Question 1160519: You want to get the temperature of your pool up to 80 degrees. There are 30 thousand gallons of water in your pool, but it can hold more water. At this point the temperature of the pool water is 78 degrees. You will be adding
water that is at a temperature of 90 degrees. How many gallons of 90 degree water do you need to add to bring the pool temperature up to 80 degrees?
Give your answer to the nearest gallon.
Found 3 solutions by solver91311, Boreal, greenestamps:
Answer by solver91311(24713)   (Show Source):
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! use units of gallon-deg
have 30000*78=2,340,000 gal-deg now
want to have (30000+x)*80=2,400,000+80x gal-deg
and the initial amount is increased by 90x gal-deg, where x is the number of gallons of 90 degree water.
so 2340000+90x=2400000+80x
10x=60000
x=6000 gallons
Another way is that there is a 2 degree deficit for 30K gallons, or 60K deficit of gal-deg
you are adding water that is a 10 degree surplus over what you want, and 6000 gallons of that is 60K gal-deg.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
Here is an informal method for solving mixture problems like this that will get you to an answer much faster and with less effort than an algebraic method.
You are starting with water at 78 degrees; you are adding water at 90 degrees, until the mixture reaches 80 degrees.
80 degrees is 1/6 of the way from 78 degrees to 90 degrees. (78 to 90 is 12; 78 to 80 is 2; 2/12 = 1/6).
That means 1/6 of the final mixture is what you are adding.
So the 30,000 gallons originally in the pool is 5/6 of the final mixture; that means 1/6 of the final mixture -- the 90 degree water you are adding -- is 30,000/5 = 6,000 gallons.
ANSWER: 6000 gallons of 90 degree water
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