Previous Year Questions

Ten people A, B, C, D, E, F, G, H, I and J live in a society.

The society consists of three colonies viz., East colony, West colony and North colony such that atleast two and atmost four people live in each colony.

No two persons were born in the same year.

Their present ages lie between 21 years and 35 years.

J lives in the same colony as D and H. F is 26 years old and lives in the west colony.

C lives neither in the west nor in the north colony.

B lives with C but with neither A nor E.

G lives in the East colony.

B lives with C but with neither A nor E.

A and H live in the different colonies.

I, who is 27 years old, lives with neither A nor E.

The sum of the ages of D and I is 61 years.

E is 31 years old.

The difference between the ages of C and F is 2 years.

H's age is an even number and only H is younger than C.

Only two people are younger than J.

The ages of the people living in the west colony are consecutive even numbers.

Here, J's age is an odd number. So, case 1 is invalid.

Therefore, we get:

The sum of the ages of all persons living in the north colony is equal to the sum of the ages of all persons living in the east colony.

People living in the north colony = J+D+H+E = 25+34+22+31 = 112.

People living in the east colony = B+C+G+I = B+24+G+27 = B+G+51

So, 112 = B+G+51

So, B+G = 112 - 51 = 61

So, B+G = 61 years.

B is older than G.

From the illustration, we can conclude that 29+32 = 61

So, G is 29 years old and B is 32 years old.

Therefore, the final arrangement is:

None of the persons is 33 years old.