Difference between revisions of "1985 AJHSME Problems/Problem 14"

(Undo revision 31685 by Edward130603 (Talk))
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<math>\text{(D)}</math> <dollar/><math>1</math>  
 
<math>\text{(D)}</math> <dollar/><math>1</math>  
  
<math>\text{(E)}</math> <dollar/><math>abefore tax is
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<math>\text{(E)}</math> <dollar/><math>10</math>
  
</math>\text{(A)}<math> <dollar/></math>.01<math>
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==Solution==
  
</math>\text{(B)}<math> <dollar/></math>.10<math>
+
The most straightforward method would be to calculate both prices, and subtract. But there's a better method...
 
</math>\text{(C)}<math> <dollar/></math>.50<math>
 
  
</math>\text{(D)}<math> <dollar/></math>1<math>
+
Before we start, it's always good to convert the word problems into [[Expression|expressions]], we can solve.
  
</math>\text{(E)}<math> <dollar/></math>abefore tax is
+
So we know that the price of the object after a <math>6.5\% </math> increase will be <math>20 \times 6.5\% </math>, and the price of it after a <math>6\% </math> increase will be <math>20 \times 6\% </math>. And what we're trying to find is <math>6.5\% \times 20 - 6\% \times 20</math>, and if you have at least a little experience in the field of [[algebra]], you'll notice that both of the items have a common [[divisor|factor]], <math>20</math>, and we can [[factoring|factor]] the expression into
 +
<cmath>\begin{align*}
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(6.5\% - 6\% ) \times 20 &= (.5\% )\times 20 \
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&= \frac{.5}{100}\times 20 \
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&= \frac{1}{200}\times 20 \
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&= .10 \
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\end{align*}</cmath>
  
<math>\text{(A)}</math> <dollar/><math>.01</math>  
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<math>.10</math> is choice <math>\boxed{\text{B}}</math>
  
<math>\text{(B)}</math> <dollar/><math>.10</math>
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==See Also==
 
<math>\text{(C)}</math> <dollar/><math>.50</math>
 
  
<math>\text{(D)}</math> <dollar/><math>1</math>
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{{AJHSME box|year=1985|num-b=13|num-a=15}}
 
+
[[Category:Introductory Algebra Problems]]
<math>\text{(E)}</math> <dollar/><math>abefore tax is
 
 
 
</math>\text{(A)}<math> <dollar/></math>.01<math>
 
 
 
</math>\text{(B)}<math> <dollar/></math>.10<math>
 
 
</math>\text{(C)}<math> <dollar/></math>.50<math>
 
 
 
</math>\text{(D)}<math> <dollar/></math>1<math>
 
 
 
</math>\text{(E)}<math> <dollar/></math>abefore tax is
 
 
 
<math>\text{(A)}</math> <dollar/><math>.01</math>
 
 
 
<math>\text{(B)}</math> <dollar/><math>.10</math>
 
 
<math>\text{(C)}</math> <dollar/><math>.50</math>
 
 
 
<math>\text{(D)}</math> <dollar/><math>1</math>
 
 
 
<math>\text{(E)}</math> <dollar/>$a
 

Revision as of 17:57, 17 May 2009

Problem

The difference between a $6.5\%$ sales tax and a $6\%$ sales tax on an item priced at <dollar/>$20$ before tax is

$\text{(A)}$ <dollar/>$.01$

$\text{(B)}$ <dollar/>$.10$

$\text{(C)}$ <dollar/>$.50$

$\text{(D)}$ <dollar/>$1$

$\text{(E)}$ <dollar/>$10$

Solution

The most straightforward method would be to calculate both prices, and subtract. But there's a better method...

Before we start, it's always good to convert the word problems into expressions, we can solve.

So we know that the price of the object after a $6.5\%$ increase will be $20 \times 6.5\%$, and the price of it after a $6\%$ increase will be $20 \times 6\%$. And what we're trying to find is $6.5\% \times 20 - 6\% \times 20$, and if you have at least a little experience in the field of algebra, you'll notice that both of the items have a common factor, $20$, and we can factor the expression into \begin{align*} (6.5\% - 6\% ) \times 20 &= (.5\% )\times 20 \\ &= \frac{.5}{100}\times 20 \\ &= \frac{1}{200}\times 20 \\ &= .10 \\ \end{align*}

$.10$ is choice $\boxed{\text{B}}$

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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