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

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==Problem==
 
==Problem==
  
The [[subtraction|difference]] between a <math>6.5\% </math> sales tax and a <math>6\% </math> sales tax on an item priced at <dollar/><math>20</math> before tax is
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The [[subtraction|difference]] between a <math>6.5\% </math> sales tax and a <math>6\% </math> sales tax on an item priced at <math> \$20</math> before tax is
  
 
<math>\text{(A)}</math> <math>\$.01</math>  
 
<math>\text{(A)}</math> <math>\$.01</math>  

Latest revision as of 17:32, 30 May 2015

Problem

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

$\text{(A)}$ $$.01$

$\text{(B)}$ $$.10$

$\text{(C)}$ $$ .50$

$\text{(D)}$ $$ 1$

$\text{(E)}$ $$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


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