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# In a party with more than three children, the average age (arithmetic

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Retired Moderator
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In a party with more than three children, the average age (arithmetic  [#permalink]

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28 Jul 2015, 10:38
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63% (01:35) correct 38% (02:01) wrong based on 109 sessions

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In a party with more than three children, the average age (arithmetic mean) was 6. What is the average age of the remaining children in the party after three of the children have left?

(1) Each of the three children who left was of age 6.
(2) The average age of the three children who left was 6.

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Re: In a party with more than three children, the average age (arithmetic  [#permalink]

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28 Jul 2015, 10:51
2
reto wrote:
In a party with more than three children, the average age (arithmetic mean) was 6. What is the average age of the remaining children in the party after three of the children have left?

(1) Each of the three children who left was of age 6.
(2) The average age of the three children who left was 6.

reto, the OA is D : https://gmat.economist.com/gmat-practic ... s-the-type

Let the number of children be n

S be the sum of the ages of the children

Thus Mean = S/n= 6 --> S =6n

Statement 1, 3 children of ages, 6,6,6 left. Thus the mean of the remaining children = $$\frac{S-18}{n-3}$$ = $$\frac{6n-18}{n-3}$$ = 6. Thus sufficient.

Statement 2 is identical to statement 1 and is thus sufficient. Average of 3 children =6 ---> sum of the ages of these 3 children = 3*6 = 18.

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Re: In a party with more than three children, the average age (arithmetic  [#permalink]

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28 Jul 2015, 11:56
Engr2012 wrote:
reto wrote:
In a party with more than three children, the average age (arithmetic mean) was 6. What is the average age of the remaining children in the party after three of the children have left?

(1) Each of the three children who left was of age 6.
(2) The average age of the three children who left was 6.

reto, the OA is D : https://gmat.economist.com/gmat-practic ... s-the-type

Let the number of children be n

S be the sum of the ages of the children

Thus Mean = S/n= 6 --> S =6n

Statement 1, 3 children of ages, 6,6,6 left. Thus the mean of the remaining children = $$\frac{S-18}{n-3}$$ = $$\frac{6n-18}{n-3}$$ = 6. Thus sufficient.

Statement 2 is identical to statement 1 and is thus sufficient. Average of 3 children =6 ---> sum of the ages of these 3 children = 3*6 = 18.

Of course. My "typo"! Thanks.
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Re: In a party with more than three children, the average age (arithmetic  [#permalink]

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21 Jul 2018, 05:51
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Re: In a party with more than three children, the average age (arithmetic &nbs [#permalink] 21 Jul 2018, 05:51
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