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Re: x,y,z are negative numbers- [#permalink]
Carcass wrote:
The numbers might be integers or fractions. Therefore, we do not know which quantity is greatest

D is the answer pretty straight without any calculation

regards


Can you justify with some examples. Still trying to wrap my head around with possible numbers to arrive at a solution
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Re: x,y,z are negative numbers- [#permalink]
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Lets assume x y z are int and all equal, x = y = z = -1

A = -3 and B = -1/3. Option B is greater

Now lets assume x y z are all fractions and all equal, x = y = z = -1/2

A = -3/2 and B = -2/3. Option B is greater

I believe correct ans is option B. Carcass, could you plz advise, I have a strong feeling that D is correct but just can't derive it.
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Re: x,y,z are negative numbers- [#permalink]
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AkkuJi wrote:
Lets assume x y z are int and all equal, x = y = z = -1

A = -3 and B = -1/3. Option B is greater

Now lets assume x y z are all fractions and all equal, x = y = z = -1/2

A = -3/2 and B = -2/3. Option B is greater

I believe correct ans is option B. Carcass, could you plz advise, I have a strong feeling that D is correct but just can't derive it.


you do not know iF they are negative and all equal or different numbers. IF they are integers or not

Suppose are -1 all the numbers. a is greater

-1/3 all the numbers and B is greater

D is the answer
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x,y,z are negative numbers- [#permalink]
1
Carcass wrote:
AkkuJi wrote:
Lets assume x y z are int and all equal, x = y = z = -1

A = -3 and B = -1/3. Option B is greater

Now lets assume x y z are all fractions and all equal, x = y = z = -1/2

A = -3/2 and B = -2/3. Option B is greater

I believe correct ans is option B. Carcass, could you plz advise, I have a strong feeling that D is correct but just can't derive it.


you do not know iF they are negative and all equal or different numbers. IF they are integers or not

Suppose are -1 all the numbers. a is greater

-1/3 all the numbers and B is greater

D is the answer


B has got to be the correct answer. Here is why:

First: x, y, z are negative numbers -- we will also assume that x, y, z are of different values
Second: we try for the case of negative integers, and negative fractions

First case, \(x=-1\), \(y=-2\), \(z=-3\)
quantity A: \(x + y + z = -6\)
quantity B: \(\frac{1}{x+y+z}=\frac{-1}{6}\) -- Bigger

Second case, First case, \(x=\frac{-1}{2}\), \(y=\frac{-3}{2}\), \(z=\frac{-5}{2}\)
quantity A: \(x + y + z = \frac{-9}{2}\)
quantity B: \(\frac{1}{x+y+z}=\frac{-2}{9}\) -- Bigger

The answer is (B)
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x,y,z are negative numbers- [#permalink]
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