Question

Gavin had 20 minutes to do a three problem quiz he spent 9 3/4 minutes on problem 1 and 3 4/5 minutes on problem 2 how much time did he have left for problem 3 write the answer in minutes and seconds how many minutes did he have left

Answer 1

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.

$20-(9\frac{3}{4}+3\frac{4}{5})$

Simplify the above fraction

$\begin{gathered} 20-(9\frac{3}{4}+3\frac{4}{5}) \\ 20-\lbrack(9+3)(\frac{3}{4}+\frac{4}{5})\rbrack \\ 20-\lbrack(12)(\frac{5\cdot3+4\cdot4}{20})\rbrack \\ 20-\lbrack(12)(\frac{15+16}{20})\rbrack \\ 20-\lbrack(12)(\frac{31}{20})\rbrack \\ 20-12\frac{31}{20} \\ 20-\frac{12\cdot20+31_{}}{20} \\ 20-\frac{271_{}}{20} \\ \frac{20\cdot20-1\cdot271}{20} \\ \frac{400-271}{20} \\ \frac{129}{20} \end{gathered}$

Let us write the answer in mixed number

$\frac{129}{20}=6\frac{9}{20}$

Finally, convert the fractional part to seconds by multiplying the fractional part by 3

$6\frac{9}{20}=6\frac{27}{60}$

This means that there are 6 minutes and 27 seconds are left for problem 3.

$6\frac{27}{60}$