How much water needs to be mixed with three liters of
45% antifreeze to dilute the concentration to 35% antifreeze?

The volume of water needed=0.86 liters

Explanation:

Step 1: Form an equation

The equation can be expressed as follows;

(Ac×Va)+(Wc+Vw)=Fc(Va+Vw)

where;

Ac=initial concentration of antifreeze

Va=volume of antifreeze in liters

Wc=concentration of water

Vw=volume of the water in liters

Fc=final concentration of the antifreeze

This expression can be written as;

(concentration of antifreeze×volume of antifreeze in liters)+(concentration of water×volume of the water)=final concentration of the antifreeze(volume of antifreeze in liters+volume of the water in liters)

In our case;

Ac=45%=45/100=0.45

Va=3 liters

Wc=0

Vw=unknown

Fc=35%=35/100=0.35

Replacing;

(0.45×3)+(0×Vw)=0.35(3+Vw)

1.35=1.05+0.35 Vw

0.35 Vw=1.35-1.05

0.35 Vw=0.30

Vw=0.3/0.35=0.86

Vw=0.86 liters

The volume of water needed=0.86 liters

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