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Difference between revisions of "1969 AHSME Problems/Problem 27"

(Created page with "== Problem == A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile ...")
 
(Solution to Problem 27)
 
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== Solution ==
 
== Solution ==
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<math>\fbox{E}</math>
 
  
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== See also ==
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Let <math>d</math> be the distance already traveled and <math>s_n</math> be the speed for the <math>n^\text{th}</math> mile.  Because the speed for the second and subsequent miles varies inversely as the integral number of miles already traveled,
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<cmath>ds_{d+1} = k</cmath>
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If the second mile is traversed in <math>2</math> hours, then the speed in the second mile (<math>s_2</math>) equals <math>\tfrac{1}{2}</math> miles per hour.  Since one mile has been traveled already, <math>k = \tfrac{1}{2}</math>.  Now solve for the speed the <math>n^\text{th}</math> mile, noting that <math>n-1</math> miles have already been traveled.
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<cmath>(n-1)s_{d+1} = \frac{1}{2}</cmath>
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<cmath>s_{d+1} = \frac{1}{2(n-1)}</cmath>
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Thus, the time it takes to travel the <math>n^\text{th}</math> mile is <math>1 \div \tfrac{1}{2(n-1)} = \boxed{\textbf{(E) } 2(n-1)}</math> hours.
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== See Also ==
 
{{AHSME 35p box|year=1969|num-b=26|num-a=28}}   
 
{{AHSME 35p box|year=1969|num-b=26|num-a=28}}   
  
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[[Category: Intermediate Algebra Problems]]
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:38, 21 June 2018

Problem

A particle moves so that its speed for the second and subsequent miles varies inversely as the integral number of miles already traveled. For each subsequent mile the speed is constant. If the second mile is traversed in $2$ hours, then the time, in hours, needed to traverse the $n$th mile is:

$\text{(A) } \frac{2}{n-1}\quad \text{(B) } \frac{n-1}{2}\quad \text{(C) } \frac{2}{n}\quad \text{(D) } 2n\quad \text{(E) } 2(n-1)$

Solution

Let $d$ be the distance already traveled and $s_n$ be the speed for the $n^\text{th}$ mile. Because the speed for the second and subsequent miles varies inversely as the integral number of miles already traveled, \[ds_{d+1} = k\] If the second mile is traversed in $2$ hours, then the speed in the second mile ($s_2$) equals $\tfrac{1}{2}$ miles per hour. Since one mile has been traveled already, $k = \tfrac{1}{2}$. Now solve for the speed the $n^\text{th}$ mile, noting that $n-1$ miles have already been traveled. \[(n-1)s_{d+1} = \frac{1}{2}\] \[s_{d+1} = \frac{1}{2(n-1)}\] Thus, the time it takes to travel the $n^\text{th}$ mile is $1 \div \tfrac{1}{2(n-1)} = \boxed{\textbf{(E) } 2(n-1)}$ hours.

See Also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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
All AHSME Problems and Solutions

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