Difference between revisions of "1971 AHSME Problems/Problem 4"
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So, the interest is <cmath>P*(0.05)(2/12) = \textdollar 2.11</cmath> | So, the interest is <cmath>P*(0.05)(2/12) = \textdollar 2.11</cmath> | ||
| − | The answer is <math>\textbf{(A) }11</math> | + | The answer is <math>\boxed{\textbf{(A) }11}</math> |
-edited by Happypuma46 | -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 08:37, 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:
![\[P = \frac{A}{1+rt}\]](http://latex.artofproblemsolving.com/e/6/f/e6fe59cb005eba4a931e84590c01368751c45331.png)
Plug in all the necessary numbers:
![\[P = \frac{255.31}{1 + (0.05)(2/12)} = \textdollar 253.21\]](http://latex.artofproblemsolving.com/6/e/3/6e3ebea10b1b32f41ad62fc2cc39d7856fc8e8e2.png)
So, the interest is ![\[P*(0.05)(2/12) = \textdollar 2.11\]](http://latex.artofproblemsolving.com/b/b/f/bbff43c417ac6736476db16a0f850eef84eb97fa.png)
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. 
