anon1 wrote:
At a certain committee meeting only associate professors and assistant professors are present. Each associate professor has brought 2 pencils and 1 chart to the meeting, while each assistant professor has brought 1 pencil and 2 charts. If a total of 10 pencils and 11 charts have been brought to the meeting, how many people are present?
A. 6
B. 7
C. 8
D. 9
E. 10
A pencil-bringing rate for associate professors is combined with a pencil-bringing rate for assistant professors to yield a pencil-bringing rate for the MIXTURE of professors.
Since this is a MIXTURE problem, an alternate approach is to solve with ALLIGATION.
Let X = the number of associate professors and Y = the number of assistant professors.
Pencil-bringing rate for X \(= \frac{2-pencils}{3-items-brought} = \frac{2}{3} = \frac{14}{21}\)
Pencil-bringing rate for Y \(= \frac{1-pencil}{3-items-brought} = \frac{1}{3} = \frac{7}{21}\)
Pencil-bringing rate for the MIXTURE of X and Y \(= \frac{10-pencils}{21-items-brought} = \frac{10}{21}\)
Since the fractions have been put over the same denominator, the alligation can be performed with only the numerators.
Step 1: Plot the 3 numerators on a number line, with the numerators for X and Y on the ends and the numerator for the mixture in the middle.X 14-----------10------------ 7 Y
Step 2: Calculate the distances between the numerators. X 14-----
4-----10-----
3-----7 Y
Step 3: Determine the ratio in the mixture. The ratio of X to Y is equal to the RECIPROCAL of the distances in red:
X:Y = 3:4
Since there are 3 X's for every 4 Y's -- and 3+4 = 7 -- the total number of people must be a multiple of 7.