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

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==Video Solution by SpreadTheMathLove==
 
==Video Solution by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=Q7jIaqd9uFk
 
https://www.youtube.com/watch?v=Q7jIaqd9uFk
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===~-from dearly beloved`~===
  
 
==Video Solution by @MathTalks==
 
==Video Solution by @MathTalks==
 
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT
 
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT
  
 
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==Video Solution by WhyMath==
 
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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

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

Video Solution by WhyMath

https://youtu.be/szzcDoUZVnA

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
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

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