Difference between revisions of "2010 AMC 10A Problems/Problem 3"
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==Solution 2 == | ==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 <math>97+11</math> balls, or <math>108</math> 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 <math>2:1</math>, meaning that together added up is <math>108</math>. The two numbers add up to <math>3</math> and we divide <math>108</math> by <math>3</math>, and outcomes <math>36</math>. This is the amount of balls Eric has, and so doubling that results in the amount of balls Tyler has, which is <math>72</math>. <math>97-72</math> is <math>25</math>, and <math>36-11</math> is <math>25</math>, thus proving our statement true, and the answer is <math>\boxed{D=25}</math>. | 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 <math>97+11</math> balls, or <math>108</math> 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 <math>2:1</math>, meaning that together added up is <math>108</math>. The two numbers add up to <math>3</math> and we divide <math>108</math> by <math>3</math>, and outcomes <math>36</math>. This is the amount of balls Eric has, and so doubling that results in the amount of balls Tyler has, which is <math>72</math>. <math>97-72</math> is <math>25</math>, and <math>36-11</math> is <math>25</math>, thus proving our statement true, and the answer is <math>\boxed{D=25}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/C1VCk_9A2KE?t=145 | ||
| + | |||
| + | ~IceMatrix | ||
== See Also == | == See Also == | ||
Revision as of 05:17, 26 May 2020
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 1
Let 
be the number of marbles Tyrone gave to Eric. Then, 
. Solving for yields 
and 
. The answer is 
.
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 
balls, or 
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 
, meaning that together added up is . The two numbers add up to 
and we divide by , and outcomes 
. This is the amount of balls Eric has, and so doubling that results in the amount of balls Tyler has, which is 
. 
is 
, and 
is , thus proving our statement true, and the answer is 
.
Video Solution
https://youtu.be/C1VCk_9A2KE?t=145
~IceMatrix
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
