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Difference between revisions of "1950 AHSME Problems/Problem 22"

(Created page with "== Problem== Successive discounts of <math>10\%</math> and <math>20\%</math> are equivalent to a single discount of: <math> \textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\tex...")
 
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<math> \textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these} </math>
 
<math> \textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these} </math>
  
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==Solution==
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==Solution 1 (Kind of Lame)==
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Without loss of generality, assume something costs <math>100</math> dollars. Then with each successive discount, it would cost <math>90</math> dollars, then <math>72</math> dollars. This amounts to a total of <math>28</math> dollars off, so the single discount would be <math>\boxed{\mathrm{(D)}\ 28\%.}</math>
  
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WLOG assume something costed <math>100</math> dollars. Then with each successive discount, it would cost <math>90</math> dollars, then <math>72</math> dollars. This amounts to a total of <math>28</math> dollars off, so the single discount would be <math>\boxed{\mathrm{(D)}\ 28\%.}</math>
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==Solution 2 (Technical)==
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Let the object cost <math>x</math> dollars. After the <math>10\%</math> discount, it's worth <math>(1-10\%)x=0.9x</math> dollars. After a <math>20\%</math> discount on that, it's worth <math>(1-20\%)(0.9x)=0.72x</math> dollars. Say the single discount is of <math>k</math>. Then <math>(1-k)x=0.72x</math>. So <math>k=0.28</math>, or <math>k=28\%</math>. So select <math>\boxed{D}</math>.
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~hastapasta
  
 
==See Also==
 
==See Also==
  
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{{AHSME box|year=1950|num-b=21|num-a=23}}
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{{AHSME 50p box|year=1950|num-b=21|num-a=23}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 12:16, 31 March 2022

Problem

Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:

$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$

Solution 1 (Kind of Lame)

Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\boxed{\mathrm{(D)}\ 28\%.}$

Solution 2 (Technical)

Let the object cost $x$ dollars. After the $10\%$ discount, it's worth $(1-10\%)x=0.9x$ dollars. After a $20\%$ discount on that, it's worth $(1-20\%)(0.9x)=0.72x$ dollars. Say the single discount is of $k$. Then $(1-k)x=0.72x$. So $k=0.28$, or $k=28\%$. So select $\boxed{D}$.

~hastapasta

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AHSME Problems and Solutions

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