Difference between revisions of "1952 AHSME Problems/Problem 14"
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==Solution== | ==Solution== | ||
− | Denote the original price of the house and the store as <math> h </math> and <math> s </math>, respectively. It is given that <math> \frac{4h}{5}=\textdollar 12,000 </math>, and that <math> \frac{ | + | Denote the original price of the house and the store as <math> h </math> and <math> s </math>, respectively. It is given that <math> \frac{4h}{5}=\textdollar 12,000 </math>, and that <math> \frac{6s}{5}=\textdollar 12,000 </math>. Thus, <math> h=\textdollar 15,000 </math>, <math> s=\textdollar10,000 </math>, and <math> h+s=\textdollar25,000 </math>. This value is <math> \textdollar1000 </math> higher than the current price of the property, <math> 2\cdot \textdollar12,000 </math>. Hence, the transaction resulted in a <math> \boxed{\textbf{(B)}\ \text{loss of }\textdollar1000} </math>. |
==See also== | ==See also== | ||
{{AHSME 50p box|year=1952|num-b=13|num-a=15}} | {{AHSME 50p box|year=1952|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:09, 19 December 2017
Problem
A house and store were sold for each. The house was sold at a loss of of the cost, and the store at a gain of of the cost. The entire transaction resulted in:
Solution
Denote the original price of the house and the store as and , respectively. It is given that , and that . Thus, , , and . This value is higher than the current price of the property, . Hence, the transaction resulted in a .
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AHSME Problems and Solutions |
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