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IMO D

Let total children be N
Avg. age = 6

Avg. age = Sum of age / No. of Children = 6
Sum of age = 6N


Reg: Avg. age when 3 children left.

(1) Each of the 3 children who left was of age 6.

Sum of ages of 3 children left = 18
New Avg. = (6N - 18)/ N-3 = 6

Sufficient.

(2) The average age of the 3 children who left was 6.
Sum of ages of 3 children who left = 3x6 = 18
New Avg. = (6N - 18)/ N-3 = 6

Sufficient.

Ans D - Both statement are sufficient alone.
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In a party with more than 3 children, the average age (arithmetic mean) was 6.

Let the total children in party be Y
Y is more than 3
So, Y = 3+X -------(1)
Let total of age be T.
Such that T/Y = 6
T = 6*Y
T = 6(3+X)
T = 18+6X -------(2)
Total = 18+6X

the average age of the remaining children in the party after 3 of the children have left

Statement (1) Each of the 3 children who left was of age 6:
=> Therefore,
=> T = 18+6X-3(6) => 18-18+6X
=> T = 6X
Remaining children at party = total no. - left children
=> 3+x-3 => x
=> therefore Average of remaining children = 6x/x = 6
(Sufficient)

(2) The average age of the 3 children who left was 6.
=> Average of 3 children = 6
=> Total/3 = 6
=> Total age of 3 children = 18
Therefore,
=> T = 18+6X-3(6) => 18-18+6X
=> T = 6X
Remaining children at party = total no. - left children
=> 3+x-3 => x
=> therefore Average of remaining children = 6x/x = 6
(Sufficient)

Answer is D

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Quote:
In a party with more than 3 children, the average age (arithmetic mean) was 6. What is the average age of the remaining children in the party after 3 of the children have left?

(1) Each of the 3 children who left was of age 6.
(2) The average age of the 3 children who left was 6.

n>3
avg=6

(1) sufic
6n-6*3/n-3=6n-18/n-3,
6(n-3)/n-3=6

(2) sufic
6n-6*3/n-3=6n-18/n-3,
6(n-3)/n-3=6

Ans (D)
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IMO D

We know that for N children, the average is 6.

(1) This says that 3 children who left were each 6 years old. Means their average age was 6. Also, the existing average is 6. Therefore, there is no change in the average .

Also, by formula we get:
S/n = 6 ---> S = 6n

(S-18)/(n-3) = ?
--> (6n-18)/(n-3)
--> 6.

Sufficient.

(2) It gives us the same information as A. Not exactly same, but the average is 6. So their leaving doesn't impact the original average.

Sufficient

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Considering each statement alone, we know the average age of the students who left the group.In both the cases, average will not change as the average age of the students who are leaving is equal to 6.
So, either statement is alone sufficient to answer the question.

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In a party with more than 3 children, the average age (arithmetic mean) was 6. What is the average age of the remaining children in the party after 3 of the children have left?

Let the total children = T.

Total age = T * 6.

Three children were left so total number of children Age T - 3

We need to find \(\frac{ ((Total age before leaving ) - (total age of left students) ) }{ (T - 3 )}\)



(1) Each of the 3 children who left was of age 6.

So total age of left students = 3 * 6 .

Req Avg = \(\frac{ ((Total age before leaving ) - (total age of left students) ) }{ (T - 3 )}\)
= \(\frac{(T *6- 3*6 ) }{ (T-3 )}\) = 6
Sufficient .

(2) The average age of the 3 children who left was 6.

So total age of left students = 3 * 6 .

Req Avg = \(\frac{ ((Total age before leaving ) - (total age of left students) ) }{ (T - 3 )}\)
= \(\frac{(T *6- 3*6 ) }{ (T-3 )}\) = 6
Sufficient .

ANs is D
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In a party with more than 3 children, the average age (arithmetic mean) was 6. What is the average age of the remaining children in the party after 3 of the children have left?

(1) Each of the 3 children who left was of age 6.

(2) The average age of the 3 children who left was 6.

for example n=5
sum = 30

1) 3*6 = 18
30-18=12
so 2 children with sum 12
average=6
sufficient

2) 3*6=18
30-18=12
same as above
sufficient

Ans D
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