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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==
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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 we didn't put <math>100</math> instead of <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 <math>3+2+1+4=\boxed{\textbf{(B)}\ 10}\ \}</math>.
+
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
 +
 
 +
==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

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 (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. AMC logo.png

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