Aservice station checks mr. gittleboro's radiator and finds it contains only 30% antifreeze. if the radiator holds 10 quarts and is full, how much must be drained off and replaced with pure antifreeze in order to bring it up to a required 50% antifreeze?

2.86 quarts ( approx )

Step-by-step explanation:

Given,

The initial quantity of the Mr. Gittleboro's radiator that contains 30% antifreeze = 10 quarts,

Let x quarts of pure antifreeze replaced x quarts of Mr. Gittleboro's radiator to bring it up to a required 50% antifreeze,

So, the quantity of 30% antifreeze radiator after drained off x quarts = (10-x) quarts

Also, the quantity of final antifreeze radiator 50% antifreeze = 10 quarts

Thus, we can write,

30% of (10-x) + 100% of x = 50% of 10

30(10-x) + 100x = 500

300 - 30x + 100x = 500

300 + 70x = 500

70x = 200

x = 2.85714285714 quarts ≈ 2.86 quarts

twenty-six miles over two gallons because the proportion for the original 13 miles to one gallon would be written 13/1 and, doing the same thing to the top and bottom, i multiplied by 2 and got 26/2