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

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== Problem 29==
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== Problem ==
 
On a examination of <math>n</math> questions a student answers correctly <math>15</math> of the first <math>20</math>. Of the remaining questions he answers one third correctly.  
 
On a examination of <math>n</math> questions a student answers correctly <math>15</math> of the first <math>20</math>. 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 <math>n</math> can there be?
 
All the questions have the same credit. If the student's mark is 50%, how many different values of <math>n</math> can there be?
 
<math>\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the problem cannot be solved}    </math>
 
<math>\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the problem cannot be solved}    </math>
 
[[1959 AHSME Problems/Problem 29|Solution]]
 
  
 
==Solution==
 
==Solution==
 
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:
  
\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=50</math>, so there is <math>\boxed{\textbf{(D) }1}</math> solution.
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~Goldroman
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== See also ==
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{{AHSME 50p box|year=1959|num-b=28|num-a=30}}
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{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 14:29, 21 July 2024

Problem

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

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 $n=50$, so there is $\boxed{\textbf{(D) }1}$ solution.

~Goldroman

See also

1959 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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