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Difference between revisions of "1986 AJHSME Problems/Problem 15"
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==Problem==
 
==Problem==
  
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Sale prices at the Ajax Outlet Store are <math>50\% </math> below original prices.  On Saturdays an additional discount of <math>20\% </math> off the sale price is given.  What is the Saturday price of a coat whose original price is <dollar/><math>180</math>?
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Sale prices at the Ajax Outlet Store are <math>50\% </math> below original prices.  On Saturdays an additional discount of <math>20\% </math> off the sale price is given.  What is the Saturday price of a coat whose original price is <math>\textdollar 180</math>?
  
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<math>\text{(A)}</math> <dollar/><math>54</math>
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<math>\text{(A)}</math> <math>\textdollar 54</math>
  
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<math>\text{(B)}</math> <dollar/><math>72</math>
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<math>\text{(B)}</math> <math>\textdollar 72</math>
  
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<math>\text{(C)}</math> <dollar/><math>90</math>
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<math>\text{(C)}</math> <math>\textdollar 90</math>
  
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<math>\text{(D)}</math> <dollar/><math>108</math>
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<math>\text{(D)}</math> <math>\textdollar 108</math>
  
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<math>\text{(D)}</math> <dollar/><math>110</math>
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<math>\text{(D)}</math> <math>\textdollar 110</math>
  
 
==Solution==
 
==Solution==
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==See Also==
 
==See Also==
  
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[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=14|num-a=16}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 14:32, 9 November 2017

Problem

Sale prices at the Ajax Outlet Store are $50\%$ below original prices. On Saturdays an additional discount of $20\%$ off the sale price is given. What is the Saturday price of a coat whose original price is $\textdollar 180$?

$\text{(A)}$ $\textdollar 54$

$\text{(B)}$ $\textdollar 72$

$\text{(C)}$ $\textdollar 90$

$\text{(D)}$ $\textdollar 108$

$\text{(D)}$ $\textdollar 110$

Solution

First we need to do the first discount, which, at Ajax Outlet Store, would be there any day of the week. \begin{align*} 180 \times 50\% &= 180 \times \frac{1}{2} \\ &= 90 \\ \end{align*}

If we discount $20\%$, then $80 \%$ will be left, so after the second discount, we get \begin{align*} 90 \times 80\% &= 90 \times \frac{8}{10} \\ &= 9 \times 8 \\ &= 72 \\ \end{align*}

$\boxed{\text{B}}$

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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