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Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 3"

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==Problem==
 
==Problem==
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A cyclist drives form town A to town B with velocity <math>40 \frac{\mathrm{km}}{\mathrm{h}}</math> and comes back with velocity <math>60 \frac{\mathrm{km}}{\mathrm{h}}</math>. The [[mean]] velocity in <math>\frac{\mathrm{km}}{\mathrm{h}}</math> for the total distance is
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A cyclist drives from town A to town B with velocity <math>40 \frac{\mathrm{km}}{\mathrm{h}}</math> and comes back with velocity <math>60 \frac{\mathrm{km}}{\mathrm{h}}</math>. The [[mean]] velocity in <math>\frac{\mathrm{km}}{\mathrm{h}}</math> for the total distance is
  
 
<math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 48\qquad \mathrm{(C) \ } 50\qquad \mathrm{(D) \ } 55\qquad \mathrm{(E) \ } 100</math>
 
<math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 48\qquad \mathrm{(C) \ } 50\qquad \mathrm{(D) \ } 55\qquad \mathrm{(E) \ } 100</math>
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The time it took the cyclist to travel from <math>B</math> to <math>A</math> was <math>\frac{d}{60}</math> hours.
 
The time it took the cyclist to travel from <math>B</math> to <math>A</math> was <math>\frac{d}{60}</math> hours.
  
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The cyclist's mean velocity was <math>\frac{2d}{\frac{d}{40}+\frac{d}{60}}=\frac{2d}{\frac{100d}{2400}}=\frac{4800d}{100d}=48\frac{\mathrm{km}}{\mathrm{h}}\Rightarrow\mathrm{ B}</math>
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The cyclist's mean velocity was <math>\frac{2d}{\frac{d}{40}+\frac{d}{60}}=\frac{2d}{\frac{100d}{2400}}= \frac{4800d}{100d}=48\frac{\mathrm{km}}{\mathrm{h}} \Longrightarrow\mathrm{ B}</math>
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2007|l=Lyceum|num-b=2|num-a=4}}
 
{{CYMO box|year=2007|l=Lyceum|num-b=2|num-a=4}}

Latest revision as of 15:01, 24 May 2009

Problem

A cyclist drives from town A to town B with velocity $40 \frac{\mathrm{km}}{\mathrm{h}}$ and comes back with velocity $60 \frac{\mathrm{km}}{\mathrm{h}}$. The mean velocity in $\frac{\mathrm{km}}{\mathrm{h}}$ for the total distance is

$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 48\qquad \mathrm{(C) \ } 50\qquad \mathrm{(D) \ } 55\qquad \mathrm{(E) \ } 100$

Solution

Let the distance from town A to town B, in kilometers, be $d$.

The time it took the cyclist to travel from $A$ to $B$ was $\frac{d}{40}$ hours.

The time it took the cyclist to travel from $B$ to $A$ was $\frac{d}{60}$ hours.

The cyclist's mean velocity was $\frac{2d}{\frac{d}{40}+\frac{d}{60}}=\frac{2d}{\frac{100d}{2400}}= \frac{4800d}{100d}=48\frac{\mathrm{km}}{\mathrm{h}} \Longrightarrow\mathrm{ B}$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 2 		Followed by
Problem 4
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
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