If |x| 0. Which of the following could

Question

If |x| < |y + 2| < |z| , y > 0 and xz > 0 . Which of the following could be true ? I. 0 < y < x < z II. 0 < x < y < z III. 0 < y + 1.5 < x < z A. I only B. II only C. III only D. I and III only E. I, II and III

LeoN88 · Answer

let's see if we can put values to get the statements true.

I Possible
0 < y < x < z
0 < 1 < 2 < 5 Satisfying all conditions.

II Possible
0 < x < y < z
0<1<3<10 Satisfying all conditions.

III Possible

0 < y + 1.5 < x < z(z can be any value no problem), also both x and z has to be +ve.

0< (0.1 + 1.5)< x <10
0< (1.6)< x <10 But x has to be less than 2.1... From equation |x| < |y + 2| so, let's put x=2
0 < (0.1 + 1.5)< 2 <10

I II III possible.

sharat1001 · Answer

What if x=1 y=2 z=5
Statement 1 is false

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LeoN88 · Answer

The question is asking could be true- so you have to find ways to make it true and move on.

sharat1001

Mo2men · Answer

The question asks what    could be true

xz > 0 means that x & z have same sign

Let x=2 & y =1 & z=4

|2| < |3| < |4|

I.  0 < 1 < 2 < 4 .......could be true.

Eliminate B & C

Let x=1/2 & y =1 & z=4

|1/2| < |3| < |4|

0 < 1/2 < 1 < 4 ..........could be true

Eliminate A & D...No need to check III as there is only one answer left

Answer: E

CrackverbalGMAT · Answer

This is a moderately difficult, ‘could be’ kind of question. In a ‘could be’ knid of question, the strategy is always to find ONE case to make a statement true, ONCE. After all, we are not trying to prove that the statement IS true, but we are only trying to figure out if there are cases which CAN make it true.

So, the trick is to take simple values in and around Zero to try and see if the statements can be made true.

We know that y is a positive number. xz>0 means that x and z are either both positive or both negative. However, in all the three statements, x and z are shown to be greater than ZERO. So, it would not be wrong for us to consider that x and z are both positive.

But while we consider positive values, we will have to be careful to pick values from all ranges, especially from the 0 to 1 range which contains proper fractions.
Also, let’s keep in mind that |x| < |y+2| < |z| , so the values that we pick should also satisfy the above inequality.

To test statement I, let’s take y = ½, x = 1 and z = 3. For these values, xz>0 and |1| < | 2.5| < |3|. So, statement I could be true. Therefore, options B and C can be eliminated.

To test statement II, let’s take x = 1, y = 2 and z = 5. For these values, xz>0 and |1| < | 4 | < |5|. Statement II could be true. Now, options A and D can be eliminated.

The only option left is E and this has to be the answer. However, let’s test statement III by taking x = 2, y = 1/10 and z = 3. For these values, xz>0 and |2| < |2.1| < |3|. Statement III could be true.

Taking values and proving statements true, is the way forward in ‘could be’ type of questions. Also look at the statements to see if they have a few clues on offer.

Hope this helps!

HoudaSR · Answer

I solved this problem using the number line: We were told that y>0 and that xz>0, which means that x and z have the same sign. So if we consider that both x and z are positive, which is the case provided in all given statements, we get the following compound inequality: x

kanikaa9 · Answer

While solving I realised 1 and 2 is surely possible and then I saw there's no option like 1 & 2 is correct so definitely it had to be option E without solving for 3rd one.

napolean92728 · Answer

Given Conditions: ∣x∣<∣y+2∣<∣z∣ y>0 xz>0 (i.e., x and z have the same sign) Step-by-step Analysis: From ∣x∣<∣y+2∣<∣z∣ , we get an order of absolute values: ∣x∣<∣y+2∣<∣z∣ This means x and z are nonzero, and their absolute values follow this inequality. Given y>0 , we know y+2>2 , so ∣y+2∣=y+2 The condition xz>0 implies x and z must have the same sign. That is, either both are positive or both are negative. Checking Each Option: Option I: 00 . We need to check if y0 ), then y+2>y , and since ∣x∣y . If x,z>0 , the inequality y0 , so y+1.5>1.5 Given ∣x∣<∣y+2∣ , and y+2>y+1.5 , it's possible that x>y+1.5 . ✔️ This is a valid possibility. Final Answer: All three conditions are possible, so the correct choice is: E. I, II, and III

sujoykrdatta · Answer

We know: |x| < |y + 2| < |z|, y > 0 and xz > 0 |x| < |z| => z is further from 0 compared to x Also, xz > 0 => both x and z are of the same sign i.e. both positive or both negative |x| < |y + 2| => (y+2) is further from 0 compared to x |y + 2| < |z| => (y+2) is closer to 0 compared to x Thus, we have the possible cases (the relative gaps between the numbers and 0 is shown correctly): (A) ----(z)----(y+2)-----(x)-----(0)------ (B) ----(0)----(x)-----(y+2)-----(z)------ (C) ----(z)------(x)---(0)------(y+2)----- (D) ----(y+2)------(0)---(x)------(z)----- Let us look at the statements: I. 0 < y < x < z - possible in (B). Since y+2 is greater than x, y can become smaller than x. Thus, you will get: ---- (0) ---- (y) --- (x) -----(y+2)----- (z) ---- II. 0 < x < y < z - possible in (B). Here, y+2 is greater than x, y can still remain greater than x. Thus, you will get: ---- (0) --- (x) ---- (y) ---(y+2)------ (z) ------ III. 0 < y + 1.5 < x < z - possible in (I) above. Since y is between 0 and x, y+1.5 can also be between 0 and x. Thus, you will get: ---- (0) --- (y+1.5) --- (x) ---(y+2)---- (z) ---- Thus, all statements can be true Ans E

luisdicampo · Answer

Deconstructing the Question We are given |x| < |y+2| < |z| y>0 xz>0 Since y>0 , we know y+2>0 so |y+2|=y+2 Also, xz>0 means x and z have the same sign The question asks which statements could be true , we only need one valid example for each statement. Step-by-step Check statement I 0

If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could

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