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A bag of 216 caramels is divided equally among a group of children.

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A bag of 216 caramels is divided equally among a group of children.  [#permalink]

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New post 09 May 2019, 18:39
1
3
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A
B
C
D
E

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  65% (hard)

Question Stats:

61% (02:45) correct 39% (02:57) wrong based on 77 sessions

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A bag of 216 caramels is divided equally among a group of children. If 6 children were added to the group, each child would receive 3 fewer caramels. How many children are in the group?

A) 36
B) 24
C) 20
D) 18
E) 12
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Re: A bag of 216 caramels is divided equally among a group of children.  [#permalink]

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New post 09 May 2019, 19:36
If there are n children then following equation can be formed.
[(216/n)-3] = [216/(n+6)]

This will give equation which can be solved to get n=18

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A bag of 216 caramels is divided equally among a group of children.  [#permalink]

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New post 16 Sep 2019, 07:46
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The answer choices in the question represent information directly asked for by the question, and the question could be solved by writing an equation, so Plug in the Answers.

The answer choices represent the number of children in the group. Start with choice C, 20. Because the question says that the bag of 216 caramels is to be divided equally among a group of children, divide 216 by 20. However, 20 does not divide 216 evenly, so the bag cannot be divided evenly among the children. Eliminate choice C.

Try one of the other two middle choices. Try choice B, 24. If there are 216 caramels to be divided amount 24 children, divide 216 by 24 to get. If 6 children were added to the group, each child would receive 6 fewer caramels. 6 more children would make 24 + 6 = 30 children. However, 30 does not divide 216 evenly, so eliminate choice B.

Three choices remain, so try the middle of the remaining choices, which is choice D, 18. If there are 18 children, then each child would receive caramels. If 6 children were to be added, there would be 18 + 6 = 24 children, each of whom would receive caramels.

Therefore, each child would receive 12 – 9 = 3 fewer candies, which is consistent with the information in the question.

The correct answer is choice D.


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Re: A bag of 216 caramels is divided equally among a group of children.  [#permalink]

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New post 09 Oct 2019, 12:09
energetics wrote:
A bag of 216 caramels is divided equally among a group of children. If 6 children were added to the group, each child would receive 3 fewer caramels. How many children are in the group?

A) 36
B) 24
C) 20
D) 18
E) 12

Given:
\(\frac{216}{n} = c\) and \(\frac{216}{n+6} = c - 3\)

Number properties solution:
1) 216 prime factored is 2^3 * 3^3, so our AC must have only factors of 2 and 3 (eliminate C)

2) Add n + 6:
A) 36 + 6 = 42
B) 24 + 6 = 30
D) 18 + 6 = 24
E) 12 + 6 = 18

n+6 must also have only factors of 2 and 3 to divide 216 evenly (eliminate A, B)

3) Notice in E) when n = 12, remaining factors are 2*3^2 = 18 (this is the number c and also n+6), HOWEVER 18 - 3 = 15, not 12, so E is out

4) Check D) if you have time:
n = 18 is 2*(3^2), remaining factors are (2^2)*3, c = 12
n+6 = 24, 3*(2^3), remaining factors are 3^2, c - 3 = 9, so 3^2, matches.

Might be easier to just check 216/9 = 24?

Algebra solution:
\(\frac{216}{n} = c\) and \(\frac{216}{n+6} = c - 3\)

1) \(\frac{216}{n+6} = \frac{216}{n - 3}\)

2) \(\frac{216}{n+6} = \frac{216 - 3n}{n}\)

3) \(216n = 216n + 6*216 - 3(n^{2}+6n)\)

4) \(0 = 6*216 - 3(n^{2}+6n)\)

5) \(3(n^{2}+6n) = 6*216\)

6) \(n^{2}+6n = 2*216\)

7) \(n^{2}+6n - 432 = 0\)

8) Factor, notice we ignore (n + root1) because n must be positive, also root 1 > root 2 because r1+r2=6 (positive) and r1*r2 = -432 (negative)

9) 432 prime factored --> (4^4)(3^3), notice 6 and -432 are even so we must have 2 even factors (multiples of 2 in each)

10) Try root 1 = 4*9 --> 36, root 2 = 4*3 --> 12, 36-12 = 24, too big, remove factor of 3 and add factor of 2 to first root

11) r1 = 8*3 --> 24, r2 = 9*2 --> 18, 24-18 = 6, so (n + 24)(n - 18) = 0

n = 18
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Re: A bag of 216 caramels is divided equally among a group of children.   [#permalink] 09 Oct 2019, 12:09
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