Difference between revisions of "1971 AHSME Problems/Problem 4"
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Plug in all the necessary numbers: | Plug in all the necessary numbers: | ||
<cmath>P = \frac{255.31}{1 + (0.05)(2/12)} = \textdollar 253.21</cmath> | <cmath>P = \frac{255.31}{1 + (0.05)(2/12)} = \textdollar 253.21</cmath> | ||
+ | So, the interest is <cmath>P*(0.05)(2/12) = \textdollar 2.11</cmath> | ||
− | The answer is <math>\textbf{( | + | The answer is <math>\boxed{\textbf{(A) }11}</math> |
− | -edited by | + | -edited by Happypuma46 |
If there are any mistakes, feel free to edit so that it is correct. | If there are any mistakes, feel free to edit so that it is correct. | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 35p box|year=1971|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 09:37, 1 August 2024
Problem
After simple interest for two months at % per annum was credited, a Boy Scout Troop had a total
of
in the Council Treasury. The interest credited was a number of dollars plus the following number of cents
Solution
We can use the formula for simple interest, where
is the principal (initial amount),
is the total,
is the interest rate per year, and
is the time in years.
We are solving for so we can rearrange the equation:
Plug in all the necessary numbers:
So, the interest is
The answer is
-edited by Happypuma46
If there are any mistakes, feel free to edit so that it is correct.
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.