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Rate problem

by ktn11 » Sat Aug 20, 2011 5:12 am
1. At its constant rate, sluice X drained off 1/2 amount of water in four hours. Then, X and Y together, at their constant rates, drained off the remaining water in three hours. How many hours will it take for Y to drain off all the water, independently?

can anyone solve this?
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by knight247 » Sat Aug 20, 2011 5:35 am
Let the amount of water to be drained be 120 Gallons

In t1=4 hrs Sluice x drained off 60 Gallons
Rate=Work/Time
R1=60/4=15 Gallons/hr


Let Rate and Time for Y be R2 and T2 respectively

For Sluice X and Y combined, Let Time and Rate be T3 and R3 respectively

Sluice X and Y combined, in T3=3 hours drained the remaining 60 gallons

R3=60/3=20 Gallons/hr

Now we can infer that

R1+R2=R3

15+R2=20
R2=5 Gallons/hr

Now if we have to drain off 120 gallons of water using only Sluice y which drains at the rate of 5 Gallons/hr
Time=work/Rate=120/5=24 hrs
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by Anurag@Gurome » Sat Aug 20, 2011 5:41 am
ktn11 wrote:1. At its constant rate, sluice X drained off 1/2 amount of water in four hours. Then, X and Y together, at their constant rates, drained off the remaining water in three hours. How many hours will it take for Y to drain off all the water, independently?
In 1 hour X drains off (1/2)/4 = 1/8 amount of water
In 1 hour X and Y together drain off (1/2)/3 = 1/6 amount of water

Hence, in 1 hour Y drains off (1/6 - 1/8) = 1/24 amount of water

Therefore, Y will take 24 hours to drain off all the water.
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by akhilsuhag » Sat Aug 20, 2011 5:54 am
Let the total work be x.
The rate of x be r1 and that of y r2.

Now, (1/2)x = r1*4 ------(A)
and (1/2)x = (r1+r2)*3 -----(B)

Equating A and B we find that, 4*r1 = 3r1 + 3r2, giving us r2 = (1/3)r1 ---- (c)

Now for just machine Y, the equation is:

x = r2 * t (where t is the time).
Now using (A) and (C)

8r1 = (1/3)*r1 * t;

This gives us [spoiler]t = 24.[/spoiler]
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by akhilsuhag » Sat Aug 20, 2011 6:01 am
Hi Ktn,

I just thought how I go about rate problems and I hope it helps you in conquering them.

I write all the rate equations possible ( the work = rate * time) for each machine/ person or combination of machines. Now in these question you will find something that u can equate.

In most cases the work done can be equated or in some you will be given a relation b/t the two rates or times or work. You can use these relations to plug in and equate the two equations and then the questions usually do solve themselves.

I hope it helps.
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by ktn11 » Sat Aug 20, 2011 6:28 am
Thanks for solving and sugestions
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by GMATGuruNY » Sat Aug 20, 2011 4:33 pm
ktn11 wrote:1. At its constant rate, sluice X drained off 1/2 amount of water in four hours. Then, X and Y together, at their constant rates, drained off the remaining water in three hours. How many hours will it take for Y to drain off all the water, independently?

can anyone solve this?
Let water = 24 units.
Rate for X alone to drain 1/2 the water = w/t = 12/4 = 3 units per hour.
Rate for X and Y combined to drain the remaining water = w/t = 12/3 = 4 units per hour.
Rate for Y alone = (Rate for X and Y combined) - (Rate for X alone) = 4-3 = 1 unit per hour.
Time for Y to drain all the water = w/r = 24/1 = 24 hours.
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