Hi,
here are my two cents for this question
we have X>0, Y>0.
we are told that \(\frac{X}{Y}\)= 3Y+Z--------(I), where Y>Z\(\geqslant{0}\)
Now we are told that \(\frac{Z}{Y}\) , remainder is 2. this means Y>2
or Z= YK+2 -----(II) where K is integer
Here we have multiple ways to visualize this statement
So X-3Y=Z --------(III)
So dividing the (III) by Y
\(\frac{Z}{Y}\)=\(\frac{X-3Y}{Y}\),
So \(\frac{X}{Y}- \frac{3Y}{Y}\)
we are told that when \(\frac{Z}{Y}\)we have remainder 2. and from above, the part that gives remainder 2 is \(\frac{X}{Y}\)
Since \(\frac{X}{Y}\) gives us remainder as Z and from above we have that Z=2
Now
X= 3Y+2
X>0,Y>2,Z>0
We have if y=3 X=11
Since this leaves us with only one option which is D. Hence answer is D
or you could think of this as below
Logically we have , Y>Z\(\geqslant{0}\) & Y>2.
so whenever we have Y>Z and\(\frac{Z}{Y}\) the remainder is always Z
Eg. \(\frac{4}{9}\) remainder will be 4 .
this tells us that Z=2
so Eq(I) becomes X= 3y+2
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