Question

jayden, spencer, and miguel made paper airplanes and had a contest to see whose could fly the farthest. jaydens airplane flew $45\frac{1}{2}$ feet. spencers airplane flew $3\frac{1}{3}$ feet farther than jaydens. miguels airplane was the winner. it flew $7\frac{1}{4}$ feet farther than spencers airplane. how far did miguels airplane fly?

Explanation:

Step1: Find Spencer's distance

Jayden's distance is $45\frac{1}{3}$ feet. Spencer's airplane flew $3\frac{1}{3}$ feet farther than Jayden's. So we add $45\frac{1}{3}$ and $3\frac{1}{3}$.
First, convert mixed numbers to improper fractions: $45\frac{1}{3}=\frac{45\times3 + 1}{3}=\frac{136}{3}$, $3\frac{1}{3}=\frac{3\times3+1}{3}=\frac{10}{3}$.
Adding them: $\frac{136}{3}+\frac{10}{3}=\frac{146}{3}=48\frac{2}{3}$ feet.

Step2: Find Miguel's distance

Miguel's airplane flew $7\frac{1}{3}$ feet farther than Spencer's. So we add Spencer's distance ($48\frac{2}{3}$) and $7\frac{1}{3}$.
Convert to improper fractions: $48\frac{2}{3}=\frac{48\times3 + 2}{3}=\frac{146}{3}$, $7\frac{1}{3}=\frac{7\times3+1}{3}=\frac{22}{3}$.
Adding them: $\frac{146}{3}+\frac{22}{3}=\frac{168}{3}=56$ feet.

Answer:

56 feet