Difference between revisions of "2019 AMC 10B Problems/Problem 11"

Problem

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$, and the ratio of blue to green marbles in Jar $2$ is $8:1$. There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$?

$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$

Solution

Call the number of marbles in each jar $x$ (because the problem specifies that they each contain the same number). Thus, $\frac{x}{10}$ is the number of green marbles in Jar $1$, and $\frac{x}{9}$ is the number of green marbles in Jar $2$. Since $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$, we have $\frac{19x}{90}=95$, so there are $x=450$ marbles in each jar.

Because $\frac{9x}{10}$ is the number of blue marbles in Jar $1$, and $\frac{8x}{9}$ is the number of blue marbles in Jar $2$, there are $\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5$ more marbles in Jar $1$ than Jar $2$. This means the answer is $\boxed{\textbf{(A) } 5}$.

See Also

 2019 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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