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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.
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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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
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.
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.
Answer D
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PhD in Applied Mathematics Love GMAT Quant questions and running.
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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PhD in Applied Mathematics Love GMAT Quant questions and running.
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.
What could I be doing wrong here Bunuel? Could you please help me with that. Thanks in advance!
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.
What could I be doing wrong here Bunuel? Could you please help me with that. Thanks in advance!
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 \(2Z<A+B\).
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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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?
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.
Answer: D.
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 \(|Y-A|=Y-A\) ?
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.
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.
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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Thanks & Regards, Anaira Mitch
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Re: If A<Y<Z<B, is |Y-A|< |Y-B|? &nbs
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07 Feb 2017, 20:08
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45% (02:14) correct 
