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work-rate problem 03 Oct 2005, 12:59
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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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work-rate problem 03 Oct 2005, 18:29
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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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work-rate problem 03 Oct 2005, 18:53
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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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work-rate problem 04 Oct 2005, 02:32
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fresinha12
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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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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work-rate problem 04 Oct 2005, 05:33
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krisrini
fresinha12
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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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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work-rate problem 05 Oct 2005, 01:38
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Vikramm, thanks much for your reply, it really helped identify what I needed to do.

Thanks



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!