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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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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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..

Consider Kudos if it helped

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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Here, you have to think about extreme values like:
r can be 0.001, or 0.999 and
s can be 1.001 or 1.99 etc.
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Bunuel

Since 0<r<1 and 1<s<2 can we multiply the inequalities to prove (II)?

(1)*(0)<R*S<(1)*(2)

0<RS<2

Thanks

Since this a MUST be question, we use the constraints to set up cases, disprove the statements and eliminate the answer options.

0 < r < 1 < s < 2.

Let r = ½ and s = \(\frac{3}{2}\).

Clearly, s- r =\( \frac{3}{2}\) – ½ = 1. So, we found a case where s - r is not less than 1. Statement III does not represent an expression that is always less than 1.

Eliminate answer options containing statement III viz., option C and option E.

Let r = \(\frac{3}{5}\) and s = \(\frac{5}{3}\).

Clearly, r * s = \(\frac{3}{5}\) * \(\frac{5}{3}\) = 1. Statement II does not represent an expression that is always less than 1.

Eliminate answer options containing statement II viz., option B and option D.

The answer option left out i.e. option A MUST be true.
The correct answer option is A.
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