Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 26"

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<math>g=\frac{b}3</math> and <math>t=\frac{b}{27}</math>
 
<math>g=\frac{b}3</math> and <math>t=\frac{b}{27}</math>
  
<math>S=b+\frac{b}{3}+\frac{b}{27}=\frac{37b}{27}\Rightarrow\mathrm{ B}</math>
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<math>S=b+\frac{b}{3}+\frac{b}{27}=\frac{37b}{27}\Longrightarrow\mathrm{ B}</math>
  
 
==See also==
 
==See also==
*[[2007 Cyprus MO/Lyceum/Problems]]
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{{CYMO box|year=2007|l=Lyceum|num-b=25|num-a=27}}
  
*[[2007 Cyprus MO/Lyceum/Problem 25|Previous Problem]]
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[[Category:Introductory Algebra Problems]]
 
 
*[[2007 Cyprus MO/Lyceum/Problem 27|Next Problem]]
 

Latest revision as of 20:21, 6 May 2007

Problem

The number of boys in a school is 3 times the number of girls and the number of girls is 9 times the number of teachers. Let us denote with $b$, $g$ and $t$, the number of boys, girls and teachers respectively. Then the total number of boys, girls and teachers equals to

$\mathrm{(A) \ } 31b\qquad \mathrm{(B) \ } \frac{37b}{27}\qquad \mathrm{(C) \ } 13g\qquad \mathrm{(D) \ } \frac{37g}{27}\qquad \mathrm{(E) \ } \frac{37t}{27}$

Solution

$g=\frac{b}3$ and $t=\frac{b}{27}$

$S=b+\frac{b}{3}+\frac{b}{27}=\frac{37b}{27}\Longrightarrow\mathrm{ B}$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 25
Followed by
Problem 27
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