Difference between revisions of "2010 AMC 8 Problems/Problem 7"
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<math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99</math> | <math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99</math> | ||
==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>2</math> dimes, and <math>4</math> pennies. Thus, the correct answer is | + | 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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| + | ==Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=Q7jIaqd9uFk | ||
| + | ===~-from dearly beloved`~=== | ||
| + | |||
| + | ==Video Solution by @MathTalks== | ||
| + | https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/szzcDoUZVnA | ||
==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}} | ||
Latest revision as of 10:26, 18 November 2024
Contents
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 
.
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=Q7jIaqd9uFk
~-from dearly beloved`~
Video Solution by @MathTalks
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT
Video Solution by WhyMath
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. 
