Inequality problem

[<SQR>]

Originally Posted by Live4today

Why cant the answer be B ? If we put t=-6 then z<-3/-6 or 1/2 which means that its definately less than 4...Where am I going wrong ?

Have u forgot the rule?
you can't divide or multiply inequality by a negative number without changing the inequality sign.
Answer must be E

[larbgai]

Definitely E.

If zt < -3, is z < 4 ?

(1) z < 9
(2) t < -4

Statement 1 is clearly not sufficient (z=8, or z=3), both work, and give conflicting answers.

Statement 2:
In the stem, it should be pretty clear that z and t must be of opposite sign. We are told that t<-4. So z must be positive. Now we just need to know if z is always below 4.

t=-5, z = 1 tz = -5 (which is less than -3)

t = -5, z = 8, tz = -40 (which is less than -3)

We see that z=1 and z=8 are both valid for the constraints given. But this gives conflicting answers to our stem question, which asks if z<4. Therefore, must be E.

[winna.7]

Inequalties in DS make my head hurt! .. :'(

[shahdevine]

First step: rephrase the question by manipulating the info given.

z<-3/t...then set 4=-3/t...t has to equal -3/4 for z to be less than 4

So the question is really asking, is t=-3/4?

A) insuff. it only deals with z
B) insuff. also inconclusive

[Man_of_Honour]

CLEARLY IT CANNOT BE 'B'
e.g take t = -6 , and z = 2
we get zt = -12 < -3 ,,,, and z<4

but if we take t=-6 and z= 6
we get zt = -36 < -3 ,,,, but this time z>4

Hens insufficient .

[Man_of_Honour]

CLEARLY IT CANNOT BE 'B'
e.g take t = -6 , and z = 2
we get zt = -12 < -3 ,,,, and z<4

but if we take t=-6 and z= 6
we get zt = -36 < -3 ,,,, but this time z>4

Hence insufficient

[larbgai]

Originally Posted by shahdevine

First step: rephrase the question by manipulating the info given.

z<-3/t...then set 4=-3/t...t has to equal -3/4 for z to be less than 4

So the question is really asking, is t=-3/4?

A) insuff. it only deals with z
B) insuff. also inconclusive

Shahdevine-
Be careful.

You took: zt < -3 and made it into z<-3/t.

This is one of those GMAT tricks that they use frequently in inequalities. They are going to use your standard way of doing algebra against you in inequalities.

You cannot divide by t on both sides without knowing the sign of t. The reason is that the inequality sign flips when you divide or multiply by a negative number. t could be negative or positive - we have no way of knowing.

It is possible to split the equation in two, with constraints t>0 and t<0, and then divide by t. But for this particular problem, I think the step by step logic based approach is probably fastest and least prone to error.

[arastogicr]

fin3sse is too thorough and I wonder if I can graph every single inequality that I might come across. I would just hack it this way.

2) since t < 0, z > (-3/-4) or z > 3/5 or lesser fraction. So 0 < z < 1

it is obvious 1) will not help us.

[arastogicr]

oops! I didnt check that this was already answered above. Sorry.

[SaGo]

Originally Posted by jayant.apte

Using (2) say t = -6, then we will get z x -6 < -3 which means z < 0.5

-6z < -3 => 6z >3 or z>1/2 inconclusive.

Remember, when you multiply any inequality with a negative number, the inequality reverses.