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If x + z > y + z, then which of the following must be true?

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If x + z > y + z, then which of the following must be true?  [#permalink]

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New post 02 Apr 2019, 04:32
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A
B
C
D
E

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Question Stats:

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If x + z > y + z, then which of the following must be true?  [#permalink]

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New post Updated on: 02 Apr 2019, 05:10
X + Z > Y + Z
Therefore,
X > Y.

1. X - Z > Y - Z
X > Y - Z + Z
X > Y
Statement 1 is correct and sufficient.

2. XZ > YZ
If X = 1 , Y= -1 , Z = -2
1-1 > -1-2
0 > -3
However if we substitute the values into statement 2 :
-2 < 2
Hence, statement 2 is not true.

3. X/Z > Y/Z
If X = 1 , Y= -1 , Z = -2
1-1 > -1-2
0 > -3
However if we substitute the values into statement 3 :
1/-2 < -1/-2
Hence, statement 3 is not true.

Answer : A

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Originally posted by justus23 on 02 Apr 2019, 05:05.
Last edited by justus23 on 02 Apr 2019, 05:10, edited 1 time in total.
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Re: If x + z > y + z, then which of the following must be true?  [#permalink]

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New post 02 Apr 2019, 05:06
If \(x+z>y+z\), then which of the following must be true?

Simplifying the question stem
\(x+z>y+z\) z cancels
\(x>y\)

Must be true questions are usually best done with algebra.

I. \(x–z>y–z\) z cancels
\(x>y\) So it must be true

P.s ( This is actually a fundamental rule in inequalities: adding (or subtracting) the same value to (a>b) will not change the inequality.

II. \(xz>yz\) --> \(xz-yz>0\) --> \(z(x-y) > 0\) (Z is unknown, it could be negative or 0) must not be true


III. \(\frac{x}{z}>\frac{y}{z} => \frac{x}{z}-\frac{y}{z} > 0 => \frac{(x-y)}{z} > 0\)

(z could be negative, therefore changing the inequality sign ) must not be true

Only A
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Re: If x + z > y + z, then which of the following must be true?  [#permalink]

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New post 02 Apr 2019, 05:12
1
justus23 wrote:
X + Z > Y + Z
Therefore,
X > Y.

1. X - Z > Y - Z
X > Y - Z + Z
X > Y
Statement 1 is correct and sufficient.

2. XZ > YZ
X(Z) > Y(Z)
X > Y
Statement 2 is correct and sufficient.

3. X/Z > Y/Z

If X = 1 , Y= -1 , Z = -2
1-1 > -1-2
0 > -3
However if we substitute the values into statement 3 :
1/-2 < -1/-2
Hence, statement 3 is not true.

Answer : D

Posted from my mobile device


Hi
Another rule in inequality
You cannot divide by an unknown (i.e., a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality

Consider

-3<1 ( multiple by -1) becomes 3<-1 (which is not true )
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Re: If x + z > y + z, then which of the following must be true?  [#permalink]

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New post 02 Apr 2019, 06:22
Top Contributor
Bunuel wrote:
If \(x + z > y + z\), then which of the following must be true?

I. \(x – z > y – z\)

II. \(xz > yz\)

III. \(\frac{x}{z} > \frac{y}{z}\)

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only


Let's check each statement

(I) x – z > y – z
It is given that: x + z > y + z
If we subtract 2z from both sides, we get: x - z > y - z
Perfect!
Statement I must be true.

Check the answer choices . . . ELIMINATE B, C and E since they suggest that statement 1 is NOT true.


(II) xz > yz
This statement need not be true.
It is given that: x + z > y + z
So, one possible case is that x = 5, y = 4 and z = -1, since 5 + (-1) > 4 + (-1)
When we plug these values into Statement II, we get: (5)(-1) > (4)(-1)
Simplify to get: -5 > -4, which is NOT TRUE
So, Statement II is not true
Check the remaining answer choices . . . ELIMINATE D

By the process of elimination, the correct answer is A

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Re: If x + z > y + z, then which of the following must be true?  [#permalink]

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New post 14 May 2019, 11:26
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X has to be greater than Y for X + Z > Y + Z to be true!
Hence, "X > Y".

Now let's have a look at the individual cases to solve this algebraically-->

i.) X - Z > Y - Z --> which is equivalent to --> X - Z + Z > Y --> which implies --> X > Y

i.) correct


ii.) XZ > YZ --> suppose Y is negative and Z is negative --> X(-Z) > (-Y)(-Z) --> X>(-Y)

This does not satisfy our original equation.

ii.) incorrect


iii.) X/Z > Y/Z --> suppose Y is negative and Z is negative --> X/(-Z) > (-Y)/(-Z) --> X>(-Y)

This does not satisfy our original equation.

iii.) incorrect


Thus, only i.) is correct

Hence, Answer = A
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Re: If x + z > y + z, then which of the following must be true?  [#permalink]

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New post 16 May 2019, 19:51
Bunuel wrote:
If \(x + z > y + z\), then which of the following must be true?

I. \(x – z > y – z\)

II. \(xz > yz\)

III. \(\frac{x}{z} > \frac{y}{z}\)

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only


Simplifying, we have:

x > y

Now let’s look at each Roman Numeral:

I.

x - z > y - z

x > y

I is true.

II.

xz > yz

Since we don't know the sign of z, we don't know whether II is true.

For example, if z is positive, then xz > yz, but if z is negative, then we will have xz < yz.

II is not necessarily true.

III.

x/z > y/z

If z is positive, then the inequality holds, but if z is negative, we will have x/z < y/z.

III is not necessarily true.

Answer: A
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Re: If x + z > y + z, then which of the following must be true?  [#permalink]

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New post 03 Aug 2019, 14:28
Bunuel wrote:
If \(x + z > y + z\), then which of the following must be true?

I. \(x – z > y – z\)

II. \(xz > yz\)

III. \(\frac{x}{z} > \frac{y}{z}\)

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only


I can solve with algebra but I avoid doing that if Possible. Since there are no restrictions on x y z being positive or negative or anything else, Lets pick numbers
x=4,y=3,z=2
Given condition satisfies so we are good.
Turns out I,II,III all three are true. Good, there is no option which says all 3. Even if there was you should always try at least one more pair.

Let's try negative z. z=-1,x=4,y=3. Given condition satisfies so we are good.
This time both II and III are not true.
Hence A. I only.
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Re: If x + z > y + z, then which of the following must be true?   [#permalink] 03 Aug 2019, 14:28
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