Difference between revisions of "2023 AMC 8 Problems/Problem 14"

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<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 55</math>
 
<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 55</math>
  
==Written Solution==
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==Solution 1==
  
 
Most stamps make <math>7.10.</math> You have 20 of each coin, nickles, dimes and quarters.
 
Most stamps make <math>7.10.</math> You have 20 of each coin, nickles, dimes and quarters.
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==Solution==
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==Solution 2==
The maximum amount of money in postage he can make with his collection of stamps is 20(5+10+25) = 800 cents, or 8 dollars. To make 7.10 in postage, he should subtract 90 cents, while removing the lowest number of stamps. To do this, he should remove as many 25 cent stamps as possible. He can remove at most 3 such stamps. After that, he will have exactly 7.25 left, and the way to remove 15 cents from that while removing the least number of stamps is by removing 1 10 cent stamp and 1 5 cent stamp. That will make 7.10 in postage, and he will have removed 5 total stamps: 3 25 cent stamps, 1 10 cent stamp, and 1 5 cent stamp. Since he has 20(3) = 60 stamps in his entire collection, the number of stamps used is 20(3) - 5 = 55, so the answer is E.
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The value of his entire stamp collect is <math>8</math> dollars. To make <math>$7.10</math> with stamps, he should remove <math>90</math> cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many <math>25</math> cent stamps as possible, as they have the greatest denomination. He can remove at most <math>3</math> of these stamps. He still has to remove <math>90-25\cdot3=15</math> cents worth of stamps. This can be done with one <math>5</math> and <math>10</math> cent stamp. In total, he has <math>20\cdot3=60</math> stamps in his entire collect. As a result, the maximum number of stamps he can use is <math>20\cdot3-5=\boxed{\textbf{(E) }55}</math>.
  
 
pianoboy
 
pianoboy
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~MathFun1000 (Rewrote for clarity and formatting)
  
  

Revision as of 22:11, 24 January 2023

Problem

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of 20 of each of 5 cent, 10 cent, and 25 cent stamps. What is the GREATEST number of stamps that Nicolas can use to make exactly 7.10 in postage?

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 55$

Solution 1

Most stamps make $7.10.$ You have 20 of each coin, nickles, dimes and quarters.

If we want to have the most amount of stamps we have to have the most amount of smaller value coins. We can use 20 nickels and 29 dimes to bring our total cost to $7.10 - 3.00 = 4.10$. However when we try to use quarters the 25 cents don’t fit evenly, so we have to give back 15 cents in order to make the quarter amount 4.25 the most efficient way to do this is give back a dime and a nickel to have 38 coins used so far. Now we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{\text{(E)}55}$

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat


Solution 2

The value of his entire stamp collect is $8$ dollars. To make $$7.10$ with stamps, he should remove $90$ cents worth of stamps with as few stamps as possible. To do this, he should start by removing as many $25$ cent stamps as possible, as they have the greatest denomination. He can remove at most $3$ of these stamps. He still has to remove $90-25\cdot3=15$ cents worth of stamps. This can be done with one $5$ and $10$ cent stamp. In total, he has $20\cdot3=60$ stamps in his entire collect. As a result, the maximum number of stamps he can use is $20\cdot3-5=\boxed{\textbf{(E) }55}$.

pianoboy

~MathFun1000 (Rewrote for clarity and formatting)


Animated Video Solution

https://youtu.be/XP_tyhTqOBY

~Star League (https://starleague.us)