The condition for which the equations $3x–2y=5\text{and}5x–ky=4$ are consistent with unique solution is

The correct option is B $k\ne \frac{10}{3}$

Condition of equation $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_2+c_2=0$ for unique solution is $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}.$
$\text{Given :}3x-2y=5\text{and}5x-ky=4$
$\text{Here,}{a}_1=3,b_2=-2\text{and}{c}_1=-5$
$\text{Also ,}{a}_2=5,b_2=-k\text{and}{c}_2=-4$
Thus, for the equations 3x – 2y = 5 and 5x – ky = 4 to be consistent with unique solution,
$\frac{3}{5}\ne \frac{-2}{-k}$
$\Rightarrow \frac{3}{5}\ne \frac{2}{k}$
$\Rightarrow k\ne \frac{10}{3}$
Hence, the correct answer is option (2).