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

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Bunuel
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If the reciprocal of the negative integer x is greater than the sum of [#permalink] 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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PKN
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Re: If the reciprocal of the negative integer x is greater than the sum of [#permalink] 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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PKN
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Re: If the reciprocal of the negative integer x is greater than the sum of [#permalink] New post  14 Aug 2018, 06:40
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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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AmitD90
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Re: If the reciprocal of the negative integer x is greater than the sum of [#permalink] New post  02 Oct 2018, 10:53
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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) \(\frac{1}{x} > z – y\)


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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BrentGMATPrepNow
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If the reciprocal of the negative integer x is greater than the sum of [#permalink] New post  Updated on: 17 May 2021, 07:16
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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: 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

Answer: D

Cheers,
Brent

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Originally posted by BrentGMATPrepNow on 20 Feb 2019, 10:45.
Last edited by BrentGMATPrepNow on 17 May 2021, 07:16, edited 1 time in total.
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JeffTargetTestPrep
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Re: If the reciprocal of the negative integer x is greater than the sum of [#permalink] New post  18 Aug 2018, 19:27
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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) \(\frac{1}{x} > z – y\)


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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apurv09
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If the reciprocal of the negative integer x is greater than the sum of [#permalink] 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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vishnu90
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Re: If the reciprocal of the negative integer x is greater than the sum of [#permalink] New post  02 Oct 2018, 11:14
C
Since , when we rearrange the option , we will get 1/x > (y+z)
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kiranteja93
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Re: If the reciprocal of the negative integer x is greater than the sum of [#permalink] New post  07 Oct 2018, 23:55
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] New post  08 Oct 2018, 09:41
vishnu90 wrote:
C
Since , when we rearrange the option , we will get 1/x > (y+z)
Posted from my mobile device


Hi vishnu90,

Could you please show your steps?
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KSBGC
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Re: If the reciprocal of the negative integer x is greater than the sum of [#permalink] New post  13 Oct 2018, 03:22
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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