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One type of granola has
30% nuts, by mass. A second type of granola has
15% nuts. What mass of each type needs to be mixed to make
600g of granola that will have
21% nuts?

One type of granola has $30 \%$ nuts, by mass. A second type of granola has $15 \%$ nuts. What mass of each type needs to be mixed to make $600 \mathrm{~g}$ of granola that will have $21 \%$ nuts?

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Weighted averages: word problems

Full solution

Q. One type of granola has

30 \%

nuts, by mass. A second type of granola has

15 \%

nuts. What mass of each type needs to be mixed to make

600 \mathrm{~g}

of granola that will have

21 \%

nuts?

Define variables: Let $x$ be the mass of the first type of granola ( $30\%$ nuts) and $y$ be the mass of the second type of granola ( $15\%$ nuts). We know the total mass of the mixture should be $600g$ . So, the equation is: $\newline$ $x + y = 600$
Set up equations: Next, we set up the equation for the nut content. The total nut content in the final mixture should be $21\%$ of $600g$ , which is $126g$ . The nut content from the first type is $0.30x$ and from the second type is $0.15y$ . The equation is: $\newline$ $0.30x + 0.15y = 126$
Solve equations: We can solve these equations simultaneously. From the first equation, express $y$ in terms of $x$ : $y = 600 - x$
Substitute values: Substitute $y$ in the second equation: $0.30x + 0.15(600 - x) = 126$ $0.30x + 90 - 0.15x = 126$ $0.15x = 36$ $x = 240$
Final solution: Substitute $x = 240$ back into the equation for $y$ : $y = 600 - 240$ $y = 360$

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