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

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== Problem 5 ==
 
== Problem 5 ==
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Mr. <math>A</math> owns a home worth <math>\</math>10,000.  He sells it to Mr. <math>B</math> at a 10% profit based on the worth of the house. Mr. <math>B</math> sells the house back to Mr. <math>A</math> at a 10% loss.  Then:
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Mr. <math>A</math> owns a home worth <nowiki>$</nowiki><math>10,000</math>.  He sells it to Mr. <math>B</math> at a <math>10\%</math> profit based on the worth of the house. Mr. <math>B</math> sells the house back to Mr. <math>A</math> at a <math>10\%</math> loss.  Then:
  
 
<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>
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==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><dollar/><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> 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>10,000</math> dollars <math>=</math> <math>11,000</math> dollars 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>.
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==See Also==
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{{AHSME 50p box|year=1951|num-b=4|num-a=6}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 12:58, 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 $=$ $1.1$ $\cdot$ $10,000$ dollars $=$ $11,000$ dollars 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
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

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