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Re: 3 Years ago, the average of a family of 5 members was [#permalink]
Here's my solution to this one =>
Mean 1=> 17
Mean 2=> 17+3 => 20
Mean 3=> 17
Hence Baby's age => 20-3*6 => 20-18=2

Hence the Baby is 2 years old today.
Hence C
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3 Years ago, the average of a family of 5 members was [#permalink]
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shriramvelamuri wrote:
3 Years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is

A. 1 year.

B. 1.5 Years.

C. 2 Years.

D. 3 Years.

E. 4 Years.


let b=age of baby
b+(5*17)+(5*3)=6*17
b=2 years
C
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Re: 3 Years ago, the average of a family of 5 members was [#permalink]
3 years Ago:

S/5 = 17
s= 17*5 ..... (1)

current age: A baby is born and avg age is same therefore: (S +15 +x)/6 = 17.... (2)
15 because all the members of the family would have grown older by 3 years hence 3+3+3+3+3 = 15

From 1 and 2
(17*5+15+x) = 17*6
x = 17(6-5) -15 = 2
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Re: 3 Years ago, the average of a family of 5 members was [#permalink]
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Solution:

Total age of the family 3 years ago was = 5 * 17 =85 years

Total age of the six members including the baby after 3 years = 17 *6 =102 years

The age of the family without the baby currently, should have been = 85 + 3*5 = 100 but due to the baby the it is 102

=>Age of the baby = 102-100=2 years (option c)

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Re: 3 Years ago, the average of a family of 5 members was [#permalink]
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Re: 3 Years ago, the average of a family of 5 members was [#permalink]
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