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25 Feb 2012, 02:06
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64% (01:24) correct
36%
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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
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
25 Feb 2012, 02:16
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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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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.
20 Jul 2016, 21:26
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nk18967
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?
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.
