Difference between revisions of "1959 AHSME Problems/Problem 29"
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| − | == Problem | + | == 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> | ||
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==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: | ||
| − | <math>\frac{\frac{n-20}{3} + 15}{n} = \frac{1}{2}</math>. Simplify to get <math>n= | + | <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 == | ||
| + | {{AHSME 50p box|year=1959|num-b=28|num-a=30}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 14:29, 21 July 2024
Problem
On a examination of 
questions a student answers correctly 
of the first 
. 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 can there be?
Solution
To calculate the student's score in terms of , you can write the following equation:
. Simplify to get 
, so there is 
solution.
~Goldroman
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
