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Math Expert V
Joined: 02 Sep 2009
Posts: 61189
If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 56% (02:08) correct 44% (02:03) wrong based on 383 sessions

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If x and y are positive integers such that $$x > y > 1$$ and $$z = \frac{x}{y}$$, which of the following must be true?

I. $$z > \frac{(x − 1)}{(y − 1)}$$

II. $$z < \frac{(x − 1)}{(y − 1)}$$

III. $$z > \frac{(x + 1)}{(y + 1)}$$

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

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Math Expert V
Joined: 02 Aug 2009
Posts: 8283
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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2
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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Manager  Joined: 09 Jul 2013
Posts: 107
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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1
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: 11 Sep 2015
Posts: 4318
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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Top Contributor
1
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

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Re: If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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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  B
Joined: 25 Jan 2013
Posts: 28
Location: United States
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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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.
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1141
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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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. Intern  B
Joined: 11 Sep 2017
Posts: 33
Schools: IMD '21
GMAT 1: 740 Q50 V40 Re: If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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If I plug in nos as x=100 and y=99 then both of these nos satisfy the inequality as per the question stem but both statement II and III stand out to be incorrect. Also in the qstn its written that both x, y are integers but nothing about z is mentioned. Z can or cant be an integer. Can someone pl explain. @brunuel
Math Expert V
Joined: 02 Sep 2009
Posts: 61189
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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hassu13 wrote:
If x and y are positive integers such that $$x > y > 1$$ and $$z = \frac{x}{y}$$, which of the following must be true?

I. $$z > \frac{(x − 1)}{(y − 1)}$$

II. $$z < \frac{(x − 1)}{(y − 1)}$$

III. $$z > \frac{(x + 1)}{(y + 1)}$$

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

If I plug in nos as x=100 and y=99 then both of these nos satisfy the inequality as per the question stem but both statement II and III stand out to be incorrect. Also in the qstn its written that both x, y are integers but nothing about z is mentioned. Z can or cant be an integer. Can someone pl explain. @brunuel

If x = 100 and y = 99 both II and III ARE correct:

II. $$(\frac{100}{99}=1.0101...) < (\frac{(100 − 1)}{(99 − 1)} =1.02...)$$

III. $$(\frac{100}{99}=1.0101...) > (\frac{(100 + 1)}{(99 + 1)}=1.01)$$

As for your second question, yes z may or may not be an integer but this does not matter, we know that z = x/y and we can substitute this in the options and solve accordingly.
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Intern  B
Joined: 10 Mar 2018
Posts: 11
Re: If x and y are positive integers such that x > y > 1 and z=x/y, which  [#permalink]

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I tried to solve this algebraically

Question says X and Y and Positive + Integer and greater than 1 i.e. least value of Y can be 2 and least of X can be 3

Also z=x/y i.e. Z will always be +ve and will be greater than 1

Also, ZY = X

option I

z (y-1)> (x-1)
Zy -z > (ZY-1)
-Z> -1...
Since Z is greater than 1, negative value of Z cannot be greater than 1. Not true

Note: here though we dont know the value of x,y,z but we can multiply Y-1 both since values are positive and Y-1 is also positive.

OPTION II

By solving equation as mentioned above, result is:

-Z < -1...which is true

OPTION III

Final result is:

Z>1...which is true

Posted from my mobile device Re: If x and y are positive integers such that x > y > 1 and z=x/y, which   [#permalink] 27 May 2018, 16:22
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