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

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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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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
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
Bunuel wrote:
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.

Hope it's clear.

When you take the sq rt of 4 in this, shouldn't the result be +/- 2?
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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ailintan wrote:
Bunuel wrote:
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.

Hope it's clear.

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

t^2 < 4

|t| < 2 (which is the same as -2 < t < 2).
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
Bunuel wrote:
ailintan wrote:
Bunuel wrote:
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.

Hope it's clear.

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

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.

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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
How do I check the difficult level in terms of 660-650-700-750 level? It just shows easy/ medium on the timer :/
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
kritikalal wrote:
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.
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
Bunuel wrote:
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.

Hope it's clear.

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!
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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Suruchim12 wrote:
Bunuel wrote:
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.

Hope it's clear.

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.
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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Re: Three points T, U and V on the number line have coordinates t, u, and [#permalink]
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