how many liters of antifreeze should be added to 5 L of water of produce a solution that is no less than 60% antifreeze and no more than 80% antifreeze?

Answer 1

Suppose you add x liters antifreeze

Then you have x liters antifreeze out of 5 + x total.

So percentage antifreeze = ( x / (5+X)) x 100

So we want 60 < (x / 5+x) x 100 < 80

Divide through by 100

0.6 < x / 5+x < 0.8

5+x > 0, so multiply through by this

0.6 (5+x) < x < 0.8 (5+x)

3 + 0.6x < x < 4 + 0.8x

Subtract 0.6 x

3 < 0.4x < 4 + 0.2x

So 0.4x > 3, so x > 3 / 0.4 = 7.5

x < 4 + 0.8 x

0.2 x < 4

x < 4 / 0.2 = 20

So 7.5 <= X <= 20

Mike