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Re: If x, y and z are positive numbers, is z between x and y?
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02 Dec 2014, 17:07
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



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Re: If x, y and z are positive numbers, is z between x and y?
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03 Dec 2014, 02:45
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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). Check for more here: inequalitiestipsandhints175001.htmlHope it helps.
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Re: If x, y and z are positive numbers, is z between x and y?
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06 Dec 2014, 11:27
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



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Re: If x, y and z are positive numbers, is z between x and y?
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30 Jul 2015, 17:53
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. Answer: C. Very ingenious answer! I never thought of chaining the inequalities!



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Re: If x, y and z are positive numbers, is z between x and y?
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21 Sep 2015, 17:15
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, 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!
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Re: If x, y and z are positive numbers, is z between x and y?
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21 Sep 2015, 17:29
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, \(x^2 > 4z^2 or 2z^2\)



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Re: If x, y and z are positive numbers, is z between x and y?
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21 Sep 2015, 18:28
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!
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Re: If x, y and z are positive numbers, is z between x and y?
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22 Sep 2015, 03:19
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



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Re: If x, y and z are positive numbers, is z between x and y?
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04 Nov 2015, 12:45
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. 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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22 Mar 2017, 16:26
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. 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.



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Re: If x, y and z are positive numbers, is z between x and y?
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Re: If x, y and z are positive numbers, is z between x and y?
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24 Apr 2017, 03:56
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 / 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
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Re: If x, y and z are positive numbers, is z between x and y?
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07 Jul 2017, 04:42
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. 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....



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Re: If x, y and z are positive numbers, is z between x and y?
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Re: If x, y and z are positive numbers, is z between x and y?
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11 Jul 2018, 07:41
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: 0xz2zy......we get YES. Case 2: 0zx2zy.....we get NO. Statement 1 is Not Sufficient. Statement 2: 2x < z < 2y Again visualize the number line Case 1: 0x2xzy2y......we get YES. Case 2: 0x2xyz2y.....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. Answer C. Thanks, GyM
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Re: If x, y and z are positive numbers, is z between x and y?
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11 Jul 2018, 08:27
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 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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09 Oct 2018, 10:25
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.
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