Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 1"

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==Problem==
 
==Problem==
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A diary industry, in a quantity of milk with <math>4\%</math> fat adds a quantity of milk with <math>1\%</math> fat and produces <math>1200</math>kg of milk with <math>2\%</math> fat.
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The quantity of milk with <math>1\%</math> fat, that was added is (in kg)
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<math>\mathrm{(A)}\ 1000\qquad\mathrm{(B)}\ 600\qquad\mathrm{(C)}\ 800\qquad\mathrm{(D)}\ 120\qquad\mathrm{(E)}\ 480</math>
  
 
==Solution==
 
==Solution==
{{solution}}
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Let <math>x</math> be the number of kg of <math>4\%</math> milk and let <math>y</math> be the number of kg of <math>1\%</math> milk.
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We then have the equations <math>x+y=1200</math> and <math>\frac{4x+y}{100}=1200*.02=24</math>.
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Solving, we find <math>X=400</math>, and <math>y=1200-400=800\Rightarrow\mathrm{(C)}</math>.
  
 
==See also==
 
==See also==
 
{{CYMO box|year=2006|l=Lyceum|before=First Question|num-a=2}}
 
{{CYMO box|year=2006|l=Lyceum|before=First Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 10:54, 27 April 2008

Problem

A diary industry, in a quantity of milk with $4\%$ fat adds a quantity of milk with $1\%$ fat and produces $1200$kg of milk with $2\%$ fat. The quantity of milk with $1\%$ fat, that was added is (in kg)

$\mathrm{(A)}\ 1000\qquad\mathrm{(B)}\ 600\qquad\mathrm{(C)}\ 800\qquad\mathrm{(D)}\ 120\qquad\mathrm{(E)}\ 480$

Solution

Let $x$ be the number of kg of $4\%$ milk and let $y$ be the number of kg of $1\%$ milk.

We then have the equations $x+y=1200$ and $\frac{4x+y}{100}=1200*.02=24$.

Solving, we find $X=400$, and $y=1200-400=800\Rightarrow\mathrm{(C)}$.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
First Question
Followed by
Problem 2
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