Question 1200059: How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 85% antifreeze
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39616)   (Show Source):
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
Let V be the total volume of the mixture (antifreeze + water).
Then 0.85V should be 1 gallon of the pure antifreeze
0.85V = 1 gallon.
From this equation, V =  = 1.176 gallon (rounded).
It means that the volume 1.176 - 1 = 0.176 gallon of water should be added. ANSWER
Solved.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
(1) A standard formal algebraic solution....
x gallons of 0% antifreeze, plus 1 gallon of 100% antifreeze, yields (x+1) gallons of 85% antifreeze:
0(x)+1.00(1)=0.85(1+x)
1=.85+.85x
.85x=.15
x=.15/.85 = 15/85 = 3/17
ANSWER: 3/17 gallons
(2) One quick and easy alternative for an algebraic solution....
1 gallon of 100% antifreeze; x gallons of water; the ratio of gallons of antifreeze to gallons of water is 1:x
The ratio of antifreeze to water in the mixture is 85:15 = 17:3, so
1/x = 17/3
17x = 3
x = 3/17
ANSWER: 3/17 gallons
(3) And a quick and easy informal solution method....
85% is 85/100 = 17/20 of the way from 0% to 100%, so 17/20 of the mixture is the original 1 gallon of antifreeze.
That means 3/20 of the mixture is the added water; so the amount of water added is 3/17 of the amount of antifreeze. 3/17 of 1 gallon is 3/17 gallons, so
ANSWER: 3/17 gallons
Assuming the student is learning how to solve problems using formal algebra, you want to understand method (1) above.
But if the speed of obtaining the answer is important and formal algebra is not required (as in a timed math competition), being able to use one of the other methods is to your advantage.
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