Question 1203331
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I thought of several different ways to attack this problem and couldn't decide easily which one would be easier, so I tried different ones.<br>
That in itself is a good lesson in problem solving, whether it is a math problem or a problem in real life.  Always be open to trying different ways of doing things. If we didn't do that, we would all still be living in caves.<br>
Method 1...<br>
let b = # of butter buns
let c = # of cheese buns<br>
The number we are to find is the fraction of the buns that are butter buns.  That fraction is b/(b+c).<br>
She sold 1/3 of the cheese buns, so the number she had left was (2/3)c.<br>
She sold 2/5 of the butter buns, so the number she had left was (3/5)b.<br>
The number of butter buns she had left was half the number of cheese buns; i.e., the number of cheese buns she had left was twice the number of butter buns.<br>
{{{(2/3)c=2((3/5)b)}}}
{{{10c=18b}}}
{{{b/c=10/18=5/9}}}
{{{b/(b+c)=5/(5+9)=5/14}}}<br>
ANSWER: 5/14<br>
Method 2...<br>
let x = # of butter buns she had left
then 2x= # of cheese buns she had left<br>
She sold 1/3 of the cheese buns, so she was left with 2/3 of them.  She was left with 2x cheese buns, so the number she started with was (3/2)(2x) = 3x.<br>
She sold 2/5 of the butter buns, so she was left with 3/5 of them. She was left with x butter buns, so the number she started with was (5/3)x.<br>
She started with (5/3)x butter buns and 3x cheese buns.  The fraction of buns that were butter buns was<br>
{{{((5/3)x)/((5/3)x+3x)=5x/(5x+9x)=(5x)/(14x)=5/14}}}<br>
ANSWER: 5/14<br>
Method 3...<br>
let x = fraction of the buns that were butter buns
then 1-x = fraction that were cheese buns<br>
She was left with (3/5)x butter buns and (2/3)(1-x) cheese buns; and the number of cheese buns left was twice the number of butter buns left:<br>
{{{(2/3)(1-x)=2(3/5)x}}}
{{{10(1-x)=18x}}}
{{{10-10x=18x}}}
{{{10=28x}}}
{{{x=5/14}}}<br>
ANSWER: 5/14<br>
Having solved the problem three different ways, I see tricky parts in each of the methods, so I don't have a strong preference for any one of them....<br>
Perhaps another tutor will present a different method for solving the problem that is easier than any of the above.<br>