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 gives some of his balls away, the ratio between the balls is $2:1$ , and added together is $108$ . The two numbers add up to $3$ and we divide $108$ by $3$ , and outcome is $36$ . 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, and the answer is $\boxed{D=25}$ .

Video Solution

https://youtu.be/C1VCk_9A2KE?t=145

~IceMatrix

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

Contents

Problem 3

Solution 1

Solution 2

Video Solution

See Also