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If 0<r<1<s<2. Which of the following must be less than 1.

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If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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If 0<r<1<s<2. Which of the following must be less than 1.

I. r/s
II. rs
III. s-r

A. I only
B. II Only
C. III Only
D. I and II
E. I and III
[Reveal] Spoiler: OA

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Re: less than 1 [#permalink]

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If 0 < r < 1 < s < 2, which of the following must be less than 1?
I. r/s
II. rs
III. s - r

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

Notice that we are asked "which of the following MUST be lees than 1, not COULD be less than 1. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

Given: \(0 < r < s\) --> divide by \(s\) (we can safely do this since we know that \(s>0\)) --> \(\frac{r}{s}<1\), so I must be true;

II. rs: if \(r=\frac{9}{10}<1\) and \(1<(s=\frac{10}{9})<2\) then \(rs=1\), so this statement is not alway true;

III. s-r: if \(r=0.5<1\) and 1\(<(s=1.5)<2\) then \(s-r=1\), so this statement is not alway true.

Answer: A (I only).

Check our new Must or Could be True Questions to practice more: search.php?search_id=tag&tag_id=193

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Re: less than 1 [#permalink]

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New post 25 Feb 2012, 02:32
So the best strategy is to pick numbers in these ques?

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Re: less than 1 [#permalink]

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New post 25 Feb 2012, 02:43
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devinawilliam83 wrote:
So the best strategy is to pick numbers in these ques?


Number picking is a good strategy for such kind of questions, though as you can see we proved that I is true with algebra (so to prove that a statement MUST be true you might need an algebraic or logical/conceptual approach).

Generally it really depends on the problem to pick the way of handling it. Check the link in my previous post to see bunch of Must or Could be True Questions solved using different approaches.
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Re: If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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New post 15 Sep 2013, 03:25
Hello Experts,

Can we evaluate the value of rs by below method

Given:
0<r<1 ---(1)
1<s<2 ---(2)

It can be seen that both r and s are positives, hence multiplying r and s will not have any effect on inequality

Hence,
Multiplying 1 and 2
0<rs<2 ---(3)

Am I right in concluding equation (3). Please clarify.

Thanks
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Re: If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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New post 15 Sep 2013, 03:44
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imhimanshu wrote:
Hello Experts,

Can we evaluate the value of rs by below method

Given:
0<r<1 ---(1)
1<s<2 ---(2)

It can be seen that both r and s are positives, hence multiplying r and s will not have any effect on inequality

Hence,
Multiplying 1 and 2
0<rs<2 ---(3)

Am I right in concluding equation (3). Please clarify.

Thanks


Though I am not an expert, the answer to your question is Yes,
And, in fact we can solve all the equations in I, II, III by this method.

If 0<r<1 and 1<s<2
then 0<r/s<1 ----- I) True

Again as you suggested 0<rs<2 --- II) Hence False since it can take values more than 1 as well.

For III) multiply r inequality by -1
we get -1 < -r < 0 and 1<s<2
adding the inequalities

0 < s-r < 2 ------ Again this inequality can take value more than 1 hence False..

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Re: If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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Re: If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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New post 20 Jul 2016, 18:14
Curious if this can be solved this way:

Given: 0<r<1<s<2

1. r/s <1
= r<s (yes)

2. rs<1
= r<1/s (no)

3. s-r<1
= s<r (no)

So only 1 must be true?

Bunuel wrote:
If 0 < r < 1 < s < 2, which of the following must be less than 1?
I. r/s
II. rs
III. s - r

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

Notice that we are asked "which of the following MUST be lees than 1, not COULD be less than 1. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

Given: \(0 < r < s\) --> divide by \(s\) (we can safely do this since we know that \(s>0\)) --> \(\frac{r}{s}<1\), so I must be true;

II. rs: if \(r=\frac{9}{10}<1\) and \(1<(s=\frac{10}{9})<2\) then \(rs=1\), so this statement is not alway true;

III. s-r: if \(r=0.5<1\) and 1\(<(s=1.5)<2\) then \(s-r=1\), so this statement is not alway true.

Answer: A (I only).

Check our new Must or Could be True Questions to practice more: search.php?search_id=tag&tag_id=193

Hope it helps.

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Re: If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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New post 20 Jul 2016, 21:26
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nk18967 wrote:
Curious if this can be solved this way:

Given: 0<r<1<s<2

1. r/s <1
= r<s (yes)

2. rs<1
= r<1/s (no)

3. s-r<1
= s<r (no)

So only 1 must be true?



Not sure how you arrive at this: = r<1/s (no)

Also, s-r<1 is not the same as s < r. It is the same as s < 1 + r.

To prove 2, think of a case such as r = 3/4 and s = 4/3. The product is 1, not less than 1.
To prove 3, think what happens when r is very close to 0 and s is very close to 2. Then s - r is almost 2, not less than 1.
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Re: If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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New post 21 Jul 2016, 06:22
Thank you Karishma.
Testing numbers seems like a much better way to solve these.
Also, thanks for pointing out mistakes in my calculations! :)

VeritasPrepKarishma wrote:
nk18967 wrote:
Curious if this can be solved this way:

Given: 0<r<1<s<2

1. r/s <1
= r<s (yes)

2. rs<1
= r<1/s (no)

3. s-r<1
= s<r (no)

So only 1 must be true?



Not sure how you arrive at this: = r<1/s (no)

Also, s-r<1 is not the same as s < r. It is the same as s < 1 + r.

To prove 2, think of a case such as r = 3/4 and s = 4/3. The product is 1, not less than 1.
To prove 3, think what happens when r is very close to 0 and s is very close to 2. Then s - r is almost 2, not less than 1.

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Re: If 0<r<1<s<2. Which of the following must be less than 1. [#permalink]

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New post 27 Aug 2017, 13:11
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Re: If 0<r<1<s<2. Which of the following must be less than 1.   [#permalink] 27 Aug 2017, 13:11
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