# how many oz of mixed nuts that contain 65% peanuts must Ryan add to 8.6 oz of mixed nuts that contain 20% peanuts to make a mixture with 50% peanuts?

Step-by-step solution-:

The Given that:

8.6~oz of mixed nuts that contains 20% peanuts:

Let be amount of peanuts be a :

By using the formula of Percent Proportion:

The percent proportion representing that x is  p% of  y is given by:

\frac{x}{y}=\frac{P}{100}

Substitute the y=8.6 oz and p=20% into the Percent Proportion:

\frac{a}{\textcolor{primary}{8.6}}=\frac{\textcolor{secondary}{20}}{100}

After Cross multiplication:

100a=8.6\times20

100a=172=>a=\frac{172}{100}=>a=1.72

Let b be the mass of mixed nuts that contains 65% peanuts:

Similarly,

By using the formula of Percent Proportion:

\frac{x}{y}=\frac{P}{100}

Substitute the y=b oz and p=65 into the Percent Proportion:

\frac{x}{\textcolor{primary}{b}}=\frac{\textcolor{secondary}{65}}{100}

After solving we get,

x=\frac{13}{20}b

The total amount of peanuts=(1.72+(13/20)b) oz:

The Total amount will be original 8.6 oz plus the added amount of peanuts

The total amount of nuts =(8.6+b) oz:

Substitute the y=(8.6+b) oz, x=(1.72+\13/20b)oz and p=50 into the Percent Proportion:

\frac{\textcolor{tertiary}{(1.72+\frac{13}{20}b)}}{\textcolor{primary}{(8.6+b)}}=\frac{\textcolor{secondary}{50}}{100}

After solving we get b:

b=17.2

Therefore, The value of b is 17.2 oz:

The Ry was added to 17.2 oz of mixed nu that contain 65% peanuts.

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