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# Given that ab< 0 and a > b, which of the following must be true?

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Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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18 Aug 2017, 23:20
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Given that $$ab< 0$$ and $$a > b$$, which of the following must be true?

I. $$a>0$$

II. $$b>0$$

III. $$\frac{1}{a} > \frac{1}{b}$$

A. I only
B. II Only
C. I and III Only
D. II and III Only
E. I, II and III

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Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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19 Aug 2017, 02:36
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susheelh wrote:
Given that $$ab< 0$$ and $$a > b$$, which of the following must be true?

I. $$a>0$$

II. $$b>0$$

III. $$\frac{1}{a} > \frac{1}{b}$$

A. I only
B. II Only
C. I and III Only
D. II and III Only
E. I, II and III

$$ab< 0$$ means that a and b have different signs. Since also given that a > b, then a > 0 > b (a and b have different signs).

So, I is true and II is not.

Next, 1/a = 1/(positive) = (positive) and 1/b = 1/(negative) = (negative), thus 1/a > 1/b. III is true too.

Hope it's clear.
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Re: Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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06 Feb 2020, 04:40
Bunuel wrote:
susheelh wrote:
Given that $$ab< 0$$ and $$a > b$$, which of the following must be true?

I. $$a>0$$

II. $$b>0$$

III. $$\frac{1}{a} > \frac{1}{b}$$

A. I only
B. II Only
C. I and III Only
D. II and III Only
E. I, II and III

$$ab< 0$$ means that a and b have different signs. Since also given that a > b, then a > 0 > b (a and b have different signs).

So, I is true and II is not.

Next, 1/a = 1/(positive) = (positive) and 1/b = 1/(negative) = (negative), thus 1/a > 1/b. III is true too.

Hope it's clear.

Hi Bunuel

I agree with the same answer but then I had a thought about the above highlighted 3rd option.
that suppose if we plugin values for a = 2 and b= -1; then 1/a= 1/2 will also be -2 if denominator and numerator shuffle and similarly RHS will become +ve too.

So how can we evidently say that 3rd statement MUST BE TRUE as well?
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Re: Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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06 Feb 2020, 04:45
Anurag06 wrote:
Bunuel wrote:
susheelh wrote:
Given that $$ab< 0$$ and $$a > b$$, which of the following must be true?

I. $$a>0$$

II. $$b>0$$

III. $$\frac{1}{a} > \frac{1}{b}$$

A. I only
B. II Only
C. I and III Only
D. II and III Only
E. I, II and III

$$ab< 0$$ means that a and b have different signs. Since also given that a > b, then a > 0 > b (a and b have different signs).

So, I is true and II is not.

Next, 1/a = 1/(positive) = (positive) and 1/b = 1/(negative) = (negative), thus 1/a > 1/b. III is true too.

Hope it's clear.

Hi Bunuel

I agree with the same answer but then I had a thought about the above highlighted 3rd option.
that suppose if we plugin values for a = 2 and b= -1; then 1/a= 1/2 will also be -2 if denominator and numerator shuffle and similarly RHS will become +ve too.

So how can we evidently say that 3rd statement MUST BE TRUE as well?

Not following you... If a = 2 and b= -1, then $$(\frac{1}{a} =\frac{1}{2})> (\frac{1}{b}=-1)$$. Anyway, as it's eplxained in the solution: 1/a is positive, since a is positive, and 1/b is negative, since b is negative: positive>negative.
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Re: Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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06 Feb 2020, 06:53
Bunuel I mean to say that inverse of fraction would result in opposite signs in both LHS and RHS when values a and b are assumed 2 and -3.
1/a = 1/2 and 1/b = 1/-3....if you reverse both sides it will be -2<3 instead of 1/2 > -1/3.. So it may not necessarily be MUST but CAN situation rather right?
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Re: Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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06 Feb 2020, 06:58
Anurag06 wrote:
Bunuel I mean to say that inverse of fraction would result in opposite signs in both LHS and RHS when values a and b are assumed 2 and -3.
1/a = 1/2 and 1/b = 1/-3....if you reverse both sides it will be -2<3 instead of 1/2 > -1/3.. So it may not necessarily be MUST but CAN situation rather right?

Could you please show values of a and b for which III is not true? I still cannot understand what you mean.
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Re: Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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06 Feb 2020, 07:14
1
inverse of fraction: there is a swap in denominator and numerator. No change in sign.
if 1/a = 2, then a = 2 not -2.

Anurag06 wrote:
Bunuel I mean to say that inverse of fraction would result in opposite signs in both LHS and RHS when values a and b are assumed 2 and -3.
1/a = 1/2 and 1/b = 1/-3....if you reverse both sides it will be -2<3 instead of 1/2 > -1/3.. So it may not necessarily be MUST but CAN situation rather right?

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Joined: 02 Sep 2009
Posts: 61396
Re: Given that ab< 0 and a > b, which of the following must be true?  [#permalink]

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06 Feb 2020, 09:03
1
1
GMATBusters wrote:
inverse of fraction: there is a swap in denominator and numerator. No change in sign.
if 1/a = 2, then a = 2 not -2.

Anurag06 wrote:
Bunuel I mean to say that inverse of fraction would result in opposite signs in both LHS and RHS when values a and b are assumed 2 and -3.
1/a = 1/2 and 1/b = 1/-3....if you reverse both sides it will be -2<3 instead of 1/2 > -1/3.. So it may not necessarily be MUST but CAN situation rather right?

Also, if you cross-multiply 1/2 > -1/3, you'll get -3 < 2 (flipping the sign because we multiplied by -3.)
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Re: Given that ab< 0 and a > b, which of the following must be true?   [#permalink] 06 Feb 2020, 09:03
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# Given that ab< 0 and a > b, which of the following must be true?

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