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 == | ||
+ | |||
+ | {{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}} |
Revision as of 19:35, 1 January 2012
Problem 3
Tyrone had marbles and Eric had 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?
Solution
Let be the number of marbles Tyrone gave to Eric. Then, . Solving for yields and . The answer is .
See Also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
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 |