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If A<Y<Z<B, is YA< YB?
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If A<Y<Z<B, is YA< YB? (1) ZA < ZB (2) YA < ZB
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Originally posted by yezz on 18 Aug 2009, 17:47.
Last edited by Bunuel on 09 Aug 2012, 00:11, edited 1 time in total.
Edited the question and added the OA.




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Re: If A<Y<Z<B, is YA< YB?
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09 Aug 2012, 00:26
If A<Y<Z<B, is YA< YB? Since given that \(A<Y\) (\(YA>0\)), then \(YA=YA\); Since given that \(Y<B\) (\(YB<0\)), then \(YB=BY\); So, the question becomes: is \(YA<BY\)? > is \(2Y<A+B\)? (1) ZA < ZB. The same way as above: Since given that \(A<Z\) (\(ZA>0\)), then \(ZA=ZA\); Since given that \(Z<B\) (\(ZB<0\)), then \(ZB=BZ\); So, we have that: \(ZA<BZ\) > \(2Z<A+B\). Now, since \(Y<Z\), then \(2Y<2Z\), so \(2Y<2Z<A+B\). Sufficient. (2) YA < ZB. The same way as above: Since given that \(A<Y\) (\(YA>0\)), then \(YA=YA\); Since given that \(Z<B\) (\(ZB<0\)), then \(ZB=BZ\); So, we have that: \(YA<BZ\) > \(Y+Z<A+B\). Now, since \(Y<Z\), then \(2Y<Y+Z\) (\(2Y=Y+Y<Y+Z\) ), so \(2Y<Y+Z<A+B\). Sufficient. Answer: D. Hope it's clear.
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Re: If A<Y<Z<B, is YA< YB?
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04 Sep 2018, 18:38
yezz wrote: If A<Y<Z<B, is YA< YB?
(1) ZA < ZB
(2) YA < ZB ab = the distance between a and b. Since a < y < z < b, we get the following number line: a..........y..........z..........b Statement 1: za < zb Since the distance between y and a is less than the distance between z and a, we get: ya < za Since the distance between z and b is less than the distance between y and b, we get: zb < yb Linking the inequalities above to the inequality in Statement 1, we get: ya < za < zb < yb ya < yb Thus, the answer to the question stem is YES. SUFFICIENT. Statement 2: ya < zb Since the distance between z and b is less than the distance between y and b, we get: zb < yb Linking the inequality above to the inequality in Statement 2, we get: ya < zb < yb ya < yb Thus, the answer to the question stem is YES. SUFFICIENT.
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Re: If A<Y<Z<B, is /YA/< /YB/?
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18 Aug 2009, 19:26
yezz wrote: If A<Y<Z<B, is /YA/< /YB/?
1) /ZA/ < /ZB/
2) /YA/ < /ZB/ Consider A,Y,Z,B lying on the straight line Given that A<Y<Z<B 1) statement 1 talks nothing about Y but however it clearly brings out the fact that the distance between B&Z is greater than the distance between A&Z. Now we know that since Y<Z, Y lies closer to A than Z and farther from B than Z. hence answers our question 2) Statement 2 : it says distance between Y&A is less than the distance between Z&B. As already stated Y lies closer to A than Z and farther to B than Z, hence this answers the question as well So choice is Either statement alone is sufficient




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Re: If A<Y<Z<B, is /YA/< /YB/?
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28 Oct 2009, 16:44
I think both are insufficient Take an example A = 5 , Y= 1 , Z = 2 , B = 3 Y A <  YB 1+5 < 13 i.e. 6 < 2 ... no A = 1 , Y= 2 , Z = 3 , B = 4 21 < 24 i.e. 1 < 2 ... yes So, A not suff similarly B  not suff I think i have mistaken somewhere , can some one correct me
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Re: If A<Y<Z<B, is /YA/< /YB/?
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28 Oct 2009, 19:41
D
I drew a number line for this which helped me conceptualize the problem (4 points A, Y, Z and B). (A) Given distance between Z and B is greater than Z and A is sufficient as Y lies between A and Z. (B) Similar approach as A. Sufficient.
Please correct this if wrong.



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Re: If A<Y<Z<B, is /YA/< /YB/? 1) /ZA/ < /ZB/
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08 Aug 2012, 20:14
Bunuel
can you please clarify is this makes sense
If a < y < z < b, is y – a < y – b? (1) z – a < z – b (2) y – a < z – b
a < y < z < b a< y implies
0< ya or ya >0
2) y < B (yb ) < 0
Implies
yb  0yb
Yes proved
2) If a < y < z < b
y – a < z – b
0 < ya ya >0
Z<b zb< 0 Y< zb implies
yb < 0
(ya) 0yb
D



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Re: If A<Y<Z<B, is YA< YB?
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09 Aug 2012, 09:54
yezz wrote: If A<Y<Z<B, is YA< YB?
(1) ZA < ZB
(2) YA < ZB Use the meaning of the absolute value: distance between two points. Rephrasing the question: we are given 4 distinct points on the number line, A, Y, Z, B, from left to right (in increasing order). The question is: is the distance between A and Y smaller than the distance between Y and B? (1) AYZB would visualize a typical situation, distance between A and Z less than the distance between Z and B. Since Y is between A and Z, the answer to the question "is the distance between A and Y smaller than the distance between Y and B" is definitely YES. Sufficient. (2) AYZB this would be a typical situation, distance between A and Y, less than the distance between Z and B. Now Z is between Y and B, so again, the distance between A and Y is smaller than the distance between Y and B. Sufficient. Answer D
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Re: If A<Y<Z<B, is /YA/< /YB/?
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09 Aug 2012, 10:31
gmattokyo wrote: D
I drew a number line for this which helped me conceptualize the problem (4 points A, Y, Z and B). (A) Given distance between Z and B is greater than Z and A is sufficient as Y lies between A and Z. (B) Similar approach as A. Sufficient.
Please correct this if wrong. You are absolutely right. Sorry, I missed your post and I have just presented a similar solution.
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Re: If A<Y<Z<B, is YA< YB?
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18 Oct 2012, 04:21
Bunuel,
For statement 1, everything works until za<bz = 2z<a+b. y<z so 2y<2z ok but this does not mean 2z<a+b. we know that a<y<z<b. Now if we plug in values say 3<4<5<6 then 2(5) is not < 3+6 i.e 2z<a+b.
What could I be doing wrong here Bunuel? Could you please help me with that. Thanks in advance!



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Re: If A<Y<Z<B, is YA< YB?
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23 Oct 2012, 05:52



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Re: If A<Y<Z<B, is /YA/< /YB/?
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06 Apr 2013, 09:56
alwynjoseph wrote: yezz wrote: If A<Y<Z<B, is /YA/< /YB/?
1) /ZA/ < /ZB/
2) /YA/ < /ZB/ Consider A,Y,Z,B lying on the straight line Given that A<Y<Z<B 1) statement 1 talks nothing about Y but however it clearly brings out the fact that the distance between B&Z is greater than the distance between A&Z. Now we know that since Y<Z, Y lies closer to A than Z and farther from B than Z. hence answers our question 2) Statement 2 : it says distance between Y&A is less than the distance between Z&B. As already stated Y lies closer to A than Z and farther to B than Z, hence this answers the question as well So choice is Either statement alone is sufficient I've struggled with this question for 10 min., then I looked at first sentence and the light bulb turned on. It becomes quite, quite a simple task the minute you draw a straight line... thank you!
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Re: If A<Y<Z<B, is YA< YB?
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06 Apr 2013, 19:30
By drawing a line with the given condition, it is easy to see ya < yb



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Re: If A<Y<Z<B, is YA< YB?
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25 Jun 2013, 17:33
If A<Y<Z<B, is YA< YB?
Given the above, YA< YB (YA) < (YB) YA < Y+B is 2Y < B+A
(1) ZA < ZB (ZA) < (ZB) ZA <Z+b 2Z < B+A
A<Y<Z<B. So, Z is greater than Y. if 2Z < B+A then 2Y (Y being less than Z) must also be less than B+A SUFFICIENT
(2) YA < ZB (YA) < (ZB) YA < Z + B Y + Z < B + A
A<Y<Z<B. Y + Z is less than B + A. If Y + Z is less than B + A then 2Y must be less than B + A. 2Y is Y + Y but Y + Z is Y + a number greater than Y. If Y + a Number greater than Y is less than B + A then Y + Y must be less than Y.
I noticed it is a bit different from the way others have solved this problem. Is my way correct?
Thanks!



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Re: If A<Y<Z<B, is YA< YB?
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14 Jun 2015, 11:07
Bunuel wrote: If A<Y<Z<B, is YA< YB?
Since given that \(A<Y\) (\(YA>0\)), then \(YA=YA\); Since given that \(Y<B\) (\(YB<0\)), then \(YB=BY\);
So, the question becomes: is \(YA<BY\)? > is \(2Y<A+B\)?
(1) ZA < ZB. The same way as above: Since given that \(A<Z\) (\(ZA>0\)), then \(ZA=ZA\); Since given that \(Z<B\) (\(ZB<0\)), then \(ZB=BZ\);
So, we have that: \(ZA<BZ\) > \(2Z<A+B\). Now, since \(Y<Z\), then \(2Y<2Z\), so \(2Y<2Z<A+B\). Sufficient.
(2) YA < ZB. The same way as above: Since given that \(A<Y\) (\(YA>0\)), then \(YA=YA\); Since given that \(Z<B\) (\(ZB<0\)), then \(ZB=BZ\);
So, we have that: \(YA<BZ\) > \(Y+Z<A+B\). Now, since \(Y<Z\), then \(2Y<Y+Z\) (\(2Y=Y+Y<Y+Z\) ), so \(2Y<Y+Z<A+B\). Sufficient.
Answer: D.
Hope it's clear. Hi Bunuel , I am learning concepts of Mod . Just need one small clarification : given that \(A<Y\) (\(YA>0\)), then \(YA=YA\) I want to know if Y is positive and A is negative then still Is \(YA=YA\) ? Thanks



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If A<Y<Z<B, is YA< YB?
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14 Jun 2015, 11:24
abhinav008 wrote: Hi Bunuel , I am learning concepts of Mod . Just need one small clarification : given that \(A<Y\) (\(YA>0\)), then \(YA=YA\)
I want to know if Y is positive and A is negative then still Is \(YA=YA\) ?
Thanks
Hello abhinav008Yes if Y is positive and A is negative then \(YA=YA\) more general rule (work in described case and in case when Y and A are both positive): if \(YA > 0\) then \(YA=YA\)
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If A<Y<Z<B, is YA< YB?
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22 Apr 2016, 23:33
Bunuel wrote: If A<Y<Z<B, is YA< YB?
Since given that \(A<Y\) (\(YA>0\)), then \(YA=YA\); Since given that \(Y<B\) (\(YB<0\)), then \(YB=BY\);
So, the question becomes: is \(YA<BY\)? > is \(2Y<A+B\)?
(1) ZA < ZB. The same way as above: Since given that \(A<Z\) (\(ZA>0\)), then \(ZA=ZA\); Since given that \(Z<B\) (\(ZB<0\)), then \(ZB=BZ\);
So, we have that: \(ZA<BZ\) > \(2Z<A+B\). Now, since \(Y<Z\), then \(2Y<2Z\), so \(2Y<2Z<A+B\). Sufficient.
(2) YA < ZB. The same way as above: Since given that \(A<Y\) (\(YA>0\)), then \(YA=YA\); Since given that \(Z<B\) (\(ZB<0\)), then \(ZB=BZ\);
So, we have that: \(YA<BZ\) > \(Y+Z<A+B\). Now, since \(Y<Z\), then \(2Y<Y+Z\) (\(2Y=Y+Y<Y+Z\) ), so \(2Y<Y+Z<A+B\). Sufficient.
Answer: D.
Hope it's clear. Bunuel, your explanations of 'absolute value questions' using number lines have taken my ability to solve such questions to amazing heights. I could solve this question easily using number line this leaves me wondering why you used algebraic approach instead of the number line approach. I want to know whether there are scenarios in which one should prefer algebraic approach to number line approach.



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Re: If A<Y<Z<B, is YA< YB?
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21 May 2016, 10:46
yezz wrote: If A<Y<Z<B, is YA< YB?
(1) ZA < ZB
(2) YA < ZB An algebric way is nice, but if you think about the given equations as intervals on a line, it becomes much easier to solve. notice that in my solution, since a,c,b are intervals they are all positive.
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Re: If A<Y<Z<B, is YA< YB?
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07 Feb 2017, 20:08
Because we know that a < y < z < b, we know abs(y  a) = y  a abs(y  b) = b  y (since y  b is negative) We can rephrase the question: Is y  a < b  y? or Is 2y < a + b? Statement (1) can be rephrased: z  a < b  z, so 2z < a + b. We also know that since y < z, then 2y < 2z. So 2y < 2z < a + b. (1) is sufficient. Statement (2) can be rephrased: y  a < b  z, so y + z < a + b. Since y < z, we can add y to both sides: 2y < y + z. So 2y < y + z < a + b, so (2) is sufficient as well. Hence D.
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