SOLUTION: A large truck has a radiator capacity of 10 gallons. Currently the radiator is 20% antifreeze. How much fluid must be replaced with a 75% solution to bring the radiator up to 50% a
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Question 268552: A large truck has a radiator capacity of 10 gallons. Currently the radiator is 20% antifreeze. How much fluid must be replaced with a 75% solution to bring the radiator up to 50% antifreeze? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of fluid that must be replaced with 75% antifreeze solution
Now we know that the amount of pure antifreeze that remains after the x amount of fluid is drained (0.20(10-x)) plus the amount of pure antifreeze in the x amount that is needed (0.75x) has to equal the amount of pure antifreeze that exists in the radiator after the 75% antifreeze is added (0.50(10)), so our equation to solve is:
0.20(10-x)+0.75x=0.50(10) get rid of parens
2-0.20x+0.75x=5 subtract 2 from each side
2-2-0.20x+0.75x=5-2 collect like terms
0.55x=3 divide each side by 0.5
x=5.5 gal--------------------amount of fluid that must be drained and replaced with a 75% solution
CK
0.20(10-5.5)+0.75*5.5=0.50*10
0.9+4.125=5
5.025~~~~~5
Does this help?---ptaylor
