iishim wrote:
Of the members in a certain club, 70 percent are women. Is the average (arithmetic mean) age of the members in the club greater than 45 years?
(1) The average age of the men in the club is 68 years.
(2) The average age of the women in the club is 65 years.
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Assume that the number of members of the club = \(x\)
Therefore
- the number of women in the club = \(0.7x\)
- the number of men in the club = \(0.3x\)
Statement 1(1) The average age of the men in the club is 68 years.
Assume that the average age of the women in the club = \(y\)
Total Age of men = 68 * 0.3 x = \(20.4x\)
Total Age of women = 0.7x * y = \(0.7xy\)
Average age = \(\frac{20.4x + 0.7xy}{x} = 20.4 + 0.7\frac{y}{x }\)
Hence, the average age depends on the ratio \(\frac{y}{x}\).
If \(\frac{y}{x}\) = 1 ⇒
Is the average (arithmetic mean) age of the members in the club greater than 45 years? →
NoIf \(\frac{y}{x}\) = 50 ⇒
Is the average (arithmetic mean) age of the members in the club greater than 45 years? →
YesStatement 1 alone is not sufficient to answer the question. Hence, we can eliminate A and D.
Statement 2(2) The average age of the women in the club is 65 years.
Assume that the average age of the men in the club = \(y\)
Total Age of men = y * 0.3 x = \(0.3xy\)
Total Age of women = 0.7x * 65 = \(45.5x\)
Average age = \(\frac{0.3xy + 45.5x}{x} = 45.5 + 0.3\frac{y}{x}\)
As \(\frac{y}{x}\) is always positive, we can conclude that the average age of the group is always greater than 45.5
Is the average (arithmetic mean) age of the members in the club greater than 45 years? →
Yes
Option B