Which of the following must be true if x/y and x/z > y/z?

\(\frac{x}{z} > \frac{y}{z}\)
--> \(\frac{(x - y)}{z} > 0\)
Since, \(x < y\),
--> \(x - y < 0\)
--> \(z\) < 0 ALWAYS (Since for \(\frac{(x - y)}{z} > 0\), ONLY negative/negative > 0)

IMO Option A

Which of the following must be true if x<y and x/z >y/z?

The eqs. x<y and x/z >y/z will only be valid when z<0.

x<y
and x/z > y/z

i.e x/z - y/z > 0
x-y/z > 0
Since x-y <0, so z has to be < 0 to make the equation > 0
So z must be < 0

OA:A

z(x-y)>0

given x-y<0 therefore Z<0

option A
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Which of the following must be true if x < y and \(\frac{x}{z} > \frac{y}{z}\) ?

A. z < 0

B. z > 0

C. \(\frac{x}{y}\) > 0

D. x > 0

E. y > 0

Since x < y and inequality \(\frac{x}{z} > \frac{y}{z}\) is with opposite sign, z must be an entity that must be causing the change of sign.

Among the choices D and E are straightforward out since both can have both + and - signs.
C is also ruled out since x and y can have opposite signs.

Only possibility left is A because z > 0 would not lead to change in sign from < to >.

Answer A.
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x<y…x-y<0
x/z>y/z…x/z-y/z>0…x-y/z>0
x-y<0 so z<0

Ans. (A)

Given x<y, and x/z > y/z,
this can only be true if x>0 and y>0 and z<0
or x<0 and y<0 and z<0.
In both cases, z<0 must be true.

To illustrate, lets consider the case where x>0 and y>0 and z<0
Let x=2 and y=3, we know x=2 < y=3. if z=-1, then x/z=-2 and y/z=-3. Since -2(x/z) > -3(y/z), then the given condition is satisfied.

In the other case where x<0 and y<0, let x=-3 and y=-2. -3(x) < -2(y). x/z can only be greater than y/z if z<0, so let z=-1. Then x/z=3 and y/z=2. Since 3(x/z) > 2(y/z), then the given condition is satisfied.

In both instances, z<0.

Hence the answer is option A

I cross multiplied, to get zx > zy which would indicate that a sign change has occurred so z < 0

A IMO

the given condition x<y and x/z >y/z can be true if x & z are -ve and y is +ve
so IMO A ; z<0 should be correct

Which of the following must be true if x<y and x/z >y/z?

A. z<0

B. z>0

C. xy>0

D. x>0

E. y>0

Given x<y but \frac{x}{z}>\frac{y}{z}. Note that the sign is changed when divided by Z.

Theory : Multiplication/division with a negative number can change the inequality sign. Thefore Z has to be negative for this to hold true.

Answer is A.

Which of the following must be true if x < y and \((\frac{x}{z}) > (\frac{y}{z})\)?

\(\frac{(x —y)}{z} > 0\)
x—y < 0.

—> in order \(\frac{(x —y)}{z}\) to be positive, z must be less than zero. (z < 0)

The answer is A

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Z should be negative. For example:
x=2 and y=3 and z=-3 then we have 2<3 so x<y and also -2/3 > -3/3 so x/z > y/z.

A is my answer

Given,
(X/z)>(y/z)
Here, z can not be zero.
So, z must be positive or negative.
If z is positive, x>y.
If z is negative, x<y

it is also given that x<y. So, Z must be negative.

Imo. A

Which of the following must be true if x<y and x/z >y/z?

Let's try from options.

A. z<0, Z is negative, x/z >y/z, If z = -1, -x>-y, y>x and the same condition x<y is mentioned. Hence, Z must be true to suffice the condition.

B. z>0, Z is positive, x/z >y/z, If z = 1, x>y, which contradicts the give condition x<y. incorrect

C. x/y>0, We don't know the signs of any variables, so there can be multiple options, which could be true, are available.

D. x>0, If x is positive, Y is also positive. But no more information about Z. So multiple values for inequality can be obtained

E. y>0. It implies that y = +ve and x can be +ve/-ve. So, again multiple values for inequality can be obtained

We see that the inequality sign flips when both x and y are divided by z. This happens when z is negative, i.e., z < 0.

Alternate Solution:

Using the second inequality, we have:

x/z - y/z > 0

(x - y)/z > 0

Since (x - y)/z is positive, (x - y) and z are either both positive or both negative. On the other hand, from the first inequality, we have:

x - y < 0

Since (x - y) is negative, it follows that z must also be negative; i.e. z < 0.

Answer: A
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