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

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

sharat1001

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

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!

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<y+2<z

On the number line this coumpound inequality can be presented as following:

0...........x.......y+2...........z

y can be either to the right of x or to the left of x (both cases are possible, therefore both statements 1 and 2 can be true).

0..........y.....x.......y+2..........z or
0........x.....y.....y+2..............z

Now for the 3rd statement y+1,5 will be to the left of y+2, either to the left or to the right of x

0...........x......y+1,5........y+2........z or
0...........y+1,5.......x.......y+2........z (case presented in statement 3)

Therefore, statement 3 can also be true.

Correct answer is E