Difference between revisions of "2013 AMC 12A Problems/Problem 16"

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Note that <math>\frac{B}{B + C} < 1</math>, and the maximal value of this factor occurs when <math>C = 1</math>
 
Note that <math>\frac{B}{B + C} < 1</math>, and the maximal value of this factor occurs when <math>C = 1</math>
  
Also note that <math>\frac{46}{3}</math> must cancel to give an integer value, and the only fraction that satisfies these conditions is <math>\frac{45}{46}</math>
+
Also note that <math>\frac{46}{3}</math> must cancel to give an integer value, and the only fraction that satisfies both these conditions is <math>\frac{45}{46}</math>
  
 
Plugging in, we get
 
Plugging in, we get
  
 
<math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math>
 
<math>44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59</math>

Revision as of 05:06, 7 February 2013

Let pile $A$ have $A$ rocks, and so on.

The mean weight of $A$ and $C$ together is $44$, so the total weight of $A$ and $C$ is $44(A + C)$

To get the total weight of $B$ and $C$, we need to add the total weight of $B$ and subtract the total weight of $A$

$44A + 44C + 50B - 40A = 4A + 44C + 50B$

And then dividing by the number of rocks $B$ and $C$ together, to get the mean of $B$ and $C$,

$\frac{4A + 44C + 50B}{B + C}$

Simplifying,

$\frac{4A + 44C + 44B + 6B}{B + C}$


$44 + \frac{4A + 6B}{B + C}$

Now, to get rid of the $A$ in the numerator, we use two definitions of the total weight of $A$ and $B$

$40A + 50B = 43A + 43B$

$3A = 7B$

$A = \frac{7}{3}B$

Substituting back in,

$44 + \frac{4(\frac{7}{3}B) + 6B}{B + C}$


$44 + \frac{46}{3}*\frac{B}{B + C}$

Note that $\frac{B}{B + C} < 1$, and the maximal value of this factor occurs when $C = 1$

Also note that $\frac{46}{3}$ must cancel to give an integer value, and the only fraction that satisfies both these conditions is $\frac{45}{46}$

Plugging in, we get

$44 + \frac{46}{3} * \frac{45}{46} = 44 + 15 = 59$

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