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One of the toughest I encountered:Algebra + DS

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One of the toughest I encountered:Algebra + DS

by harsh.champ » Thu Feb 04, 2010 5:36 am
Mark (A) if the question can be answered by using statement (1) alone.
Mark (B) if the question can be answered by using statement (2) alone.
Mark (C) if the question can be answered by using either of the statements alone.
Mark (D) if the question can be answered by using both the statements together.
Mark (E) if the question cannot be answered even by using both the statements together.

If 'x', 'y' and 'z' are real numbers, then is the value of [(9/x) + 14] > [(14/y) + (20/z)]


1. [(1/x) + (2/y) + (4/z)] > 17


2. [(3/x) - (2/y)] > [(9 + 2/z)]
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by Ian Stewart » Thu Feb 04, 2010 10:52 am
harsh.champ wrote:Mark (A) if the question can be answered by using statement (1) alone.
Mark (B) if the question can be answered by using statement (2) alone.
Mark (C) if the question can be answered by using either of the statements alone.
Mark (D) if the question can be answered by using both the statements together.
Mark (E) if the question cannot be answered even by using both the statements together.

If 'x', 'y' and 'z' are real numbers, then is the value of [(9/x) + 14] > [(14/y) + (20/z)]


1. [(1/x) + (2/y) + (4/z)] > 17


2. [(3/x) - (2/y)] > [(9 + 2/z)]
The fractions might make this confusing; we don't need to work with them. Let's replace (1/x) with a, (1/y) with b and (1/z) with c. Then our question becomes:

Is 9a + 14 > 14b + 20c?

1. a + 2b + 4c > 17
2. 3a - 2b > 9 + 2c

What I notice first here is that there appears to be no direct relationship between the statements and the question, making it unlikely the statements will be sufficient; because of this, I'd start looking for example numbers to prove this. Further, if we add the inequalities, b will vanish, which means we have a lot of freedom in choosing a value for b. If a is a lot larger than b or c, then both statements will be true; for example, if:

a = 1000
b = 1
c = 1

both statements are true, and the answer to the question is 'yes'. On the other hand, if we make b large as well, we can make both statements true and get a 'no' answer:

a = 1000
b = 1000
c = 1

So the answer is E.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by harsh.champ » Thu Feb 04, 2010 3:39 pm
Ian Stewart wrote:
harsh.champ wrote:Mark (A) if the question can be answered by using statement (1) alone.
Mark (B) if the question can be answered by using statement (2) alone.
Mark (C) if the question can be answered by using either of the statements alone.
Mark (D) if the question can be answered by using both the statements together.
Mark (E) if the question cannot be answered even by using both the statements together.

If 'x', 'y' and 'z' are real numbers, then is the value of [(9/x) + 14] > [(14/y) + (20/z)]


1. [(1/x) + (2/y) + (4/z)] > 17


2. [(3/x) - (2/y)] > [(9 + 2/z)]
The fractions might make this confusing; we don't need to work with them. Let's replace (1/x) with a, (1/y) with b and (1/z) with c. Then our question becomes:

Is 9a + 14 > 14b + 20c?

1. a + 2b + 4c > 17
2. 3a - 2b > 9 + 2c

What I notice first here is that there appears to be no direct relationship between the statements and the question, making it unlikely the statements will be sufficient; because of this, I'd start looking for example numbers to prove this. Further, if we add the inequalities, b will vanish, which means we have a lot of freedom in choosing a value for b. If a is a lot larger than b or c, then both statements will be true; for example, if:

a = 1000
b = 1
c = 1

both statements are true, and the answer to the question is 'yes'. On the other hand, if we make b large as well, we can make both statements true and get a 'no' answer:

a = 1000
b = 1000
c = 1

So the answer is E.
_____________________
Thanks.
The method of replacing the reciprocals of the variable with another variable was fantastic.When I was solving this question,I ended up getting terms like [(yz+2xz+4xy)/xyz].I racked my brains but to no avail.
Applying the shortcut technique,the question seems quite easy.
Also,I wanted to ask:Here you have solved the question using the hit-and-trial technique(assuming values of a,b and c).
But earlier I was warned about using that technique as in some questions it can prove to be counter-productive.
Isn't there any formal method to solve the question??
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by sanju09 » Fri Feb 05, 2010 12:37 am
Isn't there any formal method to solve the question??
Taking Ian's reference, if we take it when all's said and done, we end up with the following inequality

2 a + c > 13

We can pay unnecessary time by setting up more and more inequalities in a, b, and c to no avail, if we ignore the basic algebraic fact of having at least n number of different inequalities in hand in order to answer an inequality problem that involves n unknown variables. The obtained inequality above cannot be treated as a different (third) inequality as it's been derived from the already known two inequalities. Secondly, there's nothing magical about the coefficients used with the variables that could make us figure out something concrete about the left and right hand sides of the main question, "Is 9 a + 14 > 14 b + 20 c?".

Ian must have used some arbitrary numbers just to prove that things won't remain same for all real numbers if we are not supplied more information.

Hence [spoiler]E[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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