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work-rate problem

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work-rate problem [#permalink] New post 03 Oct 2005, 11:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

this question would take me some time..close to 3 min....but good for testing your guessing strategies...

BTW: how do you guys feel about CAMBRIDGE CATs? are they any good?
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 [#permalink] New post 03 Oct 2005, 17:29
I get b) 3.0

P-> 6 hrs or 1 run is 6hrs
Q -> 4 hrs or 1 run in 4 hrs
P&Q -> 2.4 hrs or 1 run in 2.4 hrs

Day's run = Half the time by Q + Half the time by both
1/T = (1/2)*(1/4) + (1/2)*(1/2.4)
1/T = 1/8 + 10/48 = 16/48 = 1/3
=> T = 3 hrs
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 [#permalink] New post 03 Oct 2005, 17:53
Went through a round about route, but eventually arrived at the answer. It is, no surprises here, B!!!
Day's run = 6*30*60
Time taken = 2X
45*60*X + 75*60*X = 6* 30*60
Solving for X, we get 2X=3.0hrs, 8-)
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Re: work-rate problem [#permalink] New post 04 Oct 2005, 01:32
fresinha12 wrote:
this question would take me some time..close to 3 min....but good for testing your guessing strategies...

BTW: how do you guys feel about CAMBRIDGE CATs? are they any good?


My answer is E - 5 . Here is my working. Please correct me where I am wrong.

P - 1 hour - 1800 g ; so 6 hours - 10800 G
Q - 1 hour - 2700 g; so 4 hours - 10800 G
P&Q - 1 hour - 4500 g; so 2.4 hours - 10800 G

The question says Q alone half the time, and both machines for half the time.

Q half the time - 2 hours - 5400G
Since Q has already been running half its time, the remaining machine is P which in half the time would be 3 hours - 5400 G

So 2 hours + 3 hours = 5 hours.

Thanks
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Re: work-rate problem [#permalink] New post 04 Oct 2005, 04:33
krisrini wrote:
fresinha12 wrote:
this question would take me some time..close to 3 min....but good for testing your guessing strategies...

BTW: how do you guys feel about CAMBRIDGE CATs? are they any good?


My answer is E - 5 . Here is my working. Please correct me where I am wrong.

P - 1 hour - 1800 g ; so 6 hours - 10800 G
Q - 1 hour - 2700 g; so 4 hours - 10800 G
P&Q - 1 hour - 4500 g; so 2.4 hours - 10800 G

The question says Q alone half the time, and both machines for half the time.

Q half the time - 2 hours - 5400G
Since Q has already been running half its time, the remaining machine is P which in half the time would be 3 hours - 5400 G

So 2 hours + 3 hours = 5 hours.

Thanks


Sreeni,

To begin with, question says for a run -> half the time machine Q was run. 2 hours is half of Q's run (alone) not the day's run. If you assume that, then you should take half the time both ran too... which would mean 2.4/2 + 3 = 3.2 hrs (not an option).

A better way to clarify maybe is to backsolve here (its a longer method for this problem):
Say total time is 5 hours (your answer) and a run is 10800 gallons -> 1/2 the time Machine Q was run at 45 gpm. 2.5hrs, 45gpm gives a total of 6750 gallons. The other 2.5 hours both machines were run with a combined rate of 75gpm (30+45) to get 11250 gallons. So, 11250+6750 > 10800. Not possible.

Lets take 3 hours now. Half the time Q was run => 45*1.5*60 = 4050 g. The other half both machines were run => 75*1.5*60 = 6750 g.
4050 + 6750 = 10800 g!

In short, the equation you need to form is (see anandsebastin's post)
45*60*T/2 + 75*60*T/2 = 10800 and solve for T (3 hrs)

Shorter way - the rates are not required at all. 1/T = (1/2)(1/4) + (1/2)(1/2.4)
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Re: work-rate problem [#permalink] New post 05 Oct 2005, 00:38
Vikramm, thanks much for your reply, it really helped identify what I needed to do.

Thanks
Re: work-rate problem   [#permalink] 05 Oct 2005, 00:38
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