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Question 1203331: Betty baked some cheese buns and butter buns. After she sold 1/3 of the cheese buns and 2/5 of the butter buns, she had 50% as many butter buns as cheese buns left. What fraction of the buns baked was butter buns?
Found 3 solutions by greenestamps, MathTherapy, josgarithmetic:
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
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.
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.
Method 1...
let b = # of butter buns
let c = # of cheese buns
The number we are to find is the fraction of the buns that are butter buns. That fraction is b/(b+c).
She sold 1/3 of the cheese buns, so the number she had left was (2/3)c.
She sold 2/5 of the butter buns, so the number she had left was (3/5)b.
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.
ANSWER: 5/14
Method 2...
let x = # of butter buns she had left
then 2x= # of cheese buns she had left
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.
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.
She started with (5/3)x butter buns and 3x cheese buns. The fraction of buns that were butter buns was
ANSWER: 5/14
Method 3...
let x = fraction of the buns that were butter buns
then 1-x = fraction that were cheese buns
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:
ANSWER: 5/14
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....
Perhaps another tutor will present a different method for solving the problem that is easier than any of the above.
Answer by MathTherapy(10555)  (Show Source):
Answer by josgarithmetic(39625)  (Show Source):
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