Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1950 AHSME Problems/Problem 50"
(soln)
Line 10: Line 10:
  
 
==Solution==
 
==Solution==
−
{{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>
  
 
==See Also==
 
==See Also==

Revision as of 15:16, 20 June 2012

Problem

A privateer discovers a merchantman $10$ miles to leeward at 11:45 a.m. and with a good breeze bears down upon her at $11$ mph, while the merchantman can only make $8$ 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 $17$ miles while the merchantman makes $15$. The privateer will overtake the merchantman at:

$\textbf{(A)}\ 3\text{:}45\text{ p.m.} \qquad \textbf{(B)}\ 3\text{:}30\text{ p.m.} \qquad \textbf{(C)}\ 5\text{:}00\text{ p.m.} \qquad \textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad \textbf{(E)}\ 5\text{:}30\text{ p.m.}$

Solution

Assume that the two boats are traveling along the positive real number line, with the merchantman starting at the number $10$ and the privateer starting at the number $0$. After two hours the merchantman is at $26$ while the privateer is at $22$. When the top sail of the privateer is carried away, the speed of the merchantman is unaffected; he still travels at $8$ miles per hour. Therefore the privateer travels at $17\cdot \frac{8}{15}=\frac{136}{15}=9\frac{1}{15}$ miles per hour. The remaining time $t$ in hours it takes for the privateer to catch up to the merchantman satisfied the equation

\[26+8t=22+\frac{136}{15}t\]

Simplification yields the equation $\frac{16}{15}t=4$, which shows that $t=\frac{15}{4}=3\frac{3}{4}$. It therefore takes a total of $5\frac{3}{4}$ hours for the privateer to catch the merchantman, so this will happen at $\boxed{\textbf{(E)}\ 5\text{:}30\text{ p.m.}}$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_50&oldid=47430"