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

(New page: ==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 problem...)
 
 
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==Problem==
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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
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<math>\text{(A)}</math> <math>\$.01</math>
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<math>\text{(B)}</math> <math>\$.10</math>
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<math>\text{(C)}</math> <math>\$ .50</math>
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<math>\text{(D)}</math> <math>\$ 1</math>
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<math>\text{(E)}</math> <math>\$10</math>
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==Solution==
 
==Solution==
  
 
The most straightforward method would be to calculate both prices, and subtract. But there's a better method...
 
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 formulas, or at least expressions, we can solve.
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Before we start, it's always good to convert the word problems into [[Expression|expressions]], we can solve.
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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
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<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>
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<math>.10</math> is choice <math>\boxed{\text{B}}</math>
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==Video Solution==
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https://youtu.be/DTx0CwhOU0o
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~savannahsolver
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==See Also==
  
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So we know that the price of the object after a 6.5% increase will be <math>20 \times 6.5%</math>, and the price of it after a 6% 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 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...
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{{AJHSME box|year=1985|num-b=13|num-a=15}}
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[[Category:Introductory Algebra Problems]]
  
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<math>(6.5% - 6%) \times 20</math>, which simplifies to <math>.5%</math>. We know that <math>.5 = \frac{1}{20}</math>, so <math>.5% = \frac{1}{2000}</math>. <math>20 \times \frac{1}{2000} = \frac{1}{100} = .01</math>, so the answer is 1 cent.
 
  
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1 cent = <math>\</math>.01$ = (D)
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{{MAA Notice}}

Latest revision as of 07:13, 13 January 2023

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 $\boxed{\text{B}}$

Video Solution

https://youtu.be/DTx0CwhOU0o

~savannahsolver

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


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