SOLUTION: A radiator holds 8 liters. Suppose you have a bottle of pure antifreeze and a bottle of 10% antifreeze mixture. How much of each must you need to make enough of a 77.5% mixture to

Question 998782: A radiator holds 8 liters. Suppose you have a bottle of pure antifreeze and a bottle of 10% antifreeze mixture. How much of each must you need to make enough of a 77.5% mixture to fill the radiator?
How liters of pure antifreeze?______________
How many liters of 10% antifreeze?__________
Found 3 solutions by mananth, josgarithmetic, ikleyn:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
percent ---------------- quantity
SoLution I 100.00% ---------------- x Oz
Solution II 10.00% ------ 8 - x Oz
Mixture 77.50% ---------------- 8
8
100.00% x + 10.00% ( 8 - x ) = 77.50% * 8
100 x + 10 ( 8 - x ) = 620
100 x + 80 - 10 x = 620
100 x - 10 x = 620 - -80
90 x = 540
/ 90
x = 6 L 100.00% SoLution I
2 L 77.50% Solution II

m.ananth@hotmail.ca

Answer by josgarithmetic(39629)   (Show Source): You can put this solution on YOUR website!
The description is poor. What sizes are those two "bottles"? Given is ONE bottle of pure antifreeze and ONE bottle of 10% antifreeze; making available only TWO bottles.

On the other hand, the question asks for 8 liters of 77.5% antifreeze. Maybe each bottle is big enough to carry about 4 liters each? The question asks for how many liters, and not how many bottles.

v, how many liters of the pure antifreeze
8-v, how many... 10%

------Six liters of the pure antifreeze
therefore 2 liters of the 10% antifreeze.

Now again, HOW BIG IS EACH BOTTLE?
Answer by ikleyn(52873)   (Show Source): You can put this solution on YOUR website!
.
Let x be the volume of pure antifreeze (in liters) and y be the volume of 10% antifreeze mixture (in liters). Then you have two equations to determine x and y. First one is x + y = 8. The second equation is x + 0.1y = 0.775*(x+y). It equalize the volume of the pure antifreeze before and after the mixing. To solve the system, express x from the first equation, x = 8-y, and substitute it into the second equation. You will get (8-y) + 0.1y = 0.775*((8-y) + y). Simplify and solve: 8 - 0.9y = 0.775*8, 0.9y = 8 - 0.775*8, 0.9 y = 1.8, y = = 2. How many liters of pure antifreeze? 6 liters. How many liters of 10% antifreeze? 2 liters.

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