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a^3b^4c^7>0

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a^3b^4c^7>0 [#permalink] New post 18 Jul 2018, 17:00
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49% (00:42) correct 49% (00:50) wrong based on 28 sessions
\(a^3b^4c^7>0\)

Which of the following statements must be true?

Indicate all such statements.

A. \(ab\) is negative

B. \(abc\) is positive

C. \(ac\) is positive
[Reveal] Spoiler: OA

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Re: a^3b^4c^7>0 [#permalink] New post 18 Jul 2018, 19:53
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Carcass wrote:
\(a^3b^4c^7>0\)

Which of the following statements must be true?

Indicate all such statements.

A. \(ab\) is negative

B. \(abc\) is positive

C. \(ac\) is positive



the above statement will be true if any of the following statement are true,

1. a , b , c are all positive numbers

2. if a is +ve, b is -ve and c is +ve

3. if a is -ve, b is +ve and c is -ve

Therefore only option C is always true
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Re: a^3b^4c^7>0 [#permalink] New post 29 Aug 2018, 07:38
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Expert's post
Carcass wrote:
\(a^3b^4c^7>0\)

Which of the following statements must be true?

Indicate all such statements.

A. \(ab\) is negative

B. \(abc\) is positive

C. \(ac\) is positive


Key Concept: (any number)^(EVEN INTEGER) ≥ 0


First, since \(a^3b^4c^7>0\), we know that a ≠ 0, b ≠ 0, and c ≠ 0

Next, since \(b^4\) must be POSITIVE, we can safely divide both sides of the inequality by \(b^4\) to get: \(a^3c^7>0\)
Also, since \(c^6\) must be POSITIVE, we can safely divide both sides of the inequality by \(c^6\) to get: \(a^3c>0\)
Finally, since \(a^2\) must be POSITIVE, we can safely divide both sides of the inequality by \(a^2\) to get: \(ac>0\)

So, the ONLY relevant conclusion we can make is that \(ac>0\)

Answer: C

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Brent
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Re: a^3b^4c^7>0 [#permalink] New post 16 Nov 2018, 18:20
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Answer: C
A: if ab <0 then:
a<0 and b>0 : then c should be negative to have a^3*b^4*c^7 >0
a>0 and b<0 : then it’s ok because b’s power is even.
So A is possible but not obligatory.

B: it’s possible, But it’s not obligatory.

C: ac is positive: True
Because power of a and c are odd, they both should have same sign, both negative or positibve to have the whole expression positive.
Re: a^3b^4c^7>0   [#permalink] 16 Nov 2018, 18:20
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