how much water is needed to add to a half-pint of syrup with 60% sugar to obtain a drink with 5% sugar?

0.6*8 = 4.8 0z of sugar.

8-4.8 = 3.2 oz of water.

Sugar = 5%.
%Water = 100-5 = 95.

(3.2+w)/(4.8 + (w+3.2)) = 0.95
Cross multiply:
3.2+w = 4.56+0.95w+3.04
w-.95w = 4.56+3.04
0.05w = 7.6
W = 88 oz. of water added.

To calculate the amount of water needed to achieve a drink with 5% sugar, we can use the concept of a dilution equation. The dilution equation is as follows:

C₁V₁ = C₂V₂

Where:
C₁ = initial concentration of the syrup (60% sugar)
V₁ = initial volume of the syrup (half-pint)
C₂ = final concentration of the drink (5% sugar)
V₂ = final volume of the drink

Let's solve the equation step by step:

1. Convert the initial volume of the syrup from a half-pint to an appropriate unit. A half-pint is equivalent to 8 fluid ounces.

2. Determine the initial concentration of the syrup in terms of a decimal. The 60% sugar concentration is equivalent to 0.6.

3. Convert the final concentration of the drink to a decimal. The 5% sugar concentration is equivalent to 0.05.

4. Substitute the known values into the dilution equation:

(0.6)(8 oz) = (0.05)(V₂)

5. Solve for V₂:

4.8 oz = 0.05V₂

V₂ = 4.8 oz / 0.05
V₂ = 96 oz

So, to obtain a drink with 5% sugar, you will need to add 96 ounces of water to the half-pint (or 8 ounces) of syrup.