work problem.....part 2!!!!!!!!!?

Question

a swimming pool can be filled by a pipe in 10 hours if the drain is opened. if the pipe was allowed to run with the drain for 2 hours, and the drain is closed afterwards, it would require the pipe 4 more hours to fill the tank. how long would it take the pipe to make the tank full if the drain is closed from the start?


i need step by step procedure, again..

pls answer....NOW!!!!!
PASS TOMORROW DEC.4 2007...!!!!!!

ARIGATO GUSAYMASU!!!

DOMO ARIGATO!!!

Answer 1

[04]
When the pipe+drain is opened ,it takes 10 hrs to fo fill the pool
Therefore if both are opened in 1 hr ,1/10 part of the pool can be filled
So,when both are opened for 2 hrs,1*2/10 or 1/5 part of the pool is filled
Part remaining to be filled=1-1/5=4/5
4/5 part of the pool is filled by the pipe in 4 hrs
Therefore,the pipe itself can fill the pool in
4 divided by 4/5 or 5 hrs.if the drain is closed from the beginning

Answer 2

I can show you this but you will not pass the test tomorrow unless you do a lot of problems between now and test time.

Let V = volume of pool; let t = time to fill it; let F = flow rate of the fill line, d = flow rate of the drain line.

to fill the pool fully to the volume in t time with both fill and drain in use we know

V = t (F-d)

we further know that this will define the volume level at any point in time if both the fill and drain are in use. Understand the above and upon what it is based before proceeding as the rest is just fundamental algebra.

Now we fill it to full volume V in 10 hours with both F and d on, which states

10(F-d) = V (understand this from above)

If we fill to full volume, V, with both F and d on for 2 hours AND then shut d off and continue with F on for 4 more hours to reach fullness we also know

2(F-d) + 4F = V (understand this from above)

now since the full volume V is the same

2(F-d) + 4F = 10(F-d)

now do the algebra and you see once this is done that

F = 2 d

now back to 10(F-d)=V and substitute

10 (2d-d) = v or
10d = v or d = V/10 and
10(F -F/2) = V or
5F = V or F = V/5

now since we are not draining, d is shut off and
t(F-0) = V
t(V/5) = V
t = 5

Answer 3

5 hrs

it takes 2 hours to fill the pool up 20% with the drain open then 4 hours with the drain closed to fill it up the other 80%. If it takes 4 hours to fill the pool 80% with the drain closed, it will take 5 hours to fill the pool 100% with the drain closed.

Answer 4

if when the drain is open it fills the pool in ten hours then it fills 1/10 per hour, so in 2 hours it fills 1/5 of the pool. if then the drain is closed then it takes 4 hours to fill 4/5 of the pool, meaning 1/5 of the pool per hour, then to fill the whole pool it must take 5 hours.

Answer 5

Let the pipe be p and the drain,d.
When both p and d are on,it takes 10 hours i.e
p+d=10 hrs.
that means 100%=10hrs
so in 1hr,1/10 will be done,that is 10%

In the second case,p and d are on for 2hours then p alone for 4hours.
If in 1 hour,p and d fill 10%,they will fill 20% in two hours.p works alone for 4hours to fill the remaining 80%
if it takes 4hours for p to fill80%,it means it will fills 20% in 1hour.

If p fills 20% in one hour,it will take 5hours to fill 100%.

you answer is therefore 5hours.