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Difference between revisions of "1959 AHSME Problems/Problem 29"

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To calculate the student's score in terms of <math>n</math>, you can write the following equation:
 
To calculate the student's score in terms of <math>n</math>, you can write the following equation:
  
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\frac{\frac{n-20}{3} + 15}{n} = \frac{1}{2}<math>. Simplify to get </math>n=55$, so there is one solution.
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<math>\frac{\frac{n-20}{3} + 15}{n} = \frac{1}{2}</math>. Simplify to get <math>n=55</math>, so there is one solution.

Revision as of 13:15, 16 July 2024

Problem 29

On a examination of $n$ questions a student answers correctly $15$ of the first $20$. Of the remaining questions he answers one third correctly. All the questions have the same credit. If the student's mark is 50%, how many different values of $n$ can there be? $\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the problem cannot be solved}$

Solution

Solution

To calculate the student's score in terms of $n$, you can write the following equation:

$\frac{\frac{n-20}{3} + 15}{n} = \frac{1}{2}$. Simplify to get $n=55$, so there is one solution.

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