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Today Rose is twice as old as Sam and Sam is 3 years younger [#permalink]

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28 Apr 2012, 15:01

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Today Rose is twice as old as Sam and Sam is 3 years younger than Tina. If Rose, Sam, and Tina are all alive 4 years from today, which of the following must be true on that day.

I. Rose is twice as old as Sam II. Sam is 3 years younger than Tina III. Rose is older than Tina

A. I only B. II only C. III only D. I and II E. II and III

Today Rose is twice as old as Sam and Sam is 3 years younger than Tina. If Rose, Sam, and Tina are all alive 4 years from today, which of the following must be true on that day?

I. Rose is twice as old as Sam II. Sam is 3 years younger than Tina III. Rose is older than Tina

A. I only B. II only C. III only D. I and II E. II and III

The official answer is C .

After reading the explanation also i am unable to understand how C[3 only] can be answer.

Answer cannot be C, it's B.

Notice that we are asked which of the following MUST be true not COULD be true.

Stem says that "Sam is 3 years younger than Tina" which naturally be exactly so ANY number of years from today. So, II is always true. Which means that the answer is B (II only), D (I and II only) or E (II and III only).

Analyze options I and III I. Rose is twice as old as Sam. III. Rose is older than Tina

Now, if today Rose is 4 years old, Sam is 2 years old and Tina is 5 years old then after 4 years Rose will be 8 years old, Sam is will be 6 years old and Tina will be 9 years old, so neither option I nor option III holds true.

Re: Today Rose is twice as old as Sam and Sam is 3 years younger [#permalink]

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28 Apr 2012, 16:35

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Thanks for responding.Apologies,answer is B.

Here is where I am not getting it:-

Let Tina's current age be X years. As per the question:- Tina = X yrs Sam= (X-3) yrs Rose = 2*Sam = 2(X-3)

After 4 yrs

Tina = X+4 yrs Sam = [(X-3)+4] yrs Rose = [2(X-3)+4] yrs

Now,I understand and agree option 1 is not correct in 4+ yr scenario. Option 2 is also fine,infact that is the most evident option.

The problem is option 3 "Rose is older than Tina". Just like in any case a person who is younger will always remain younger in age(case in point option 2); a person older will be older always. Also if you look at my equations abv. I have relation b/w Rose and Tina's age.For all values of X the end result will be +4 in Tina's age.

Let Tina's current age be X years. As per the question:- Tina = X yrs Sam= (X-3) yrs Rose = 2*Sam = 2(X-3)

After 4 yrs

Tina = X+4 yrs Sam = [(X-3)+4] yrs Rose = [2(X-3)+4] yrs

Now,I understand and agree option 1 is not correct in 4+ yr scenario. Option 2 is also fine,infact that is the most evident option.

The problem is option 3 "Rose is older than Tina". Just like in any case a person who is younger will always remain younger in age(case in point option 2); a person older will be older always. Also if you look at my equations abv. I have relation b/w Rose and Tina's age.For all values of X the end result will be +4 in Tina's age.

So,how option 3 is not correct?

Thanks a lot!

There is an example given in my post above proving that III in not always true.

In your algebraic approach in 4 years Rose is 2x-2 years old and Tina is x+4 years old. How is 2x-2 always more than is x+4? 2x-2>x+4 holds true for x>6, so if Tina is now 7 years old or older.
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Re: Today Rose is twice as old as Sam and Sam is 3 years younger [#permalink]

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15 Jun 2013, 21:18

My thoughts :

Clearly B is Correct, eg: think about siblings, your brother/sister is always older than you by x yrs, when u were born and also in 2013 Evaluating C : Rose is older than Tina in 4 yrs? , going by previous example, In 4 years, nothing will change. Rose is older than Tina if she is older than Tina today? So today, is Rose older than Tina? 2(Tina-3)>Tina? => is Tina>6? we dont know, so its not always true!

Re: Today Rose is twice as old as Sam and Sam is 3 years younger [#permalink]

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21 Sep 2013, 11:10

Bunuel wrote:

arkle wrote:

Thanks for responding.Apologies,answer is B.

Here is where I am not getting it:-

Let Tina's current age be X years. As per the question:- Tina = X yrs Sam= (X-3) yrs Rose = 2*Sam = 2(X-3)

After 4 yrs

Tina = X+4 yrs Sam = [(X-3)+4] yrs Rose = [2(X-3)+4] yrs

Now,I understand and agree option 1 is not correct in 4+ yr scenario. Option 2 is also fine,infact that is the most evident option.

The problem is option 3 "Rose is older than Tina". Just like in any case a person who is younger will always remain younger in age(case in point option 2); a person older will be older always. Also if you look at my equations abv. I have relation b/w Rose and Tina's age.For all values of X the end result will be +4 in Tina's age.

So,how option 3 is not correct?

Thanks a lot!

There is an example given in my post above proving that III in not always true.

In your algebraic approach in 4 years Rose is 2x-2 years old and Tina is x+4 years old. How is 2x-2 always more than is x+4? 2x-2>x+4 holds true for x>6, so if Tina is now 7 years old or older.

Sorry Bunnel, I got stuck in the algebra regarding Statement 3.

So if T = S+3 in 4 years then T = S+7

But in the case of R= 2S, in 4 years we cannot establish a proper relation right? Cause we are not given any real difference, just the ratio. So algebraically, how could we compare Tina and Rose to see what age does Sam have to be in order for one to be older than the other?

Let Tina's current age be X years. As per the question:- Tina = X yrs Sam= (X-3) yrs Rose = 2*Sam = 2(X-3)

After 4 yrs

Tina = X+4 yrs Sam = [(X-3)+4] yrs Rose = [2(X-3)+4] yrs

Now,I understand and agree option 1 is not correct in 4+ yr scenario. Option 2 is also fine,infact that is the most evident option.

The problem is option 3 "Rose is older than Tina". Just like in any case a person who is younger will always remain younger in age(case in point option 2); a person older will be older always. Also if you look at my equations abv. I have relation b/w Rose and Tina's age.For all values of X the end result will be +4 in Tina's age.

So,how option 3 is not correct?

Thanks a lot!

There is an example given in my post above proving that III in not always true.

In your algebraic approach in 4 years Rose is 2x-2 years old and Tina is x+4 years old. How is 2x-2 always more than is x+4? 2x-2>x+4 holds true for x>6, so if Tina is now 7 years old or older.

Sorry Bunnel, I got stuck in the algebra regarding Statement 3.

So if T = S+3 in 4 years then T = S+7

But in the case of R= 2S, in 4 years we cannot establish a proper relation right? Cause we are not given any real difference, just the ratio. So algebraically, how could we compare Tina and Rose to see what age does Sam have to be in order for one to be older than the other?

Thanks

Today Rose is twice as old as Sam --> R = 2S. Sam is 3 years younger than Tina --> T = S + 3

In four years: Tina = T + 4 = S + 7; Rose = R + 4 = 2S + 4.

The third option says: Rose is older than Tina --> 2S + 4 > S + 7 --> S > 3. So, the third option says that Sam is more than 3 years old. Do we know that? No. Thus this option is not always true.

Today Rose is twice as old as Sam and Sam is 3 years younger [#permalink]

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05 Aug 2014, 16:10

I: If we add the same number to both numerator and denominator, the ratio changes. So after 4 years the ratio of R/S= 2 will chnage. I is Not True. II: is quite clear. III: Let Sam be 1 yr old. Therefore T= 4 and R = 2. Thus III need not be true

Ans: B
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Re: Today Rose is twice as old as Sam and Sam is 3 years younger [#permalink]

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19 May 2016, 19:35

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Re: Today Rose is twice as old as Sam and Sam is 3 years younger [#permalink]

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19 May 2016, 19:53

R= 2S S= T-3

Proportion of their current age R:S:T= 2S:S:S+3

After 4 years R:S:T= 2S+4 :S+4 : S+7

Looking at the statements I. Rose is twice as old as Sam Not true II. Sam is 3 years younger than Tina True III. Rose is older than Tina What if S=1, 2 or 3 then no. But if S >3 then yes _________________

I welcome critical analysis of my post!! That will help me reach 700+

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Re: Today Rose is twice as old as Sam and Sam is 3 years younger
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19 May 2016, 19:53

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