How in the world?

pick the specific numbers . OA is 3 only

you have got to pick nos. and check

case 1 : both x&y are positive
assume x=6;y=-2.
then x-y=6-(-2)=8

Hence 1 is not the answer.

Case 2 : If x is positive ,y must be positive
Not necessary.. from the example above

Case 3: If x is negative y must be negative
Assume x=-7;then y has to be negative for the answer to be positive.
since only then will the sins cancel out for the no. to be positive.
So in this case if y=-15
x-y= -7-(-15) = 8

Hence option 3 is right

I agree with the above - the best way to do this (and most other problems involving variables) is to pick numbers.

This question tells us that x-y=8. So, we can pick numbers such as:

x=10, y =2
x = 4, y = -4
x = -2, y=-10

etc. etc.

Then, plug these numbers into the answer choices and eliminate ones that come out "false". You may have to do multiple iterations of this before you arrive at the correct answer.

Take case I : Both x and y are positive ----- Need not be, Y can be negative eg. x=4 and y=-4 => x-y = 4+4 = 8

Take Case II : If x is positive, y must be positive ----- Disproved above

Case III - If x is negative, y must be negative ----- If x is negative and y is positive, then -x-y will give a negative result. This shows that if x is negative, y must be negative so that the eqn becomes -x+y which gives us a possibility of having positive result.

Right answer is C. III only

HTH

Amit

Thanks Guys! time consuming for sure

notgoodinmath wrote:Thanks Guys! time consuming for sure

Most roman numeral questions are susceptible to a strategic attack - GMAC wants to reward strategic thinkers!

Start by looking at the choices and (usually) begin with the roman numeral that appears most frequently. As you deal with each roman numeral, eliminate choices as you go.

In this question, II is the most frequent among the choices. Once we see that II is not a MUST be true, we eliminate (b), (d) and (e) - worst case, we have a 50/50 shot.

The remaining choices are:

(a) I only; and
(c) III only.

Since they appear an equal number of times, we just work with the one that seems easier, which in this case will be I, since in all likelihood the numbers you picked to show that II doesn't have to be true will also show that I doesn't have to be true.

Once we elminate I (and (a)), the only answer choice left is (c) III only. We can confidently pick (c) without even thinking about III.

So, if you attacked this question strategically, you probably only needed to pick 1 set of numbers.

This is such an important strategy for saving time.
Thanks Stuart!! I've been spending awful lot of time in such problems.