GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Jun 2019, 21:58 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If x, y and z are positive numbers, is z between x and y?

Author Message
TAGS:

Hide Tags

Manager  Joined: 08 Feb 2014
Posts: 204
Location: United States
Concentration: Finance
GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking)
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Bunuel wrote:
(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

If the inequality signs were not all neatly facing the same directions, would we still be able to add?

Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 55618
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

1
JackSparr0w wrote:
Bunuel wrote:
(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

If the inequality signs were not all neatly facing the same directions, would we still be able to add?

Thanks

No.

1. You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

2. You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Check for more here: inequalities-tips-and-hints-175001.html

Hope it helps.
_________________
Manager  Joined: 07 Dec 2009
Posts: 88
GMAT Date: 12-03-2014
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Hi Bunuel,

I have been trying to solve some DS questions lately and am very confused. Its probably a silly question but its been bothering me...

There are certain DS questions where one needs to manipulate the equations to decide whether the option is Sufficient or not where as some other questions like this one where we directly plug in values (usually 1,-1,0,-1/2,1/2). Is there a way you decide which approach to take before solving it ? Is it just a matter of practice ?

Cheers
Manager  Joined: 10 Mar 2013
Posts: 186
GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

gdediegoi wrote:
vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 4, 2z = 5, y = 6 --> 4 < 5 < 6 BUT z < x < y -- NO
If x = 4, 2z = 10, y = 12 --> 4 < 10 < 12 and x < z < y -- YES

Not sufficient.

(2) 2x < z < 2y.

If 2x = 4, z = 5, 2y = 6 --> 4 < 5 < 6 BUT x < y < z --> NO
If 2x = 4, z = 5, 2y = 18 --> 4 < 5 < 9 and x < z < y --> YES

Not sufficient.

(1)+(2)

x < 2x (always, for positive numbers)
2x < z (from 2)
z < 2z (always, for positive numbers)
2z < y (from 1)

Which gives x < 2x < z < 2z < y

Therefore x < z < y

Sufficient.

Very ingenious answer! I never thought of chaining the inequalities!
Manager  Joined: 10 May 2014
Posts: 138
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Bunuel wrote:
jlgdr wrote:
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.

Hi Bunuel,

Would it have been the same if instead of adding the inequalities we would have multiplied them?
I assume that since we know the variables are positive and the signs are always in the same direction, we can do so.

My reasoning is
- First Inequality: x < 2z < y
- Second Inequality: 2x < z < 2y

Inequalities multiplied: 2$$x^2$$ < 2$$z^2$$ < 2$$y^2$$
Square root each: 2x < 2z < 2y
Divide each by 2: x < z < y

I know adding inequalities is way simpler to solve this particular problem, but I just want to know if this path is also feasible.

Thank you so much!
_________________
Consider giving me Kudos if I helped, but don´t take them away if I didn´t! What would you do if you weren´t afraid?
CEO  S
Joined: 20 Mar 2014
Posts: 2622
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

minwoswoh wrote:

Hi Bunuel,

Would it have been the same if instead of adding the inequalities we would have multiplied them?
I assume that since we know the variables are positive and the signs are always in the same direction, we can do so.

My reasoning is
- First Inequality: x < 2z < y
- Second Inequality: 2x < z < 2y

Inequalities multiplied: 2$$x^2$$ < 2$$z^2$$ < 2$$y^2$$
Square root each: 2x < 2z < 2y
Divide each by 2: x < z < y

I know adding inequalities is way simpler to solve this particular problem, but I just want to know if this path is also feasible.

Thank you so much!

Multiplying or dividing 2 variables in inequalities is ONLY allowed when you know the signs of the 2 variables involved. In your question, it must not be done. First of, if you square 2z you will get $$4z^2$$ and NOT $$2z^2$$ as you have mentioned above.

Secondly, if lets say, x<2z with x=2 and z=3, then yes, $$x^2<4z^2$$or$$2z^2$$ but if x=-3 and z=-0.25 then in this case, $$x^2 > 4z^2 or 2z^2$$
Manager  Joined: 10 May 2014
Posts: 138
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Engr2012 wrote:
minwoswoh wrote:

Hi Bunuel,

Would it have been the same if instead of adding the inequalities we would have multiplied them?
I assume that since we know the variables are positive and the signs are always in the same direction, we can do so.

My reasoning is
- First Inequality: x < 2z < y
- Second Inequality: 2x < z < 2y

Inequalities multiplied: 2$$x^2$$ < 2$$z^2$$ < 2$$y^2$$
Square root each: 2x < 2z < 2y
Divide each by 2: x < z < y

I know adding inequalities is way simpler to solve this particular problem, but I just want to know if this path is also feasible.

Thank you so much!

Multiplying or dividing 2 variables in inequalities is ONLY allowed when you know the signs of the 2 variables involved. In your question, it must not be done. First of, if you square 2z you will get $$4z^2$$ and NOT $$2z^2$$ as you have mentioned above.

Secondly, if lets say, x<2z with x=2 and z=3, then yes, $$x^2<4z^2$$or$$2z^2$$ but if x=-3 and z=-0.25 then in this case, $$x^2 > 4z^2 or 2z^2$$

Hi Engr2012,
But in my previous post I was not squaring the first inequality.
- I was multiplying the first and the second inequality together --> x(2x) < z(2z) < y(2y) --> 2$$x^2$$ < 2$$z^2$$ < 2$$y^2$$
- I was then dividing each term by 2, yielding --> $$x^2$$ < $$z^2$$ < $$y^2$$
- And finally square rooting each term, yielding --> x < z < y

In my previous post, I mistakenly mixed the order of these last 2 steps.

Regarding your point of "Multiplying or dividing 2 variables in inequalities is ONLY allowed when you know the signs of the 2 variables involved" the question stem tells us that x, y, and z are positive numbers.

Is my reasoning correct then?

Thanks!
_________________
Consider giving me Kudos if I helped, but don´t take them away if I didn´t! What would you do if you weren´t afraid?
CEO  S
Joined: 20 Mar 2014
Posts: 2622
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

minwoswoh wrote:
Engr2012 wrote:
minwoswoh wrote:

Hi Bunuel,

Would it have been the same if instead of adding the inequalities we would have multiplied them?
I assume that since we know the variables are positive and the signs are always in the same direction, we can do so.

My reasoning is
- First Inequality: x < 2z < y
- Second Inequality: 2x < z < 2y

Inequalities multiplied: 2$$x^2$$ < 2$$z^2$$ < 2$$y^2$$
Square root each: 2x < 2z < 2y
Divide each by 2: x < z < y

I know adding inequalities is way simpler to solve this particular problem, but I just want to know if this path is also feasible.

Thank you so much!

Multiplying or dividing 2 variables in inequalities is ONLY allowed when you know the signs of the 2 variables involved. In your question, it must not be done. First of, if you square 2z you will get $$4z^2$$ and NOT $$2z^2$$ as you have mentioned above.

Secondly, if lets say, x<2z with x=2 and z=3, then yes, $$x^2<4z^2$$or$$2z^2$$ but if x=-3 and z=-0.25 then in this case, $$x^2 > 4z^2 or 2z^2$$

Hi Engr2012,
But in my previous post I was not squaring the first inequality.
- I was multiplying the first and the second inequality together --> x(2x) < z(2z) < y(2y) --> 2$$x^2$$ < 2$$z^2$$ < 2$$y^2$$
- I was then dividing each term by 2, yielding --> $$x^2$$ < $$z^2$$ < $$y^2$$
- And finally square rooting each term, yielding --> x < z < y

In my previous post, I mistakenly mixed the order of these last 2 steps.

Regarding your point of "Multiplying or dividing 2 variables in inequalities is ONLY allowed when you know the signs of the 2 variables involved" the question stem tells us that x, y, and z are positive numbers.

Is my reasoning correct then?

Thanks!

As I said before, for inequalities I would be very wary of multiplying or dividing by variables as nature of numbers change when you square them especially if lets say a variable x, 0<x<1

If x = 0.2 , then x^2 < x but if x>1, then x^2>x. Your approach might have worked for this question but will not be applicable for scenarios wherein the variables lie between 0 and 1.

This question was straightforward by adding the 2 inequalities to give you x<z<y
Intern  Joined: 21 Jul 2015
Posts: 1
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

I ENVY you for having such a systematic approach to questions EVERY single time! Beautiful!

Bunuel wrote:
jlgdr wrote:
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.
Manager  S
Joined: 13 Dec 2013
Posts: 150
Location: United States (NY)
Schools: Cambridge"19 (A)
GMAT 1: 710 Q46 V41 GMAT 2: 720 Q48 V40 GPA: 4
WE: Consulting (Consulting)
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Bunuel wrote:
jlgdr wrote:
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.

For 1)&2) is this rationale also correct?

From 1), x<2z<y and from 2), x<z/2<y, therefore z must lie between x and y?

Thanks.
Math Expert V
Joined: 02 Sep 2009
Posts: 55618
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Cez005 wrote:
For 1)&2) is this rationale also correct?

From 1), x<2z<y and from 2), x<z/2<y, therefore z must lie between x and y?

Thanks.

_____________
Yes, that's true.
_________________
Director  V
Joined: 06 Jan 2015
Posts: 644
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Bunuel wrote:
jlgdr wrote:
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.

Hi Bunuel / Experts,

How to quickly choose the numbers to satisfy

(1) x < 2z < y
(2) 2x < z < 2y

I'm Getting Stuck to choose the numbers. Please throw some light
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution S
Joined: 06 Mar 2017
Posts: 194
Concentration: Operations, General Management
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Bunuel wrote:
jlgdr wrote:
vcbabu wrote:
x,y,z>0, is y between x and z?
1). x<(y/2)<z
2). x<2y<z

Hey all, was just checking this question out.
Wanted to know what is the OA for this one. I got B. Let me walk you through the way I solved it
So is y between x and z?
Ok, let's do it!

Statement 1
y/2 is actually between x and z, but we don't really know about y, only that it is positive but still this is not sufficient

Statement 2
Ok, now this says that 2y is between x and z, so algebraically x<2y<z, now if we divide all terms by 2 we get x/2<y<z/2
Now if they are all positive and y is between half of x and half of z, then y must definetely also be in the range x<y<z, If i'm not mistaken
So IMO, Answer should be (B)

Please let me know if this works
Kudos if you like!
Cheers
J No B is not correct. Check below. Notice that it's basically the same exact question.

If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y.

If x = 1, y = 10, and z = 1 --> answer NO.
If x = 1, y = 10, and z = 2 --> answer YES.

Not sufficient.

(2) 2x < z < 2y.

If x = 1, y = 2, and z = 3 --> answer NO.
If x = 1, y = 10, and z = 3 --> answer YES.

Not sufficient.

(1)+(2) Add the inequalities: 3x < 3z < 3y --> divide by 3: x < z < y. Sufficient.

Hope it's clear.

Bunuel,
Pls give the adding and subtracting rule for inequalities.
I saw it on one of your posts.
Same sign you add and different sign you subtract...Something like that....
Math Expert V
Joined: 02 Sep 2009
Posts: 55618
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

1
Director  P
Joined: 14 Dec 2017
Posts: 520
Location: India
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

Given x, y, z > 0

Question: is z between x & y?

Statement 1: x < 2z < y

Visualize the number line

Case 1: 0---------x---------z-------------------2z------y......we get YES.

Case 2: 0-------z--x----2z-------------------------------y.....we get NO.

Statement 1 is Not Sufficient.

Statement 2: 2x < z < 2y

Again visualize the number line

Case 1: 0----x-----2x---------z---y----------------------2y......we get YES.

Case 2: 0----x-----2x---------y----z------------------2y.....we get NO.

Statement 2 is not Sufficient.

Combining both statements, we can add the inequalities

hence 3x < 3z < 3y, removing 3 we get

x < z < y

Combining is Sufficient.

Thanks,
GyM
_________________
Manager  S
Joined: 07 Feb 2017
Posts: 183
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

Concept:
b > 2a must be true that b > a

b > a/2 could be true that b > a
Ex: a=4; b=3 or a=4; b=999999999999

(1) z > x/2 could be true z > x
Must be true y > z
Insufficient

(1) Must be true z > x
z/2 < y could be true y > z
Insufficient

(1) and (2)
y > z and z > x so the result is x < z < y
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 936
Re: If x, y and z are positive numbers, is z between x and y?  [#permalink]

Show Tags

vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?

(1) x < 2z < y
(2) 2x < z < 2y

$$x,y,z\,\,\, > 0\,\,\,\left( * \right)$$

$$?\,\,\,\,:\,\,\,\,z\,\,{\text{between}}\,\,x\,\,{\text{and}}\,\,y\,\,\,$$

$$\left( 1 \right)\,\,\,x < 2z < y\,\,\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,2,5} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \end{gathered} \right.$$

$$\left( 2 \right)\,\,2x < z < 2y\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\ \,{\text{Take}}\,\,\left( {x,z,y} \right) = \left( {1,3,4} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\ \end{gathered} \right.$$

$$\left( {1 + 2} \right)\,\,\,\,x\,\,\mathop < \limits^{\left( * \right)} \,\,\,2x\,\,\mathop < \limits^{\left( 2 \right)} \,\,z\,\,\mathop < \limits^{\left( * \right)} \,\,\,2z\,\,\mathop < \limits^{\left( 1 \right)} \,\,\,y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x < z < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net Re: If x, y and z are positive numbers, is z between x and y?   [#permalink] 09 Oct 2018, 11:25

Go to page   Previous    1   2   [ 37 posts ]

Display posts from previous: Sort by

If x, y and z are positive numbers, is z between x and y?  