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# If a < y < z < b, is |y - a | < |y - b | ?

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Re: If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
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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
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Re: If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
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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]
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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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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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]
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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Re: If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
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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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Re: If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
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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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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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If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
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the first thing you should do here is rephrase the question.

big takeaway:
if you see the ABSOLUTE VALUE OF A DIFFERENCE, you should read it as the DISTANCE BETWEEN THE TWO THINGS on the number line.

therefore, |y - a| is the distance between y and a, and so on.

hence:

QUESTION: is y closer to a than to b ?
(1) z is closer to a than to b
(2) y is closer to a than z is to b

statement (1):
note that the distance y-a is less than the distance z-a, because y is placed between a and z.
also, note that the distance y-b is greater than the distance z-b, since z lies between y and b.
therefore:
distance y-a < distance z-a < distance z-b < distance y-b (note that these are color-coded to the statements above)
so, distance y-a < distance y-b
"yes"
SUFFICIENT

statement (2):
same thing as statement (1), except for the second term of the inequality above isn't there anymore.
i.e., distance y-a < distance z-b < distance y-b.
SUFFICIENT

ans = (d)

Originally posted by RashedVai on 28 Feb 2020, 10:36.
Last edited by RashedVai on 31 Jul 2023, 13:26, edited 1 time in total.
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Re: If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
yezz wrote:
If $$a < y < z < b$$, is $$|y - a | < |y - b |$$ ?

(1) $$|z - a | < |z - b |$$

(2) $$|y - a | < |z - b |$$

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
We can figure out the followings from the original condition.
|y-a| < |z-a| and |z-b| < |y-b|.

Condition 1)
We have |y-a| < |z-a| < |z-b| < |y-b|.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
We have |y-a| < |z-b| < |y-b|.

Since condition 2) yields a unique solution, it is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
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MathRevolution wrote:
yezz wrote:
If $$a < y < z < b$$, is $$|y - a | < |y - b |$$ ?

(1) $$|z - a | < |z - b |$$

(2) $$|y - a | < |z - b |$$

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
We can figure out the followings from the original condition.
|y-a| < |z-a| and |z-b| < |y-b|.

Condition 1)
We have |y-a| < |z-a| < |z-b| < |y-b|.

Since condition 1) yields a unique solution, it is sufficient.

Condition 2)
We have |y-a| < |z-b| < |y-b|.

Since condition 2) yields a unique solution, it is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

No offense, your approach don't work in most of the questions. You only selectivly post to few questions. I have been following your posts. GMAT is not that easy. So don't ALWAYS request us to "forget about conventional approach".

The variable approach will work on maximum 2 questions out of 14/15 questions in the entire test. How do I get a Q51 depending on your approach?

But our conventional approach like algebra, number picking and visualization can do all the DS questions on the test.
Bunuel generis

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Re: If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
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 Bunnel,
after taking out the absolute sign, we're moving Y from RHS to LHS but what if Y was -ve ?
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Re: If a < y < z < b, is |y - a | < |y - b | ? [#permalink]
(1) |z-a| < |z-b|

a<y<z<b

The expression therefore should be:

z-a < b-z

Now going back to expression in question:

|y-a| < |y-b| ?
y-a < b-y?

Comparing (b-y) with (b-z), we can easily verify that (b-y) > (b-z)
Now how about (z-a) and (y-a)

We can see that (z-a) > (y-a)

We can see that expression 1 says that:

LARGER than (y-a) < SMALLER THAN (b-y)

Therefore we can easily say that since (y-a) is smaller than the condition provided, and since (b-y) is larger than the condition provided, the inequality will hold true always.

SUFFICIENT

(2) |y-a| < |z-b|

y-a < b-z

Again (b-z) is smaller than (b-y)

Since y-a is still smaller than a value smaller than (b-y); therefore, (b-y) is greater than (y-a) always.

SUFFICIENT

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