riteshpatnaik wrote:If a, b and c are integers, is abc an even integer?

(1) b is halfway between a and c

(2) a=b−c

I chose E. But the answer is B. The explanation given is

a=b−c . Re-arrange: a+c=b. Since it's not possible the sum of two odd integers to be odd, then the case of 3 odd numbers is ruled out, hence at least one of them must be even. Sufficient.

Please help

• 1. a and c are both ODD => ODD + ODD = EVEN => c is EVEN => ODD*ODD*EVEN = EVEN
  
  2. a and c are both EVEN => EVEN + EVEN = EVEN => c is EVEN => EVEN*EVEN*EVEN = EVEN
  
  3. One of a and c is ODD and the other is EVEN => ODD + EVEN = ODD => c is ODD => ODD*EVEN*ODD = EVEN

riteshpatnaik wrote:
Is xy≤1/2?

(1) x2+y2=1

(2) x2−y2=0

here also I chose E but the answer is A

Explanation given is (1) x2+y2=1. Recall that (x−y)2≥0(x−y)2≥0 (square of any number is more than or equal to zero). Expand: x2−2xy+y2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.

(2) x2−y2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.
Please help