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

Revision as of 16:38, 20 December 2010

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 $\boxed{D}$ .

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

Revision as of 16:38, 20 December 2010

Problem 3

Solution

@@ Line 1: / Line 1: @@
+== Problem 3 ==
+Tyrone had <math>97</math> marbles and Eric had <math>11</math> 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?
+<math>
+\mathrm{(A)}\ 3
+\qquad
+\mathrm{(B)}\ 13
+\qquad
+\mathrm{(C)}\ 18
+\qquad
+\mathrm{(D)}\ 25
+\qquad
+\mathrm{(E)}\ 29
+</math>
+==Solution==
+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>.