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

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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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.

So, 4/2<t<6/2 --> c/2<t<h/2.

Hope it's clear.
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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.

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.
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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.

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.
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t = ch/(c + h)

Option 1 : 0 < t < h

ch/(c + h) < h
i.e c< c + h
h > 0

ch/(c + h) > 0
i.e ch > 0 ( c and h not -ve)
i.e both c & h > 0
Satisfies option 1

Option 2 : c < t < h
c < ch/(c + h)
c + h < h
c < 0 ..False
(No need to check other inequality)

Option 3 : c/2 < t/2 < h/2

c/2 < ch/(c + h)
c + h < 2h
c < h (true ,as per question)

ch/(c + h) < h/2
2c <c + h
c<h (true,as per question)..Hence always true

Ans 1 & 3

We know that c < h, so the cold tap fills the tub more quickly than the hot tap.

If you turn on both taps, you'll fill the tub more quickly than if you only turn on one tap, so t < c < h, and I is true and II is false (I think there's a typo in the original post - if I recall correctly, in the GMATPrep version of this question, the first roman numeral item reads "t < c < h").

If you had two of the fast taps (the cold taps), they'd fill the tub in c/2 hours. If you had two of the slow taps (the hot taps), they'd fill the tub in h/2 hours. We have one fast tap and one slow tap, so the time they take together must be somewhere in between c/2 and h/2, which is what III says, so III must also be true.
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1.Take an example. C= 1 hr, h= 2 hrs, t=1/(1/1+1/2)=2/3
2.Take a totally different example c=1 hr, h=100hrs, t=1/(1/1+1/100)=100/101
In both the cases I and III are true.
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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:

GMAT assassins aren't born, they're made,
Rich

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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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

This problem can be solved easily by plugging in numbers for example C = 2hrs and H = 3 hrs.

T becomes 4/3hrs therefore option 1 and 3 holds true.

My 2 cents on this:

Personally more comfortable taking numbers and so here is my attempt with that

C --> 4 hrs and H --> 7 hrs given that C<H

Rate of c = 1/4 ; Rate of h = 1/7

1/4 + 1/7 = 1/t

t = 28/11 which is approx 2.5

C=4, H=7, T = 2.5

I and III satisfy, hence E.

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