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 "1953 AHSME Problems/Problem 30"

(Created page with "==Problem 30== A house worth \$ <math>9000</math> is sold by Mr. A to Mr. B at a <math>10</math> % loss. Mr. B sells the house back to Mr. A at a <math>10</math> % gain. The...")
 
m (→‎Solution)
 
(3 intermediate revisions by one other user not shown)
Line 10: Line 10:
  
 
==Solution==
 
==Solution==
−
When Mr.A sells the house at a <math>10</math>% loss, he sells it for <math>9000(1 - .1) = 8100</math>. When Mr.B sells the house back to Mr. A at a <math>10</math> % gain he sells it for <math>8100(1 + .1) = 8910</math>. Therefore Mr. A has lost <math>8100-8910 = 900</math> dollars, so the answer is <math>\boxed{D}</math>.
+
When Mr.A sells the house at a <math>10</math>% loss, he sells it for <math>9000(1 - .1) = 8100</math>. When Mr.B sells the house back to Mr. A at a <math>10</math> % gain he sells it for <math>8100(1 + .1) = 8910</math>. Therefore Mr. A has lost <math>8100-8910 = 810</math> dollars, so the answer is <math>\boxed{D}</math>.
  
 
==See Also==
 
==See Also==
  
−
{{AHSME 50p box|year=1953|before=First Question|num-a=2}}
+
{{AHSME 50p box|year=1953|num-b=29|num-a=31}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:33, 22 April 2020

Problem 30

A house worth $ $9000$ is sold by Mr. A to Mr. B at a $10$ % loss. Mr. B sells the house back to Mr. A at a $10$ % gain. The result of the two transactions is:

$\textbf{(A)}\ \text{Mr. A breaks even} \qquad \textbf{(B)}\ \text{Mr. B gains }$900 \qquad \textbf{(C)}\ \text{Mr. A loses }$900\\ \textbf{(D)}\ \text{Mr. A loses }$810\qquad \textbf{(E)}\ \text{Mr. B gains }$1710$

Solution

When Mr.A sells the house at a $10$% loss, he sells it for $9000(1 - .1) = 8100$. When Mr.B sells the house back to Mr. A at a $10$ % gain he sells it for $8100(1 + .1) = 8910$. Therefore Mr. A has lost $8100-8910 = 810$ dollars, so the answer is $\boxed{D}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME 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=1953_AHSME_Problems/Problem_30&oldid=121533"