1993 AJHSME Problems/Problem 5

Revision as of 01:54, 2 June 2011 by Eaglehawk7 (talk | contribs) (Created page with '==Solution== Since the white portion of the graph is twice as much as the black and gray, its bar in the bar graph will need to be twice as much as the black and gray. Note that…')
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Solution

Since the white portion of the graph is twice as much as the black and gray, its bar in the bar graph will need to be twice as much as the black and gray. Note that the black and gray portions are equal. This cancels choices $A, B, D, E$. The final answer is now $\boxed{C}$.