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

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Joined: 02 Sep 2009
Posts: 49271
If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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13 Aug 2018, 04:41
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45% (medium)

Question Stats:

63% (01:07) correct 37% (01:28) wrong based on 67 sessions

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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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Posts: 852
WE: Supply Chain Management (Energy and Utilities)
Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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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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PKN

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Joined: 10 Oct 2017
Posts: 11
If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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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
Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 852
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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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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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PKN

Rise above the storm, you will find the sunshine

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

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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

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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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