154.825
154.825
154.825
154.825
<span>30 hours
For this problem, going to assume that the actual flow rate for both pipes is constant for the entire duration of either filling or emptying the pool. The pipe to fill the pool I'll consider to have a value of 1/12 while the drain that empties the pool will have a value of 1/20. With those values, the equation that expresses how many hour it will take to fill the pool while the drain is open becomes:
X(1/12 - 1/20) = 1
Now solve for X
X(5/60 - 3/60) = 1
X(2/60) = 1
X(1/30) = 1
X/30 = 1
X = 30
To check the answer, let's see how much water would have been added over 30 hours.
30/12 = 2.5
So 2 and a half pools worth of water would have been added. Now how much would be removed?
30/20 = 1.5
And 1 and half pools worth would have been removed. So the amount left in the pool is
2.5 - 1.5 = 1
And that's exactly the amount needed.</span>
300=360=380 cause the 8 brings up the 6 and
200=300=100 it will stay the same so
A=future amount
P=present amount
r=rate in decimal
t=time in years
so
when will A=double of original?
or when will A=2P
r=3%=0.03
t=t
divide both sides by P
take ln of both sides
ln(2)=0.03t
divide both sides by 0.03
(ln(2))/0.03=t
use calculator
23.1049=t
so a little over 23 years, about 23 years and 38.2 days
