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 Post subject: OG - DS - #139, #140
 Post Posted: Fri Aug 03, 2007 9:00 pm 
I read the solutions for DS questions 139 and 140. I solved using other techniques and can't figure out where I'm going wrong. If any one can explain. Here are my techniques

Problem 139 Page 289

I rephrased the question as below
Is (x -y) > (x+y)
Is -y > y ?
Statement 2 tells us that y is negative, hence -y is positive. So according to my method the ans is B, which is wrong per OG11.


Problem 140 Page 289
If points (r,s) and (u,v) are equidistance from the origin then, by midpoint formula
(r+u)/2 = 0 and (s+v)/2 = 0 Because origin (0,0) will be the midpoint.
Which means (r+u) = 0 and (s+v) = 0.
But, stem 2 tells us that r+u = 1 and s+v = 1. Hence the midpoint is (1,1). So the points are not equidistant. Ans B. which is wrong per OG11.

Please explain wot I'm doing wrong. Thanks!


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 Post subject:
 Post Posted: Sat Aug 04, 2007 10:49 am 
Regarding Q139:

I made a mistake once in a similar situation, where I converted the inequality x/y > 1 as x > y.
It is absolutely correct to make such a conversion, but in the process we need to make sure that both the conditions x/y > 1 and x > y is satisfied.

Example: Let us assume x = -3 and y = -4. In this case x > y, but x/y is < 1.

So when we simplify inequalities, you always have to be very careful and consider all the possible combinations of Positive and negative values for variables.

In your problem, take Statement (1), x > 0.

We can have multiple situations for variable y, which is as below:
a) y > 0 (y > x)
b) y > 0 (y <= x)
c) y = 0
d) y < 0 (with |y| > x)
e) y < 0 (with |y| < x)).

Please note you cannot have a situation where |x| = |y| when y is negative (In the question it is given x is not equal to -y)

If you work out examples for above situations, we can find that we will get inconsistent results for (x - y)/(x + y). In certain cases it could be > 1, but not in all cases. So INSUFFICIENT.

Statement (2): We can do the same thing by assuming various conditions for variable x, and we will get inconsistent results. So INSUFFICIENT.

If you combine (1) and (2), you are narrowing it down to situations (d) and (e) mentioned above. So you can work out an example for that.

Situation (d): Take x = 3 and y = -4. x - y = 7, x + y = -1 So (x - y) / (x + y) is -7 which is not greater than 1.
Situation (e): Take x = 3 and y = -2 x - y = 5, x + y = 1 So (x - y)/(x + y) is 5, which is greater than 1.


So even after combining (1) and (2), we cannot get a definite answer. So answer is (E), I guess.

I will give my explanation for Q140 in my next post, as my explanations are getting longer :)


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 Post subject:
 Post Posted: Sat Aug 04, 2007 11:01 am 
The mid point formula that you mentioned is for the Straight line that is formed by 2 points (r,s) and (u,v). But the question doesn't say that the points form a straight line. Hence your approach will not work in this situation.

Example: We can have a point (3,4) in I Quadrant and (-3,4) in II Quadrant, If you include the origin, then you are forming a shape like V, with 2 lines from origin (0,0) forming the two arms of the V. In fact the above points are equidistant from the origin, but (0,0) is not the mid point of the line joining (3,4) and (-3,4).

Hope the above example clarifies.

In this question, it is better to use the distance formula to figure out the answer. In fact, OG does a much better explanation, and hence I would request you to refer that.


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