CUET UG 2025 — General Test Quantitative Reasoning

For two linear equations to have no solution, the lines must be parallel.

For equations:

• $a_1x + b_1y + c_1 = 0$a1​x+b1​y+c1​=0
• $a_2x + b_2y + c_2 = 0$a2​x+b2​y+c2​=0

The condition for no solution is:

$$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$$a2​a1​​=b2​b1​​=c2​c1​​

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From $3x - ky - 20 = 0$3x−ky−20=0:

$$a_1 = 3$$a1​=3

$$b_1 = -k$$b1​=−k

$$c_1 = -20$$c1​=−20

From $6x - 10y + 40 = 0$6x−10y+40=0:

$$a_2 = 6$$a2​=6

$$b_2 = -10$$b2​=−10

$$c_2 = 40$$c2​=40

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Calculate $\dfrac{a_1}{a_2}$a2​a1​​:

$$\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}$$a2​a1​​=63​=21​

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For no solution, $\dfrac{b_1}{b_2} = \dfrac{a_1}{a_2}$b2​b1​​=a2​a1​​:

$$\dfrac{b_1}{b_2} = \dfrac{-k}{-10} = \dfrac{k}{10}$$b2​b1​​=−10−k​=10k​

Setting equal to $\dfrac{a_1}{a_2}$a2​a1​​:

$$\dfrac{k}{10} = \dfrac{1}{2}$$10k​=21​

$$k = 10 \times \dfrac{1}{2}$$k=10×21​

$$k = 5$$k=5

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Check $\dfrac{c_1}{c_2}$c2​c1​​:

$$\dfrac{c_1}{c_2} = \dfrac{-20}{40} = -\dfrac{1}{2}$$c2​c1​​=40−20​=−21​

Since $\dfrac{1}{2} = \dfrac{1}{2} \neq -\dfrac{1}{2}$21​=21​=−21​, the condition for no solution is satisfied.

Therefore, $k = 5$k=5