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# If r, s, and t are integers such that r^2 > s^2 > t^2, which of the

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If r, s, and t are integers such that r^2 > s^2 > t^2, which of the  [#permalink]

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Updated on: 05 May 2017, 03:57
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If r, s, and t are integers such that r^2 > s^2 > t^2, which of the following must be true?

I. r > s
II. |r| > |s|
III. (t + 1)^2 ≥ r^2

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

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Originally posted by Bunuel on 30 Jan 2017, 10:46.
Last edited by Bunuel on 05 May 2017, 03:57, edited 1 time in total.
Edited the question.
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Re: If r, s, and t are integers such that r^2 > s^2 > t^2, which of the  [#permalink]

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30 Jan 2017, 11:12
Bunuel wrote:
If r, s, and t are integers such that r^2 > s^2 > t^2, which of the following must be true?

I. r > s
II. | r | > | s |
III. (t + 1)2 ≥ r2

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

Case 1 r=-5, s = -4, t=-3 r2> s2> t2 r<s so I is out
But II will always be true.
III is also negated with this example. So, only II is correct, thus A.
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Re: If r, s, and t are integers such that r^2 > s^2 > t^2, which of the  [#permalink]

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30 Jan 2017, 11:27
Bunuel wrote:
If r, s, and t are integers such that r^2 > s^2 > t^2, which of the following must be true?

I. r > s
II. | r | > | s |
III. (t + 1)2 ≥ r2

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

$$r^2 > s^2 > t^2$$

Let $$r = 5$$ , $$s = 4$$ & $$t = 3$$

So, $$r^2 > s^2 > t^2$$ = 25 > 16 > 9

Again , $$r^2 > s^2 > t^2$$

Let $$r = -5$$ , $$s = -4$$ & $$t = -3$$

So, $$r^2 > s^2 > t^2$$ = 25 > 16 > 9

Now, check the options -

I. $$r > s$$ - Not true if $$r = -5$$ & $$s = -4$$

II. | r | > | s | - True for both +ve and -ve values of r & s

III. $$(t + 1)^2 ≥ r^2$$ - Not true

Hence, answer will be (A) only II.

PS : Assuming $$(t + 1)2 ≥ r2$$ as $$(t + 1)^2 ≥ r^2$$
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Abhishek....

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Re: If r, s, and t are integers such that r^2 > s^2 > t^2, which of the  [#permalink]

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30 Jan 2017, 12:48
What does the third case III. (t + 1)2 ≥ r2 mean?

2(t+1) >= 2r or (t+1)^2 >= r^2?
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Re: If r, s, and t are integers such that r^2 > s^2 > t^2, which of the  [#permalink]

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20 May 2017, 02:24
Abhishek009 wrote:
Bunuel wrote:
If r, s, and t are integers such that r^2 > s^2 > t^2, which of the following must be true?

I. r > s
II. | r | > | s |
III. (t + 1)2 ≥ r2

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

$$r^2 > s^2 > t^2$$

Let $$r = 5$$ , $$s = 4$$ & $$t = 3$$

So, $$r^2 > s^2 > t^2$$ = 25 > 16 > 9

Again , $$r^2 > s^2 > t^2$$

Let $$r = -5$$ , $$s = -4$$ & $$t = -3$$

So, $$r^2 > s^2 > t^2$$ = 25 > 16 > 9

Now, check the options -

I. $$r > s$$ - Not true if $$r = -5$$ & $$s = -4$$

II. | r | > | s | - True for both +ve and -ve values of r & s

III. $$(t + 1)^2 ≥ r^2$$ - Not true

Hence, answer will be (A) only II.

PS : Assuming $$(t + 1)2 ≥ r2$$ as $$(t + 1)^2 ≥ r^2$$

I really find it difficult to figure out when to apply the mod properties? Like in this case when we are asked that |r|>|s|, then it holds true only if r and s are positive. In case they are negative then |r| equals -r and |s| equals -s, so -r<-s. Is this not the case?

I understand by mod we mean only positive values. But i get confuse whenever i see |x| equals -x,x<0. I do not understand when to consider only positive values and when to consider |x| equals -x,x<0 in case of mod.

It will be really helpful if you can help
PS: I have read the theory also.
Re: If r, s, and t are integers such that r^2 > s^2 > t^2, which of the &nbs [#permalink] 20 May 2017, 02:24
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