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Difference between revisions of "2019 AMC 10B Problems/Problem 11"
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==Solution==
 
==Solution==
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Call the amount of marbles in each jar <math>x</math>, because they are equivalent. Thus, <math>\frac{x}{10}</math> is the amount of green marbles in <math>1</math>, and <math>\frac{x}{9}</math> is the amount of green marbles in <math>2</math>. <math>\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}</math>, <math>\frac{19x}{90}=95</math>, and <math>x=450</math> marbles in each jar. Because the <math>\frac{9x}{10}</math> is the amount of blue marbles in jar <math>1</math>, and <math>\frac{8x}{9}</math> is the amount of blue marbles in jar <math>2</math>, <math>\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90}</math>, so there must be <math>5</math> more marbles in jar <math>1</math> than jar <math>2</math>. The answer is <math>(\boxed{A})</math>
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Call the amount of marbles in each jar <math>x</math>, because they are equivalent. Thus, <math>\frac{x}{10}</math> is the amount of green marbles in <math>1</math>, and <math>\frac{x}{9}</math> is the amount of green marbles in <math>2</math>. <math>\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}</math>, <math>\frac{19x}{90}=95</math>, and <math>x=450</math> marbles in each jar. Because the <math>\frac{9x}{10}</math> is the amount of blue marbles in jar <math>1</math>, and <math>\frac{8x}{9}</math> is the amount of blue marbles in jar <math>2</math>, <math>\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90}</math>, so there must be <math>5</math> more marbles in jar <math>1</math> than jar <math>2</math>. The answer is <math>\boxed{A}</math>
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(Edited by Lcz)
 
  
 
==See Also==
 
==See Also==

Revision as of 12:17, 15 February 2019

Problem

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar 1 the ratio of blue to green marbles is 9:1, and the ratio of blue to green marbles in Jar 2 is 8:1. There are 95 green marbles in all. How many more blue marbles are in Jar 1 than in Jar 2?

$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$

Solution

Call the amount of marbles in each jar $x$, because they are equivalent. Thus, $\frac{x}{10}$ is the amount of green marbles in $1$, and $\frac{x}{9}$ is the amount of green marbles in $2$. $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$, $\frac{19x}{90}=95$, and $x=450$ marbles in each jar. Because the $\frac{9x}{10}$ is the amount of blue marbles in jar $1$, and $\frac{8x}{9}$ is the amount of blue marbles in jar $2$, $\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90}$, so there must be $5$ more marbles in jar $1$ than jar $2$. The answer is $\boxed{A}$

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

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