2010 AMC 10A Problems/Problem 3

Problem 3

Tyrone had $97$ marbles and Eric had $11$ marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?

$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 18 \qquad \mathrm{(D)}\ 25 \qquad \mathrm{(E)}\ 29$

Solution 1

Let $x$ be the number of marbles Tyrone gave to Eric. Then, $97-x = 2\cdot(11+x)$ . Solving for $x$ yields $75=3x$ and $x = 25$ . The answer is $\boxed{D}$ .

Solution 2

Since the amount of balls Tyler and Eric have a specific ratio when Tyler give some of his balls to Eric, we can divide the balls to their respective ratios. Altogether, they have $97+11$ balls, or $108$ balls, and it does not change when Tyler give some of his to Eric, since none are lost in the process. After Tyler give a few of his balls away, the ratio between the balls is $2:1$ , meaning that together added up is $108$ . The two numbers add up to $3$ and we divide $108$ by $3$ , and outcomes $36$ . This is the amount of balls Eric has, and so doubling that results in the amount of balls Tyler has, which is $72$ . $97-72$ is $25$ , and $36-11$ is $25$ , thus proving our statement true.

2010 AMC 10A (Problems • Answer Key • Resources)
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2010 AMC 10A Problems/Problem 3

Contents

Problem 3

Solution 1

Solution 2

See Also