Difference between revisions of "2017 AMC 8 Problems/Problem 9"

(Created page with "==Problem 9== All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the sma...")
 
m (Problem)
 
(19 intermediate revisions by 7 users not shown)
Line 1: Line 1:
==Problem 9==
+
==Problem==
All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?
+
All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?
  
 
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math>
 
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math>
 +
 +
==Solution 1==
 +
 +
The <math>6</math> green marbles and yellow marbles form <math>1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}</math> of the total marbles. Now suppose the total number of marbles is <math>x</math>. We know the number of yellow marbles is <math>\frac{5}{12}x - 6</math> and a positive integer. Therefore, <math>12</math> must divide <math>x</math>. Trying the smallest multiples of <math>12</math> for <math>x</math>, we see that when <math>x = 12</math>, we get there are <math>-1</math> yellow marbles, which is impossible. However when <math>x = 24</math>, there are <math>\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}</math> yellow marbles, which must be the smallest possible.
 +
 +
==Solution 2==
 +
The 6 green and yellow marbles make up <math>1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}</math> of the total marbles, just like as in the previous solution. Now we know that there are <math>\frac{5}{12}(\text{total}) - 6</math> yellow marbles. Now, because 12 marbles for the total doesn't work (there would be -1 yellow marbles), we multiply the 12 by 2, to find out there are <math>\frac{5}{12}(24) - 6=\boxed{\textbf{(D) }4}</math> yellow marbles.
 +
 +
-[[User:elbertpark|elbertpark]]
 +
 +
==Video Solution==
 +
https://youtu.be/rQUwNC0gqdg?t=770
 +
 +
==See Also==
 +
{{AMC8 box|year=2017|num-b=8|num-a=10}}
 +
 +
{{MAA Notice}}

Latest revision as of 16:24, 13 February 2021

Problem

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Solution 1

The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now suppose the total number of marbles is $x$. We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$. Trying the smallest multiples of $12$ for $x$, we see that when $x = 12$, we get there are $-1$ yellow marbles, which is impossible. However when $x = 24$, there are $\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}$ yellow marbles, which must be the smallest possible.

Solution 2

The 6 green and yellow marbles make up $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles, just like as in the previous solution. Now we know that there are $\frac{5}{12}(\text{total}) - 6$ yellow marbles. Now, because 12 marbles for the total doesn't work (there would be -1 yellow marbles), we multiply the 12 by 2, to find out there are $\frac{5}{12}(24) - 6=\boxed{\textbf{(D) }4}$ yellow marbles.

-elbertpark

Video Solution

https://youtu.be/rQUwNC0gqdg?t=770

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 AJHSME/AMC 8 Problems and Solutions

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

Invalid username
Login to AoPS