Fraction Reduction 8C820A

1. The problem is to reduce each fraction to its simplest form by dividing numerator and denominator by their greatest common divisor (GCD). 2. The formula to reduce a fraction $\frac{a}{b}$ is: $$\frac{a}{b} = \frac{\cancel{d} \times a'}{\cancel{d} \times b'} = \frac{a'}{b'}$$ where $d = \gcd(a,b)$ and $a' = \frac{a}{d}$, $b' = \frac{b}{d}$. 3. We find the GCD for each fraction and simplify: - $\frac{5}{13}$: GCD(5,13)=1, so fraction is $\frac{5}{13}$. - $\frac{3}{7}$: GCD(3,7)=1, so fraction is $\frac{3}{7}$. - $\frac{4}{9}$: GCD(4,9)=1, so fraction is $\frac{4}{9}$. - $\frac{11}{14}$: GCD(11,14)=1, so fraction is $\frac{11}{14}$. - $\frac{60}{135}$: GCD(60,135)=15 $$\frac{60}{135} = \frac{\cancel{15} \times 4}{\cancel{15} \times 9} = \frac{4}{9}$$ - $\frac{60}{156}$: GCD(60,156)=12 $$\frac{60}{156} = \frac{\cancel{12} \times 5}{\cancel{12} \times 13} = \frac{5}{13}$$ - $\frac{110}{140}$: GCD(110,140)=10 $$\frac{110}{140} = \frac{\cancel{10} \times 11}{\cancel{10} \times 14} = \frac{11}{14}$$ - $\frac{45}{105}$: GCD(45,105)=15 $$\frac{45}{105} = \frac{\cancel{15} \times 3}{\cancel{15} \times 7} = \frac{3}{7}$$ - $\frac{22}{28}$: GCD(22,28)=2 $$\frac{22}{28} = \frac{\cancel{2} \times 11}{\cancel{2} \times 14} = \frac{11}{14}$$ - $\frac{27}{63}$: GCD(27,63)=9 $$\frac{27}{63} = \frac{\cancel{9} \times 3}{\cancel{9} \times 7} = \frac{3}{7}$$ - $\frac{28}{63}$: GCD(28,63)=7 $$\frac{28}{63} = \frac{\cancel{7} \times 4}{\cancel{7} \times 9} = \frac{4}{9}$$ - $\frac{55}{143}$: GCD(55,143)=11 $$\frac{55}{143} = \frac{\cancel{11} \times 5}{\cancel{11} \times 13} = \frac{5}{13}$$ - $\frac{42}{98}$: GCD(42,98)=14 $$\frac{42}{98} = \frac{\cancel{14} \times 3}{\cancel{14} \times 7} = \frac{3}{7}$$ - $\frac{55}{70}$: GCD(55,70)=5 $$\frac{55}{70} = \frac{\cancel{5} \times 11}{\cancel{5} \times 14} = \frac{11}{14}$$ - $\frac{6}{14}$: GCD(6,14)=2 $$\frac{6}{14} = \frac{\cancel{2} \times 3}{\cancel{2} \times 7} = \frac{3}{7}$$ - $\frac{10}{26}$: GCD(10,26)=2 $$\frac{10}{26} = \frac{\cancel{2} \times 5}{\cancel{2} \times 13} = \frac{5}{13}$$ - $\frac{40}{90}$: GCD(40,90)=10 $$\frac{40}{90} = \frac{\cancel{10} \times 4}{\cancel{10} \times 9} = \frac{4}{9}$$ - $\frac{99}{126}$: GCD(99,126)=9 $$\frac{99}{126} = \frac{\cancel{9} \times 11}{\cancel{9} \times 14} = \frac{11}{14}$$ - $\frac{15}{35}$: GCD(15,35)=5 $$\frac{15}{35} = \frac{\cancel{5} \times 3}{\cancel{5} \times 7} = \frac{3}{7}$$ - $\frac{20}{52}$: GCD(20,52)=4 $$\frac{20}{52} = \frac{\cancel{4} \times 5}{\cancel{4} \times 13} = \frac{5}{13}$$ - $\frac{12}{28}$: GCD(12,28)=4 $$\frac{12}{28} = \frac{\cancel{4} \times 3}{\cancel{4} \times 7} = \frac{3}{7}$$ - $\frac{121}{154}$: GCD(121,154)=11 $$\frac{121}{154} = \frac{\cancel{11} \times 11}{\cancel{11} \times 14} = \frac{11}{14}$$ - $\frac{15}{39}$: GCD(15,39)=3 $$\frac{15}{39} = \frac{\cancel{3} \times 5}{\cancel{3} \times 13} = \frac{5}{13}$$ - $\frac{33}{77}$: GCD(33,77)=11 $$\frac{33}{77} = \frac{\cancel{11} \times 3}{\cancel{11} \times 7} = \frac{3}{7}$$ 4. After reducing, the fractions correspond to the bins as follows: - Bin $\frac{5}{13}$: $\frac{5}{13}$, $\frac{60}{156}$, $\frac{55}{143}$, $\frac{10}{26}$, $\frac{20}{52}$, $\frac{15}{39}$ - Bin $\frac{3}{7}$: $\frac{3}{7}$, $\frac{45}{105}$, $\frac{27}{63}$, $\frac{42}{98}$, $\frac{6}{14}$, $\frac{15}{35}$, $\frac{12}{28}$, $\frac{33}{77}$ - Bin $\frac{4}{9}$: $\frac{4}{9}$, $\frac{60}{135}$, $\frac{28}{63}$, $\frac{40}{90}$ (twice) - Bin $\frac{11}{14}$: $\frac{11}{14}$, $\frac{110}{140}$, $\frac{22}{28}$, $\frac{55}{70}$, $\frac{99}{126}$, $\frac{121}{154}$ This sorting helps clean up the grounds by placing each paper scrap in the correct bin based on its reduced fraction. Final answer: All fractions are reduced and sorted into bins labeled $\frac{5}{13}$, $\frac{3}{7}$, $\frac{4}{9}$, and $\frac{11}{14}$ accordingly.