The value of k, for which the system of equations $3x-ky-20= 0$, and $6x-10y+40= 0$ has no solution, is:

5

The correct answer is Option (3) → 5

We are asked to find k such that the system has no solution.

The system of equations:

1. $3x - ky - 20 = 0$
2. $6x - 10y + 40 = 0$

Step 1: Condition for no solution

Two lines have no solution if they are parallel but not coincident:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Where the equations are in the form: $a x + b y + c = 0$

• Equation 1: $3x - k y - 20 = 0 \Rightarrow a_1 = 3, b_1 = -k, c_1 = -20$
• Equation 2: $6x - 10y + 40 = 0 \Rightarrow a_2 = 6, b_2 = -10, c_2 = 40$

Step 2: Parallel condition

$\frac{a_1}{a_2} = \frac{b_1}{b_2}$

$\frac{3}{6} = \frac{-k}{-10} \Rightarrow \frac{1}{2} = \frac{k}{10} \Rightarrow k = 5$