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.
In a school outing with only adults and young children, the
[#permalink]
Updated on: 22 Oct 2022, 08:06
Updated on: 22 Oct 2022, 08:06
5
1
Bookmarks
00:00
Question Stats:
31% (02:01) correct
68% (01:50) wrong based on 16 sessions
Hide Show timer Statistics
In a school outing with only adults and young children, the average weight of the adults is 60 kg and the average weight of the children is 24 kg. If the average weight of everyone on the school outing is 30 kg, what is the ratio of adults to children on this outing?
A) \(\frac{1}{12}\)
B) \(\frac{1}{6}\)
C) \(\frac{1}{5}\)
D) \(\frac{1}{4}\)
E) \(\frac{1}{3}\)
A) \(\frac{1}{12}\)
B) \(\frac{1}{6}\)
C) \(\frac{1}{5}\)
D) \(\frac{1}{4}\)
E) \(\frac{1}{3}\)
Re: In a school outing with only adults and young children, the
[#permalink]
07 Apr 2020, 00:28
07 Apr 2020, 00:28
2
We need to find the ratio and the average is given as 30.
Start with C, so that we can get an idea whether the number is more or less than that.
Here, there is 1 adult and 5 children.
1(60)+5(24) / 6
60+120 / 6
180 / 6 = 30, which is the mean weight.
Thus the answer is C.
Start with C, so that we can get an idea whether the number is more or less than that.
Here, there is 1 adult and 5 children.
1(60)+5(24) / 6
60+120 / 6
180 / 6 = 30, which is the mean weight.
Thus the answer is C.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In a school outing with only adults and young children, the
[#permalink]
07 Apr 2020, 05:25
07 Apr 2020, 05:25
2
workout wrote:
In a school outing with only adults and young children, the average weight of the adults is 60 kg and the average weight of the children is 24 kg. If the average weight of everyone on the school outing is 30 kg, what is the ratio of adults to children on this outing?
A) \(\frac{1}{12}\)
B) \(\frac{1}{6}\)
C) \(\frac{1}{5}\)
D) \(\frac{1}{4}\)
E) \(\frac{1}{3}\)
A) \(\frac{1}{12}\)
B) \(\frac{1}{6}\)
C) \(\frac{1}{5}\)
D) \(\frac{1}{4}\)
E) \(\frac{1}{3}\)
Let's solve this question using weighted averages
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...
Let \(A\) = the number of adults
Let \(C\) = the number of children
This means \(A+C\) = the total number of people
Also \(\frac{A}{A+C}\) = the proportion of adults and the group
And \(\frac{C}{A+C}\) = the proportion of children and the group
Substitute values into our equation to get: \(30 = (\frac{A}{A+C})(60) + (\frac{C}{A+C})(24)\)
Multiply both sides of the equation by \((A+C)\) to get: \(30(A+C) = 60A + 24C\)
Expand to get: \(30A+30C = 60A + 24C\)
Subtract \(30A\) from both sides: \(30C = 30A + 24C\)
Subtract \(24C\) from both sides: \(6C = 30A\)
Divide both sides by \(C\) to get: \(6 = \frac{30A}{C}\)
Divide both sides by \(30\) to get: \(\frac{6}{30} = \frac{A}{C}\)
Simplify: \(\frac{1}{5} = \frac{A}{C}\)
Answer: C
Cheers,
Brent
