Rates problem

[lightbulb]

Guys,

Here is another one from the GMAT Prep that I couldn't figure completely.

In a certain bath tub, both the cold water and hot water fixuters leak. The cold water leak alone would fill an empty bucket in c hours, and the hot water leak 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

o I only
o II only
o III only
o I and II
o I and III

Ans: E

[earth@work]

IMO it's E= I and III are true..what is OA

[Ian Stewart]

In a certain bath tub, both the cold water and hot water fixuters leak. The cold water leak alone would fill an empty bucket in c hours, and the hot water leak 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

This is a good 'conceptual' rates problem. Let's look at a concrete example, with numbers, to see what must be true. Say you have a cold tap that will fill a sink in 10 hours, and a hot tap that's slower- it will fill the sink in 12 hours. If t is the time it takes the hot and cold taps together, what must be true?

Well, firstly, t must be less than 10 hours, since the cold tap alone takes 10 hours, and so with the help of the hot tap it must take less than 10 hours.

Secondly, if you had two taps working together which were identical to the cold tap, they would take 10/2 = 5 hours together to fill the sink. You don't have two taps like the cold tap, though; the hot tap is slower, so t must be greater than 5 hours. If you had two taps just like the hot tap, it would take 12/2 = 6 hours, but you don't have two taps just like the hot tap; the cold tap is faster, so it must take less than 6 hours.

Applying the same logic to the above problem, we can see that t must be less than h, t must be less than c, and c/2 < t < h/2. So I and III must be true, and II must be false (c is certainly not less than t).

[earth@work]

Wow Ian, your method is much simpler than the calculations i was doing.

Just wanted to know if my method was correct for III. c/2 < t < h/2
Taking speed of Cold leakage as S1 & hot as S2
Since volume is same = S1*c = S2*h =(S1+S2)*t
Taking S1*c=(S1+S2)*t......(1)
also we know S1>S2, putting S1 in place of S2 in (1) we get : S1*c<2S1*t, which gives us c<t/2
similarly, for S2*h =(S1+S2)*t, we get S2*h>2*S2*t, implies h/2>t

which gives us c/2 < t < h/2 true, is this method correct?

[lightbulb]

Nice response Ian, thanks.
I wasn't able to figure out the rational for III, but now it makes sense. I got all wound up in the math, trying to prove/disprove III using the following relationship and lost a lot of time.

ch/(c + h) = t

What should be the strategy if you get hung up like this on the test?

[logitech]

Wow Ian, your method is much simpler than the calculations i was doing.

Just wanted to know if my method was correct for III. c/2 < t < h/2
Taking speed of Cold leakage as S1 & hot as S2
Since volume is same = S1*c = S2*h =(S1+S2)*t
Taking S1*c=(S1+S2)*t......(1)
also we know S1>S2, putting S1 in place of S2 in (1) we get : S1*c<2S1*t, which gives us c<t/2
similarly, for S2*h =(S1+S2)*t, we get S2*h>2*S2*t, implies h/2>t

which gives us c/2 < t < h/2 true, is this method correct?

Remember, GMAC people put a lot of time to design the answer choices. And nothing is in answer choices without a reason.

c/2 < t < h/2 AHA!! :twisted:

So we should actually see this signal and try to understand why c/2 or h/2 is here...

just my 2 cents.

IAN YOU ROCK!

[dmateer25]

Nice response Ian, thanks.
I wasn't able to figure out the rational for III, but now it makes sense. I got all wound up in the math, trying to prove/disprove III using the following relationship and lost a lot of time.

ch/(c + h) = t

What should be the strategy if you get hung up like this on the test?

I got to the same point you did with this problem and then it clicked in my head to plug in numbers for c and h. We know h > c so:

h=4
c=2

(2)(4)/(2+4) = T
8/6 = 4/3 = T

Now just fill in:

I. 0 < 4/3 < 4.............yes
II. 2 < 4/3 < 4............no
III. 2/2 < 4/3 < 4/2.....yes

Answer would have to be E

[logitech]

Nice response Ian, thanks.
I wasn't able to figure out the rational for III, but now it makes sense. I got all wound up in the math, trying to prove/disprove III using the following relationship and lost a lot of time.

ch/(c + h) = t

What should be the strategy if you get hung up like this on the test?

I got to the same point you did with this problem and then it just clicked in my head to just plug in numbers for c and h. We know h > c so:

h=4
c=2

(2)(4)/(2+4) = T
8/6 = 4/3 = T

Now just fill in:

I. 0 < 4/3 < 4.............yes
II. 2 < 4/3 < 4............no
III. 2/2 < 4/3 < 4/2.....yes

Answer would have to be E

I WOULD BE CAREFUL WITH PLUGGING NUMBERS IN "MUST BE TRUE" questions!

[pablo1]

can anybody explain the name of the forumla given above.. ch/c+h=t