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

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13 Aug 2018, 03:41
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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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13 Aug 2018, 08:39
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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) $$\frac{1}{x} > z – y$$

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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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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14 Aug 2018, 05:23
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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14 Aug 2018, 05:40
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

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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, 18:27
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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02 Oct 2018, 09:53
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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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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02 Oct 2018, 10:14
C
Since , when we rearrange the option , we will get 1/x > (y+z)
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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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07 Oct 2018, 22:55
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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Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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08 Oct 2018, 08:41
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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13 Oct 2018, 02:22
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 .

The best answer is D.
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Posts: 4340
Re: If the reciprocal of the negative integer x is greater than the sum of  [#permalink]

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20 Feb 2019, 09:45
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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Re: If the reciprocal of the negative integer x is greater than the sum of   [#permalink] 20 Feb 2019, 09:45
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