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

(Solution)
(Solution 2)
 
(8 intermediate revisions by 4 users not shown)
Line 20: Line 20:
  
 
==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.
+
Since the number of balls Tyrone and Eric have a specific ratio when Tyrone gives some of his balls to Eric, we can divide the total amount of balls by their respective ratios. Altogether, they have <math>97+11</math> balls, or <math>108</math> balls, and it does not change when Tyrone gives some of his to Eric, since none are lost in the process. After Tyrone gives some of his balls away, the ratio between the balls is <math>2:1</math>, and added together is <math>108</math>. The two numbers add up to <math>3</math> and we divide <math>108</math> by <math>3</math>, and the outcome is <math>36</math>. This is the number of balls Eric has, and so doubling that results in the number of balls Tyrone 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 ==

Latest revision as of 19:50, 5 March 2023

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 1

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 $\boxed{D}$.


Solution 2

Since the number of balls Tyrone and Eric have a specific ratio when Tyrone gives some of his balls to Eric, we can divide the total amount of balls by their respective ratios. Altogether, they have $97+11$ balls, or $108$ balls, and it does not change when Tyrone gives some of his to Eric, since none are lost in the process. After Tyrone gives some of his balls away, the ratio between the balls is $2:1$, and added together is $108$. The two numbers add up to $3$ and we divide $108$ by $3$, and the outcome is $36$. This is the number of balls Eric has, and so doubling that results in the number of balls Tyrone has, which is $72$. $97-72$ is $25$, and $36-11$ is $25$, thus proving our statement true, and the answer is $\boxed{D=25}$.

Video Solution

https://youtu.be/C1VCk_9A2KE?t=145

~IceMatrix

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png