Question

Hector had 30 minutes to do a three-problem quiz. He spent 11 3 4 minutes on question A and 6 1 2 minutes on question B. How much time did he have left for question C?

Answer 1

Answer:

$11 \frac{3}{4}$ min

Step-by-step explanation:

In order to solve this problem, we must subtract the times he has already used from the time he has available, so we get:

$30-11 \frac{3}{4} - 6 \frac{1}{2}$

we can start solving this subtraction by turning all of the numbers into improper fractions so we get:

$\frac{30}{1}-\frac{11*4+3}{4}- \frac{6*2+1}{2}$

$\frac{30}{1}-\frac{47}{4}- \frac{13}{2}$

next, we can find a least common denominator. In this case it will be 4, so we need to turn all denominators into a 4, so we get:

$\frac{30*4}{1*4}-\frac{47}{4}- \frac{13*2}{2*2}$

$\frac{120}{4}-\frac{47}{4}- \frac{26}{4}$

and now we can subtract the numerators and copy the denominators so we get:

$\frac{120-47-26}{4}$

$\frac{47}{4}$

which can now be turned into a mixed number by dividing 47/4 which yields 11 with a remainder of 3, so the mixed number is:

$11 \frac{3}{4}$ min

and this is the amount of time he has available to solve the las problem.