Difference between revisions of "1955 AHSME Problems/Problem 33"
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He arrives at his destination between <math>2</math> p.m. and <math>3</math> p.m. when the hands of the clock are exactly <math>180^\circ</math> apart. The trip takes: | He arrives at his destination between <math>2</math> p.m. and <math>3</math> p.m. when the hands of the clock are exactly <math>180^\circ</math> apart. The trip takes: | ||
| − | <math> \textbf{(A)}\ \text{6 hr.}\qquad\textbf{(B)}\ \text{6 hr. 43 | + | <math> \textbf{(A)}\ \text{6 hr.}\qquad\textbf{(B)}\ \text{6 hr. } 43\frac{7}{11} \text{min.}\qquad\textbf{(C)}\ \text{5 hr. } 16\frac{4}{11} \text{min.}\qquad\textbf{(D)}\ \text{6 hr. 30 min.}\qquad\textbf{(E)}\ \text{none of these} </math> |
| − | [[1955 AHSME Problems | + | ==Solution== |
| + | |||
| + | ==See Also== | ||
| + | In order to get back to the problem set, click [[1955 AHSME Problems|here]] | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 20:10, 15 July 2025
Problem 33
Henry starts a trip when the hands of the clock are together between 
a.m. and 
a.m.
He arrives at his destination between 
p.m. and 
p.m. when the hands of the clock are exactly 
apart. The trip takes:
Solution
See Also
In order to get back to the problem set, click here
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
