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Difference between revisions of "2023 AMC 12B Problems/Problem 4"
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{{duplicate|[[2023 AMC 10B Problems/Problem 4|2023 AMC 10B #4]] and [[2023 AMC 12B Problems/Problem 4|2023 AMC 12B #4]]}}
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==Problem==
 
==Problem==
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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?  
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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==
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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>.  
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<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>.  
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~Failure.net
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==Solution 2 (Scientific Form)==
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<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>.
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~darrenn.cp
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==Remark==
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A good way to remember metric conversions is with the phrase ''' "King Henry Died By Drinking Chocolate Milk" ''' (Kilo, Hecto, Deka, Basic units, Deci, Centi, Milli). As you read from left to right you divide by 10. So 1 meter (which is a basic unit) is equivalent to <math>1000</math> millimeters, or <math>100</math> centimeters.
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~[https://artofproblemsolving.com/wiki/index.php/User:grogg007 grogg007]
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==Video Solution 1 by SpreadTheMathLove==
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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
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 +
~Math-X
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==Video Solution==
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https://youtu.be/BZ1zeFvw5hU
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution by Interstigation==
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https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
  
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~Failure,net
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==See also==
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{{AMC10 box|year=2023|ab=B|num-b=3|num-a=5}}
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{{AMC12 box|year=2023|ab=B|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 00:41, 25 June 2025

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

Remark

A good way to remember metric conversions is with the phrase "King Henry Died By Drinking Chocolate Milk" (Kilo, Hecto, Deka, Basic units, Deci, Centi, Milli). As you read from left to right you divide by 10. So 1 meter (which is a basic unit) is equivalent to $1000$ millimeters, or $100$ centimeters.

~grogg007

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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