CAT 2022 Set-3 | Quantitative Aptitude | Question: 2

Question

If $c=\frac{16 x}{y}+\frac{49 y}{x}$ for some non-zero real numbers $x$ and $y$, then $c$ cannot take the value

• A. $-60$ 
• B. $-50$ 
• C. $60$ 
• D. $-70$ 

Answer 1

Let z=(x/y). Now, c = 16z + 49/z  => 16z$^{2}$ – cz + 49 = 0. Since x & y are non zero real number. So, z is also a non zero real number. So, the equation 16z$^{2}$ – cz + 49 = 0 satisfies for some real number z, if b$^{2}$ – 4ac >= 0  => c$^{2}$ – 56$^{2}$ >= 0  => |c| >= 56.

Therefore the absolute value of c is always >= 56. So, value of c can’t be -50.

Ans is (B).