Are negative integers s and t both less than t? 1) s

Question

Are negative integers  s  and  t  both less than  r  ?
1)  s < r+t 
2) \frac{s}{r}  <t

niks18 · Answer

Statement 1: s < r+t=>r>s-t if s=-1 and t=-4 then r>-1+4=>r>3 . hence we have a Yes for our question stem but if s=-5 and t=-2 , then r>-5+2=>r>-3 , so r could be -2 . in this case r=t so we have a NO . Insufficient Statement 2: \frac{s}{r} \frac{s}{rt}>1 , now s & t are negative. hence \frac{s}{t} is positive. so we have \frac{positive}{r}>1 => r>0 . Hence we have a Yes for our question stem. Sufficient Option B

EgmatQuantExpert · Answer

Solution

Given:
• Both s and t are negative integers

To find:
• Whether the value of s and t less than r

Analysing Statement 1
• As per the information given in Statement 1, s < r + t

o Or, r > s – t• As it is given that both s and t are negative integers, if we assume that s is less than t, we can have the following scenario possible for r:

Depending on the values of s and t, r can belong to any of the regions indicated.
For example, if s = -7 and t = -1, r > (-7) – (-1) or r > -6
Now, r > -6 means r belong to any of the 3 regions indicated above (red, blue, green)

Hence, statement 1 is not sufficient to answer

Analysing Statement 2
• As per the information given in Statement 2, \(\frac{s}{r}\) < t
• We know that t is negative. Hence \(\frac{s}{r}\) should also be negative.
• We know that s is already negative. Thus, to ensure that \(\frac{s}{r}\) is negative. r MUST be positive.

Hence, from the given statement \(\frac{s}{r}\) < t, we can say r > 0 and \(\frac{s}{t}\) > 0

Hence, statement 2 is sufficient to answer

Hence, the correct answer is option B.

Answer: B

Bunuel · Answer

Similar question to practice: https://gmatclub.com/forum/are-negative ... 25947.html

niks18 · Answer

Hi  EgmatQuantExpert

May be I am missing something here. kindly explain the highlighted part. how are we arriving at  r>\frac{s}{t} . we know  t  is negative and we know nothing about  r . so if you are cross-multiplying  t  then sign of inequality should change.

Eventually  r  will be positive.

GMATBusters · Answer

Hi

The best approach to consider statement 2 is as follows:

2) s/r <t
as we know that s and t are negative,
So if r is also negative, then s/r would be positive and this inequality s/r <t would not be valid .
 So r is positive. hence s and t are both less than r  .

SUFFICIENT.

GMATBusters · Answer

Hi  niks18

Your Query: May be I am missing something here. kindly explain the highlighted part. how are we arriving at  r>\frac{s}{t} . we know  t  is negative and we know nothing about  r . so if you are cross-multiplying  t  then
  sign of inequality should change. -  Since we don't know the nature of r, we can not predict the nature of sign of inequality at this stage of time, without determining the nature of sign of r..

Hi  EgmatQuantExpert

May be I am missing something here. kindly explain the highlighted part. how are we arriving at  r>\frac{s}{t} . we know  t  is negative and we know nothing about  r . so if you are cross-multiplying  t  then sign of inequality should change.

Eventually  r  will be positive.

EgmatQuantExpert · Answer

Hi niks,
You are correct! We cannot infer anything about the range of r. Ideally we can only say that:
 • As both s and t are negative integers, the right-hand side of the given expression is negative. But  \frac{s}{t}  is positive.
• Given that s is also negative, it means r is positive 
Hence, from the given statement  \frac{s}{r}  < t, we can say r > 0 and  \frac{s}{t}  > 0

Have edited the solution. Apologies for the confusion.

push12345 · Answer

1) s<r+t

Remember we don't know sign of r 
So multiple cases arise
1) -6<-2-3
s=-6 r=-2 t=-3. S and t are less than r

2)-6<-3-2
s=-6 r=-3 t=-2. S is less than r but t is not

So Insufficient

2)s/r<t

S and t are given as negative integers

R cannot be negative here as it would result in positive value on left side
Whereas on the left side we have negative value

So r is positive and t and s are less than r as they are negative

Sufficient

B is answer

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bumpbot · Answer

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