Difference between revisions of "2023 AMC 12B Problems/Problem 4"
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==Problem== | ==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? | + | 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 centimeters. 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 | ~Failure.net | ||
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| − | ==Solution 2( | + | ==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 | + | <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 | ~darrenn.cp | ||
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==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=SUnhwbA5_So | https://www.youtube.com/watch?v=SUnhwbA5_So | ||
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| + | ==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== | ==See also== | ||
Latest revision as of 16:39, 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.
Contents
Problem
Jackson's paintbrush makes a narrow strip with a width of 
millimeters. Jackson has enough paint to make a strip 
meters long. How many square centimeters of paper could Jackson cover with paint?
Solution 1
millimeters is equal to 
centimeters. meters is 
centimeters. The answer is 
, so the answer is 
.
~Failure.net
Solution 2 (Scientific Form)
millimeters can be represented as 
centimeters. meters is 
centimeters. Multiplying out these results in 
, which is 
making the answer .
~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
~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. 
