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 Post subject: Applied Problems
PostPosted: Fri Jan 25, 2008 6:42 am 
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Joined: Sun Oct 14, 2007 6:41 am
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Anne is twice as old as Beth and Beth is four years younger to Cathy. Which of the following must be true five years hence?

I. Anne is older than Cathy.
II. Anne is twice as old as Beth.
III. Beth is four years younger to Cathy.

(1) Statement I only.
(2) Statement II only.
(3) Statement III only
(4) Statements II & III.
(5) Statements I & II.


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PostPosted: Sun Feb 17, 2008 4:37 pm 
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Joined: Wed Nov 21, 2007 5:33 am
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Anne is twice as old as Beth and Beth is four years younger to Cathy. Which of the following must be true five years hence?

I. Anne is older than Cathy.
II. Anne is twice as old as Beth.
III. Beth is four years younger to Cathy

Five years hence Beth will still be four years younger to Cathy. Statement III is definitely true.

Five years hence Anne will certainly not be twice as old as Beth, if Anne at present is twice as old as Beth. So, statement II is definitely false.

mmmm... statement I. If Anne is older than Cathy now, Anne will be older than Cathy five years hence. Is Anne older than Cathy. Not necessarily. If Anne is 2 years old now, then Beth is 1 year old and Cathy is 5 years old. So, Anne is younger to Cathy.

On the contrary if Anne is 10 years old now, then Beth is 5 years old now and Cathy is 9 years old now. Then Anne is older than Cathy. So, we cannot conclude if Statement I is true or not.

Correct answer according to me is choice (3).

What is the OA?


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PostPosted: Wed Mar 05, 2008 2:29 pm 
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Joined: Wed Jan 02, 2008 7:52 am
Posts: 29
Initially,
A = 2B
B+4 = C
=> A = 2 * (C-4) -----> EQ: 1

This implies that --- A = 2C - 8 and therefore the minimum value that C can take is 5, as for any value less than this, A will have a ZERO / Negative age which is not possible.

So, For C= 5, we have A = 2 which proves that A < C => Anne is NOT elder than Cathy now and thus after 5 yrs also.

So, stmt I is wrong, so is stmt II.

Thus, stmt III alone is TRUE. Hope this is clear.


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