Difference between revisions of "1971 AHSME Problems/Problem 4"
Happypuma46 (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{(A) }11</math> | The answer is <math>\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. |
Revision as of 10:52, 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: 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.