In a certain bathtub, both the hot and cold water fixtures : GMAT Problem Solving (PS)
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# In a certain bathtub, both the hot and cold water fixtures

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In a certain bathtub, both the hot and cold water fixtures [#permalink]

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29 Jan 2010, 00:49
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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 tool 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Feb 2013, 00:49, edited 1 time in total.
Edited the question and added OA.
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29 Jan 2010, 02:44
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If both fixtures leak then it take to fill the bucket in t time which will be less than the time taken by any of fixture.
so, 1 is right

so 2 may not be right.

3 is also right

t=ch/(c+h)

doing back calculation

as 3 say

(h/2)<t<(c/2)

h/2<t
=> h/2< [ch/(c+h)]
=> c+h<2c
=>h<c which is right as per question
similarly we can prove that t<c/2

so iii is also correct
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01 Feb 2010, 12:23
I think OA is 1 (I Only)
Reason: Both condition 2 and 3 imply that c< h which is opposite to the problem statement. 1 is correct because t will be always less than either c or h by itself (sum of flow rates always grater than an individual flow rate) and will be greater than zero, as time that takes to fill up the bath tab has to be positive number.
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01 Feb 2010, 16:32

I is always true since the T value of C+H will always be smaller than H alone.

II may not always be true as it is dependent on the values for C & H.

III is always true given the fact that C > H.

Sanjay already did the math for the last part to prove that the result of C + H < 2C must always be true.

I took the algebra-free approach. Given it takes C less time to fill the tub than it does H, then you know that C/2 is less than H/2. Additionally, the two working together (C+H) would have to be somewhere between the two due to their unequal rates of flow.
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01 Feb 2010, 20:51
(c x h)/ (c + h) = t

c > h

Thus by putting C = 1hr and h = 2hr

clearly (i) and (ii) are correct. Therefore only answer choice which seems possible is (4)
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02 Feb 2010, 05:19
LJ wrote:

I is always true since the T value of C+H will always be smaller than H alone.

II may not always be true as it is dependent on the values for C & H.

III is always true given the fact that C > H.

Sanjay already did the math for the last part to prove that the result of C + H < 2C must always be true.

I took the algebra-free approach. Given it takes C less time to fill the tub than it does H, then you know that C/2 is less than H/2. Additionally, the two working together (C+H) would have to be somewhere between the two due to their unequal rates of flow.

How can c/2 < h/2, while we have c>h, and we are talking about positive whole numbers? Please explain.
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03 Feb 2010, 17:42
alexBLR wrote:
LJ wrote:

I is always true since the T value of C+H will always be smaller than H alone.

II may not always be true as it is dependent on the values for C & H.

III is always true given the fact that C > H.

Sanjay already did the math for the last part to prove that the result of C + H < 2C must always be true.

I took the algebra-free approach. Given it takes C less time to fill the tub than it does H, then you know that C/2 is less than H/2. Additionally, the two working together (C+H) would have to be somewhere between the two due to their unequal rates of flow.

How can c/2 < h/2, while we have c>h, and we are talking about positive whole numbers? Please explain.

I see the way I wrote it could be confusing.

C/2 would be described as half the time it takes C to fill the tub alone.
H/2 would be described as half the time it takes H to fill the tub alone.
T, aka (H + C) is the time it takes for the faucets to fill the tub together.

Since C alone fills the tub faster than H, you can see that C working twice as fast (C/2) would fill the tub faster than H+C, therefore it is less than T. The opposite is true for H.
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04 Feb 2010, 05:16
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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 tool $$t$$ hours to fill the bucket, which of the following must be true?

1. 0 < t < h
2. c < t <h
3. c/2 < t < h/2

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

There is NO WAY (E) can be the correct answer. The answer for the question above must be (A) I only.

Given $$c>h$$:
2. is never correct as $$c < t <h$$, means $$c<h$$ and that contradicts the stem;

3. is never correct as $$\frac{c}{2}<t<\frac{h}{2}$$, means $$\frac{c}{2}<\frac{h}{2}$$ or $$c<h$$ and that contradicts the stem.

1. is always correct, as 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.

Guess the original question had $$c<h$$ (not $$c>h$$). In this case yes E is the correct answer.

1. remains correct as explained above.

2. can not be correct: $$t$$, 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.

3. To prove that this is always correct we can use pure logic or algebra.

Logic:
If both fixtures were leaking at identical rate then c/2=h/2=t but as the rate of cold water is higher (because it needs less time) then c/2<t and as the rate of hot water is lower then t<h/2.

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 3 is also always true. Answer E (in case we change the stem).

Hope it's clear.
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28 Mar 2010, 10:01
[Reveal] Spoiler:
OA--E

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28 Mar 2010, 11:04
Very tricky problem. The math itself requires some insight, then they stick it to you with an inequalities and wording trick!

The cold faucet rate is 1 bucket in c hours = 1/c (buckets per hour)
The hot faucet rate is 1 bucket in h hours = 1/h (buckets per hour)

The combined rate can be computed as the sum of the individual rates = [1/c + 1/h] (buckets per hour).
The combined rate is also given to us directly as 1 bucket in t hours = 1/t (buckets per hour).

Relating t to c and h:
[1/c + 1/h] = 1/t
[(h+c)/ch] = 1/t
t = ch/(h+c)

Seeing that all the answers had inequalities, and that c<h was given, I wrote the following on paper:

t = c*c'/(c'+c), where the mark (') indicates "a little more than."
t = (c^2)'/(2c)' = (c/2)'

t = h"*h/(h + h"), where the mark (") indicates "a little less than."
t = (h^2)"/(2h)" = (h/2)"

Put it together: c/2 < t < h/2, which corresponds to III directly.

But the final trick is that since c is positive (i.e. the cold faucet leak doesn't fill the bucket in literally no time.), 0< c/2 < t.
Since h is positive (same reason), h > h/2 > t.

Put all that together:
0 < c/2 < t < h/2 < h
0 < t < h, so I also "must be true."
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27 Feb 2011, 22:23
esledge wrote:
Very tricky problem. The math itself requires some insight, then they stick it to you with an inequalities and wording trick!

The cold faucet rate is 1 bucket in c hours = 1/c (buckets per hour)
The hot faucet rate is 1 bucket in h hours = 1/h (buckets per hour)

The combined rate can be computed as the sum of the individual rates = [1/c + 1/h] (buckets per hour).
The combined rate is also given to us directly as 1 bucket in t hours = 1/t (buckets per hour).

Relating t to c and h:
[1/c + 1/h] = 1/t
[(h+c)/ch] = 1/t
t = ch/(h+c)

Seeing that all the answers had inequalities, and that c<h was given, I wrote the following on paper:

t = c*c'/(c'+c), where the mark (') indicates "a little more than."
t = (c^2)'/(2c)' = (c/2)'

t = h"*h/(h + h"), where the mark (") indicates "a little less than."
t = (h^2)"/(2h)" = (h/2)"

Put it together: c/2 < t < h/2, which corresponds to III directly.

But the final trick is that since c is positive (i.e. the cold faucet leak doesn't fill the bucket in literally no time.), 0< c/2 < t.
Since h is positive (same reason), h > h/2 > t.

Put all that together:
0 < c/2 < t < h/2 < h
0 < t < h, so I also "must be true."

Once you have found the value of t=ch/c+h, Try plugging in the values of t in choices:

(1) 0 < t < h

0<ch/c+h<h we get c>0 true & c<c+h also true hence Chose 1 is correct.

(2) c < t < h

c<ch/c+h<h, we get c+h<h (from solving the L.H.S of the inequality) - False & c<c+h (from solving the R.H.S of the inequality) - True - Hence Choice 2 is incorrect.

(3) (c/2) < t < (h/2)

c/2<ch/c+h<h/2, we get c<h (from solving the L.H.S of the inequality) - True & c<h (from solving the R.H.S of the inequality) - True - Hence Choice 3 is incorrect.

Therefore, Choice 5 is the correct Answer.
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Re: In a certain bathtub,both the hot and cold water fixtures [#permalink]

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02 Sep 2012, 12:17
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Question asked is "which option MUST BE true", so if i can prove any option wrong , i can easily select the correct answer.
So i preferred to use value for these variables as i am trying to prove few options are incorrect
As c>h
let c = 4 hrs
h = 2 hrs
Thus t = 8/6 = 4/3 hr

Now if put these value only option A is true

Thus as per me the answer has to be A
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Re: In a certain bathtub,both the hot and cold water fixtures [#permalink]

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14 Nov 2012, 05:33
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Let c = 4
Let h = 5
$$\frac{1}{c}+\frac{1}{h}=\frac{1}{t}$$
$$\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$$

I. 0 < $$\frac{20}{9}$$ < 5 TRUE
II. 4 < $$\frac{20}{9}$$ < 5 FALSE
III. 2 < 20/9 < 5/2 TRUE!

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

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21 Feb 2013, 21:37
The original posted stem says c>h. In which case E can not be the answer, only A. In the correct version (posted later) the stem says c<h. In this case E is the answer. Can someone please change the original posted stem to match the question they were trying to copy in order to clear up the confusion.
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Re: In a certain bathtub,both the hot and cold water fixtures [#permalink]

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22 Feb 2013, 00:51
coreyrnichols wrote:
The original posted stem says c>h. In which case E can not be the answer, only A. In the correct version (posted later) the stem says c<h. In this case E is the answer. Can someone please change the original posted stem to match the question they were trying to copy in order to clear up the confusion.

Edited the question and added OA.
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Re: In a certain bathtub, both the hot and cold water fixtures [#permalink]

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19 Sep 2014, 12:50
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Re: In a certain bathtub, both the hot and cold water fixtures [#permalink]

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08 May 2016, 03:42
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Re: In a certain bathtub, both the hot and cold water fixtures   [#permalink] 08 May 2016, 03:42
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