Difference between revisions of "2010 AMC 8 Problems/Problem 7"
(→Solution) |
MRENTHUSIASM (talk | contribs) m (Undo revision 160329 by Raina0708 (talk) I PM'ed Raina0708 for suggesting NOT to erase LaTeX. So, I will restore the LaTeX code. LaTeX makes solutions more professional.) (Tag: Undo) |
||
| Line 4: | Line 4: | ||
<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 2 dimes, 1 nickel, and 4 pennies for the first 25 cents. From 26 cents to 50 cents, you only need to add 1 quarter. From 51 cents to 75 cents, you also only need to add 1 quarter. The same for 76 cents to 99 cents. Notice that instead of 100, it is 99 We are left with 3 quarters, 1 nickel, 2 dimes, and 4 pennies. | + | 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>. |
==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 14:23, 15 August 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. 
