SOLUTION: A 10-quart radiator contains a 30% antifreeze solution. How much of the solution needs to be drained out and replaced with pure antifreeze in order to raise the solution to 65% ant
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Question 1039181: A 10-quart radiator contains a 30% antifreeze solution. How much of the solution needs to be drained out and replaced with pure antifreeze in order to raise the solution to 65% antifreeze? Found 2 solutions by addingup, ikleyn:Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website! So you want:
0.3x+1x = 0.65(10)
1.3x = 6.5
x = 5
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Check:
5 quarts pure antifreeze, represented by the number 1, mixed with 5 quarts of a 30% antifreeze solution, and we get:
5(1)+5(0.3)= 6.5 Correct.
You can put this solution on YOUR website! .
A 10-quart radiator contains a 30% antifreeze solution. How much of the solution needs to be drained out and replaced
with pure antifreeze in order to raise the solution to 65% antifreeze?
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Let "x" be (an unknown) volume of the 30% antifreeze solution which need to be drained out and replaced,
in accordance with the condition.
After draining, you will have (10-x) quarts of the antifreeze solution in the radiator
(assuming that initially the radiator was full and contained 10 quarts of the solution).
The amount of the pure antifreeze in the (10-x) quarts of the 30% solution is 0.3*(10-x).
After replacing of "x" quarts by the pure antifreeze by "x" quarts of the pure antifreeze
the amount of the pure antifreeze in the radiator is 0.3*(10-x) + x quarts.
To get the antifreeze concentration of 65%, the volume "x" should satisfy the equation
= . (1)
First step to solve this equation is to multiply both sides by 10 to rid of the denominator. You will get
0.3(10-x) + x = 6.5, or
3 - 0.3x + x = 6.5 ---> 0.7x = 6.5 - 3 ---> 0.7x = 3.5 ---> x = 5.
Answer. The amount of the 65% antifreeze solution that needs to be drained and replaced by the pure antifreeze is 5 quarts.
