MathRevolution wrote:
DHAR wrote:
Is |a| - |b| < |a-b| ?
Statement 1: \(b^a\)<0
Statement 2: |b|>|a|
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.
|a| - |b| < |a-b|
⇔ |a| < |a-b| + |b|
⇔ |a-b + b| < |a-b| + |b|
⇔ |a-b + b|^2 < (|a-b| + |b|)^2 since both sides are non-negative in the above inequality.
⇔ (a-b + b)^2 < (|a-b| + |b|)^2
⇔ (a-b)^2 + 2b(a-b) + b^2 < |a-b|^2 + 2|a-b||b| + |b|^2
⇔ (a-b)^2 + 2b(a-b) + b^2 < (a-b)^2 + 2|a-b||b| + b^2
⇔ 2b(a-b) < 2|(a-b)b|
⇔ b(a-b) < |(a-b)b|
⇔ b(a-b) < 0
Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider 1) & 2) first.
b = -2, a = 1 : Yes
b = -4, a = -3 : No
Thus, both conditions 1) & 2) are not sufficient.
Therefore, E is the answer.
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Dear
MathRevolution,
My colleague, with all due respect, I am going to disagree on a few points.
You first made three opening statements with which I disagree
1)
Forget conventional ways of solving math questions.
2)
For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem.
3)
Remember that equal numbers of variables and independent equations ensure a solution.
In response to #1, I would assert that the heart of advanced problem-solving is elegance. In response to #2, I would say that the balance of your post seems a rebuttal to this statement. In response to #3, I would say--yes, this is a good rule, but the caveat should be mention that if two equations are dependent, we are not guaranteed a solution.
I studied your algebra carefully. I don't find a single step that's off: your inequality may be equivalent to the prompt inequality.
b(a-b) < 0
Your claim is that this is equivalent to the prompt inequality. I cannot gainsay that claim.
Then you picked two pairs of numbers:
Pair #1:
b = -2, a = 1 Pair #2:
b = -4, a = -3 But here's the thing. Both of these pairs satisfy your inequality. In fact, I would submit that it's impossible to find a pair of numbers consistent with Statement #2 that doesn't satisfy your inequality.
Both of these pairs definitely satisfy the prompt inequality.
Prompt Inequality: |a| - |b| < |a-b|
Pair #1:
|a| - |b| = |1| - |-2| = -1
|a-b| = |1 - (-2)| = |3| = +3
Prompt inequality works.
Pair #1:
|a| - |b| = |-3| - |-4| = 3 - 4 = -1
|a-b| = |-3 - (-4)| = |1| = +1
Prompt inequality works.
If both pairs satisfy both your inequality and the prompt inequality, they do not constitute counterexamples. I make the claim that counterexamples consistent with Statement #2 do not exist. Please provide one if I am mistaken in this claim.
Finally, I will say that the great irony is that this problem can be resolved an astonishingly elegant solution.
Statement 1 can be dismissed with numerical counterexamples
Statement 2: |b|>|a|
Since these are not equal, we know a ≠ b, so |a - b| > 0
|b|>|a| is equivalent to the inequality 0 > |a| - |b|
Combining, we get |a| - |b| < 0 < |a - b|, which directly implies |a| - |b| < |a - b|.
With remarkable elegance, Statement #2 directly implies the prompt inequality and is sufficient. OA =
(B).
My colleague, I am open to anything you would like to say in reply.
With sincere respect,
Mike McGarry
I agree with your mentioned points.