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# In a school outing with only adults and young children, the

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Joined: 22 Oct 2018
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In a school outing with only adults and young children, the [#permalink]  07 Apr 2020, 00:19
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Question Stats:

87% (01:42) correct 12% (02:38) wrong based on 8 sessions
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) $$\frac{1}{3}$$
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Joined: 09 Mar 2020
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Kudos [?]: 150 [1] , given: 16

Re: In a school outing with only adults and young children, the [#permalink]  07 Apr 2020, 00:28
1
KUDOS
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.
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Kudos [?]: 4647 [1] , given: 70

Re: In a school outing with only adults and young children, the [#permalink]  07 Apr 2020, 05:25
1
KUDOS
Expert's post
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) $$\frac{1}{3}$$

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: $$\frac{1}{5} = \frac{A}{C}$$

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.

Re: In a school outing with only adults and young children, the   [#permalink] 07 Apr 2020, 05:25
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