Difference between revisions of "2010 AMC 8 Problems/Problem 7"

Latest revision as of 15:42, 1 July 2024

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?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99$

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. Thus, the correct answer is $3+2+1+4=\boxed{\textbf{(B)}\ 10}$ .

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

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.

@@ Line 1: / Line 1: @@
 ==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?
+<math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99</math>
+==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>.
+==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
-<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 99 </math>
-==Solution==
-You have to have 4 pennies so you can have 4 cents, 9 cents, 14, etc. Then you start with 1 dime, 1 nickel, and 1 quarter. You go through 1-9, 10-19, etc. until 90-99, and along the way, you add coins with the most value. You end up with 4 pennies, 1 nickel, 2 dimes, and 3 quarters. <math>4+1+2+3=10</math>, so our answer is <math>\boxed{B(10)}</math>
 ==See Also==
 {{AMC8 box|year=2010|num-b=6|num-a=8}}
 {{MAA Notice}}

Page

Toolbox

Search