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

Revision as of 13:33, 28 December 2023

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 8: / Line 8: @@
 ==Video Solution by SpreadTheMathLove==
 https://www.youtube.com/watch?v=Q7jIaqd9uFk
+==~-from dearly beloved`~==
 ==Video Solution by @MathTalks==

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