2010 AMC 10A Problems/Problem 3

Revision as of 08:30, 10 August 2017 by Progamexd (talk | contribs) (Solution 2)

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, and the answer is $\boxed{D=25}$.

See Also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

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