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Three members of a family Raina, Mike and Peter were born on the same

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Three members of a family Raina, Mike and Peter were born on the same  [#permalink]

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New post 06 Jan 2019, 10:05
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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
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Re: Three members of a family Raina, Mike and Peter were born on the same  [#permalink]

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New post 06 Jan 2019, 19:45
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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



So, what all do we know..
1) the ages are in GP
2) R=4P
3) P is the youngest..

So there can be two cases for a GP
A) R>M>P...4P>2P>P.
GP with r as 2..
B) M>R>P....16P>4P>P
GP with r as 4

Let us check the options

I. Mike is 16 times as old as Peter.
M can be 16 times P or 2 times P
Need not be true

II. Raina is the eldest
Mike can also be the eldest..
Need not be true

III. Sum of the ages of all three members is atleast 7.
P is atleast 1, so minimum sum of ages = 4P+2P+1P=7P=7*1=7
Yes, this will always be true

C
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Re: Three members of a family Raina, Mike and Peter were born on the same  [#permalink]

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New post 07 Jan 2019, 10:05
chetan2u wrote:
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



So, what all do we know..
1) the ages are in GP
2) R=4P
3) P is the youngest..

So there can be two cases for a GP
A) R>M>P...4P>2P>P.
GP with r as 2..
B) M>R>P....16P>4P>P
GP with r as 4

Let us check the options

I. Mike is 16 times as old as Peter.
M can be 16 times P or 2 times P
Need not be true

II. Raina is the eldest
Mike can also be the eldest..
Need not be true

III. Sum of the ages of all three members is atleast 7.
P is atleast 1, so minimum sum of ages = 4P+2P+1P=7P=7*1=7
Yes, this will always be true

C


Thanks for the detailed explanation. :thumbup: :heart
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Re: Three members of a family Raina, Mike and Peter were born on the same  [#permalink]

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New post 07 Jan 2019, 10:22
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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Re: Three members of a family Raina, Mike and Peter were born on the same &nbs [#permalink] 07 Jan 2019, 10:22
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