How to pick values for such problems

gmatrant

Can you give me suggestions as to how to deal with such problems
if d>0 and 0<1-(c/d)<1, which of the following must be true
I.c>0
II. c/d <1
III. c^2 +d^2>1

A)I only
B) II only
C) I and II only
D) II and III only
E) I ,II, and III

OA is C

I am really stumped when I encounter such problems( more so when such problems come in DS). How to I pick values intelligently and attack such problems? Crying or Very sad

naren_nayak · Posted: Sun Oct 28, 2007 9:54 am Post subject:

I don't know if this is the easiest method, but I try to focus just on being compliant with the original conditions with values that go against the conditon mentioned in the individual questions.
In this case, the conditions given are that d is positive &
0 < 1 - (c/d) < 1
i.e. 0 < (d-c)/d < 1

For the first one, they're asking if c must be > 0.
Going against this condition, c can be 0 or negative.
If c=0, then (d-c)/d = 0 and the condition is not satisfied.
Choosing a negative value of c, our goal is to prove that there is a value of c that is compliant with the original condition ((d-c)/d)).
If we choose d=5 and c=-1, then (d-c)/d = (5-(-1))/5 = 6/5 which is greater than 1. So any negative value of c will end up with the numerator greater than the denominator, giving you a number always greater than 1. Now you might find a value of c that is positive and is not compliant with the original condition (e.g. d=5 and c=10 => (5-10)/5, which is negative), but that has nothing to do with the question. If there is a value of c to satisfy the original conditon, it must be positive. So I is true.

For the second condition, they're asking if c/d < 1, i.e. if c must be < than d.
So if we choose values such that c=d and c>d, we can check if the condition is still valid.
If c=d, (d-c)/d becomes 0 so the condition is not valid
If c>d, since d is positive, the end result is negative and the condition is not valid.
So c must be < than d, and II is true as well.

For the third condition, they're asking if c^2 + d^2 > 1. This one is a little tricky.
We're looking for values of c & d such that this condition is not true, but the original conditon is. Positive & negative values of c & d that are integers will not work. If we choose d=0.5 and c=0.25, c^2 + d^2 < 1 and (d-c)/d = 0.25/0.5, which is between 0 and 1, so we've found values of c & d such that III is not true, but the original condition is, So this condition does not have to be true.

Short answer => the answer is C Smile

samirpandeyit62 · Posted: Sun Oct 28, 2007 10:08 am Post subject:

Can you give me suggestions as to how to deal with such problems
if d>0 and 0<1-(c/d)<1, which of the following must be true
I.c>0

1-(c/d) d > 0 +ve

so here if c is -ve then c/d will get added to 1 making it greater than 1 NOT Possible, also if c=0 then 1-c/d = 1 which again voilates the inequality.

hence c>0 must be true

II. c/d <1

if c/d = 1 or greater then 1-c/d = 0 or -ve which voilates the ineuqlity so this also needs to be true.

III. c^2 +d^2>1

c/d < 1 now this will be true even if value of c > d or c is -ve i.e voilation condition + this will be false if both c & d are decmimal between 0 & 1