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

(Video Solution by Interstigation)
(22 intermediate revisions by 13 users not shown)
Line 7: Line 7:
 
==Solution 1==
 
==Solution 1==
  
Let's use the most stamps to make <math>7.10.</math> We have <math>20</math> of each stamp, <math>5</math>-cent (like nickels), <math>10</math>-cent (like dimes), and <math>25</math>-cent (like quarters).
+
Let's use the most stamps to make <math>7.10.</math> We have <math>20</math> of each stamp, <math>5</math>-cent (nickels), <math>10</math>-cent (dimes), and <math>25</math>-cent (quarters).
  
If we want to have the highest number of stamps, we have to have the highest number of the smaller value stamps (like the coins above). We can use <math>20</math> nickels and <math>20</math> dimes to bring our total cost to <math>7.10 - 3.00 = 4.10</math>. However, when we try to use quarters, the <math>25</math> cents don’t fit evenly, so we have to give back <math>15</math> cents in order to make the quarter amount <math>4.25</math>. The most efficient way to do this is to give back a  <math>10</math>-cent (dime) stamp and a <math>5</math>-cent (nickel) stamp to have <math>38</math> stamps (coins) used so far. Now, we just use <math>\frac{425}{25} = 17</math> quarters to get a grand total of <math>38 + 17 = \boxed{\textbf{(E)}\ 55}</math>.
+
If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use <math>20</math> nickels and <math>20</math> dimes to bring our total cost to <math>7.10 - 3.00 = 4.10</math>. However, when we try to use quarters, the <math>25</math> cents don’t fit evenly, so we have to give back <math>15</math> cents to make the quarter amount <math>4.25</math>. The most efficient way to do this is to give back a  <math>10</math>-cent (dime) stamp and a <math>5</math>-cent (nickel) stamp to have <math>38</math> stamps used so far. Now, we just use <math>\frac{425}{25} = 17</math> quarters to get a grand total of <math>38 + 17 = \boxed{\textbf{(E)}\ 55}</math>.
  
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
+
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, InterstellerApex, mahika99
  
 
==Solution 2==
 
==Solution 2==
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>.
+
The value of his entire stamp collection 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 collection. As a result, the maximum number of stamps he can use is <math>20\cdot3-5=\boxed{\textbf{(E)}\ 55}</math>.
  
 
~pianoboy
 
~pianoboy
  
 
~MathFun1000 (Rewrote for clarity and formatting)
 
~MathFun1000 (Rewrote for clarity and formatting)
 +
 +
~vadava_lx (minor edits)
 +
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/vq3voJZ-hvw
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by Math-X (Smart and Simple)==
 +
https://youtu.be/Ku_c1YHnLt0?si=GHLp1q_7Le68a4rJ&t=2454 ~Math-X note from InterstellerApex: this is wrong, he didn’t choose the correct answer. :(
  
 
==Animated Video Solution==
 
==Animated Video Solution==
Line 28: Line 38:
 
https://youtu.be/-N46BeEKaCQ?t=4280
 
https://youtu.be/-N46BeEKaCQ?t=4280
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
https://youtu.be/1bA7fD7Lg54?t=1078
+
https://youtu.be/DBqko2xATxs&t=1435
  
==Video Solution (CREATIVE THINKING!!!)==
+
==Video Solution by harungurcan==
https://youtu.be/vq3voJZ-hvw
+
https://www.youtube.com/watch?v=VqN7c5U5o98&t=449s
 +
 
 +
~harungurcan
 +
 
 +
==Video Solution (Solve under 60 seconds!!!)==
 +
https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=633
 +
 
 +
~hsnacademy
  
~Education, the Study of Everything
+
==Video Solution by Dr. David==
 +
https://youtu.be/bVA6Sx4mbdM
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=13|num-a=15}}
 
{{AMC8 box|year=2023|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:10, 27 October 2024

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 $5$-cent, $10$-cent, and $25$-cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $$7.10$ in postage? (Note: The amount $$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55$

Solution 1

Let's use the most stamps to make $7.10.$ We have $20$ of each stamp, $5$-cent (nickels), $10$-cent (dimes), and $25$-cent (quarters).

If we want the highest number of stamps, we must have the highest number of the smaller value stamps (like the coins above). We can use $20$ nickels and $20$ 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 to make the quarter amount $4.25$. The most efficient way to do this is to give back a $10$-cent (dime) stamp and a $5$-cent (nickel) stamp to have $38$ stamps used so far. Now, we just use $\frac{425}{25} = 17$ quarters to get a grand total of $38 + 17 = \boxed{\textbf{(E)}\ 55}$.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, InterstellerApex, mahika99

Solution 2

The value of his entire stamp collection 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 collection. 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)

~vadava_lx (minor edits)

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/vq3voJZ-hvw

~Education, the Study of Everything

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=GHLp1q_7Le68a4rJ&t=2454 ~Math-X note from InterstellerApex: this is wrong, he didn’t choose the correct answer. :(

Animated Video Solution

https://youtu.be/XP_tyhTqOBY

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

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=4280

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=1435

Video Solution by harungurcan

https://www.youtube.com/watch?v=VqN7c5U5o98&t=449s

~harungurcan

Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=633

~hsnacademy

Video Solution by Dr. David

https://youtu.be/bVA6Sx4mbdM

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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