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

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20 Feb 2012, 15:58

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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. 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?

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.

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

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23 Feb 2012, 14:09

I am sorry I do not understand your explanation for statement 3. I do not quite sure how got the inequalities in the algebraic approach and cant seem to understand the logic approach either.

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

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23 Feb 2012, 15:09

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Expert's post

gmatpunjabi wrote:

I am sorry I do not understand your explanation for statement 3. I do not quite sure how got the inequalities in the algebraic approach and cant seem to understand the logic approach either.

Consider this: say the cold-water leak needs 4 hours to fill an empty bucket and the hot-water leak needs 6 hours to fill an empty bucket.

Now, if both leaks needed 4 hours (so if hot-water were as fast as cold-water) then working together they would take 4/2=2 hours to fill the bucket, but we don't have two such fast leaks, so total time must be more than 2 hours. Similarly, if both leaks needed 6 hours (so if cold-water were as slow as hot-water) then working together they would take 6/2=3 hours to fill the bucket, but we don't have two such slow leaks, so total time must be less than 3 hours.

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

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25 Apr 2012, 01:03

Expert's post

LalaB wrote:

let c=2; h=3 ,

then 1/2+1/3=1/t

t=6/5

now plug in this numbers in every case only cases I and III satisfies

my question -is this method not ideal? I mean what are the flaws of this method?

For must be true questions number picking is a good strategy to discard an option. Your example showed that II is not true, so it's out.

As for I and III, you have that they are true for these particular numbers but they might not be true for others, so based on only one example you cannot say that I and III are ALWAYS true.

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

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25 Apr 2012, 02:39

Quote:

Hope it's clear.

Sorry, but I will annoy you one more time

Bunuel wrote:

As for I and III, you have that they are true for these particular numbers but they might not be true for others, so based on only one example you cannot say that I and III are ALWAYS true.

Yep! that is why I am asking this question. But how to be sure that for any number these two cases are true? is only the method shown by you a 100% guarantee of a true answer?

I mean when we use the 'plug-in' method in another type of questions we just check numbers from the ranges --(-1)--0--(1)-- . is there any range for this particular question? or it is safer not to use this method at all? _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

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

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25 Apr 2012, 02:53

Expert's post

LalaB wrote:

Quote:

Hope it's clear.

Sorry, but I will annoy you one more time

Bunuel wrote:

As for I and III, you have that they are true for these particular numbers but they might not be true for others, so based on only one example you cannot say that I and III are ALWAYS true.

Yep! that is why I am asking this question. But how to be sure that for any number these two cases are true? is only the method shown by you a 100% guarantee of a true answer?

I mean when we use the 'plug-in' method in another type of questions we just check numbers from the ranges --(-1)--0--(1)-- . is there any range for this particular question? or it is safer not to use this method at all?

"MUST BE TRUE" questions: These questions ask which of the following MUST be true, or which of the following is ALWAYS true no matter what set of numbers you choose. Generally for such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

So, for "MUST BE TRUE" questions plug-in method is good to discard an option but not 100% sure thing to prove that an option is ALWAYS true.

As for "COULD BE TRUE" questions: The questions asking which of the following COULD be true are different: if you can prove that a statement is true for one particular set of numbers, it will mean that this statement could be true and hence is a correct answer.

So, for "COULD BE TRUE" questions plug-in method is fine to prove that an option could be true. But here, if for some set of numbers you'll see that an option is not true, it won't mean that there does not exist some other set which will make this option true. _________________

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

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25 Apr 2012, 03:43

Bunuel wrote:

LalaB wrote:

Quote:

So, for "COULD BE TRUE" questions plug-in method is fine to prove that an option could be true. But here, if for some set of numbers you'll see that an option is not true, it won't mean that there does not exist some other set which will make this option true.

hm, my question is different. Perhaps I cant explain properly what I want to know or just do not understand what you are trying to say.

I used "must be/could be true" method to reject case 2. If my problem was in "must be/could be true", I could not cancel out even case 2.

My question is which numbers (!) should I plug in to be 100% sure that cases 1 and 3 are right? that is why I mentioned ranges.

but does ur answer mean that there is no such ranges to get 'must be true condition'? _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

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

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22 Oct 2012, 14:59

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?

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

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22 Oct 2012, 15:43

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carcass 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. \(\frac{c}{2}\) \(< t <\) \(\frac{h}{2}\)

(A) I only

(B) II only

(C) III only

(D) I and II

(E) I and III

The two rates for the cold water and that of the hot water fixture are 1/c and 1/h, respectively. We can write the following equation:

(1/c+1/h)t = 1, or t/c + t/h = 1.

I. t > 0 (represents time to fill the bucket), and from the above inequality it follows that t/h < 1, or t < h. TRUE

II. If t > c then t/c > 1, which is impossible according to the above equality. FALSE

III. Since t/c + t/h = 1 and c < h, t/c > t/h, necessarily t/h < 1/2 < t/c, or c/2 < t < h/2. TRUE

Answer E. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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22 Oct 2012, 22:24

EvaJager wrote:

carcass 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. \(\frac{c}{2}\) \(< t <\) \(\frac{h}{2}\)

(A) I only

(B) II only

(C) III only

(D) I and II

(E) I and III

The two rates for the cold water and that of the hot water fixture are 1/c and 1/h, respectively. We can write the following equation:

(1/c+1/h)t = 1, or t/c + t/h = 1.

I. t > 0 (represents time to fill the bucket), and from the above inequality it follows that t/h < 1, or t < h. TRUE

II. If t > c then t/c > 1, which is impossible according to the above equality. FALSE

III. Since t/c + t/h = 1 and c < h, t/c > t/h, necessarily t/h < 1/2 < t/c, or c/2 < t < h/2. TRUE

Answer E.

My approach was by just plugging in numbers. I put in \(1\) for c and \(2\) for h and \(\frac{2}{3}\)

\(\frac{1}{1} + \frac{1}{2} = \frac{3}{2}\)

so.,

\(c = 1, h = 2, t = \frac{2}{3}, \frac{c}{2} = \frac{1}{2}, \frac{h}{2} = \frac{2}{2} = 1\)

\(0<t<c<h\) \(&\) \(\frac{c}{2} < t < \frac{h}{2}\)

and I got the answer as E. Did I just get lucky or would it work with any two numbers. _________________

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

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22 Oct 2012, 22:37

1

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MacFauz wrote:

EvaJager wrote:

carcass 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. \(\frac{c}{2}\) \(< t <\) \(\frac{h}{2}\)

(A) I only

(B) II only

(C) III only

(D) I and II

(E) I and III

The two rates for the cold water and that of the hot water fixture are 1/c and 1/h, respectively. We can write the following equation:

(1/c+1/h)t = 1, or t/c + t/h = 1.

I. t > 0 (represents time to fill the bucket), and from the above inequality it follows that t/h < 1, or t < h. TRUE

II. If t > c then t/c > 1, which is impossible according to the above equality. FALSE

III. Since t/c + t/h = 1 and c < h, t/c > t/h, necessarily t/h < 1/2 < t/c, or c/2 < t < h/2. TRUE

Answer E.

My approach was by just plugging in numbers. I put in \(1\) for c and \(2\) for h and \(\frac{2}{3}\) and I got the answer as E. Did I just get lucky or would it work with any two numbers.

From the above solution, you can see that any triplet such that c < h and t which fulfill the equation (1/c + 1/h)t = 1 will work. The conclusion doesn't depend on the numbers themselves, but on the relationships between them. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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23 Oct 2012, 02:29

Expert's post

carcass 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. \(\frac{c}{2}\) \(< t <\) \(\frac{h}{2}\)

(A) I only

(B) II only

(C) III only

(D) I and II

(E) I and III

Merging similar topics. Please see the the solutions above. _________________

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

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23 Oct 2012, 20:47

LalaB wrote:

Bunuel wrote:

Anyway, for must be true questions you CANNOT be 100% sure that an option is always true just by number plugging.

ah, that is what I wanted to hear (to read, actually heh)! thanks a lot.

but in number properties questions u can be 100% sure if u use numbers from the ranges (infitiniy; -1) (-1;0) (0;1) (1;infinity)

That is why I asked whether there is any range for this kind of problem

ok, thanks a lot. sorry for annoying questions

LalaB, though I am not sure but I think there is still some clarification needed here.

First, I agree with you that in number properties questions which have finite ranges like natural numbers from 1 to 100 or integers from -7 to 8 etc, we can be 100% sure of the properties if they hold for each and everyone of the member of these ranges. We can check if the property is satisfied or not, by plugging in each of these numbers in the property.

However, for number property questions which have infinite ranges like natural numbers belonging to (1,infinity) or real numbers between 0 & 1 etc, we can never be 100% sure if the number properties are satisfied. Why? Because to check this, we would need to plug in each and every number (belonging to the range) in the property to see its validity. Since there are infinite numbers, we cannot do so.

Second, when Bunuel said "Anyway, for must be true questions you CANNOT be 100% sure that an option is always true just by number plugging", he is referring here to infinite range problem, like the way we are given; both c & h can have infinite real number values from anything greater than 0 to infinity. But suppose, if we were given that c & h are both prime numbers less than 20, then since range becomes finite, we could have used your number plugging method to verify the validity of the statements.

Third, there is always a range for any real number. It could be (minus infinity,infinity) or anything less than this range but range is always there. But having a range doesn't help, if the range has infinite members.

Fourthly, a friendly suggestion is that even when an expert says that one thing cannot be done, don't just accept it, ask questions, go deeper and don't accept till you are satisfied with the logic.

I hope my post is useful and I am sorry if I have written something wrong or unwanted.

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