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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>
  
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The answer is <math>\textbf{(A) }11</math>
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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.
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=3|num-a=5}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 09:37, 1 August 2024

Problem

After simple interest for two months at $5$% per annum was credited, a Boy Scout Troop had a total of $\textdollar 255.31$ in the Council Treasury. The interest credited was a number of dollars plus the following number of cents

$\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }21\qquad \textbf{(E) }31$

Solution

We can use the formula $A = P(1+rt)$ for simple interest, where $P$ is the principal (initial amount), $A$ is the total, $r$ is the interest rate per year, and $t$ is the time in years.

We are solving for $P,$ so we can rearrange the equation: \[P = \frac{A}{1+rt}\] Plug in all the necessary numbers: \[P = \frac{255.31}{1 + (0.05)(2/12)} = \textdollar 253.21\] So, the interest is \[P*(0.05)(2/12) = \textdollar 2.11\]

The answer is $\boxed{\textbf{(A) }11}$

-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

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