Correct Answer:
C
To find branches with a net decrease in costs it will help to understand what is represented by each quadrant. Quadrant I: these branches had an increase over both periods (3 branches) Quadrant II: these branches had a decrease during the first period but an increase during the second (6 branches) Quadrant III: these branches had a decrease during both periods (1 branch) Quadrant IV: these branches had an increase during the first period but a decrease during the second (5 branches) None of the branches that have an increase during both periods will have a net overall decrease, so we can ignore the branches in Quadrant I (3 branches). All branches that have a decrease during both periods will have a net overall decrease, so we count all branched in Quadrant III (1 branch). Now we must look at the more complicated situations – an increase in one period but a decrease in the other. To get a net overall decrease, we need a larger dollar decrease than increase. For Quadrant II (decrease in first period but increase in second), we need the X-coordinate (period 1) to be larger than the Y-coordinate (period). This appears to be the case for 3 points: (-7, 5.5), (-15.5, 5.5), and (-16, 0.5). Finally, for Quadrant IV (decrease in the second period but increase in the first), we need the Ycoordinate (period 2) to have larger magnitude than the X-coordinate (period 1). This appears to be the case for 3 points (2.5, -4.5), (0.5, -7.5), and (11, -17). For all these points, the decrease outweighs the increase. So in total, we have 1+3+3=7 branches out of the 15, a number slightly less than ½ or 50%. Plug it into the calculator to get 7/15 = 0.4667 = 47%. As an alternative, you could find all the points satisfying the expression x+y<0, or y<-x. Mentally sketch the downward-sloping line y=-x, which runs through the origin. What points fall below the line? You'll find 7 points there. From here, the final calculation is the same.