Difference between revisions of "2010 AMC 8 Problems/Problem 15"

(Problem)
(See Also)
 
(4 intermediate revisions by 3 users not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
We do <math>100-30-20-15-10</math> to find the percent of gumdrops that are green. We find that <math>25\%</math> of the gumdrops are green. That means there are <math>120</math> gumdrops. If we replace all blue gumdrops with green gumdrops, then <math>35\%</math> of the jar's gumdrops are brown. <math>\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}</math>
+
We do <math>100-30-20-15-10</math> to find the percent of gumdrops that are green. We find that <math>25\%</math> of the gumdrops are green. That means there are <math>120</math> gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then <math>15\%</math> of the jar's gumdrops are brown. <math>\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}</math>
 +
 
 +
==Video by MathTalks==
 +
 
 +
https://www.youtube.com/watch?v=6hRHZxSieKc
 +
 
 +
 
 +
 
  
 
==See Also==
 
==See Also==
 +
 
{{AMC8 box|year=2010|num-b=14|num-a=16}}
 
{{AMC8 box|year=2010|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:55, 22 January 2023

Problem

A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?

$\textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$

Solution

We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then $15\%$ of the jar's gumdrops are brown. $\dfrac{35}{100} \cdot 120=42 \Rightarrow \boxed{\textbf{(C)}\ 42}$

Video by MathTalks

https://www.youtube.com/watch?v=6hRHZxSieKc



See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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