Difference between revisions of "2010 AMC 10A Problems/Problem 3"

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Let <math>x</math> be the number of marbles Tyrone gave to Eric. Then, <math>97-x = 2\cdot(11+x)</math>. Solving for <math>x</math> yields <math>75=3x</math> and <math>x = 25</math>. The answer is <math>\boxed{D}</math>.
 
Let <math>x</math> be the number of marbles Tyrone gave to Eric. Then, <math>97-x = 2\cdot(11+x)</math>. Solving for <math>x</math> yields <math>75=3x</math> and <math>x = 25</math>. The answer is <math>\boxed{D}</math>.
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== See Also ==
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{{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}}

Revision as of 19:35, 1 January 2012

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

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 $\boxed{D}$.


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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