akurathi12 wrote:

Three members of a family Raina, Mike and Peter were born on the same date in different years. This year on their birthday, the ages of Raina, Mike and Peter will form a geometric progression series, not necessarily in the same order. Raina is 4 times as old as Peter whereas Peter is the youngest amongst the three. Which of the following must be true?

I. Mike is 16 times as old as Peter.

II. Raina is the eldest

III. Sum of the ages of all three members is atleast 7.

A. I only

B. I&II only

C. III only

D. I,II&III

E. None of these

Given, Raina = R, Mike = M and P= Peter. R, M and P are three members of a family.

R = 4P

P is the youngest of all

Now the series Can be in the order of P M R (1) or P R M (2)

To Elaborate on the series, values can be from a GP in which common ratio is 2 or 4

Now series (1) can be 1(P), 2(M) , 4(R) and series (2) can be 1(P), 4(R), 16(M)

Now to answer this question

I. Mike is 16 times as old as Peter. - Can be true from (2), cannot be true from (1)

II. Raina is the eldest -> Can be true from (1), cannot be true from (2)

III. Sum of the ages of all three members is atleast 7 -> This will always be true Series (1), Sum = 7 and Series(2), Sum = 21

Making the

Correct Answer as C

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