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

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==Solution==
 
==Solution==
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Mr. <math>A</math> earns <math>1.1\cdot <dollar/> 10,000= <dollar/> 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 <dollar/>11,000=<dollar/>9,900</math>, so <math>\boxed{\textrm{(B)}\ \text{A makes 1100 on the deal}}</math>.
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Mr. <math>A</math> sells his home for  <math>(1 + 10</math>%<math>)</math> <math>\cdot</math> <math>10,000</math> dollars <math>=</math> <math>1.1</math> <math>\cdot</math> <math>11,000</math> dollars after he sells it to Mr. <math>B</math>. Then, Mr. <math>B</math> sells it at a price of <math>(1-10</math>%<math>)</math> <math>\cdot</math> <math>11,000</math> dollars <math>=</math> <math>0.9</math> <math>\cdot</math> <math>11,000</math> dollars <math>=</math> <math>9,900</math> dollars, thus <math>11,000 - 9,900</math> <math>=</math> <math>\boxed{\textrm{(B)}\ \text{A makes 1100 on the deal}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 11:57, 19 February 2016

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. $A$ sells his home for $(1 + 10$%$)$ $\cdot$ $10,000$ dollars $=$ $1.1$ $\cdot$ $11,000$ dollars after he sells it to Mr. $B$. Then, Mr. $B$ sells it at a price of $(1-10$%$)$ $\cdot$ $11,000$ dollars $=$ $0.9$ $\cdot$ $11,000$ dollars $=$ $9,900$ dollars, thus $11,000 - 9,900$ $=$ $\boxed{\textrm{(B)}\ \text{A makes 1100 on the deal}}$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AHSME Problems and Solutions

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