marwahshubham wrote:
AbdurRakib wrote:
If y ≠2xz, what is the value of (2xz + yz)/(2xz – y)?
1) 2x + y = 3
2) z = 2
I understand that statement 1 & 2 alone are not sufficient but I opted for C in this way
From statement 2 we get z=2
denominator becomes 4x-y
=> 2x^2 -1y^2
=> (2x+y) (2x-y)
And numerator becomes 2(2x+y)
2x+y cancels out from both numerator and denominator
now we are left with 2/2x-y
from statement 1 we get 2x+y = 3
=> y=3-2x
putting value of y in 2/2x-y
we get 2/2x-3-2x
which comes up to 2/-3I know this must be one of the silliest question you have encountered but can`t help myself with this.
your help will be highly appreciated.
Thanks in advance!
Hello, everyone. I came across this question today and noticed that, as of this writing, 13 percent of people have opted for (C). I wanted to discuss why this
cannot be the answer without resorting to self-evident logic. By factoring and substituting, you can indeed work out quite a bit of the expression:
\(\frac{z(2x+y)}{(2xz-y)}\)
\(\frac{(2)(3)}{(2x(2)-y)}\)
\(\frac{6}{4x-y}\)
Now, from statement (1), you can isolate the
y and substitute into the above expression:
\(2x+y=3\)
\(y=3-2x\)
But you have to be careful when substituting, or you could flip a sign, a mistake that a few people may be making. (I have
highlighted such an error in the quoted post above.)
\(\frac{6}{4x-(3-2x)}\)
\(\frac{6}{4x-3+2x}\)
\(\frac{6}{6x-3}\)
Or, if you prefer the reduced fraction,
\(\frac{2}{2x-1}\)
That is it. Since we cannot determine the value of
x, we will not be able to determine the value of the quotient even with the two statements together. Thus,
the answer must be (E).
I hope this post may prove helpful to those who may have gone wayward but not understood why.
- Andrew