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Question

In 2001 there were 6 members in Binod's family and their average age was 28 years. He got married between 2001 and 2004 and in 2004 there was an addition of a child in his family. In 2006, the average age of his family is 32 years.

What is the present age (in 2006) of Vinod wife (in years) is:

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What is the present age (in 2006) of Vinod wife (in years) is:

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fauji · Answer

Please post your question with options and also add the source of the question.

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BrushMyQuant · Answer

In 2001 there were 6 members in Binod's family and their average age was 28 years 
Using Sum = Average * Count
Sum of ages of 6 members = 28*6 = 168

He got married between 2001 and 2004 and in 2004 there was an addition of a child in his family 
Age of child in 2004=0

In 2006, the average age of his family is 32 years. 
Let Age of wife is x in 2006
Age of child in 2006 = 2006 - 2004 = 2 yrs
Sum of Ages of 6 members excluding wife and child in 2006 = Sum of their age in 2001 + (2006-2001)*6  
Sum of Ages of 6 members excluding wife and child in 2006 = 168 + 5*6 = 168 + 30 = 198
Sum of Ages of all family members of Binod in 2006 = Sum of Ages of 6 members excluding wife and child in 2006 + Age of wife in 2006 + Age of child in 2006
= 198 + x + 2 = 200 + x
=> In 2006, the average age of his family =  \frac{Sum}{Count}  =  \frac{200+x}{8}   
=> In 2006, the average age of his family =  \frac{200+x}{8}  = 32  
=> 200 + x = 32*8
=> x = 256-200 = 56

So, answer will be 56  
Hope it helps!

preeti95 · Answer

In the year 2006, we have the equation for ages:

(28+5) x 6 + W + 2 = 32 x 8

Solving this, we get W = 56

Hope this helps!

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