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by vipulgoyal » Wed Sep 30, 2015 10:24 pm
X,Y,Z ARE NUMBERS ON THE NUMBER LINE. IS X<Y<Z?
(1) | Y - X | +| Y - Z | = | Z - X |
(2) X < Z


Oa n/a

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by GMATGuruNY » Thu Oct 01, 2015 5:40 am
vipulgoyal wrote:X,Y,Z ARE NUMBERS ON THE NUMBER LINE. IS X<Y<Z?
(1) | Y - X | +| Y - Z | = | Z - X |
(2) X < Z
|a-b| = the DISTANCE between a and b.

Note the following:
The distance between a and b is the same as the distance between b and a.
In math terms:
|a-b| = |b-a|.

Statement 1, in words:
(the distance between x and y) + (the distance between y and z) = (distance between x and z).
This relationship will hold true if x≤y≤z.

Case 1: x=0, y=0, z=10
Case 2: x=0, y=1, z=10

Cases 1 and 2 satisfy both statements.
In Case 1, x=y.
In Case 2, x<y<z.
INSUFFICIENT.

The correct answer is E.
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by vipulgoyal » Sat Oct 03, 2015 10:10 am
X,Y,Z ARE NUMBERS ON THE NUMBER LINE. IS X<Y<Z?
(1) | Y - X | +| Y - Z | = | Z - X |
(2) X < Z

Thanks Mitch, Got the point

I did, this way..

lets say IS X<Y<Z? is true

now | Y - X | +| Y - Z | = | Z - X | = y-x (bcz y>x) - y+z = z-x
therefore y= y, which is true , considering the given inequality in 1. we get the true result , hence i marked A ans Ans,,... as you explained is wrong

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by Max@Math Revolution » Fri Oct 09, 2015 8:51 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

X,Y,Z ARE NUMBERS ON THE NUMBER LINE. IS X<Y<Z?
(1) | Y - X | +| Y - Z | = | Z - X |
(2) X < Z


There are 3 variables from the original condition, but only 2 equations are provided from the 2 conditions, so there is high chance that (E) is going to be our answer.
Looking at the conditions together,
The answer to the question is 'yes' for x=1, y=2, z=3, but 'no' for x=y=1,z=3, so the conditions are insufficient and the answer becomes (E).

Normally for cases where we need 3 more equations, such as original conditions with 3 variables, or 4 variables and 1 equation, or 5 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore E has a high chance of being the answer (especially about 90% of 2by2 questions where there are more than 3 variables), which is why we attempt to solve the question using 1) and 2) together. Here, there is 80% chance that E is the answer, while C has 15% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer according to DS definition, we solve the question assuming E would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

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