Difference between revisions of "2010 AMC 8 Problems/Problem 15"
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==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 | + | 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 brown 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> |
==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}} | ||
Revision as of 18:43, 8 November 2015
Problem
A jar contains 
different colors of gumdrops. 
are blue, 
are brown, 
are red, 
are yellow, and other 
gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
Solution
We do 
to find the percent of gumdrops that are green. We find that 
of the gumdrops are green. That means there are 
gumdrops. If we replace all blue gumdrops with brown gumdrops, then 
of the jar's gumdrops are brown. 
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
