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Difference between revisions of "1951 AHSME Problems/Problem 5"

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<math> \mathrm{(A) \ A\ comes\ out\ even  } \qquad</math> <math>\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}</math> <math> \qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ deal } \qquad</math> <math>\mathrm{(D) \ A\ loses\ 900\ on\ the\ deal }</math> <math>\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal }  </math>
 
<math> \mathrm{(A) \ A\ comes\ out\ even  } \qquad</math> <math>\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}</math> <math> \qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ deal } \qquad</math> <math>\mathrm{(D) \ A\ loses\ 900\ on\ the\ deal }</math> <math>\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal }  </math>
 
==Solution==
 
==Solution==
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Mr. <math>A</math> earns <math>1.1\cdot</math> <dollar/> <math>10,000=</math> <dollar/>\ <math>11,000</math> after he sells it to Mr. <math>B</math>. Then, Mr. <math>B</math> sells it at a price of <math>(1-0.1)\cdot</math> \ <math>11,000=</math><dollar/><math>9,900</math>, so <math>\boxed{\textbf{(B)}\ \text{A makes 1100 on the deal}}</math>.
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Mr. <math>A</math> earns <math>1.1\cdot</math> <dollar/> <math>10,000=</math> <dollar/> <math>11,000</math> after he sells it to Mr. <math>B</math>. Then, Mr. <math>B</math> sells it at a price of <math>(1-0.1)\cdot</math> <dollar/>\ <math>11,000=</math><dollar/><math>9,900</math>, so <math>\boxed{\textbf{(B)}\ \text{A makes 1100 on the deal}}</math>.

Revision as of 23:49, 25 April 2011

Problem 5

Mr. $A$ owns a home worth $$10,000. He sells it to Mr. $B$ at a 10% profit based on the worth of the house. Mr. $B$ sells the house back to Mr. $A$ at a 10% loss. Then:

$\mathrm{(A) \ A\ comes\ out\ even } \qquad$ $\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}$ $\qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ deal } \qquad$ $\mathrm{(D) \ A\ loses\ 900\ on\ the\ deal }$ $\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal }$

Solution

Mr. $A$ earns $1.1\cdot$ <dollar/> $10,000=$ <dollar/> $11,000$ after he sells it to Mr. $B$. Then, Mr. $B$ sells it at a price of $(1-0.1)\cdot$ <dollar/>\ $11,000=$<dollar/>$9,900$, so $\boxed{\textbf{(B)}\ \text{A makes 1100 on the deal}}$.

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