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# If x and y are positive integers such that x > y > 1 and z=x/y, which

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Joined: 02 Sep 2009
Posts: 44287
If x and y are positive integers such that x > y > 1 and z=x/y, which [#permalink]

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05 Apr 2016, 08:36
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If x and y are positive integers such that x > y > 1 and z=x/y, which of the following must be true?

I. z > (x − 1)/(y − 1)
II. z < (x − 1)/(y − 1)
III. z > (x + 1)/(y + 1)

A. I​ only
B. I and II​
C. II and III
D. II only
E. I​ and III
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Aug 2009
Posts: 5715
If x and y are positive integers such that x > y > 1 and z=x/y, which [#permalink]

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05 Apr 2016, 22:22
1
KUDOS
Expert's post
Bunuel wrote:
If x and y are positive integers such that x > y > 1 and z=x/y, which of the following must be true?

I. z > (x − 1)/(y − 1)
II. z < (x − 1)/(y − 1)
III. z > (x + 1)/(y + 1)

A. I​ only
B. I and II​
C. II and III
D. II only
E. I​ and III

Hi,

lets see the choices

I. $$z > \frac{(x − 1)}{(y − 1)}$$
z=x/y
so $$\frac{x}{y} >\frac{(x-1)}{(y-1)}$$....$$\frac{x}{y} -\frac{(x-1)}{(y-1)}>0$$.....$$\frac{(xy-x-xy+y)}{y*(y-1)}>0$$..
we know y>1 so y*(y-1)>0...so xy-x-xy+y>0
OR y>x
NEVER true

II. $$z < \frac{(x − 1)}{(y − 1)}$$..
$$z=\frac{x}{y}$$..
so $$\frac{x}{y} <\frac{(x-1)}{(y-1)}$$.....$$\frac{x}{y} -\frac{(x-1)}{(y-1)}<0$$.....$$\frac{(xy-x-xy+y)}{y*(y-1)}<0$$..
we know y>1 so y*(y-1)>0...so xy-x-xy+y<0
OR y<x
Always TRUE

III. $$z > \frac{(x + 1)}{(y + 1)}$$
z=x/y
so $$\frac{x}{y} >\frac{(x+1)}{(y+1)}$$.....$$\frac{x}{y} -\frac{(x+1)}{(y+1)}>0$$.....$$\frac{(xy+x-xy-y)}{y*(y+1)}>0$$..
we know y>1 so y*(y+1)>0...so xy+x-xy-y>0
OR x>y
Always TRUE

II and III are true
ans C
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Joined: 09 Jul 2013
Posts: 110
If x and y are positive integers such that x > y > 1 and z=x/y, which [#permalink]

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06 Apr 2016, 08:34
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Since we're given some solid restrictions on what x and y could be, picking numbers is a quick and easy strategy.

x>y>1, and x and y are integers. Let's try x=3 and y=2. So z=x/y = 3/2

Looking at statements I and II, we can see that they are mutually exclusive. z cannot be both > and < a number at the same time. So right away, we know that either I or II must be true, but never both. So B is out.

Now, using our numbers for x and y, we can set up a little chart to see what happens when we add or subtract 1 from the numerator and denominator.

x-----y-----z-----(x+1)/(y+1)-----(x-1)/(y-1)
3-----2----3/2--------4/3--------------2/1
------------------------<z---------------->z

We can see that when 1 is added to the numerator and denominator, the value decreases, and when it is subtracted, the value increases.

Looking at the statements then, we can see that II and III are true, and I is not.

Now, in general, it is good to know what happens to fractions when you add or subtract a constant from both the numerator and denominator. As long as we're dealing with positive numbers, adding a constant on top and bottom of a fraction will cause the value of the fraction to move closer to 1, and subtracting a constant from the top and bottom a fraction will cause the value of the fraction to get farther from 1. Notice I didn't say increase or decrease, because that will depend on whether the fraction is greater than or less than 1.

Examples:
$$\frac{20-10}{100-10} < \frac{20}{100} < \frac{20+100}{100+100}$$

$$\frac{50+50}{10+50}< \frac{50}{10} < \frac{50-5}{10-5}$$

So in general:

If x>y --> $$\frac{x+a}{y+a} < \frac{x}{y} < \frac{x-a}{y-a}$$

And if x<y --> $$\frac{x-a}{y-a} < \frac{x}{y} < \frac{x+a}{y+a}$$

In our case, x>y, so adding a constant will decrease the fraction (statement III) and subtracting a constant will increase the fraction (statement II).
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Dave de Koos

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Joined: 12 Sep 2015
Posts: 2140
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which [#permalink]

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08 Jun 2017, 15:17
Expert's post
Top Contributor
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Bunuel wrote:
If x and y are positive integers such that x > y > 1 and z=x/y, which of the following must be true?

I. z > (x − 1)/(y − 1)
II. z < (x − 1)/(y − 1)
III. z > (x + 1)/(y + 1)

A. I​ only
B. I and II​
C. II and III
D. II only
E. I​ and III

Notice that statements I and II say the exact OPPOSITE thing.
Also, notice that either I or II appear in ALL 5 answer choices. So, one of them must be true.
Given this, let's see what happens when we assign some values to x and y.

Say x = 4 and y = 2
So, z = x/y = 4/2 = 2

Now test (x − 1)/(y − 1) by plugging in x = 4 and y = 2
We get: (4 - 1)/(2 - 1) = 3/1 = 3
Since z = 2, we can see that z < (x − 1)/(y − 1)
So, statement II is true, and statement I is false.

This means the correct answer is C or D

Now test statement III: z > (x + 1)/(y + 1)
Since z = x/y, we can rewrite III as x/y > (x + 1)/(y + 1)
Let's find out if this statement is true.
Since y is POSITIVE, we can safely multiply both sides by y to get: x > (y)(x + 1)/(y + 1)
y+1 is also POSITIVE, we can safely multiply both sides by y+1 to get: x(y + 1) > (y)(x + 1)
Expand both sides: xy + x > xy + y
Subtract xy from both sides to get: x > y
Is this true? YES! It's given in the question
So statement III is true

[Reveal] Spoiler:
C

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Joined: 31 Oct 2013
Posts: 29
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which [#permalink]

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05 Mar 2018, 16:55
chetan2u wrote:
Bunuel wrote:
If x and y are positive integers such that x > y > 1 and z=x/y, which of the following must be true?

I. z > (x − 1)/(y − 1)
II. z < (x − 1)/(y − 1)
III. z > (x + 1)/(y + 1)

A. I​ only
B. I and II​
C. II and III
D. II only
E. I​ and III

Hi,

lets see the choices

I. $$z > \frac{(x − 1)}{(y − 1)}$$
z=x/y
so $$\frac{x}{y} >\frac{(x-1)}{(y-1)}$$....$$\frac{x}{y} -\frac{(x-1)}{(y-1)}>0$$.....$$\frac{(xy-x-xy+y)}{y*(y-1)}>0$$..
we know y>1 so y*(y-1)>0...so xy-x-xy+y>0
OR y>x
NEVER true

II. $$z < \frac{(x − 1)}{(y − 1)}$$..
$$z=\frac{x}{y}$$..
so $$\frac{x}{y} <\frac{(x-1)}{(y-1)}$$.....$$\frac{x}{y} -\frac{(x-1)}{(y-1)}<0$$.....$$\frac{(xy-x-xy+y)}{y*(y-1)}<0$$..
we know y>1 so y*(y-1)>0...so xy-x-xy+y<0
OR y<x
Always TRUE

III. $$z > \frac{(x + 1)}{(y + 1)}$$
z=x/y
so $$\frac{x}{y} >\frac{(x+1)}{(y+1)}$$.....$$\frac{x}{y} -\frac{(x+1)}{(y+1)}>0$$.....$$\frac{(xy+x-xy-y)}{y*(y+1)}>0$$..
we know y>1 so y*(y+1)>0...so xy+x-xy-y>0
OR x>y
Always TRUE

II and III are true
ans C

i don't understand how 3 one has become true. i used plunging method. i found 3 one is wrong. kindly, make me clear and tell me where make mistake. thanks.
Intern
Joined: 25 Jan 2013
Posts: 28
Concentration: General Management, Entrepreneurship
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which [#permalink]

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09 Mar 2018, 07:50
selim wrote:
chetan2u wrote:
Bunuel wrote:
If x and y are positive integers such that x > y > 1 and z=x/y, which of the following must be true?

i don't understand how 3 one has become true. i used plunging method. i found 3 one is wrong. kindly, make me clear and tell me where make mistake. thanks.

When you plug in number, make sure x > y.
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Joined: 25 Feb 2013
Posts: 1001
Location: India
GPA: 3.82
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which [#permalink]

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11 Mar 2018, 03:30
Bunuel wrote:
If x and y are positive integers such that x > y > 1 and z=x/y, which of the following must be true?

I. z > (x − 1)/(y − 1)
II. z < (x − 1)/(y − 1)
III. z > (x + 1)/(y + 1)

A. I​ only
B. I and II​
C. II and III
D. II only
E. I​ and III

This is a great question to test few properties of a fraction. The properties are -

I) If $$x>y$$ and given $$a$$ is a positive integer, then

$$\frac{x+a}{y+a}<\frac{x}{y}$$ and $$\frac{x−a}{y−a}>\frac{x}{y}$$

II) If $$x<y$$ and given that $$a$$ a positive integer, then

$$\frac{x+a}{y+a}>\frac{x}{y}$$ and $$\frac{x−a}{y−a}<\frac{x}{y}$$

So if you are aware of these properties then you will get the answer directly to the question Option C

-----------------------------------------------------------------

Now, let's use this question to derive the options

Given $$x>y$$, multiply both sides by $$1$$ (because in the choice given each fraction is increased/decreased by 1)

So we get $$x*1>y*1$$, now add $$xy$$ to both sides of the inequality

$$=>x+xy>y+xy=>x(y+1)>y(x+1)$$

or $$\frac{x}{y}>\frac{x+1}{y+1} => z>\frac{x+1}{y+1}$$. Hence Statement II must be true

now multiply the inequality $$x>y$$ by $$-1$$

$$=>-1*x<-1*y$$, now add $$xy$$ to both sides of the inequality

$$xy-x<xy-y=>x(y-1)<y(x-1)=>\frac{x}{y}<\frac{x-1}{y-1}=>z<\frac{x-1}{y-1}$$. Hence Statement III must be true

Hi Bunuel,

May I request to add these properties to the Number Properties article in GC forum or book because this can be very useful for tricky questions. If its already present there, then kindly ignore.
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which   [#permalink] 11 Mar 2018, 03:30
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