voodoo_child wrote:what is the shortcut to do such problems in < 2 minutes? My mind started spinning while doing this problem under test conditions. I missed it because Choice A) seemed good to me. This was a killer problem.
Dear
Voodoo-Child
First of all, I'll say the way this post began, with the graph of |y|>|x|, is leagues away from the fastest way to answer this question.
If ab≠0 and |a|<|b|, which of the following must be negative?
A. (a/b)− (b/a)
B. (a−b)/(a+b)
C. a^b−b^a
D. a(b/(a−b))
E. (b−a)/b
The tricky thing about moving through this problem quickly is --- it involves
number-sense, a certain facility with jumping around among the possibilities -- here, shuffling back and forth quickly between positive and negative values.
For example, let a = +/-3, and b = +/-5. We will swap back and forth between positive and negative possibilities as it serves our interests.
In choice (a), if both are positive, then (3/5 - 5/3), small minus big, is negative. Therefore, I could make either a or b negative, and then it would be (-3/5 + 5/3), minus small plus big, which is positive. Be aware of when changing the sign of one or the other or both changes the sign of the answer. Choice (a) is out.
With (b), I notice that if b is positive in the numerator, then -b is negative in the denominator, and if b is negative, then -b is positive. The bigger positive plus or minus the smaller one (a) is still positive. The bigger negative plus or minus the smaller one (a) is still negative. This seems like it's going to be positive/negative or negative/positive no matter how we swap around the signs.
(c) is easy to make positive ===> make b an positive even number and a a negative number ------ negative to an even will be a big positive number, and even to a negative will be a fraction less than one. For example, let's say a = -2 and b = 4
Then a^b−b^a = (-2)^4 - 4^(-2) = 16 - 1/16, which is positive. Choice (c) is out.
For choice (d):
a(b/(a−b)) = (ab)/(a-b)
If both a & b are positive, or if both are negative, then the numerator is positive and the denominator is negative. If I make a positive and b negative --- remember, the opposite sign possibility which many test takers will overlook --- then
[(3)*(-5)]/[(3) + (-5)} = (-15)/(-2)
That gives negative/negative, which is positive. Choice (d) is out.
For choice (e):
(b−a)/b
Here if both are positive, a = 3 and b = 5, the answer is positive, so we can eliminate this right away. (e) is out.
It took me more than two minutes to type all that out, but I could see each one of those steps very quickly. The more you develop your number sense ---- which is
really what this question is testing --- the more fluent you will be in similar quick manipulations of possible choices.
Does this make sense?
Mike
