This question comes from the Quest for 750 Quant Homework Set. I have a question about the answer explanation and if anyone has any suggestions on how to solve this problem quickly (possibly by plugging in values?). (The answer is C)
If a and b are distinct integers and their product is not equal to zero, is a > b?
(1) (a(^3)b - b(^3)a)/(a(^3)b + b(^3)a - 2a(^2)b(^2)) < 0
(2) b < 0
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(1) Rephrased: (a + b)/(a - b) < 0
The answer explanation then instructs the following computations:
So statement (1) simplifies to (a + b)/(a - b) < 0. If a - b is positive (in other words, if a > b), we can simplify even further:
IF a > b:
a + b < a - b (MY MAIN CONCERN: HOW IS IT POSSIBLE TO MULTIPLY (a-b) by 0?)
b < -b
2b < 0
b < 0
This tells us that b < 0 whenever a > b. The converse of this statement is also true. Whenever b < 0, a must be greater than b for the inequality to hold (try plugging sample values into (a + b)/(a - b) < 0 to verify that when b is negative, an a value smaller than b fails to satisfy the inequality).
However, if a - b is negative (in other words, if a < b), we must flip the sign when multiplying through by a - b:
IF a < b:
a + b > a - b (MY MAIN CONCERN: HOW IS IT POSSIBLE TO MULTIPLY (a-b) by 0?)
b > -b
2b > 0
b > 0
This tells us that b > 0 whenever a < b. The converse of this statement is also true. Whenever b > 0, a must be less than b for the inequality to hold (try plugging sample values into (a + b)/(a - b) < 0 to verify that when b is positive, a values greater than b fail to satisfy the inequality).