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In a certain bathtub, both the cold-water and the hot-water

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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 17 Nov 2016, 15:43
First ever post, hope it's somewhat of a helpful one

The way I thought about S#3 was this...

Cold water leak will fill a bucket (B) in C hours

Hot water leak will fill a bucket (B) in H hours.

Together, the two leaks will fill two buckets (2B).

C+H=2B ----> C/2 + H/2 = B

Got it wrong but after the fact, this is what helped me understand that the combined time "t" was between C/2 and H/2
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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 23 Aug 2018, 14:09
we can treat as t/2 is harmonic mean of c and h which leaves III is true

t= ch/(c+h) , if t/2 then t = 2hc/(c+h) which is formula for harmonic mean of c and h and given c<h, which leaves c<t/2<h
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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 14 Oct 2018, 13:42
Any mean of two numbers is always between the two numbers, so c<2ch/c+h<h and c/2<ch/c+h<h/2.
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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 11 Jan 2019, 11:57
Hi All,

We're told that In a certain bathtub, both the hot and cold water fixtures leak. The cold water leak alone would fill an empty bucket in C hours, and the hot water leak alone will 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. We're asked which of the following MUST be true. Based on the five answer choices, we know AT LEAST one of the three Roman Numerals is always true - and we can TEST VALUES to define which are always true and which are not always true.

To start, this is an example of a Work Formula question, so we can use the Work Formula:

(A)(B)/(A+B) = time it takes to complete the task together, where A and B are the individual times needed to complete the task. In the prompt, we're told that C < H, so we can TEST C = 3 hours, H = 6 hours... meaning that the TOTAL time to fill the bucket would be (3)(6)/(3+6) = 18/9 = 2 hours... so T = 2. With those three values, we can check the Roman Numerals...

I. 0 < T < H

With our values, T = 2 and H = 6... and 0 < 2 < H, so Roman Numeral 1 appears to be true. Logically, we can also deduce that Roman Numeral 1 will ALWAYS be true, since when BOTH fixtures leak, the amount of time needed to fill the bucket would obviously be SMALLER than if just one of the fixtures was leaking. This means that T < H and T < C will ALWAYS be true and all of those variables will be greater than 0.
Eliminate Answers B and C.

II. C < T < H

With our values, C=3, T = 2 and H = 6... but 3 < 2 < H is NOT true, so Roman Numeral 2 is NOT true
Eliminate Answer D.

III. C/2 < T < H/2

With our values, C=3, T = 2 and H = 6... and 3/2 < 2 < 6/2 IS true, so Roman Numeral 3 appears to be true. Roman Numeral 3 will also ALWAYS be true, but you would have to do a bit more work to prove it. With ANY pair of C and H that fits the given parameters, we'll end up with a T that is less than both. Since C is the faster rate (in this example, 3 hours to fill a bucket is faster than 6 hours to fill a bucket), if we divide C by 2 and H by 2, then those rates DOUBLE (it would then take just 1.5 hours and 3 hours, respectively, to fill the bucket). That clearly gives us one values that is LESS than the current value of T and one that is MORE than the current value of T.
Eliminate Answer A.

Final Answer:

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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 17 Jan 2019, 19:13
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gmatpunjabi wrote:
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. c/2 < t < h/2

A) I only
B) II only
C) III only
D) I and II
E) I and III


We can create the equation:

1/c + 1/h = 1/t

If we let c = 2 and h = 3, we have:

1/2 + 1/3 = 3/6 + 2/6 = 5/6.

So we see that t = 1/(5/6) = 6/5 = 1.2.

So we have t < c < h, so statement I is correct and statement II is not.

Let’s now analyze statement III.

c/2 = 2/2 = 1

h/2 = 3/2 = 1.5

Thus:

c/2 < t < h/2

Answer: E
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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 24 Jun 2019, 00:40
BUNUEL can I plug in values in questions like these?



Bunuel wrote:
gmatpunjabi wrote:
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. c/2 < t < h/2

A) I only
B) II only
C) III only
D) I and II
E) I and III - Answer

Can someone please provide a comprehensive explanation as to why statement III is also valid?


I. 0 < t < h. That is always correct, as the time needed for both fixtures leaking (working) together to fill the bucket, \(t\), must always be less than time needed for either of fixture leaking (working) alone to fill the bucket;

II. c < t < h. That cannot be correct: \(t\), the time needed for both fixtures leaking (working) together to fill the bucket, must always be less than time needed for either of fixture leaking (working) alone to fill the bucket. So \(c<t\) not true.

III. c/2 < t < h/2. To prove that this is always correct we can use pure logic or algebra.

Logic:
If both fixtures were leaking at identical rate then \(\frac{c}{2}=\frac{h}{2}=t\) but since \(c<h\) then \(\frac{c}{2}<t\) (as the rate of cold water is higher) and \(t<\frac{h}{2}\) (as the rate of hot water is lower).

Algebraic approach would be:

Given: \(c<h\) and \(t=\frac{ch}{c+h}\)

\(\frac{c}{2}<\frac{ch}{c+h}<\frac{h}{2}\)? break down: \(\frac{c}{2}<\frac{ch}{c+h}\)? and \(\frac{ch}{c+h}<\frac{h}{2}\)?

\(\frac{c}{2}<\frac{ch}{c+h}\)? --> \(c^2+ch< 2ch\)? --> \(c^2<ch\)? --> \(c<h\)? Now, this is given to be true.

\(\frac{ch}{c+h}<\frac{h}{2}\)? --> \(2ch<ch+h^2\)? --> \(ch<h^2\)? --> \(c<h\)? Now, this is given to be true.

So III is also always true.

Answer: E.

Hope it's clear.
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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 24 Jun 2019, 16:30
Hi ishanivarma,

YES - you can TEST VALUES in this question - but in a way that you are probably not used to thinking about that Tactic. My explanation (two posts above yours) shows how you can do it.

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Re: In a certain bathtub, both the cold-water and the hot-water  [#permalink]

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New post 04 Aug 2019, 05:49
gmatpunjabi wrote:
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. c/2 < t < h/2

A) I only
B) II only
C) III only
D) I and II
E) I and III

Can someone please provide a comprehensive explanation as to why statement III is also valid?


If 2 c are working, T = c/2
If 2 h are working, T = h/2

Therefore, if both c and h are working, then c/2 < T < h/2

1) Covers the whole range. MBT
2) Doesn't cover c/2 to c. Not MBT
3) Covers the whole range as calculated above. MBT

ANSWER: E
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Re: In a certain bathtub, both the cold-water and the hot-water   [#permalink] 04 Aug 2019, 05:49

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