ueh55406 wrote:
Hi
chetan2u,
Bunuel,
VeritasKarishmaWhere exactly did I go wrong?
If xy + z = x(y + z), which of the following must be true?
\((y+z)= xy+z/x\)
\((y+z)=y+z/x\)
D. x=1 or y=0
0+z=0+z/1
z=z
Super confused :/
ueh55406Note what is given to you and what is asked. The approach of plugging in values using the options is not correct in such "MUST BE TRUE" questions.
You are given that xy + z = x(y + z).
You are asked whether (D) is implied by this equation. Note that even if after plugging in values using (D) you find that the equation holds, it does not mean that (D) is correct.
Take a simpler example to understand this:
Given: x > 0
Which of the following must be true:
(A) x is not 0
(B) x > 2
(A) x is not 0
Put x = 2. Inequality x > 0 holds.
Put x = -1. Inequality x > 0 does not hold.
Does this mean option (A) must not be true? No.
If x is positive as given to us, it must be true that x is not 0.
(B) x > 2
Again, put x = 3. Inequality x > 0 holds.
Put x = 4. Inequality x > 0 holds.
Put x = 10. Inequality x > 0 holds.
For all values of x that are greater than 2. inequality will hold. Does this mean that option (B) must be true?
No.
If x is positive, it does not necessarily imply that x > 2. x could be 1 too.
What does this tell us? That we need to focus on what is given and what is implied from it.
Check this post:
https://www.veritasprep.com/blog/2017/0 ... -question/Hope this point is clear.
Now, note that in our original question, option (D) gives x = 0 or y = 1. This means that at least one of them must be true. So either x = 0 or y = 1. You are taking them both to be true simultaneously (which is possible but not necessary).
In any case, as we discussed above, even if (D) satisfied the equation, it DOES NOT mean that (D) must be true.
Then how do we solve such questions?
We start with the given inequality and solve that to see what must be true.
xy + z = x(y + z)
xy + z = xy + xz
x (z - 1) = 0
So either x = 0 or z = 1 (at least one of them must be true)
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Karishma
Owner of Angles and Arguments at https://anglesandarguments.com/
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