Quantity I The sum of the ages of three friends A B and C is 42 years and the respective ratio of ag...........

Quantitative Aptitude ( CCS) Arithmatic inequalities

Quantity I: The sum of the ages of three friends A,B and C is 42 years, and the respective ratio of age of A and B is 3:2. If B is 7 years older than C. Find the age of C 3 years ago.

Quantity II: The average age of family of five members five years before was 32 years. In the meantime, a new baby was born in the family and present average age of family is half year less than the average age of family of five members, five years before. Find the present age of newborn baby.

I > II

I =< II

I >=II

I = II

I < II

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Explanation:

The average age of family of five members, five years before. Find the present age of newborn baby.
From I: Let the ages of A and B be 3x and 2x years resp.
Age of C = (2x-7) years
A+B+C = 42 years 3x + 2x + (2x-7) = 42
=> x = 7
Age of C = (2*7-7) = 7 years Age of C, 3 years before = 7 – 3 = 4 years
From II: Sum of age of family five years before = 32*5 = 160 years
Sum of age of present age of family (without baby) = 160 + 5*5 = 185 years
Sum of age of present age of family (with baby) = (32 – 0.5)*(5+1) = 189 years
Age of newborn baby = 189 – 185 = 4 years
I = II

Hence, option 4 is the correct answer.