Difference between revisions of "1985 AJHSME Problems/Problem 14"
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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> | + | <math>\text{(E)}</math> <dollar/><math>10</math> |
| − | + | ==Solution== | |
| − | + | The most straightforward method would be to calculate both prices, and subtract. But there's a better method... | |
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| − | + | Before we start, it's always good to convert the word problems into [[Expression|expressions]], we can solve. | |
| − | </math>\ | + | 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*} | ||
| + | (6.5\% - 6\% ) \times 20 &= (.5\% )\times 20 \\ | ||
| + | &= \frac{.5}{100}\times 20 \\ | ||
| + | &= \frac{1}{200}\times 20 \\ | ||
| + | &= .10 \\ | ||
| + | \end{align*}</cmath> | ||
| − | <math> | + | <math>.10</math> is choice <math>\boxed{\text{B}}</math> |
| − | + | ==See Also== | |
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| − | + | {{AJHSME box|year=1985|num-b=13|num-a=15}} | |
| − | + | [[Category:Introductory Algebra Problems]] | |
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Revision as of 18:57, 17 May 2009
Problem
The difference between a 
sales tax and a 
sales tax on an item priced at <dollar/>
before tax is
<dollar/>
<dollar/>
<dollar/>
<dollar/>
<dollar/>
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 increase will be 
, and the price of it after a increase will be 
. And what we're trying to find is 
, 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, , and we can factor the expression into
is choice 
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
