Gavin had 20 minutes to do a three-problem quiz.
He spent 9 3/4 minutes on problem 1.
He spent 3 4/5 minutes on problem 2.
How much time did he have left for problem 3?
We need to add the time spent on problem 1 and problem 2 and then subtract the sum from the total time.
[tex]20-(9\frac{3}{4}+3\frac{4}{5})[/tex]Simplify the above fraction
[tex]\begin{gathered} 20-(9\frac{3}{4}+3\frac{4}{5}) \\ 20-\lbrack(9+3)(\frac{3}{4}+\frac{4}{5})\rbrack \\ 20-\lbrack(12)(\frac{5\cdot3+4\cdot4}{20})\rbrack \\ 20-\lbrack(12)(\frac{15+16}{20})\rbrack \\ 20-\lbrack(12)(\frac{31}{20})\rbrack \\ 20-12\frac{31}{20} \\ 20-\frac{12\cdot20+31_{}}{20} \\ 20-\frac{271_{}}{20} \\ \frac{20\cdot20-1\cdot271}{20} \\ \frac{400-271}{20} \\ \frac{129}{20} \end{gathered}[/tex]Let us write the answer in mixed number
[tex]\frac{129}{20}=6\frac{9}{20}[/tex]Finally, convert the fractional part to seconds by multiplying the fractional part by 3
[tex]6\frac{9}{20}=6\frac{27}{60}[/tex]This means that there are 6 minutes and 27 seconds are left for problem 3.
[tex]6\frac{27}{60}[/tex]
