If 0

Question

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

Bunuel · Accepted Answer

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.

KarishmaB · Answer

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.

devinawilliam83 · Answer

So the best strategy is to pick numbers in these ques?

Bunuel · Answer

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.

imhimanshu · Answer

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

shameekv · Answer

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

NYCBS19 · Answer

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?

RashedVai · Answer

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.

amirza19 · Answer

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

GMATWhizTeam · Answer

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.

sachi-in · Answer

The best strategy here is to use limits or approximation for r and s:

r/s will have highest value when r is highest , s is smallest ( and still r will be less than 1 and s greater than 1 ) hence 0 < max( r/s ) < 1

rs max when r max and s max ( r can be a little less than 1, s a little less than 2 ) hence 2 > max( rs ) > 1

s - r max when s max and r min ( r can be little more than 0 , s a little less than 2 ) hence 0 < max(r - s) < 2

Only r/s will be always less than 1

Adarsh_24 · Answer

May be we can do this way too. 1) Split r and s 0 1/s > 1/2 ( if we take reciprocal sign flips) Max limits for r* 1/s are 0, 0, 1, 1/2 0< r/s < 1 2) 0

Vidhi12 · Answer

0.1<=r<=0.9
&
1.1<=s<=1.9

1)r/s
max value==> 0.9/1.1 <1(maximum value of r and minimum value of s)
min value==> 0.1/1.9 <1
It is less than 1, hence we can say this option is correct, but lets check other possibilities as well

0.9/1.9==> less than 1
0.1<1.1==> less than 1

2)rs
max=0.9*1.9>1
hence this option can directly be called wrong
but lets check other possibilities as well
0.9*1.1<1
0.1*1.1<1
0.1*1.9<1
since there is one value not satisfying the condition , option is wrong

3)s-r
max value=(max value of s - smallest value of r)
1.9-0.1>1
Hence this option is also incorrect

Answer:- A- 1 ONLY

devinawilliam83

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

egmat · Answer

Understanding What "Must Be Less Than 1" Really Means

First, let's be crystal clear about what the question is asking. When it says "must be less than 1," it's asking which expressions are  always  less than 1, no matter what values you pick for  r  and  s  within the given constraints. This isn't about finding expressions that  can  be less than 1 - they need to be  guaranteed  to be less than 1.

The Strategic Testing Approach

Here's the key insight: if you want to check whether something  must  be less than 1, try to make it as large as possible. If you can find even one case where it reaches or exceeds 1, then it doesn't "must be less than 1."

Let me show you what I mean. Let's pick strategic values that push these expressions toward their maximum:
- Let  r = 0.9  (close to 1, to maximize expressions involving  r )
- Let  s = 1.9  (close to 2, to maximize expressions involving  s )

Testing Each Expression

Expression I:   \frac{r}{s}

\frac{r}{s} = \frac{0.9}{1.9} \approx 0.47

Now here's where you need to see the deeper pattern: You're dividing a number less than 1 by a number greater than 1. Think about it - if  r < 1  and  s > 1 , then  \frac{r}{s} < \frac{r}{1} = r < 1 .

So  \frac{r}{s}  will  always  be less than 1. ✓

Expression II:   rs

rs = 0.9 	imes 1.9 = 1.71

Whoa! This is greater than 1. So expression II doesn't always stay below 1. ✗

You can see why this happens: when  r  is close to 1 and  s  is close to 2, their product approaches  1 	imes 2 = 2 .

Expression III:   s - r

s - r = 1.9 - 0.9 = 1.0

This equals 1, and we could even make it larger if we chose  r = 0.1  and  s = 1.9 , giving us  1.9 - 0.1 = 1.8 . So expression III can definitely exceed 1. ✗

The Answer

Only Expression I  \left(\frac{r}{s}ight)  must always be less than 1, while Expressions II and III can equal or exceed 1 under certain conditions.

Answer: A (I only)

Taking Your Understanding Further

Now, what I've shown you here is enough to solve this specific problem, but here's what you're missing: there's a systematic framework for approaching  all  "must be true" inequality questions that'll save you time and prevent errors. The complete solution on Neuron breaks down the strategic approach you need, shows you the common traps (like confusing "must be" with "can be"), and gives you an alternative method using boundary analysis that's even faster.

You can check out the  step-by-step solution on Neuron by e-GMAT  to master the inequality constraint satisfaction approach systematically. You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice  here .

Hope this helps!

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards