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 Your Progress
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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
1) Case 1: y/2=4, case 2: y/2=2, in two case x=1, z=6 In two case, we have different ans (not suf) 2) with x=3 and 2y = 10 and 4, we have different ans (not suf) Together we have, x<y/2<y<2y<z => YES Hence C
Re: x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). [#permalink]
15 Oct 2013, 15:30
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
Re: x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). [#permalink]
15 Oct 2013, 23:24
11
This post received KUDOS
Expert's post
12
This post was BOOKMARKED
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.
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
10 Nov 2013, 14:52
5
This post received KUDOS
1
This post was BOOKMARKED
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
question- is "x < z < y" or "y < z < x" ????
1) x < 2z < y since 2z<y ( and all are positive) we can infer that z<y . But we do not know weather x<z Insufficient (2) 2x < z < 2y since 2x<z , we can infer x<z , but we cant say if z<y Insufficient
Combine both 1 and 2 - x<z<y ... sufficient _________________
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
11 Nov 2013, 20:24
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
Question Basically asks whether x<z<y or y<z<x
from St 1 we have x<2z<y Consider x=4, y= 9, 2z= 7.8 --->z= 3.9 Is Z between x and y : No Consider x=4, y=9 and 2z= 8.4 ---->z =4.2. Is Z between x and y : yes
So A and D ruled out
Consider St 2, we have 2x<z<2y------> x<z/2< y
Consider x= 4, y=14, z/2= 6----> then z=12, Is Z between x and y : yes Consider x=4, y=14, z/2=8 ----> then z=16, Is Z between x and y: no
So B ruled out.
Combining we get x<2z<y and x<z/2<y. Now if if both z/2 and 2z are in between x and y then z will also between X and Y.
Ans C _________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
22 Feb 2014, 07:00
Hi Bunnel,
I have made mistakes in these questions a couple of times... I solve them incorrectly under time pressure as I jumble up the number cases to pick.
Do you have a Drill or a 'set of questions' for specifically this type of questions that test inequalities in multiple variables? Do you have some theory tips / pointers? (For eg. - I finnd that consider very close values in Case 1 and considering extremely far values in Case 2 helps...)
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
14 Apr 2014, 04:42
vcbabu wrote:
If x, y and z are positive numbers, is z between x and y?
(1) x < 2z < y (2) 2x < z < 2y
I want to know what is wrong with the following method:
1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No
Insufficient.
Statement 2 : Since all are positive I can reduce by 2 , we get x<(z/2)<y
x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes
x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no
1+2 Adding both of these
x < 2z < y x<z/2<y
we get 2x<2.5z<2y
x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1
I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
14 Apr 2014, 05:58
1
This post received KUDOS
Expert's post
qlx 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
I want to know what is wrong with the following method:
1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No
Insufficient.
Statement 2 : Since all are positive I can reduce by 2 , we get x<(z/2)<y
x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes
x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no
1+2 Adding both of these
x < 2z < y x<z/2<y
we get 2x<2.5z<2y
x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1
I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.
(1) x < 2z < y (2) 2x < z < 2y
When you add the inequalities you get:
(x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
16 Apr 2014, 12:05
Bunuel wrote:
qlx 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
I want to know what is wrong with the following method:
1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No
Insufficient.
Statement 2 : Since all are positive I can reduce by 2 , we get x<(z/2)<y
x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes
x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no
1+2 Adding both of these
x < 2z < y x<z/2<y
we get 2x<2.5z<2y
x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1
I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.
(1) x < 2z < y (2) 2x < z < 2y
When you add the inequalities you get:
(x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.
sorry I didn't get it. My question was since all are positive I can definitely reduce statement 2 by 2, right? now statement 2 becomes x<(z/2)<y why am I not getting the answer if I take statement 1 as x < 2z < y , and statement 2 as x<(z/2)<y
Adding 1 and 2 we get 2x<2.5z<2y
x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So how come this is not working out if after reducing stat.2 I add it to 1?
In the same way if we were to solve :
(Q2) x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). x<2y<z
adding 1 and 2--> 2x<2.5z<2y so here the answer is E x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
17 Apr 2014, 00:28
1
This post received KUDOS
Expert's post
qlx wrote:
Bunuel wrote:
qlx wrote:
If x, y and z are positive numbers, is z between x and y?
(1) x < 2z < y (2) 2x < z < 2y
I want to know what is wrong with the following method:
1) x= 1, z = 3, y = 8 this satisfies statement 1 and answer is Yes x=2, z= 2 , y = 8 this also satisfies statement 1 and answer is No
Insufficient.
Statement 2 : Since all are positive I can reduce by 2 , we get x<(z/2)<y
x= 1, z= 4, y = 8 This satisfies statement 2and answer is yes
x= 1, z= 4, y = 4 this satisfies statement 2 also and the answer is no
1+2 Adding both of these
x < 2z < y x<z/2<y
we get 2x<2.5z<2y
x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No So how come I am getting E, if I reduce statement 2, by 2 and then add with statement 1
I thought we could reduce statement 2 by 2 since all are positive ?If any one could point out the flaw here I would be grateful.
(1) x < 2z < y (2) 2x < z < 2y
When you add the inequalities you get:
(x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.
sorry I didn't get it. My question was since all are positive I can definitely reduce statement 2 by 2, right? now statement 2 becomes x<(z/2)<y why am I not getting the answer if I take statement 1 as x < 2z < y , and statement 2 as x<(z/2)<y
Adding 1 and 2 we get 2x<2.5z<2y
x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So how come this is not working out if after reducing stat.2 I add it to 1?
In the same way if we were to solve :
(Q2) x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). x<2y<z
adding 1 and 2--> 2x<2.5z<2y so here the answer is E x= 1 z= 2 y = 3 this satisfies the combined statement and the answer is yes x= 2, z=2 and y =3 this satisfies the combined statement too and the answer is No
So what is the answer to Q2, C or E
Thank you for your help.
1. You can reduce 2x < z < 2y by 2 no matter whether x, y, and z are positive or negative. You can always reduce/multiply an inequality by a positive value (2 in our case).
2. Reducing by 2 and adding, though legit is not a correct way to solve. You can add the two inequalities without reducing and directly get (x+2x) < (2z+z) < (y+2y) --> 3x < 3z < 3y --> x < y < z.
3. When you reduce and then add you'll get 2x < 2.5z < 2y but you cannot plug arbitrary value of x, y, and z there. The values must satisfy all the inequalities, while x=1, z=2, y=3 and x=2, z=2, y=3 does NOT satisfy neither x < 2z < y nor 2x < z < 2y.
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
17 Apr 2014, 05:11
Thank you , the fact that after addition of the 2 statements , the values we choose should satisfy all the inequalities , made it clear.
Now for the question below : (Q2) x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). x<2y<z
Simply adding 1 and 2 gives 2x<2.5y<2z for this we have to plug and test , we have to choose values that satisfy all the 3 inequalities , This way we get the answer as C .
But if we were to multiply statement 1 by 2 and then add to statement 2 we would get 3x<3y<3z which is x<y<z , this way was much easier here.
so as we can see , in some cases it is easier to get the answer without any manipulation(reducing/ multiplication), as in first case (1) x < 2z < y (2) 2x < z < 2y
and in some cases it is easier to get the answer after manipulation( reducing/ multiplication) as in this case : x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). x<2y<z
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
17 Apr 2014, 05:26
1
This post received KUDOS
Expert's post
qlx wrote:
Thank you , the fact that after addition of the 2 statements , the values we choose should satisfy all the inequalities , made it clear.
Now for the question below : (Q2) x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). x<2y<z
Simply adding 1 and 2 gives 2x<2.5y<2z for this we have to plug and test , we have to choose values that satisfy all the 3 inequalities , This way we get the answer as C .
But if we were to multiply statement 1 by 2 and then add to statement 2 we would get 3x<3y<3z which is x<y<z , this way was much easier here.
so as we can see , in some cases it is easier to get the answer without any manipulation(reducing/ multiplication), as in first case (1) x < 2z < y (2) 2x < z < 2y
and in some cases it is easier to get the answer after manipulation( reducing/ multiplication) as in this case : x,y,z>0, is y between x and z? 1). x<(y/2)<z 2). x<2y<z
So are these deductions correct?
Yes, sometimes you need some kind of manipulations before adding/subtracting, _________________
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
28 Sep 2014, 07:17
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.
Answer: C.
Hope it's clear.
Hi Bunuel, It may be a dumb question but i am really not able to understand why Statement 1 is not sufficient
In a you have mentioned that If x = 1, y = 10, and z = 1 --> answer NO.
But if i put the values in inequality i get x < 2z < y= 1 < 2 < 10 which is absolutely fine. then why the answer is No
Re: If x, y and z are positive numbers, is z between x and y? [#permalink]
28 Sep 2014, 07:38
Expert's post
kd1989 wrote:
Bunuel wrote:
jlgdr wrote:
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.
Answer: C.
Hope it's clear.
Hi Bunuel, It may be a dumb question but i am really not able to understand why Statement 1 is not sufficient
In a you have mentioned that If x = 1, y = 10, and z = 1 --> answer NO.
But if i put the values in inequality i get x < 2z < y= 1 < 2 < 10 which is absolutely fine. then why the answer is No
It seems that you don't understand how DS questions work.
The question asks whether z is between x and y.
(1) says that x < 2z < y.
If x = 1, y = 10, and z = 1 (x < 2z < y), then z is NOT between x and y --> answer NO. If x = 1, y = 10, and z = 2 (x < 2z < y), then z IS between x and y --> answer YES.
We have two different answers to the question, which means that the statement is NOT sufficient. _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...