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If x, y and z are positive numbers, is z between x and y?
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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
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Originally posted by vcbabu on 04 Jun 2009, 10:42.
Last edited by Bunuel on 15 Oct 2013, 23:15, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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Re: x,y,z>0, is y between x and z? 1). x<(y/2)<z 2).
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15 Oct 2013, 23:24
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.
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Re: If x, y and z are positive numbers, is z between x and y?
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10 Nov 2013, 14:52
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
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Re: inequality
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04 Jun 2009, 13:23
x<(y/2)<z case 1 x y z 1.0 1.1 1.2 TRUE case 1 x y z 1.0 1.2 1.1 FALSE INSUFFICIENT x<2y<z amkes it necessary for y to be between x and z SUFFICIENT IMO B
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Re: inequality
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06 Jun 2009, 16:23
Quote: x<2y<z amkes it necessary for y to be between x and z SUFFICIENT Say, x=5 y=3 z=10 x<2y<z is true ?5<6<10 but we can't say that y is b/w x and z..
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Re: inequality
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08 Jun 2009, 18:56
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



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



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Re: If x, y and z are positive numbers, is z between x and y?
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16 Oct 2013, 03:46
Hmm wonder where i went wrong in my reasonining for the second statement. Thanks Bunuel PS. I think theres a small typo in the answer think it should be x, y and z only after divided by 3 Thanks again Cheers! J Posted from my mobile device



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Re: If x, y and z are positive numbers, is z between x and y?
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11 Nov 2013, 12:32
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. Answer: C.



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Re: If x, y and z are positive numbers, is z between x and y?
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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
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Re: If x, y and z are positive numbers, is z between x and y?
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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...)
THanks for the help



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



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Re: If x, y and z are positive numbers, is z between x and y?
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14 Apr 2014, 05:58
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. Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.htmlHope this helps.
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Re: If x, y and z are positive numbers, is z between x and y?
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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. Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.htmlHope this helps. 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.



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Re: If x, y and z are positive numbers, is z between x and y?
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17 Apr 2014, 00:28
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. Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.htmlHope this helps. 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. Hope it's clear.
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Re: If x, y and z are positive numbers, is z between x and y?
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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
So are these deductions correct?



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Re: If x, y and z are positive numbers, is z between x and y?
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17 Apr 2014, 05:26
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,
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Re: If x, y and z are positive numbers, is z between x and y?
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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



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Re: If x, y and z are positive numbers, is z between x and y?
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28 Sep 2014, 07:38
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.
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Re: If x, y and z are positive numbers, is z between x and y?
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20 Oct 2014, 21:08
[quote="vcbabu"]If x, y and z are positive numbers, is z between x and y? (1) x < 2z < y (2) 2x < z < 2y 1  This would conclude that y is greater than z i.e. y > z. No information can be deduced about the relationship between x and z. x can be 3 and z can be 2 or x can be 2 and z can be 3 There are infinite options. 2  Here, we can conclude that z > x but nothing can be said of z and y. Combining both we get that y >z >x Thus C
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Re: If x, y and z are positive numbers, is z between x and y? &nbs
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