How Full Is the Larger Bucket After Pouring?

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SUMMARY

Joseph pours blue paint into two buckets, where the larger bucket's volume is three times that of the smaller bucket. The larger bucket is initially 3/4 full, while the smaller bucket is 1/3 full, equating to 1/9 of the larger bucket's capacity. After pouring the contents of the smaller bucket into the larger one, the larger bucket becomes 31/36 full. This calculation involves finding a common denominator to add the fractions accurately.

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Before Joseph paints his house, he pours blue paint into 2 buckets of different sizes. He notices that the volume of the larger bucket is 3 times the volume of the smaller bucket. At the end of the day, Joseph estimates that the larger bucket is 3/4 full and the smaller bucket is 1/3 full/ He decides to pour all of the paint from the smaller bucket into the larger bucket. After that, how full is the larger bucket?

I know the answer is 31/36, but can someone show me how to set it up?
 
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Re: ACT problem

The smaller bucket is 1/3 the size of the larger bucket, and so if the smaller bucket is 1/3 full, it contains 1/9 the capacity of the larger bucket (since 1/3 of 1/3 is 1/9). So, to determine how full the larger bucket would be if the smaller bucket's contents are added, we need to start with what it already has and add 1/9 to that:

$$F=\frac{3}{4}+\frac{1}{9}$$

Since the two denominators are co-prime (they share no common factors), their LCM is their product and so the LCD is $4\cdot9=36$:

$$F=\frac{3}{4}\cdot\frac{9}{9}+\frac{1}{9}\cdot\frac{4}{4}=\frac{27}{36}+\frac{4}{36}=\frac{27+4}{36}=\frac{31}{36}$$
 

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