Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12B Problems/Problem 4"
(→‎Solution 2 (Standard Form))
 
(11 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2023 AMC 10B Problems/Problem 4|2023 AMC 10B #4]] and [[2023 AMC 12B Problems/Problem 4|2023 AMC 12B #4]]}}
 +
 
==Problem==
 
==Problem==
−
Jackson's paintbrush makes a narrow strip with a width of 6.5 millimeters. Jackson has enough paint to make a strip long. How many square centimeters of paper could Jackson paint?  
+
Jackson's paintbrush makes a narrow strip with a width of <math>6.5</math> millimeters. Jackson has enough paint to make a strip <math>25</math> meters long. How many square centimeters of paper could Jackson cover with paint?  
  
 
<math>\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250</math>
 
<math>\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250</math>
  
 
==Solution 1==
 
==Solution 1==
−
6.5 millimeters is equal to 0.65 centimeters. 25 meters is 2500 meters. The answer is <math>0.65 \times 2500</math>, so the answer is <math>\boxed{\textbf{(C) 1,625}}</math>.  
+
<math>6.5</math> millimeters is equal to <math>0.65</math> centimeters. <math>25</math> meters is <math>2500</math> centimeters. The answer is <math>0.65 \times 2500</math>, so the answer is <math>\boxed{\textbf{(C) 1,625}}</math>.  
 +
 
 +
~Failure.net
 +
 
 +
==Solution 2 (Scientific Form)==
 +
<math>6.5</math> millimeters can be represented as <math>65 \times 10^{-2}</math> centimeters. <math>25</math> meters is <math>25 \times 10^{2}</math> centimeters. Multiplying out these results in <math>(65 \times 10^{-2}) \times (25 \times 10^{2})</math>, which is <math>65 \times 25</math> making the answer <math>\boxed{\textbf{(C) 1,625}}</math>.
 +
 
 +
~darrenn.cp
 +
 
 +
==Video Solution 1 by SpreadTheMathLove==
 +
 
 +
https://www.youtube.com/watch?v=SUnhwbA5_So
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/EuLkw8HFdk4?si=mEwHK_VyAr1h0O1A&t=808
 +
 
 +
~Math-X
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/BZ1zeFvw5hU
 +
 
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
  
−
~Failure,net
+
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=3|num-a=5}}
 +
{{AMC12 box|year=2023|ab=B|num-b=3|num-a=5}}
 +
{{MAA Notice}}

Latest revision as of 15:39, 21 November 2023

The following problem is from both the 2023 AMC 10B #4 and 2023 AMC 12B #4, so both problems redirect to this page.

Problem

Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?

$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\textbf{(E) } 16,250$

Solution 1

$6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$, so the answer is $\boxed{\textbf{(C) 1,625}}$.

~Failure.net

Solution 2 (Scientific Form)

$6.5$ millimeters can be represented as $65 \times 10^{-2}$ centimeters. $25$ meters is $25 \times 10^{2}$ centimeters. Multiplying out these results in $(65 \times 10^{-2}) \times (25 \times 10^{2})$, which is $65 \times 25$ making the answer $\boxed{\textbf{(C) 1,625}}$.

~darrenn.cp

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=SUnhwbA5_So

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/EuLkw8HFdk4?si=mEwHK_VyAr1h0O1A&t=808

~Math-X

Video Solution

https://youtu.be/BZ1zeFvw5hU


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 AMC 10 Problems and Solutions
2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12B_Problems/Problem_4&oldid=205130"