Difference between revisions of "2010 AMC 10A Problems/Problem 3"
Mariekitty (talk | contribs) (→Solution) |
|||
| Line 17: | Line 17: | ||
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>. | ||
| + | |||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}} | ||
Revision as of 18:35, 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 | ||
