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505-555 Level|   Algebra|      
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Bunuel
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If xyz = tyz, is x = t ? (1) x = 0 (2) y 0 21 Jun 2022, 01:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

15% (low)

Question Stats:

84% (00:59) correct 16% (01:07) wrong based on 32 sessions

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

(1) \(x = 0\)

(2) \(y ≠ 0\)




Project DS Butler Data Sufficiency (DS3)


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E
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If xyz = tyz, is x = t ? (1) x = 0 (2) y 0 21 Jun 2022, 02:13
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(1)Only
If x=0, t=0, then x=t
But if x=0,y=0,t=10, then x<>t
Insufficient

(2)Only
Suppose y=5
If x=t=0, then x=t
If x=0, z=0, t=5, then x<>t
Insufficient

(1)(2)Together
Cases under (2)Only also satisfies condition (1).
Insufficient

The answer is thus (E).
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If xyz = tyz, is x = t ? (1) x = 0 (2) y 0 21 Jun 2022, 22:17
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IMO E

Given, xyz = tyz
Therefore, yz(x-t)=0
Now, Statement1: x=0
If YZ = 0 then we cannot conclude whether x=t. hence, insufficient.

Statement2: y is not equal to zero
Here, Z can be zero. hence, again insufficient.

Combining also we can’t decide whether z=0 or not hence, Hence, E.
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If xyz = tyz, is x = t ? (1) x = 0 (2) y 0 21 Jun 2022, 22:38
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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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