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- If N is an integer, and 5N
^{2}is even, then which of the following must be true? Select all that apply.

- N + 5 is even
- N
^{2}–4 is even - N + 8 is even
- N/2 is even
- N/2 is odd

Answer: B, C

The only way 5N^{2} is even is if N^{2} 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 N^{2} – 4 is equal to 0 and it is even

C is also true.

- If a
^{4}b^{3}c^{7}> 0, then which of the following must be true? Select all that apply

- bc > 0
- abc > 0
- ab < 0
- b/c > 0

Answer: A, D

a^{4}b^{3}c^{7} > 0 can be written as a^{4}.(bc)^{3}.c^{4} > 0

Now, a^{4} and c^{4} 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

- x > y; xy ≠ 0

Quantity A

x^{2}

x

Quantity B

y^{2}

y

- if Quantity A is greater
- if Quantity B is greater
- if the two quantities are equal
- if the relationship between the two quantities cannot be determined

Answer: 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

- N is an integer that leaves a remainder 1 when divided by 7

Quantity A

The remainder when N is divided by 3

The remainder when N is divided by 3

Quantity B

5

- if Quantity A is greater
- if Quantity B is greater
- if the two quantities are equal
- if the relationship between the two quantities cannot be determined

Answer: B

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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- 3a + 5
- 3a – 9
- 3a + 10
- 3a – 4
- 3a + 4

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