Difference between revisions of "1950 AHSME Problems/Problem 50"
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\textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad | \textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad | ||
\textbf{(E)}\ 5\text{:}30\text{ p.m.}</math> | \textbf{(E)}\ 5\text{:}30\text{ p.m.}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | Assume that the two boats are traveling along the positive real number line, with the merchantman starting at the number <math>10</math> and the privateer starting at the number <math>0</math>. After two hours the merchantman is at <math>26</math> while the privateer is at <math>22</math>. When the top sail of the privateer is carried away, the speed of the merchantman is unaffected; he still travels at <math>8</math> miles per hour. Therefore the privateer travels at <math>17\cdot \frac{8}{15}=\frac{136}{15}=9\frac{1}{15}</math> miles per hour. The remaining time <math>t</math> in hours it takes for the privateer to catch up to the merchantman satisfied the equation | ||
| + | |||
| + | <cmath>26+8t=22+\frac{136}{15}t</cmath> | ||
| + | |||
| + | Simplification yields the equation <math>\frac{16}{15}t=4</math>, which shows that <math>t=\frac{15}{4}=3\frac{3}{4}</math>. It therefore takes a total of <math>5\frac{3}{4}</math> hours for the privateer to catch the merchantman, so this will happen at <math>\boxed{\textbf{(E)}\ 5\text{:}30\text{ p.m.}}</math> | ||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=l4lAvs2P_YA&t=437s | ||
| + | |||
| + | ~MathProblemSolvingSkills.com | ||
| + | |||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1950|num-b=49|after=Last Question}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | [[Category:Rate Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:53, 15 July 2023
Contents
Problem
A privateer discovers a merchantman 
miles to leeward at 11:45 a.m. and with a good breeze bears down upon her at 
mph, while the merchantman can only make 
mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only 
miles while the merchantman makes 
. The privateer will overtake the merchantman at:
Solution
Assume that the two boats are traveling along the positive real number line, with the merchantman starting at the number and the privateer starting at the number 
. After two hours the merchantman is at 
while the privateer is at 
. When the top sail of the privateer is carried away, the speed of the merchantman is unaffected; he still travels at miles per hour. Therefore the privateer travels at 
miles per hour. The remaining time 
in hours it takes for the privateer to catch up to the merchantman satisfied the equation
![\[26+8t=22+\frac{136}{15}t\]](http://latex.artofproblemsolving.com/2/0/9/209102dc6412bd18c339068515503950e882ab4e.png)
Simplification yields the equation 
, which shows that 
. It therefore takes a total of 
hours for the privateer to catch the merchantman, so this will happen at 
Video Solution
https://www.youtube.com/watch?v=l4lAvs2P_YA&t=437s
~MathProblemSolvingSkills.com
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 49 |
Followed by Last Question | |
| 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. 
