# CAT 2023 Question Paper | Quant Slot 1

###### CAT Previous Year Paper | CAT Quant Questions | Question 2

CAT 2023 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 2 : If $x$ and $y$ are real numbers such that $x^2+(x-2 y-1)^2=-4 y(x+y)$, then the value $x-2 y$ is

1. 0
2. 1
3. 2
4. -1

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$x ^ { 2 } + ( x - 2 y - 1 ) ^ { 2 } = - 4 y ( x + y )$
$x ^ { 2 } + ( 2 y ) ^ { 2 } + 2 $cdot x $cdot 2 y +$ x - 2 y - 1$ ^ { 2 } = 0$
$( x + 2 y ) ^ { 2 } + ( x - 2 y - 1 ) ^ { 2 } = 0$
Since $x $& y$ are real numbers, $$ x + 2 y ) $&$ x - 2 y - 1 )$ are both real.
A square of a real number is always non-negative.
For this reason, for the equation: $( x + 2 y ) ^ { 2 } + ( x - 2 y - 1 ) ^ { 2 } = 0$ to be true, both $( x + 2 y ) $&$ x - 2 y - 1 )$ must be equal to 0.
$x - 2 y - 1 = 0$
$x - 2 y = 1$
We solve the linear equations $( x + 2 y = 0 ) $&$ x - 2 y - 1 = 0 )$ to get the exact values of x & y, but that is not required to solve the question at hand.

The question is " If $x$ and $y$ are real numbers such that $x^2+(x-2 y-1)^2=-4 y(x+y)$, then the value $x-2 y$ is "

##### Hence, the answer is '1'

Choice B is the correct answer.

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