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DiveshR
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In a certain bathtub, both the cold-water and the hot-water fixtures 22 Jun 2025, 02:33
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EvaJager
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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

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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I followed up until t/h < 1/2 < t/c; But I lost you after that. How did you arrive at c/2 < t < h/2? Can you please elaborate?
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In a certain bathtub, both the cold-water and the hot-water fixtures 22 Jun 2025, 03:31
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you can split the inequality 't/h < 1/2 < t/c ' into 2 bits

t/h < 1/2 ==> multiply both sides by h ==> t < h/2
1/2 < t/c ==> multiply both sides by c ==> c/2 < t

Using the chaining rule on those two results, you can say:
c/2 < t < h/2

Hope that helps!


DiveshR
EvaJager
carcass
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.
I followed up until t/h < 1/2 < t/c; But I lost you after that. How did you arrive at c/2 < t < h/2? Can you please elaborate?
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Aboyhasnoname
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In a certain bathtub, both the cold-water and the hot-water fixtures Updated on: 24 Jul 2025, 10:35
Originally posted by Aboyhasnoname on 24 Jul 2025, 08:46.
Last edited by Bunuel on 24 Jul 2025, 10:35, edited 1 time in total.
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Let's Just focus on the Statement 3...
Don't confuse c and h with rates of cold water and hot water.
C is less than h... Means time taken by cold tap is less than hot tap...means the rate of cold tap is more than hot tap...
c<h and rate of Cold tap > rate of hot tap...


Now....c/2 < t < h/2
Suppose you open the taps....at the fastest speed of the close tap...both same...Now you have 2 taps that are working...so time taken will be c/2
No you decide, that let me put the hot water tap to the rate as it should be...less than cold water (Remember rate of h not time)...so you slow it down a little....Now you will take more time than you are would have taken if both operated at the speed of cold water tap...
so c/2 < t


Now Again..fresh start...
You open the two taps..but this time at the rate of hot water..which is slower...this will take you h/2 time....since both are at a speed of h and working simultaneously...
But now, you decided to increase the speed of cold water tap to it should be...This time it will take less time than it should have if they operated at a speed of hot water tap..

so t<h/2

Hence, III will always be true.....

Let us just take an example to solidify our understanding.....
Bucket Capacity - 12 Litres
Cold water tap takes 2 hours..means rate is is 6 litre/hr
Hot water tap takes 3 hours..means rate is 4 litres/hr....

Working together....
They would take 1.2 hours (6+4 litres to fill 12 litres)

Now...
Both hot and cold working at the speed of cold tap..means both working at 6 litre/hr speed
So it will be c/2 ... 2/2 ..only 1 hour to fill the bucket...


Now lets assume, both hot and cold working at the speed of hot tap.... means both working at 4 litre/hr ...means around 8litre/hr...they will take
h/2 that is 1.5 hours to fill the bucket (12/8)

Voila... !
c/2 is 1 which is less than t which is 1.2 which is less than h/2 which is 1.5... Hence 3 will be always true.


gmatpunjabi
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


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