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If S is a data set of three numbers x, y and z, is the range of the nu

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If S is a data set of three numbers x, y and z, is the range of the nu [#permalink]

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New post 05 Sep 2017, 00:18
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

61% (01:17) correct 39% (00:51) wrong based on 57 sessions

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Re: If S is a data set of three numbers x, y and z, is the range of the nu [#permalink]

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New post 05 Sep 2017, 00:30
Bunuel wrote:
If S is a data set of three numbers x, y and z, is the range of the numbers in S less than 2?

(1) x – y < 1
(2) y – z < 2


(1) no information abt z..not sufficient

(2) no information abt x..not sufficient

on combining if (x,y,z) are (1,1/2,1/4) then range is less than 2
however if (x,y,z) are (0,5,20) then range is greater than 2

E
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Re: If S is a data set of three numbers x, y and z, is the range of the nu [#permalink]

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New post 05 Sep 2017, 01:50
Bunuel wrote:
If S is a data set of three numbers x, y and z, is the range of the numbers in S less than 2?

(1) x – y < 1
(2) y – z < 2


Interesting question.

Statement 1 & 2 itself seem INSUFFICIENT, so this is the battle between C and E.

- Question : range [x,y,z] is less than 2, or \(z-y<2\)
- #1 : \(x-y < 1\) -> \(x-1 < y\).
- #2 : \(y-z < 2\) -> \(Y < z+2\)
- Combine 2 inequality : \(x-1 < y < z+2\).
- Plug one number as y. Let say \(y = 3\). -> \(x<4\) and \(z>2\), still insufficient.

E.

Wdyt?
Btw, the question doesn't tell us about the number x,y,z itself. But I assume that x is the smallest and z is the biggest. Am I correct?
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If S is a data set of three numbers x, y and z, is the range of the nu [#permalink]

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New post 10 Sep 2017, 01:01
I think question stem should tell us whether statements can be negative or not. Otherwise it is time-consuming.

Let's analyze

Case 1
Stmtnt 1: x-y<1 If this can only be positive, I mean between 0 and 1, then x>y. Insufficient

Stmtnt 2 : y-z<2 again, if this can only be positive, between 0 and 2, then y>z. Insufficient

If we combine both conclusions we'll get x>y>z meaning that we can write it in ascending order {z,y,x}. In this case range will be x-z. And if we solve both equations:

x-y<1
y-z<2

We add them and get x-z<3 , can be less than 2 or not. Insufficient Ans E

Case 2

If both statements can also be negative.we should check negative cases

Stmtnt 1: x-y<1 Assume that this is negative,then x<y, Insufficient

Stmtnt 2 :y-z<2 again, Assume that this is negative, then y<z, Insufficient

If we combine both we get x<y<z and write it in ascending order {x,y,z}. In this case range will be z-x. If x-z<3,then we flip it z-x>-3. Again we don't know exactly if it is less than 2 or not. Insufficient Ans E

Case 3

Now, we need to check another scenario 1st statement positive, 2nd negative.

x-y<1 can only be positive: x>y Insufficient

y-z<2, is negative: y<z Insufficient

Combining them x>y ,y<z then x>z. In ascending order it will be {y,z,x}, range will be x-y which less than 1,also less than 2. [b]Sufficient. Ans C[/b]

Case 4

1st stmnt negative, 2nd positive

x-y<1 negative, x<y Insufficient

y-z<2 positive, y>z Insufficient

Combining them x<z<y In ascending order it will be {x,z,y}, range y-x. If x-y<1 then y-x>-1. Again, we don't know the exact answer. Insufficient Ans E

I don't know what I'm missing here. :lol: But it seems one can't do this within 2 mins.
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If S is a data set of three numbers x, y and z, is the range of the nu   [#permalink] 10 Sep 2017, 01:01
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