Three points T, U and V on the number line have coordinates t, u, and

Three points T, U and V on the number line have coordinates t, u, and v respectively. Is T between points U and V?

(1) \(t^2<4<u^2<v^2\). Since all parts of the inequality are non-negative, we can safely take the square root from it: \(|t| < 2 < |u| < |v|\). It's possible that T is between U and V, for example, if t=0, u=-3 and v=4, as well as it's possible that it's not, for example, t=0, u=3 and v=4. Not sufficient.

(2) \(u<0<v\). Clearly insufficient.

(1) Since from (2) u<0<v, then from (1) we get \(|t| < 2 < -u < v\). If break it: we'll get \(u<-2\) (from 2 < -u), \(-2<t<2\) (from |t| < 2), and \(2<v\). T is between U and V. Sufficient.

Answer: C.

Hope it's clear.
_________________

St1: t^2 can take values between 0 and 4 --> -2 < t < 2
When u and v have the same sign: v > u > 2 or v < u < -2 --> t does not lie between u and v
When u and v have oppositie sign: u > 2 and v < -u < -2 or u < -2 and v > -u > 2 --> t lies between u and v

Illustrating the above statements:
When t = 1; u = 3; v = 4, t does not lie between u and v
When t = 1; u = -3; v = 4, t lies between u and v
Not Sufficient.

St2: u is -ve and v is positive --> Clearly insufficient as we have no information about t

Combining St1 and St2:
-2 < t < 2;
u is negative --> u < -2
v is positive --> v > -u > 2
As shown in St1, when u and v have opposite signs, t lies between u and v.
Sufficient

Answer: C

Tip: I found drawing the number line here instead of trying to write out inequalities got to the correct answer much quicker. ~20-30 seconds

When you take the sq rt of 4 in this, shouldn't the result be +/- 2?

Are you talking about t^2 < 4? What is your question?

t^2 < 4

|t| < 2 (which is the same as -2 < t < 2).
_________________

Yeah, It is a medium level question. I bit of guess/application work is needed to solve this question. Mentioned question can be solved only if we were given both information A and B.

So Answer C

How do I check the difficult level in terms of 660-650-700-750 level? It just shows easy/ medium on the timer :/

We have three grades of difficulty: easy, which corresponds to sub-600; medium, which corresponds to 600-700; and hard, which corresponds to 700+. You can check difficulty level of a question along with the stats on it in the first post. The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question.
_________________

Bunuel Hi, can you please help clarify my doubt below:

For st. 1 when we take the sq. root. why do we take them as the absolute values for u and v? why don't we simply say take the +ve values of u and v since u and v can't be -ve value of the sq. root if they are bigger than 2?

TIA!

Why not?

From 4 < u^2, u can be more than 2 as well as less than -2, for example, -3:

4 < (-3)^2.

Generally, \(\sqrt{x^2}=|x|\). Consider this, say we have \(\sqrt{x^2}=5\). What is the value of x? Well, x can obviously be 5: \(\sqrt{5^2}=\sqrt{25}=5\) but it can also be -5: \(\sqrt{(-5)^2}=\sqrt{25}=5\). So, as you can see \(\sqrt{x^2}=5\) means that \(\sqrt{x^2}=|x|=5\), which gives x = 5 or x = -5.

So, after taking the square root from 4 < u^2 we get 2 < |u| and 2 < |u| in turn means u < -2 or u > 2.

Hope it's clear.
_________________