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

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Joined: 19 Jul 2009
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If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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Updated on: 15 Oct 2014, 23:25
2
3
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Difficulty:

45% (medium)

Question Stats:

63% (00:51) correct 37% (00:52) wrong based on 331 sessions

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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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Joined: 02 Sep 2009
Posts: 52910
Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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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.

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Location: India
Schools: HBS '16, HEC Jan'16
GPA: 3
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Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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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..
Math Expert
Joined: 02 Sep 2009
Posts: 52910
Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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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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Status: Onward and upward!
Joined: 09 Apr 2013
Posts: 13
Location: United States
Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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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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Math Expert
Joined: 02 Sep 2009
Posts: 52910
Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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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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Joined: 10 Sep 2013
Posts: 73
Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 52910
Re: If 0 < r < 1 < s < 2, which of the following must be less  [#permalink]

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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.

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

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24 Jan 2019, 08:04
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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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