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Que: A fitness club has 40 male and 10 female members. The average (arithmetic mean) age of all of the members

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by Max@Math Revolution » Tue Nov 03, 2020 3:51 am

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Que: A fitness club has 40 male and 10 female members. The average (arithmetic mean) age of all of the members is 24 years. If the average age of the male members was 22 years, which of the following is the average age, in years, of the female members?

(A) 25
(B) 30
(C) 32
(D) 34
(E) 26
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Re: Que: A fitness club has 40 male and 10 female members. The average (arithmetic mean) age of all of the members

by Max@Math Revolution » Thu Nov 05, 2020 1:24 am
Solution: We have 40 males and 10 females. Hence, total members are: 40 + 10 = 50

Let the average age, in years, of the female members be x, then we get

=> \(\frac{40\ \cdot22\ +\ 10\ \cdot\ x}{40\ +\ 10}\ =\ 24\)

=> 40 * 22 + 10x = 24 * 50

=> 880 + 10x = 1,200

10x=1200 – 880 = 320 which means that the sum of the ages of 10 female members is 320

Also, we get 10x = 320, x = \(\frac{320}{10}=32\) , which means that the average age of these 10 female members is = 32.

C is the correct answer.

Answer C
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Re: Que: A fitness club has 40 male and 10 female members. The average (arithmetic mean) age of all of the members

by Scott@TargetTestPrep » Tue Nov 10, 2020 1:21 pm
Max@Math Revolution wrote: ↑
Tue Nov 03, 2020 3:51 am
 Que: A fitness club has 40 male and 10 female members. The average (arithmetic mean) age of all of the members is 24 years. If the average age of the male members was 22 years, which of the following is the average age, in years, of the female members?

(A) 25
(B) 30
(C) 32
(D) 34
(E) 26
Solution:

We can let f = the average age of the female members and calculate the weighted average:

(40 x 22 + 10f) / (40 + 10) = 24

(880 + 10f)/50 = 24

880 + 10f = 1200

10f = 320

f = 32

Answer: C

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