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25 Nov 2020, 16:56
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chetan2u
Can't this approach be a little risky as sometimes the variables cancel out? So wouldn't it still be worth it to try and work it out to make sure that we are in fact still only left with variables?
25 Nov 2020, 18:58
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gd93
You should always try to solve the equations when you have time.
But in case you are short of time and in the given scenarios as in this question, you can mark E.
There is no number in the equations, it just contains variable.
If one of the variables cancel out we will have the other as equal to 0 in most of the cases, but here none can be 0 as all variables are positive integers.
16 Jan 2021, 01:11
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AbdurRakib
\(Average of x+z=\frac{x+z}{2}\)
(1) \(x-y=y-z\)
\(⇒ x+z=2y\)
\(⇒ \frac{x+z}{2}=\frac{2y}{2}=y\)
(2) Will not give any value for x+z. Insufficient.
The answer is A.
