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If a < 0 and b < 0, what are the possible values for
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14 Feb 2022, 05:15

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Question Stats:

If \(a < 0\) and \(b < 0\), what are the possible values for \(\frac{a+b}{ab }\) ?

Indicate all that apply.

A 2

B 1

C 0

D -1

E -2

Indicate all that apply.

A 2

B 1

C 0

D -1

E -2

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If a < 0 and b < 0, what are the possible values for
[#permalink]
14 Feb 2022, 06:42

1

Carcass wrote:

If \(a < 0\) and \(b < 0\), what are the possible values for \(\frac{a+b}{ab }\) ?

Indicate all that apply.

A 2

B 1

C 0

D -1

E -2

Indicate all that apply.

A 2

B 1

C 0

D -1

E -2

Useful (and often-tested) fraction property: \(\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}\)

When we apply the above property, we get: \(\frac{a+b}{ab } = \frac{a}{ab } + \frac{b}{ab } = \frac{1}{b} + \frac{1}{a}\)

Since \(a\) and \(b\) are both NEGATIVE, we know that \(\frac{1}{a}\) and \(\frac{1}{b}\) are both NEGATIVE, which means \(\frac{1}{a} + \frac{1}{b}\) must be NEGATIVE.

So we can immediately and eliminate answer choices A, B, and C.

Now recognize that, if \(a = -2\) and \(b = -2\), then \(\frac{1}{a} + \frac{1}{b}=\frac{1}{-2} + \frac{1}{-2} = \frac{2}{-2} = -1\), which means D is a possible answer.

Also, if \(a = -1\) and \(b = -1\), then \(\frac{1}{a} + \frac{1}{b}=\frac{1}{-1} + \frac{1}{-1} = (-1) + (-1) = -2\), which means E is a possible answer.

Answer: D and E

Re: If a < 0 and b < 0, what are the possible values for
[#permalink]
14 Feb 2022, 09:35

1

Carcass wrote:

If \(a < 0\) and \(b < 0\), what are the possible values for \(\frac{a+b}{ab }\) ?

Indicate all that apply.

A 2

B 1

C 0

D -1

E -2

Indicate all that apply.

A 2

B 1

C 0

D -1

E -2

Since a & b are both -ve. Denominator will always be +ve. Now neither a nor b can be =0 and will start with -1. So a/b=-1: (-1-1)/-1*-1= -2

Re: If a < 0 and b < 0, what are the possible values for
[#permalink]
15 Feb 2022, 03:36

This is soo cool! Now I can say that maths is fun. Never thought I would say it.

gmatclubot

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