Respuesta :
Answer:
No.
Step-by-step explanation:
It is given that Darius is putting tiles on the roof of his house.
Darius had completed = [tex]$\frac{4}{5}$[/tex] th of the work
He takes time = [tex]$3 \frac{1}{3}$[/tex] hours to complete [tex]$\frac{4}{5}$[/tex] th of the work
The remaining work = [tex]$1 -\frac{4}{5}$[/tex]
[tex]$=\frac{5-4}{5}$[/tex]
[tex]$=\frac{1}{5}$[/tex]
∴ Darius completes [tex]$\frac{4}{5}$[/tex] th of the work in = [tex]$\frac{10}{3}$[/tex] hours
So, [tex]$\frac{1}{5}$[/tex] th of the work in = [tex]$\frac{10}{3} \times \frac{5}{4} \times \frac{1}{5}$[/tex] hours
[tex]$=\frac{10}{12}$[/tex] hours
Therefore total time taken to complete the work [tex]$=\frac{10}{3}+\frac{10}{12}$[/tex] hours
[tex]$=\frac{40+10}{12}$[/tex] hours
[tex]$=\frac{50}{12}$[/tex] hours
[tex]$=4\frac{2}{12}$[/tex] hours
[tex]$=4\frac{1}{6}$[/tex] hours
Thus, Darius continues to work at the same rate will not be able to complete the work in 4 hours.