Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎Solution)
(→‎Solution)
Line 9: Line 9:
 
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:
  
−
<math>\frac{n-20}{3} + 15 = \frac{1}{2}</math>. Simplify to get <math>n=55</math>, so there is one solution.
+
\frac{\frac{n-20}{3} + 15}{n} = \frac{1}{2}<math>. Simplify to get </math>n=55$, 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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_AHSME_Problems/Problem_29&oldid=222845"