% (02:15) wrong based on 12 sessions
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In a certain bathtub, both the cold-water and the hot-water fixtures leak. The cold-water leak alone would fill an empty bucket in c hours and the hot-water leak alone would fill the same bucket in h hours, where c < h. If both fixtures began to leak at the same time into the empty bucket at their respective constant rates and consequently it took t hours to fill the bucket, which of the following must be true?
I. \(0 < t < h\)
II. \(c < t < h\)
III. \(\frac{c}{2} < t < \frac{h}{2}\)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
I. \(0 < t < h\)
II. \(c < t < h\)
III. \(\frac{c}{2} < t < \frac{h}{2}\)
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
Re: In a certain bathtub, both the cold-water and the hot-water fixtures
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20 Nov 2021, 02:52
20 Nov 2021, 02:52
3
Let the capacity of bucket be \(L\)
rate of cold-water leak, \(r_c = \frac{L}{c}\)
rate of hot-water leak, \(r_h = \frac{L}{h}\)
Both filling bucket together -
\(t(r_c + r_h) = L\)
\(t(\frac{L}{c} + \frac{L}{h}) = L \Rightarrow t = \frac{1}{\frac{1}{c} + \frac{1}{h}} = \frac{ch}{c+h}\)
I - \(0 < t < h\)
This is so obviously true since both filling the bucket together is going to take less time than individual. Hence, true
II - \(c < t < h\)
Same point as stated in I, \(t < c,h\). Hence, false
III - \(\frac{c}{2} < t < \frac{h}{2}\)
\(\frac{c}{2} < t < \frac{h}{2}\)
\(\frac{c}{2} < \frac{ch}{c+h} < \frac{h}{2}\)
\(c\frac{1}{2} < c\frac{h}{c+h}\) and \(h\frac{c}{c+h} < h\frac{1}{2}\)
\(\frac{1}{2} < \frac{h}{c+h}\) and \(\frac{c}{c+h} < \frac{1}{2}\)
\(c+h < 2h\) and \(2c < c+h\)
\(c < h\) and \(c < h\) - this has already been given in question. Hence, true
Therefore, I and III are true
Hence, Answer is E
rate of cold-water leak, \(r_c = \frac{L}{c}\)
rate of hot-water leak, \(r_h = \frac{L}{h}\)
Both filling bucket together -
\(t(r_c + r_h) = L\)
\(t(\frac{L}{c} + \frac{L}{h}) = L \Rightarrow t = \frac{1}{\frac{1}{c} + \frac{1}{h}} = \frac{ch}{c+h}\)
I - \(0 < t < h\)
This is so obviously true since both filling the bucket together is going to take less time than individual. Hence, true
II - \(c < t < h\)
Same point as stated in I, \(t < c,h\). Hence, false
III - \(\frac{c}{2} < t < \frac{h}{2}\)
\(\frac{c}{2} < t < \frac{h}{2}\)
\(\frac{c}{2} < \frac{ch}{c+h} < \frac{h}{2}\)
\(c\frac{1}{2} < c\frac{h}{c+h}\) and \(h\frac{c}{c+h} < h\frac{1}{2}\)
\(\frac{1}{2} < \frac{h}{c+h}\) and \(\frac{c}{c+h} < \frac{1}{2}\)
\(c+h < 2h\) and \(2c < c+h\)
\(c < h\) and \(c < h\) - this has already been given in question. Hence, true
Therefore, I and III are true
Hence, Answer is E
Re: In a certain bathtub, both the cold-water and the hot-water fixtures
[#permalink]
22 Jun 2023, 03:51
22 Jun 2023, 03:51
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
