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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If the reciprocal of the negative integer x is greater than the sum of

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

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Difficulty:   55% (hard)

Question Stats: 57% (01:39) correct 43% (01:55) wrong based on 223 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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Director  D
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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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3
1
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
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If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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

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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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Regards,

PKN

Rise above the storm, you will find the sunshine
Target Test Prep Representative G
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2801
Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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1
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  [#permalink]

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1
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) $$\frac{1}{x} > z – y$$

In this case option D should be written as -1< xy +xz. In that case it will be more correct option without any confusion.
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Location: India
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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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C
Since , when we rearrange the option , we will get 1/x > (y+z)
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Intern  Joined: 30 Sep 2018
Posts: 6
Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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Kindly note that D is answer since it is given that X is negative integer and so when it is cross multiplied the inequality sign flips direction.

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

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vishnu90 wrote:
C
Since , when we rearrange the option , we will get 1/x > (y+z)
Posted from my mobile device

Hi vishnu90,

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PKN

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

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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) $$\frac{1}{x} > z – y$$

x is negative integer.

Reciprocal of x is 1/x.

Given,

1/x > y + z

1<x(y+z) ...........As we know x is negative , inequality sign must be flipped .

GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4340
Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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2
Top Contributor
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: The reciprocal of the negative integer x is greater than the sum of y and z
The reciprocal of x is 1/x

So, we can write: 1/x > y + z
Multiply both sides by x to get: 1 < x(y + z) [since x is NEGATIVE, we REVERSE the direction of the inequality symbol]
Expand the right side to get: 1 < xy + xz

Cheers,
Brent

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