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If 0 < r < 1 < s < 2, which of the following must be less

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If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post Updated on: 15 Oct 2014, 23:25
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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

Can anyone give an example of why II does not work? i just can't figure it out.

thanks

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-0-r-1-s-2-which-of-the-following-must-be-less-than-128112.html

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Originally posted by azule45 on 21 Jan 2011, 10:24.
Last edited by Bunuel on 15 Oct 2014, 23:25, edited 3 times in total.
Edited the question
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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post 21 Jan 2011, 10:39
3
2
azule45 wrote:
Can anyone give an example of why II does not work? i just can't figure it out.

thanks


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

Note that the question asks "which of the following MUST be less than 1".

Given: 0 < r < s --> divide by s (we can safely do this as we know that s>0) --> r/s<1, so I must be true;

II. rs: if r=9/10<1 and 1<s=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).
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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post 04 Jun 2013, 09:20
Bunuel, is the approach below correct for S-R :
1<S<2 ------- (A)
0<R<1
0>-R>-1 (Reversal of sign on multiplication with -1)
-1<-R<0 ----- (B)

Adding (A) & (B) as inequality sign is same

0<S-R<2

So III is not true..
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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post 04 Jun 2013, 23:57
karjan07 wrote:
Bunuel, is the approach below correct for S-R :
1<S<2 ------- (A)
0<R<1
0>-R>-1 (Reversal of sign on multiplication with -1)
-1<-R<0 ----- (B)

Adding (A) & (B) as inequality sign is same

0<S-R<2

So III is not true..


You could get the same by simply subtracting r from each side of the given inequality: -r < 0 < 1 - r < s - r < 2 - r. Now, 1 - r is between 0 and 1, not inclusive. Similarly 2 - r is between 1 and 2, not inclusive. So, s - r may or may not be less than 1.
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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post 24 Sep 2013, 14:19
Bunuel wrote:
karjan07 wrote:
Bunuel, is the approach below correct for S-R :
1<S<2 ------- (A)
0<R<1
0>-R>-1 (Reversal of sign on multiplication with -1)
-1<-R<0 ----- (B)

Adding (A) & (B) as inequality sign is same

0<S-R<2

So III is not true..


You could get the same by simply subtracting r from each side of the given inequality: -r < 0 < 1 - r < s - r < 2 - r. Now, 1 - r is between 0 and 1, not inclusive. Similarly 2 - r is between 1 and 2, not inclusive. So, s - r may or may not be less than 1.



I still did not understand the explanation here..This seems like a relatively easy problem, but I do not understand why III is not always less than one...I picked numbers and subtracted 3/4-5/4 and then 1/2 - 5/3. Can someone explain to be what I am doing wrong? Thank you!
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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post 25 Sep 2013, 01:51
TAL010 wrote:
Bunuel wrote:
karjan07 wrote:
Bunuel, is the approach below correct for S-R :
1<S<2 ------- (A)
0<R<1
0>-R>-1 (Reversal of sign on multiplication with -1)
-1<-R<0 ----- (B)

Adding (A) & (B) as inequality sign is same

0<S-R<2

So III is not true..


You could get the same by simply subtracting r from each side of the given inequality: -r < 0 < 1 - r < s - r < 2 - r. Now, 1 - r is between 0 and 1, not inclusive. Similarly 2 - r is between 1 and 2, not inclusive. So, s - r may or may not be less than 1.



I still did not understand the explanation here..This seems like a relatively easy problem, but I do not understand why III is not always less than one...I picked numbers and subtracted 3/4-5/4 and then 1/2 - 5/3. Can someone explain to be what I am doing wrong? Thank you!


Example discarding the third option is given in my post: if r=0.5<1 and 1<(s=1.5)<2 then s-r=1, so this statement is not always true.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post 25 Sep 2013, 02:21
1
TAL010 wrote:
Bunuel wrote:
karjan07 wrote:
Bunuel, is the approach below correct for S-R :
1<S<2 ------- (A)
0<R<1
0>-R>-1 (Reversal of sign on multiplication with -1)
-1<-R<0 ----- (B)

Adding (A) & (B) as inequality sign is same

0<S-R<2

So III is not true..


You could get the same by simply subtracting r from each side of the given inequality: -r < 0 < 1 - r < s - r < 2 - r. Now, 1 - r is between 0 and 1, not inclusive. Similarly 2 - r is between 1 and 2, not inclusive. So, s - r may or may not be less than 1.



I still did not understand the explanation here..This seems like a relatively easy problem, but I do not understand why III is not always less than one...I picked numbers and subtracted 3/4-5/4 and then 1/2 - 5/3. Can someone explain to be what I am doing wrong? Thank you!



If you dont like fractions, work with decimals! This is how I approached the problem:

Since o<r<1, I used extreme values: Let r = 0.1 or 0.9
Since 1<s<2, let 1.1 or 1.9

1) r/s = 0.9/1.9 = always less than one. Besides, since r<s this will always be true
2) rs = (0.9)*(1.9) = greater than 1
3)s-r = 1.9-0.1 = 1.8 also greater than 1

so, A. 1 only is your answer!
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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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New post 15 Oct 2014, 23:26
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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-0-r-1-s-2-which-of-the-following-must-be-less-than-128112.html
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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Re: If 0 < r < 1 < s < 2, which of the following must be less   [#permalink] 24 Jan 2019, 08:04
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