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Bunuel
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Bunuel
If \(xyz = tyz\), is \(x = t\) ?

(1) \(x = 0\)

(2) \(y ≠ 0\)


Lets solve this question in the traditional DS style..

Given \(xyz = tyz\) => \((x-t)yz=0\)

Statement 1 : \(x = 0\)
If x=0, the above equation will become
=> (0-t)yz=0
=> tyz=0; {Any one of them has to be zero}

If t=0; Then \(x = t\) true
If y or z = 0, then t can be any non-zero number & 0 (4,5,.. 0) \(x = t\) false

Hence A,D are eliminated.

Statement 2 : \(y ≠ 0\)
If \(y ≠ 0\),then the main equation will be
(x-t)yz=0 , Either x=t or z=0, Hence we cannot conclude any thing.

So B is eliminated.



Using both Statement 1 : \(x = 0\) & Statement 2 : \(y ≠ 0\)

tyz=0; {Statement 1}

Now as y cannot be 0 from {Statement 2}

t or z has to be zero..

If z = 0 then \(x ≠ t\)
But if z ≠ 0 then \(x = t\), No concluding answer.

Hence C is also eliminated.


E is the correct answer.
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