Sure, let's break down the solution step by step.
1. First, determine the number of cakes sold by Abigail:
Abigail sold [tex]\( 3 \frac{3}{6} \)[/tex] cakes.
- The fraction [tex]\( \frac{3}{6} \)[/tex] can be simplified. [tex]\( \frac{3}{6} = \frac{1}{2} \)[/tex].
- So, [tex]\( 3 \frac{3}{6} = 3 \frac{1}{2} \)[/tex].
2. Next, determine the number of cakes sold by Cody:
Cody sold [tex]\( 7 \frac{2}{3} \)[/tex] cakes.
3. Now, we need to find out how many more cakes Cody sold than Abigail:
- Abigail sold [tex]\( 3 \frac{1}{2} \)[/tex] cakes.
- Cody sold [tex]\( 7 \frac{2}{3} \)[/tex] cakes.
4. We'll subtract the number of cakes Abigail sold from the number of cakes Cody sold:
- Convert the mixed numbers to improper fractions to make subtraction easier.
- For Abigail: [tex]\( 3 \frac{1}{2} \)[/tex]:
[tex]\[
3 \frac{1}{2} = \frac{7}{2}
\][/tex]
- For Cody: [tex]\( 7 \frac{2}{3} \)[/tex]:
[tex]\[
7 \frac{2}{3} = \frac{23}{3}
\][/tex]
5. Find a common denominator to subtract the fractions:
- The common denominator for 2 and 3 is 6.
- Convert [tex]\( \frac{7}{2} \)[/tex] to have a denominator of 6:
[tex]\[
\frac{7}{2} = \frac{7 \times 3}{2 \times 3} = \frac{21}{6}
\][/tex]
- Convert [tex]\( \frac{23}{3} \)[/tex] to have a denominator of 6:
[tex]\[
\frac{23}{3} = \frac{23 \times 2}{3 \times 2} = \frac{46}{6}
\][/tex]
6. Subtract the fractions:
[tex]\[
\frac{46}{6} - \frac{21}{6} = \frac{46 - 21}{6} = \frac{25}{6}
\][/tex]
7. Convert the improper fraction [tex]\( \frac{25}{6} \)[/tex] into a mixed number:
- Divide 25 by 6. The quotient is 4 and the remainder is 1.
- So, [tex]\( \frac{25}{6} = 4 \frac{1}{6} \)[/tex].
Therefore, Cody sold [tex]\( 4 \frac{1}{6} \)[/tex] more cakes than Abigail.
