Art of Problem Solving
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
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2009 AMC 8 Problems/Problem 8: Difference between revisions

← Older edit
Gina (talk | contribs)
783 edits
No edit summary
Sunwl06 (talk | contribs)
editor
22 edits
 
(12 intermediate revisions by 9 users not shown)
Line 1: Line 1:
==Problem==
==Problem==


The length of a rectangle is increased by <math> 10%</math> and the width is decreased by <math> 10%</math>. What percent of the old area is the new area?
The length of a rectangle is increased by <math> 10\%</math> percent and the width is decreased by <math> 10\%</math> percent. What percent of the old area is the new area?




Line 11: Line 11:


==Solution==
==Solution==
In a rectangle with dimensions <math>10 \times 10</math>, the new rectangle would have dimensions <math>11 \times 9</math>. The ratio of the old area to the new area is <math>99/100 = \boxed{\textbf{(B)}\ 99}</math>.
In a rectangle with dimensions <math>10 \times 10</math>, the new rectangle would have dimensions <math>11 \times 9</math>. The ratio of the new area to the old area is <math>99/100 = \boxed{\textbf{(B)}\ 99}</math>.
 
===Solution 2===
If you take the length as <math>x</math> and the width as <math>y</math> then
A(OLD)= <math>xy</math>
A(NEW)= <math>1.1x\times.9y</math> = <math>.99xy</math>
<math>0.99/1=99\%</math>
 
==Video Solution==
https://youtu.be/4io4bjzgQKI
 
~savannahsolver
 
==Easier to understand Video Solution==
https://www.youtube.com/watch?v=cZUT7lYu474&t=4s


==See Also==
==See Also==
{{AMC8 box|year=2009|num-b=7|num-a=9}}
{{AMC8 box|year=2009|num-b=7|num-a=9}}
{{MAA Notice}}

Latest revision as of 20:39, 26 December 2023

Problem

The length of a rectangle is increased by $10\%$ percent and the width is decreased by $10\%$ percent. What percent of the old area is the new area?


$\textbf{(A)}\ 90 \qquad \textbf{(B)}\ 99 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 101 \qquad \textbf{(E)}\ 110$

Solution

In a rectangle with dimensions $10 \times 10$, the new rectangle would have dimensions $11 \times 9$. The ratio of the new area to the old area is $99/100 = \boxed{\textbf{(B)}\ 99}$.

Solution 2

If you take the length as $x$ and the width as $y$ then A(OLD)= $xy$ A(NEW)= $1.1x\times.9y$ = $.99xy$ $0.99/1=99\%$

Video Solution

https://youtu.be/4io4bjzgQKI

~savannahsolver

Easier to understand Video Solution

https://www.youtube.com/watch?v=cZUT7lYu474&t=4s

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_8_Problems/Problem_8&oldid=208459"