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# If A<Y<Z<B, is |Y-A|< |Y-B|?

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If A<Y<Z<B, is |Y-A|< |Y-B|?  [#permalink]

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Updated on: 09 Aug 2012, 00:11
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If A<Y<Z<B, is |Y-A|< |Y-B|?

(1) |Z-A| < |Z-B|

(2) |Y-A| < |Z-B|

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 |Y-A|< |Y-B|?  [#permalink]

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09 Aug 2012, 00:26
36
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If A<Y<Z<B, is |Y-A|< |Y-B|?

Since given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$;
Since given that $$Y<B$$ ($$Y-B<0$$), then $$|Y-B|=B-Y$$;

So, the question becomes: is $$Y-A<B-Y$$? --> is $$2Y<A+B$$?

(1) |Z-A| < |Z-B|. The same way as above:
Since given that $$A<Z$$ ($$Z-A>0$$), then $$|Z-A|=Z-A$$;
Since given that $$Z<B$$ ($$Z-B<0$$), then $$|Z-B|=B-Z$$;

So, we have that: $$Z-A<B-Z$$ --> $$2Z<A+B$$. Now, since $$Y<Z$$, then $$2Y<2Z$$, so $$2Y<2Z<A+B$$. Sufficient.

(2) |Y-A| < |Z-B|. The same way as above:
Since given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$;
Since given that $$Z<B$$ ($$Z-B<0$$), then $$|Z-B|=B-Z$$;

So, we have that: $$Y-A<B-Z$$ --> $$Y+Z<A+B$$. Now, since $$Y<Z$$, then $$2Y<Y+Z$$ ($$2Y=Y+Y<Y+Z$$ ), so $$2Y<Y+Z<A+B$$. Sufficient.

Hope it's clear.
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Re: If A<Y<Z<B, is |Y-A|< |Y-B|?  [#permalink]

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04 Sep 2018, 18:38
3
yezz wrote:
If A<Y<Z<B, is |Y-A|< |Y-B|?

(1) |Z-A| < |Z-B|

(2) |Y-A| < |Z-B|

|a-b| = the distance between a and b.

Since a < y < z < b, we get the following number line:
a..........y..........z..........b

Statement 1: |z-a| < |z-b|
Since the distance between y and a is less than the distance between z and a, we get:
|y-a| < |z-a|
Since the distance between z and b is less than the distance between y and b, we get:
|z-b| < |y-b|
Linking the inequalities above to the inequality in Statement 1, we get:
|y-a| < |z-a| < |z-b| < |y-b|
|y-a| < |y-b|
Thus, the answer to the question stem is YES.
SUFFICIENT.

Statement 2: |y-a| < |z-b|
Since the distance between z and b is less than the distance between y and b, we get:
|z-b| < |y-b|
Linking the inequality above to the inequality in Statement 2, we get:
|y-a| < |z-b| < |y-b|
|y-a| < |y-b|
Thus, the answer to the question stem is YES.
SUFFICIENT.

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Re: If A<Y<Z<B, is /Y-A/< /Y-B/?  [#permalink]

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18 Aug 2009, 19:26
5
3
yezz wrote:
If A<Y<Z<B, is /Y-A/< /Y-B/?

1) /Z-A/ < /Z-B/

2) /Y-A/ < /Z-B/

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
##### General Discussion
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Re: If A<Y<Z<B, is /Y-A/< /Y-B/?  [#permalink]

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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| < | Y-B|
|1+5| < |1-3| i.e. 6 < 2 ... no

A = 1 , Y= 2 , Z = 3 , B = 4

|2-1| < |2-4| 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 /Y-A/< /Y-B/?  [#permalink]

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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.

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Re: If A<Y<Z<B, is /Y-A/< /Y-B/? 1) /Z-A/ < /Z-B/  [#permalink]

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08 Aug 2012, 20:14
1
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< y-a or y-a >0

2) y < B  (y-b ) < 0

Implies

y-b --------------- 0-------------y-b

Yes proved

2) If a < y < z < b

|y – a| < |z – b|

0 < y-a  y-a >0

Z<b  z-b< 0
Y< z-b implies

y-b < 0

(y-a) -------------0--------------y-b

D
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Re: If A<Y<Z<B, is |Y-A|< |Y-B|?  [#permalink]

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09 Aug 2012, 09:54
3
1
yezz wrote:
If A<Y<Z<B, is |Y-A|< |Y-B|?

(1) |Z-A| < |Z-B|

(2) |Y-A| < |Z-B|

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) A---Y--Z------B 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) A---Y-Z----B 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.

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Re: If A<Y<Z<B, is /Y-A/< /Y-B/?  [#permalink]

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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.

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 |Y-A|< |Y-B|?  [#permalink]

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18 Oct 2012, 04:21
Bunuel,

For statement 1, everything works until z-a<b-z = 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.

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Re: If A<Y<Z<B, is |Y-A|< |Y-B|?  [#permalink]

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23 Oct 2012, 05:52
2
liarish wrote:
Bunuel,

For statement 1, everything works until z-a<b-z = 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.

You derived yourself that $$2Z<A+B$$ and later you are negating that (blue parts).

We know from (1) that $$2Z<A+B$$ (i) .
We also know that $$Y<Z$$, so $$2Y<2Z$$ (ii).

Now, combine (i) and (ii): 2Z is less than A+B and 2Y is less than 2Z, thus 2Y is less than A+B ($$2Y<2Z<A+B$$).

Also, numbers you chose does not satisfy $$2Z<A+B$$.

Hope it's clear.
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Re: If A<Y<Z<B, is /Y-A/< /Y-B/?  [#permalink]

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06 Apr 2013, 09:56
alwynjoseph wrote:
yezz wrote:
If A<Y<Z<B, is /Y-A/< /Y-B/?

1) /Z-A/ < /Z-B/

2) /Y-A/ < /Z-B/

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 |Y-A|< |Y-B|?  [#permalink]

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06 Apr 2013, 19:30
By drawing a line with the given condition, it is easy to see |y-a| < |y-b|
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Re: If A<Y<Z<B, is |Y-A|< |Y-B|?  [#permalink]

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25 Jun 2013, 17:33
1
If A<Y<Z<B, is |Y-A|< |Y-B|?

Given the above, |Y-A|< |Y-B|
(Y-A) < -(Y-B)
Y-A < -Y+B
is 2Y < B+A

(1) |Z-A| < |Z-B|
(Z-A) < -(Z-B)
Z-A <-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) |Y-A| < |Z-B|
(Y-A) < -(Z-B)
Y-A < -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 |Y-A|< |Y-B|?  [#permalink]

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14 Jun 2015, 11:07
Bunuel wrote:
If A<Y<Z<B, is |Y-A|< |Y-B|?

Since given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$;
Since given that $$Y<B$$ ($$Y-B<0$$), then $$|Y-B|=B-Y$$;

So, the question becomes: is $$Y-A<B-Y$$? --> is $$2Y<A+B$$?

(1) |Z-A| < |Z-B|. The same way as above:
Since given that $$A<Z$$ ($$Z-A>0$$), then $$|Z-A|=Z-A$$;
Since given that $$Z<B$$ ($$Z-B<0$$), then $$|Z-B|=B-Z$$;

So, we have that: $$Z-A<B-Z$$ --> $$2Z<A+B$$. Now, since $$Y<Z$$, then $$2Y<2Z$$, so $$2Y<2Z<A+B$$. Sufficient.

(2) |Y-A| < |Z-B|. The same way as above:
Since given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$;
Since given that $$Z<B$$ ($$Z-B<0$$), then $$|Z-B|=B-Z$$;

So, we have that: $$Y-A<B-Z$$ --> $$Y+Z<A+B$$. Now, since $$Y<Z$$, then $$2Y<Y+Z$$ ($$2Y=Y+Y<Y+Z$$ ), so $$2Y<Y+Z<A+B$$. Sufficient.

Hope it's clear.

Hi Bunuel ,
I am learning concepts of Mod . Just need one small clarification :
given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$

I want to know if Y is positive and A is negative then still
Is $$|Y-A|=Y-A$$ ?

Thanks
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If A<Y<Z<B, is |Y-A|< |Y-B|?  [#permalink]

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14 Jun 2015, 11:24
1
abhinav008 wrote:

Hi Bunuel ,
I am learning concepts of Mod . Just need one small clarification :
given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$

I want to know if Y is positive and A is negative then still
Is $$|Y-A|=Y-A$$ ?

Thanks

Hello abhinav008

Yes if Y is positive and A is negative then $$|Y-A|=Y-A$$

more general rule (work in described case and in case when Y and A are both positive):
if $$Y-A > 0$$ then $$|Y-A|=Y-A$$
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If A<Y<Z<B, is |Y-A|< |Y-B|?  [#permalink]

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22 Apr 2016, 23:33
Bunuel wrote:
If A<Y<Z<B, is |Y-A|< |Y-B|?

Since given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$;
Since given that $$Y<B$$ ($$Y-B<0$$), then $$|Y-B|=B-Y$$;

So, the question becomes: is $$Y-A<B-Y$$? --> is $$2Y<A+B$$?

(1) |Z-A| < |Z-B|. The same way as above:
Since given that $$A<Z$$ ($$Z-A>0$$), then $$|Z-A|=Z-A$$;
Since given that $$Z<B$$ ($$Z-B<0$$), then $$|Z-B|=B-Z$$;

So, we have that: $$Z-A<B-Z$$ --> $$2Z<A+B$$. Now, since $$Y<Z$$, then $$2Y<2Z$$, so $$2Y<2Z<A+B$$. Sufficient.

(2) |Y-A| < |Z-B|. The same way as above:
Since given that $$A<Y$$ ($$Y-A>0$$), then $$|Y-A|=Y-A$$;
Since given that $$Z<B$$ ($$Z-B<0$$), then $$|Z-B|=B-Z$$;

So, we have that: $$Y-A<B-Z$$ --> $$Y+Z<A+B$$. Now, since $$Y<Z$$, then $$2Y<Y+Z$$ ($$2Y=Y+Y<Y+Z$$ ), so $$2Y<Y+Z<A+B$$. Sufficient.

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 |Y-A|< |Y-B|?  [#permalink]

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21 May 2016, 10:46
yezz wrote:
If A<Y<Z<B, is |Y-A|< |Y-B|?

(1) |Z-A| < |Z-B|

(2) |Y-A| < |Z-B|

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 |Y-A|< |Y-B|?  [#permalink]

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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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Re: If A<Y<Z<B, is |Y-A|< |Y-B|?   [#permalink] 07 Feb 2017, 20:08
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