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2021 JMC 10 Problems/Problem 5

Revision as of 16:08, 1 April 2021 by Skyscraper (talk | contribs) (Created page with "==Problem== A mixture has <math>96</math> grams of aluminum and <math>4</math> grams of barium. Nir the chemist uses magic to remove some aluminum. Now, exactly <math>90\%</ma...")
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Problem

A mixture has $96$ grams of aluminum and $4$ grams of barium. Nir the chemist uses magic to remove some aluminum. Now, exactly $90\%$ of the mixture consists of aluminum. How many grams of the mixture now remain?

$\textbf{(A) } 25 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 50 \qquad\textbf{(D) } 70 \qquad\textbf{(E) } 72$

Solution

Let $x$ be the amount of aluminum removed, in grams. There are $96-x$ grams of aluminum left and $100-x$ grams of the mixture left. So, $\tfrac{96-x}{100-x} = \tfrac{9}{10} \implies x = 60.$ The mixture has $100-60 = 40$ grams.

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