Is the product st negative?

Question

(1) s^2 – s < 0

(2)  \frac{s-4}{t-3}=1

Bunuel could you shed some light on the fastest approach? I'd like to understand the mechanics. 
Thanks in advance.

Bunuel · Accepted Answer

Is the product st negative?

(1) s^2 – s < 0 --> \(s(s-1)<0\) --> \(0<s<1\) --> s is positive. We know nothing about t. Not sufficient.

(2) \(\frac{s-4}{t-3}=1\) --> \(s=t+1\). If s=10 and t=11, then st>0 but if s=0.5 ans t=-0.5, then st<0. Not sufficient.

(1)+(2) Since from (2) \(s=t+1\), then from (1) we have that \(0<t+1<1\) --> \(-1<t<0\), so we have that t is negative. Therefore st = positive*negative = negative. Sufficient.

Answer: C.

Hope it's clear.

viksingh15 · Answer

Well I did this other way around. St 1 - s^2 -s <0 ( we need both s and t ) not sufficient St 2- s-4/t-3 =1, from this we get => s-4=t-3 => s-1=t, not sufficient From both we get, from St 1 s(s-1)<0 and St 2 says s-1=t, if we replace s-1 with t, we get St<0

HarveyS · Answer

Bunuel,

I am confused with the one...

In Stm2

S-t = 1

I thought both S and T will have same sign to get 1

E.g- S= 5, t= 4

S-T = 4

S= -4, T= -5

-4-(-5) = 1

In both the case S and T are having same sign and definitely ST is not negative....

Pls explain what am I missing??

Thanks

Bunuel · Answer

In post you are quoting there is an example giving negative product:  s=0.5  ans  t=-0.5 , then  st<0 .

amz14 · Answer

I have a question..

How did s(s-1)<0 become s>0 and s<1 ..

shouldn't it be s<0 and s<1 ?

Bunuel · Answer

First of all: s(s - 1) < 0 is true for 0x2-4x-94661.html#p731476 inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html everything-is-less-than-zero-108884.html graphic-approach-to-problems-with-inequalities-68037.html All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184 All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189 700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope this helps.

amz14 · Answer

Thanks Bunuel ! I have read them now, this is very helpful.

smyarga · Answer

(1) It means that  0<s<1 . Nothing about  t . Insufficient.
(2)  t=s-1 , don't know signs of  s  and  t . Insufficient.

(1)+(2) Since  0<s<1  and  t=s-1 , then  -1<t<0 . Hence,  st<0 . Sufficient.

lastshot · Answer

For last 30 mins i am pondering on this B statement ...superb , & when i searched it over here, i was sure you must have already address it...& now i understand what i am missing !!! I want to Kill DS....10 days to Exam only !!! :oops:

grainflow · Answer

Bunuel,

This might be a stupid question but for some reason I struggle with this.

in the following step:
(s−4)/(t−3)=1   so s-4=t-3   so s=t+1

How can you multiply by an unknown variable if you don't know if the variable is zero? Does this rule only apply if the variable is the only figure in the denominator? or does it only apply when dividing by an unknown variable?

Thanks in advance!

Bunuel · Answer

We know that t-3 is not 0, because if it were, then (s−4)/(t−3) would be undefined and not equal to 1.

testcracker · Answer

hi for your ease understanding you can see the inequality as under: s(1-0)(s-1) < 0 here you can see the 2 roots easily .. hope this helps .. cheers... :lol:

achloes · Answer

Logical translation of the question: do s and t have opposite signs?

S1. s^2 < s 
We're dealing with a positive proper fraction. We know the sign of s but don't have info on t

INSUFFICIENT

S2. s-4 = t-3 
s = t+1

s and t can both be positive or negative. If s=3, then t=2. If s=-1, then t=-2. In either cases, st = positive value 
Alternatively, if s=0, then t=-1, which means st = 0

INSUFFICIENT

Statements combined: 
S1. s = positive fraction 
S2. s-1 = t

We know that s is a positive proper fraction, therefore t will have to be negative for S2 to hold true.

Let's take s to be 1/2. 
s-1 = t 
1/2 - 1 = -(1/2)

SUFFICIENT

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Is the product st negative?

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