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Question

2021-08-04

Let x be the amount of 45% fat cheese and yy be the amount of 20% fat cheese, both in grams.

The cheese mixture is 30 grams:

\(\displaystyle{x}+{y}={30}{\left({1}\right)}\)

In terms of percentage of fat, it must be 30%:

\(\displaystyle{0.45}{x}+{0.2}{y}={0.3}{\left({30}\right)}\)

\(\displaystyle{0.45}{x}+{0.2}{y}={9}{\left({2}\right)}\)

Solve by substitution. Solve for yy using (1) to obtain (3):

\(\displaystyle{y}={30}−{x}{\left({3}\right)}\)

Substitute (3) to (2) and solve for xx:

\(\displaystyle{0.45}{x}+{0.2}{\left({30}−{x}\right)}={9}\)

\(\displaystyle{0.45}{x}+{6}−{0.2}{x}={9}\)

\(\displaystyle{0.25}{x}+{6}={9}\)

\(\displaystyle{0.25}{x}={3}\)

\(\displaystyle{x}={12}\)

Solve for yy using (3):

\(\displaystyle{y}={30}−{12}\)

\(\displaystyle{y}={18}\)

So, the chef must use 12 grams of 45% fat cheese and 18 grams of 20% fat cheese.

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