Difference between revisions of "2010 AMC 8 Problems/Problem 7"
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==Solution== | ==Solution== | ||
You need <math>2</math> dimes, <math>1</math> nickel, and <math>4</math> pennies for the first <math>25</math> cents. From <math>26</math> cents to <math>50</math> cents, you only need to add <math>1</math> quarter. From <math>51</math> cents to <math>75</math> cents, you also only need to add <math>1</math> quarter. The same for <math>76</math> cents to <math>99</math> cents. Notice that instead of <math>100</math>, it is <math>99</math>. We are left with <math>3</math> quarters, <math>1</math> nickel, <math>2</math> dimes, and <math>4</math> pennies. Thus, the correct answer is <math>3+2+1+4=\boxed{\textbf{(B)}\ 10}</math>. | You need <math>2</math> dimes, <math>1</math> nickel, and <math>4</math> pennies for the first <math>25</math> cents. From <math>26</math> cents to <math>50</math> cents, you only need to add <math>1</math> quarter. From <math>51</math> cents to <math>75</math> cents, you also only need to add <math>1</math> quarter. The same for <math>76</math> cents to <math>99</math> cents. Notice that instead of <math>100</math>, it is <math>99</math>. We are left with <math>3</math> quarters, <math>1</math> nickel, <math>2</math> dimes, and <math>4</math> pennies. Thus, the correct answer is <math>3+2+1+4=\boxed{\textbf{(B)}\ 10}</math>. | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=6|num-a=8}} | {{AMC8 box|year=2010|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:24, 26 November 2021
Problem
Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?
Solution
You need dimes,
nickel, and
pennies for the first
cents. From
cents to
cents, you only need to add
quarter. From
cents to
cents, you also only need to add
quarter. The same for
cents to
cents. Notice that instead of
, it is
. We are left with
quarters,
nickel,
dimes, and
pennies. Thus, the correct answer is
.
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.