Does ts/k 0? (1) ts/(k^2) 0 (2) k t + s

Question

Does  \frac{ts}{k} >0 ?

(1)  \frac{ts}{{k^2}} >0

(2)  k > t +s

shameekv1989 · Answer

We need to find whether ts and k have same signs

1)  \frac{ts}{{k^2}} > 0   => ts>0 (Nothing about sign of k) Hence insufficient

2)  k > t +s  
Case i) :- 
When t=0 s=-4 => t+s = -4 => ts = 0 and k can be anything positive or negative =>
 \frac{ts}{k} = 0

Case ii) t=5 s=4 => t+s = 9 => ts > 0 and k > 9 i.e. positive => 
 \frac{ts}{k} > 0

Case iii) t=-5 s=4 => t+s = -1 => ts < 0 and k can be positive or negative =>
 \frac{ts}{k}  can be either positive or negative

Case iii) t=-5 s=-4 => t+s = -9 => ts > 0 and k >-9 i.e. k can be positive or negative => 
 \frac{ts}{k}  can be either positive or negative
 
Insufficient

Combining 2 statements
ts > 0 => t and s can not be 0
Case i) t = 5, s= 4 => t+s = 9 => ts>0 and k > 9 i.e. k>0 =>  \frac{t}{ks} > 0

Case ii) t=-5, s=-4 => t+s = -9 => ts>0 and k > -9 i.e. k can be positive or negative =>  \frac{t}{ks}  can be either positive or negative

Hence Insufficient

Answer - E

Archit3110 · Answer

#1
  \frac{ts}{{k^2}} >0 
k can be + or -ve ; so insufficient to say that  \frac{ts}{k} >0 
#2
 k > t +s 
here we can have two cases
if k is +ve and either of t & s is -ve so  \frac{ts}{k} >0  ; NO
and if all k,s,t are +ve then  \frac{ts}{k} >0  ; YES
insufficient
from 1 &2
nothing can be determined in common
IMO E

Does ts/k > 0? (1) ts/(k^2) > 0 (2) k > t + s

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