Difference between revisions of "2023 AMC 12B Problems/Problem 4"

Revision as of 21:40, 16 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 (Standard 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

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.

@@ Line 10: / Line 10: @@
 ~Failure.net
-~Mintylemon66
 ==Solution 2 (Standard Form)==

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