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 | + | The [[subtraction|difference]] between a <math>6.5\% </math> sales tax and a <math>6\% </math> sales tax on an item priced at <math> $20</math> before tax is |
− | <math>\text{(A)}</math | + | <math>\text{(A)}</math> <math>$.01</math> |
− | <math>\text{(B)}</math | + | <math>\text{(B)}</math> <math>$.10</math> |
− | <math>\text{(C)}</math | + | <math>\text{(C)}</math> <math>$ .50</math> |
− | <math>\text{(D)}</math | + | <math>\text{(D)}</math> <math>$ 1</math> |
− | <math>\text{(E)}</math | + | <math>\text{(E)}</math> <math>$10</math> |
==Solution== | ==Solution== | ||
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<math>.10</math> is choice <math>\boxed{\text{B}}</math> | <math>.10</math> is choice <math>\boxed{\text{B}}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/DTx0CwhOU0o | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 07:13, 13 January 2023
Contents
[hide]Problem
The difference between a sales tax and a sales tax on an item priced at before tax is
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
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.