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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]

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15 Oct 2013, 16:30

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

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]

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10 Nov 2013, 15:52

7

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]

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11 Nov 2013, 21:24

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]

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22 Feb 2014, 08:00

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]

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14 Apr 2014, 05:42

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.

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]

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16 Apr 2014, 13:05

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

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]

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17 Apr 2014, 06:11

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

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]

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28 Sep 2014, 08:17

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

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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