Difference between revisions of "2001 AMC 8 Problems/Problem 10"
(→Solution 2) |
|||
(4 intermediate revisions by 4 users not shown) | |||
Line 8: | Line 8: | ||
<math> 2000\% </math> is equivalent to <math> 20\times100\% </math>. Therefore, <math> 2000\% </math> of a number is the same as <math> 20 </math> times that number. <math> 4 </math> quarters is <math> 1 </math> dollar, so Bryden will get <math> 20\times1={20} </math> dollars, <math> \boxed{\text{A}} </math>. | <math> 2000\% </math> is equivalent to <math> 20\times100\% </math>. Therefore, <math> 2000\% </math> of a number is the same as <math> 20 </math> times that number. <math> 4 </math> quarters is <math> 1 </math> dollar, so Bryden will get <math> 20\times1={20} </math> dollars, <math> \boxed{\text{A}} </math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Since <math> 2000\% </math> is just <math>\frac{2000}{100}</math>, we can multiply that by <math>100</math>, because four quarters is a <math>100</math> cents. After the multiplication, we get <math>2000</math>. Since our answer is in cents right now, we need to convert it to dollars, which would be <math>\boxed{\text{(A) 20}}</math> dollars. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=9|num-a=11}} | {{AMC8 box|year=2001|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:41, 10 January 2023
Contents
Problem
A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
Solution
is equivalent to . Therefore, of a number is the same as times that number. quarters is dollar, so Bryden will get dollars, .
Solution 2
Since is just , we can multiply that by , because four quarters is a cents. After the multiplication, we get . Since our answer is in cents right now, we need to convert it to dollars, which would be dollars.
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.