Difference between revisions of "1971 AHSME Problems/Problem 4"
Coolmath34 (talk | contribs) (→Solution) |
Happypuma46 (talk | contribs) (→Solution) |
||
| Line 17: | Line 17: | ||
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>\textbf{(A) }11</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
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. | ||
Revision as of 11:51, 27 April 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 coolmath34
If there are any mistakes, feel free to edit so that it is correct.
