Number line product

[karthikpandian19]

On the number line, if the number "k" is left to the number "t", is the product "kt" to the right of "t"?

1. t<0
2. k<1

How to do this DS problem fast???

[anandhimanshu85]

I think its A

On the number line, if the number "k" is left to the number "t", is the product "kt" to the right of "t"?

1. t<0
2. k<1

How to do this DS problem fast???

[karthikpandian19]

OA is A

but can anyone provide the explanation????

[neelgandham]

On the number line, if the number "k" is left to the number "t", is the product "kt" to the right of "t"?

1. t<0

t<0, implies k<0 as the number "k" is left to the number "t". The value of kt > 0(-ve * -ve = +ve)
Sufficient to answer 'is the product "kt" to the right of "t"?'

2. k<1

let k = 1/2, and t = 2, kt = 1 and lies between k and t
let k = 0 and t = 2, kt = 0 and lies on k
let k = -1/2 and t = 10, kt = -5, lies to the left of k
Hence, Insufficient

IMO A

[mj78ind]

On the number line, if the number "k" is left to the number "t", is the product "kt" to the right of "t"?

1. t<0
2. k<1

How to do this DS problem fast???

Interesting!

We have to find kt > t ----- (1) , when t > k.

Solving (1), t(k-1)>0
Hence there are two cases possible for the above to hold:
1. t > 0 AND k > 1

OR

2. t < 0 AND k < 1

Now looking at the statements:
Stmt 1 - t < 0, k has to be < 1 in fact k < t < 0. This is always case 2 above. Hence Stmt 1 is sufficient.

Stmt 2 - k<1. t can be <0 or >0. For example, k = -2, t = -1 kt = 2 hence kt > t. k = 0.7 t = 0.9 kt = 0.63, hence kt <t. Hence Stmt 2 is insufficient.

Answer A