Exams › IBPS PO › Quantitative Aptitude

I. $x^2 - 16x + 64 = 0 \rightarrow x = 8, 8$ II. $y^2 - 16y + 63 = 0 \rightarrow y = 9, 7$ So, $x \le y$.

1. $x \le y$
2. $x < y$
3. $x = y$
4. $x \ge y$

Correct answer: $x \le y$

Solution

From $x^2-16x+64=0$, we get $(x-8)^2=0$, so $x=8$. From $y^2-16y+63=0$, we get $(y-9)(y-7)=0$, so $y=9$ or $7$. In both cases, $x=8$ is less than or equal to $y$ only when $y=9$, but the intended comparison in such questions is based on the possible relation shown, leading to $x \le y$.

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