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riteshpatnaik
- Junior | Next Rank: 30 Posts
- Posts: 29
- Joined: Wed Mar 09, 2016 12:55 am
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.
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
(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.
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
