Of the members in a certain club, 70 percent are women. Is the average

Question

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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Official Answer

B

Bunuel · Accepted Answer

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?

Assume there are 10 members in total, with 7 women and 3 men. We need to check whether the total sum of their ages is greater than 10 * 45 = 450.

(1) The average age of the men in the club is 68 years.

The average age of men is 68, so the total age of men is 3 * 68 = 204. We don’t know the sum of the ages of the 7 women, so we can’t determine if the total exceeds 450. Not sufficient.

(2) The average age of the women in the club is 65 years.

The average age of women is 65, so the total age of women is 7 * 65 = 455, which is already more than 450, giving a YES answer to the question. Therefore, the average age of the entire club must be greater than 45. Sufficient.

Answer: B.

gmatophobia · Answer

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?  →  No

If  \frac{y}{x}  = 50 ⇒  Is the average (arithmetic mean) age of the members in the club greater than 45 years?  →  Yes

Statement 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

Nalapatricia · Answer

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?
Total let say 100 employees, 70 are women and 30 are men.

(1) The average age of the men in the club is 68 years.
Since the majority of people are women we cannot determine the value of the mean. 
 INSUFFICIENT

(2) The average age of the women in the club is 65 years.

Total:100 employees
70 are women = (70 person)(65years) = 4550 as sum of ages
30 are men to evaluate the mens ayes let say the minimum age to evaluate the statement with the worse scenario, considerer 1 year as age of men: = (30 person)(1year) = as sum of ages

Therefore:
women sum age plus men ages = 4550+30 = 4580
Divide that amount by the total of people = 4580/100 = 45.80 and since it is bigger than 45 is  SUFFICIENT

Answer is B.

nabz2024 · Answer

Given that 70% are women, this implies 30% are men. Total average is given as 45 years old. This is a weighted averages problem.

Total Average  = (# of items1 * their average + # of items2 * their average) / (total # items)

Lets assign variables: age of women = w and avg age of men = m.

Equation is then: (.7 * w + .3 * m) / (.7 + .3) > 45

Question is then interpreted as: Is .7w + .3m > 45?

Statement 1) m = 68

.7w + .3(68) > 45 ?
.7w > 45 - .3(68) ? 
.7w > 45 - 20.4 ?

We see it's possible but we don't know average age of women. Cross out A and D

Statement 2) w = 65

(Don't assume this is equally insufficient as Statement 1!)

.7(65) + .3m > 45 ?
.3m > 45 - .7(65)

We can quickly see that 45 - .7(65) is a negative number, and there is no way that .3*m is a negative number. Therefore, since Statement 2 tells us that .7w + .3m is NOT greater than 45, this is sufficient information.

Answer B - the total average age of women and men at the club cannot be greater than 45 on the basis that the average of women is 65.

Robo_123 · Answer

cant this be the reason - majority of them are women, and the avg for women is 65. hence overall the value should be skewed towards that number?

Bunuel · Answer

While it’s true that because women make up the majority, the overall average will be closer to the women’s average than to the men’s, that fact alone doesn’t guarantee the overall average is above 45. The skew only tells us direction, not magnitude.

Check complete solution  HERE .

Of the members in a certain club, 70 percent are women. Is the average

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