GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
There are 35 females in a club having an average age of f, and 45 ma
[#permalink]
30 Sep 2024, 00:42
30 Sep 2024, 00:42
1
Expert Reply
3
Bookmarks
00:00
Question Stats:
71% (02:30) correct
28% (02:11) wrong based on 7 sessions
Hide Show timer Statistics
There are 35 females in a club having an average age of ‘f’, and 45 males in the same club having an average age of ‘m’. ‘f’ is greater than ‘m’
Quantity A |
Quantity B |
Average age of all the 80 members of the club |
\(\frac{f+m}{2} \) |
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2024 Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Re: There are 35 females in a club having an average age of f, and 45 ma
[#permalink]
09 Oct 2024, 22:52
09 Oct 2024, 22:52
1
Averages are always tricky, find the sum of each ie sum/35=f hence sum age of females= 35*f
Quantity A would be: (35f+45m )/ 80
Plug in some values taking f>m in consideration (I just take simple values like 2 and 1)
Answer is B
boom!
Quantity A would be: (35f+45m )/ 80
Plug in some values taking f>m in consideration (I just take simple values like 2 and 1)
Answer is B
boom!
Re: There are 35 females in a club having an average age of f, and 45 ma
[#permalink]
15 Nov 2024, 16:28
15 Nov 2024, 16:28
1
Answer is B
Two important pieces of information:
1) There are more males in the club than females
2) The average age of the female club goers is higher than males.
We can determine the answer without calculations by noticing that Quality A is a weighted average and Quantity B is a unweighted average.
With unweighted averages (Option B), the average ages for males and females are given the same weight regardless of how many males and females there are in each group. It doesn't matter if there are 45 males or 35 females. It's taking an average of the average (average female age + average male age)/2.
With weighted averages (Option A), the calculation accounts for the number of individuals in each group (male and female). Since there are more males in the club and males have a lower average age, the average of all club members (male and female) will be pulled closed towards the average male age since there are more of them. As the average age of the males in the club is less than women, we know Quantity B is greater than Quantity A.
Example in practice w/smaller numbers and no club-goers:
- There is 1 boy (10) and 1 girl (14). The average age and weighted age are equal (10+14)/2 = 12 (average age is right in the middle of 10 and 14)
- 1 more boy enters the group who is also 10. The total average age decreases (10+10+14)/3 = 11.333. The weighted age stays the same (10+14)/2 = 12
- 1 more boy enters the group. They are also 10. The total average age decreases even further (10+10+10+14)/4 =11. The weighted age stays the same (10+14)/4 = 12.
Two important pieces of information:
1) There are more males in the club than females
2) The average age of the female club goers is higher than males.
We can determine the answer without calculations by noticing that Quality A is a weighted average and Quantity B is a unweighted average.
With unweighted averages (Option B), the average ages for males and females are given the same weight regardless of how many males and females there are in each group. It doesn't matter if there are 45 males or 35 females. It's taking an average of the average (average female age + average male age)/2.
With weighted averages (Option A), the calculation accounts for the number of individuals in each group (male and female). Since there are more males in the club and males have a lower average age, the average of all club members (male and female) will be pulled closed towards the average male age since there are more of them. As the average age of the males in the club is less than women, we know Quantity B is greater than Quantity A.
Example in practice w/smaller numbers and no club-goers:
- There is 1 boy (10) and 1 girl (14). The average age and weighted age are equal (10+14)/2 = 12 (average age is right in the middle of 10 and 14)
- 1 more boy enters the group who is also 10. The total average age decreases (10+10+14)/3 = 11.333. The weighted age stays the same (10+14)/2 = 12
- 1 more boy enters the group. They are also 10. The total average age decreases even further (10+10+10+14)/4 =11. The weighted age stays the same (10+14)/4 = 12.
