Given: The time different students ran in a race as a team
[tex]\begin{gathered} B_{\text{arbara time}}=3\frac{3}{10}\min utes,D_{\text{ona time}}=2\frac{4}{5}\min utes \\ C_{indy\text{ time}}=x,N_{icole\text{ time}}=2\frac{1}{10}\min utes_{} \end{gathered}[/tex]To Determine:
a) Cindy's time if the total time is 11 3/5 minutes
This means that the addition of all their time gives 11 3/5 minutes. Therefore
[tex]3\frac{3}{10}+2\frac{4}{5}+x+2\frac{1}{10}=11\frac{3}{5}[/tex][tex]\begin{gathered} x+3+2+2+\frac{3}{10}+\frac{4}{5}+\frac{1}{10}=11\frac{3}{5} \\ x+7\frac{3+8+1}{10}=11\frac{3}{5} \\ x+7\frac{12}{10}=11\frac{3}{5} \\ x+8\frac{1}{5}=11\frac{3}{5} \\ x=11\frac{3}{5}-8\frac{1}{5} \\ x=11-8\frac{3}{5}-\frac{1}{5} \\ x=3\frac{3-1}{5} \\ x=3\frac{2}{5}\min utes \end{gathered}[/tex]2) The difference between the fastest student and the slowest student
It can be observed that the fastest students is Cindy and the slowest student is Nicole
Therefore, the difference would be
[tex]\begin{gathered} D_{ifference}=3\frac{2}{5}-2\frac{1}{10} \\ D_{ifference}=1\frac{4-1}{10} \\ D_{ifference}=1\frac{3}{10}\min utes \end{gathered}[/tex]3)
[tex]\begin{gathered} T_{he\text{ girls' total time }}=11\frac{3}{5}\min utes \\ S_{chool\text{ record time}}=12\frac{2}{5}\min \\ 11\frac{3}{5}\min utes<12\frac{2}{5}\min \end{gathered}[/tex]Since the girls's time is less than the record time, it can be concluded that the girls' did not break the record
The difference is
[tex]\begin{gathered} D_{ifference}=12\frac{2}{5}\min utes-11\frac{3}{5}\min utes \\ D_{ifference}=\frac{62}{5}\min utes-\frac{58}{5}\min utes \\ D_{ifference}=\frac{62-58}{5}\text{minutes} \\ D_{ifference}=\frac{4}{5}\text{minutes} \end{gathered}[/tex]Summary
1) x = 3 2/5 minutes
2) The difference between the fastest student and slowest student is 1 3/10 minutes
3) The girls' did NOT break the record, they were 4/5 minutes slower
