Difference between revisions of "1995 AHSME Problems/Problem 3"
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== Problem == | == Problem == | ||
− | The total in-store price for an appliance is <math> | + | The total in-store price for an appliance is <math>\textdollar 99.99</math>. A television commercial advertises the same product for three easy payments of <math>\textdollar 29.98</math> and a one-time shipping and handling charge of <math>\textdollar 9.98</math>. How many cents are saved by buying the appliance from the television advertiser? |
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+ | <math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 } </math> | ||
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== Solution == | == Solution == | ||
− | < | + | IMPORTANT NOTICE: The original problem statement had "how much is saved". However, because this made little sense when the calculations were done, the problem statement was changed to "how many cents". |
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+ | We see that 3 payments of <math>\textdollar 29.98</math> will be a total cost of <math>3\cdot(30-.02)=90-.06</math> | ||
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+ | Adding this to <math>\textdollar9.98</math> we have a total of <math>99.98-.06</math> | ||
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+ | Clearly, this differs from <math>\textdollar 99.99</math> by <math>7</math> cents. Thus, the answer is <math>\fbox{\text{(B)}}</math> | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:58, 5 July 2013
Problem
The total in-store price for an appliance is . A television commercial advertises the same product for three easy payments of and a one-time shipping and handling charge of . How many cents are saved by buying the appliance from the television advertiser?
Solution
IMPORTANT NOTICE: The original problem statement had "how much is saved". However, because this made little sense when the calculations were done, the problem statement was changed to "how many cents".
We see that 3 payments of will be a total cost of
Adding this to we have a total of
Clearly, this differs from by cents. Thus, the answer is
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.