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

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(Standardized answer choices; minor edits)
 
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The quantity of milk with <math>1\%</math> fat, that was added is (in kg)
 
The quantity of milk with <math>1\%</math> fat, that was added is (in kg)
  
A. <math>1000</math>
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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>
 
 
B. <math>600</math>
 
 
 
C.  <math>800</math>
 
 
 
D. <math>120</math>
 
 
 
E. <math>480</math>
 
  
 
==Solution==
 
==Solution==
Let <math>x</math> be the number of kg of 4% milk and let y be the number of kg of 1% milk. We then have the equations
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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.
 
 
<math>x+y=1200</math>
 
 
 
<math>\dfrac{4x+y}{100}=1200*.02=24</math>
 
 
 
<math>4x+y=2400</math>
 
 
 
<math>3x=1200</math>
 
  
<math>x=400</math>
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We then have the equations <math>x+y=1200</math> and <math>\frac{4x+y}{100}=1200*.02=24</math>.
  
<math>y=1200-400=800 \Rightarrow \mathrm {(C)}</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==

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)
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Problem 2
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