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

Latest revision as of 18:50, 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

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.

@@ 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 1==
+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>.
+==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 <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 ==
+{{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}}
+{{MAA Notice}}

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