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EBITDA
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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 12 Jul 2016, 01:38
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

38% (02:06) correct 62% (02:15) wrong based on 53 sessions

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If a and b are integers and 4a + b is an odd number, is a^2/b > 0?

(1) b^3 + 1 > 0
(2) a*b < 0
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C
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chisichei
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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 12 Jul 2016, 01:56
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If a and b are integers and 4a + b is an odd number, is (a^2) / b > 0?

1) b^3 + 1 > 0
This tell us that B> 0 but it doesnt tell us whether A <0 or >0 so Insufficient

2) a * b < 0
This tells us that either a or b is <0, if a is <0 then (a^2) / b > 0 but if a > 0 then (a^2) / b < 0 so insufficient

If you combine both then B>0 and A<0 then (a^2) / b > 0 sufficient
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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 12 Jul 2016, 02:35
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chisichei
If a and b are integers and 4a + b is an odd number, is (a^2) / b > 0?

1) b^3 + 1 > 0
This tell us that B> 0 but it doesnt tell us whether A <0 or >0 so Insufficient

2) a * b < 0
This tells us that either a or b is <0, if a is <0 then (a^2) / b > 0 but if a > 0 then (a^2) / b < 0 so insufficient

If you combine both then B>0 and A<0 then (a^2) / b > 0 sufficient
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from a) if we know that b is positive, then the ans should be A, because
a^2 is allways positive. positive/positive will always be greater than 0.
How can C be the answer then?

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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 12 Jul 2016, 02:39
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The question stem tells us that 4a + b is an odd integer.

Hence, either:

i) 4 * Even + Odd ---> Odd, or
ii) 4 * Odd + Odd ---> Odd

Therefore, b must be odd.

1) b^3 + 1 > 0 ---> This tells us that b must be a positive odd integer.

Given that a^2 will always be positive, regardless of the sign of a, a^2 / b must be positive. Therefore, statement 1 is sufficient.

However, the official source says that statement 1 is not sufficient.

Could someone please tell me why am I wrong?

2) a * b < 0

This can occur if either:

a is positive and b is negative ---> (a^2) / b < 0 ---> No
a is negative and b is positive ---> (a^2) / b > 0 ---> Yes

Therefore, statement 2 is not sufficient, and the answer is A.

Thank you so much.
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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 12 Jul 2016, 02:51
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EBITDA
The question stem tells us that 4a + b is an odd integer.

Hence, either:

i) 4 * Even + Odd ---> Odd, or
ii) 4 * Odd + Odd ---> Odd

Therefore, b must be odd.

1) b^3 + 1 > 0 ---> This tells us that b must be a positive odd integer.

Given that a^2 will always be positive, regardless of the sign of a, a^2 / b must be positive. Therefore, statement 1 is sufficient.

However, the official source says that statement 1 is not sufficient.

Could someone please tell me why am I wrong?

2) a * b < 0

This can occur if either:

a is positive and b is negative ---> (a^2) / b < 0 ---> No
a is negative and b is positive ---> (a^2) / b > 0 ---> Yes

Therefore, statement 2 is not sufficient, and the answer is A.

Thank you so much.
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eventhough we know that b is positive and a^2 is positive, we do not know whether a is equal to 0.
if a is 0, it would be 0/positive, which is equal to 0.
Therefore we need b, with a, to prove that a is not equal to 0 and that b is positive.
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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 12 Jul 2016, 03:59
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chisichei
If a and b are integers and 4a + b is an odd number, is (a^2) / b > 0?

1) b^3 + 1 > 0
This tell us that B> 0 but it doesnt tell us whether A <0 or >0 so Insufficient --> whetehr A<0 or >0 doesn't matter since A^2 will always be positive. What matters is taking A as 0( 0 being an even integer). if A is 0, A^2/B =0 , hence this statement is insufficient . because for A not equal to 0 AND b^3+1 >0, all a^2/B will always be >0 . hence two diff outcomes, hence insufficient .

2) a * b < 0
This tells us that either a or b is <0, if a is <0 then (a^2) / b > 0 but if a > 0 then (a^2) / b < 0 so insufficient--> good enough

If you combine both then B>0 and A<0 then (a^2) / b > 0 sufficient
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Combine both and A can never be 0, hence sufficient --> question took more than 2 mins to do but practice is the Key..
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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 12 Jul 2016, 04:11
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If a and b are integers and 4a + b is an odd number, is (a^2)/b > 0?

Notice two things:
a. 4a + b = even + b = odd, which implies that b = odd - even = odd.
b. (a^2)/b > 0 to hold true, a must not be 0 AND b must be positive.

(1) b^3 + 1 > 0 --> b^3 > -1 --> b can be 1, 3, 5, ... so, b is positive. We don't know whether a is 0 or not. Not sufficient.

(2) a * b < 0 --> a and b have different signs. Also, this statement implies that \(a \neq 0\). Not sufficient.

(1) From (1) we have that b is positive and from (2) we have that \(a \neq 0\) --> (a^2)/b > 0. Sufficient.

Answer: C.
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If a and b are integers and 4a + b is an odd number, is a^2/b > 0? 29 May 2019, 02:05
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