Difference between revisions of "1959 AHSME Problems/Problem 46"
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== Solution == | == Solution == | ||
| − | <math>\boxed{textbf{(B)}}</math> | + | |
| + | From the given information, we know that there were <math>7+5+6=18</math> halves of the day (either afternoons or mornings), so there were <math>\frac{18}{2}=9</math> days in total, which is answer choice <math>\boxed{\textbf{(B)}}</math>. | ||
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=45|num-a=47}} | {{AHSME 50p box|year=1959|num-b=45|num-a=47}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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Latest revision as of 11:49, 22 July 2024
Problem
A student on vacation for 
days observed that (1) it rained 
times, morning or afternoon (2) when it rained in the afternoon,
it was clear in the morning (3) there were five clear afternoons (4) there were six clear mornings. Then equals:
Solution
From the given information, we know that there were 
halves of the day (either afternoons or mornings), so there were 
days in total, which is answer choice 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 45 |
Followed by Problem 47 | |
| 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. 
