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If the reciprocal of the negative integer x is greater than the sum of

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If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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New post 13 Aug 2018, 04:41
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If the reciprocal of the negative integer x is greater than the sum of y and z, then which of the following must be true?


(A) \(x > y + z\)

(B) y and z are positive.

(C) \(1 > x(y + z)\)

(D) \(1 < xy + xz\)

(E) \(\frac{1}{x} > z – y\)

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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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New post 13 Aug 2018, 09:39
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Bunuel wrote:
If the reciprocal of the negative integer x is greater than the sum of y and z, then which of the following must be true?


(A) \(x > y + z\)

(B) y and z are positive.

(C) \(1 > x(y + z)\)

(D) \(1 < xy + xz\)

(E) \(\frac{1}{x} > z – y\)


Given \(\frac{1}{x}>(y+z)\), where x<0
Multiplying 'x' both sides, (inequality sign to be flipped since x is negative)
\(x*\frac{1}{x}>x(y+z)\)
Or, \(1 < xy+xz\)

Ans. (D)
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If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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New post 14 Aug 2018, 06:23
PKN wrote:
Bunuel wrote:
If the reciprocal of the negative integer x is greater than the sum of y and z, then which of the following must be true?


(A) \(x > y + z\)

(B) y and z are positive.

(C) \(1 > x(y + z)\)

(D) \(1 < xy + xz\)

(E) \(\frac{1}{x} > z – y\)


Given \(\frac{1}{x}>(y+z)\), where x<0
Multiplying 'x' both sides, (inequality sign to be flipped since x is negative)
\(x*\frac{1}{x}>x(y+z)\)
Or, \(1 < xy+xz\)

Ans. (D)



PKN
can we write the reciprocal as -1/x and then proceed. please correct me if my thought process is wrong
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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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New post 14 Aug 2018, 06:40
apurv09 wrote:
PKN wrote:
Bunuel wrote:
If the reciprocal of the negative integer x is greater than the sum of y and z, then which of the following must be true?


(A) \(x > y + z\)

(B) y and z are positive.

(C) \(1 > x(y + z)\)

(D) \(1 < xy + xz\)

(E) \(\frac{1}{x} > z – y\)


Given \(\frac{1}{x}>(y+z)\), where x<0
Multiplying 'x' both sides, (inequality sign to be flipped since x is negative)
\(x*\frac{1}{x}>x(y+z)\)
Or, \(1 < xy+xz\)

Ans. (D)



PKN
can we write the reciprocal as -1/x and then proceed. please correct me if my thought process is wrong


apurv09,
No.
Suppose x=-5, reciprocal of x is 1/x=1/(-5)=-1/5
If you say, reciprocal of x is -1/x, then -1/x=-1/-5=1/5
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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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New post 18 Aug 2018, 19:27
Bunuel wrote:
If the reciprocal of the negative integer x is greater than the sum of y and z, then which of the following must be true?


(A) \(x > y + z\)

(B) y and z are positive.

(C) \(1 > x(y + z)\)

(D) \(1 < xy + xz\)

(E) \(\frac{1}{x} > z – y\)


We are given that 1/x > y + z, and x is negative. Thus, we can multiply both sides of the inequality by x (and flip the inequality sign since x is negative) and obtain:

1 < x(y + z)

1 < xy + xz

Answer: D
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Re: If the reciprocal of the negative integer x is greater than the sum of &nbs [#permalink] 18 Aug 2018, 19:27
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