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Answer: B, C
The only way 5N2 is even is if N2 is even, which means N is even
Clearly N + 5 is odd, eliminate A
Now if N = 6 then N/2 is NOT even eliminate D
Now if N = 4 then N/2 is NOT odd eliminate E
Don’t over think B even if N = 2 then N2 – 4 is equal to 0 and it is even
C is also true.
Answer: A, D
a4b3c7 > 0 can be written as a4.(bc)3.c4 > 0
Now, a4 and c4 must be positive so (bc)3> 0
An odd power of bc is positive so bc is definitely positive. Select A
If bc is positive then b and c are both positive or negative. Select D
Solution: Don’t jump to option A
Even though x > y, if x and y are negative then
-2 > -3 but (-2)2 < (-3)2
This double case tells us we cannot be sure of the relationship between the two quantities
Solution: This is a classic misdirection tactic.
N is an integer that leaves a remainder 1 when divided by 7 so N is of the form 7k + 1
But it’s irrelevant because when you divide N by 3 you can only get remainders of 0, 1, and 2 all of which are smaller than 5
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