Question

A turtle is 20 5/6 inches below the surface of a pond. It dives to a depth of 32 1/4 inches. What is the change in the turtle’s position?

Answer 1

In this question, the turtle doing a dive from an initial position(20 5/6 inch) into a final position( 32 1/4 inch) and you asked the change in the position. Change in position will be determined by dividing final position with the initial position. The calculation would be:

distance= final position - initial position
distance = 32 1/4 inch - 20 5/6 inch= 11 5/12 inch = 11.416 inch below the surface

Answer 2

The turtle is initially 20 5/6 inches below the surface. 
After which he dives to a depth of 32 1/4 inches
then the initial position below - 20 5/6 inch
final position - 32 1/4 inch
the difference in the positions can be calculated taking the difference between initial and final position 
this can be done with a simple subtraction of fractions 
difference = $32 \frac{1}{4} - 20 \frac{5}{6}$
when subtracting mixed fractions the whole numbers are first subtracted
  =  $(32 - 20) \frac{1}{4} - \frac{5}{6}$
  = $12\frac{1}{4} - \frac{5}{6}$
 we can now bring the denominators to a common denominator which is 12
then in the first fraction 1/4, denominator is 4 , to bring it to 12 we can multiply the whole fraction by 3
$\frac{1*3}{4*3} = \frac{3}{12}$
next fraction 5/6 , denominator is 6, to bring it  to 12 we multiply whole fraction by 2
$\frac{5*2}{6*2} = \frac{10}{12}$
now since  3/12 is smaller than 10/12 so we borrow a whole number from 12 outside the brackets, then $\frac{3}{12} + \frac{12}{12} = \frac{15}{12}$
then 
 = (12 - 1) ($\frac{15}{12} - \frac{10}{12}$)
  = 11 $\frac{5}{12}$
therefore change in turtles position is 11 5/12 inches