Linear Equations Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

Given that system of equations has no solution

⇒ |A| = 0

|A| = – 30 + 6 k

But |A| = 0

⇒ 6k – 30 = 0

⇒ k = 5

∴ Option (b) is correct.

Answer: (c)

$x^2 – y^2$ = 0

(Equation of y = |x|)

$(x – a)^2 + y^2$ = 1

[Equation of a circle with centre (a, 0) and radius 1]

These have single positive solution

⇒ y = |x| to be tangent to the circle

AB = radius of circle

∠AOB = 45°

∠OAB = 90°

[tangent to the circle]

∠ABO = 45°

Length of OB = $√{AB^2 + OA^2} = √{1^2 + 1^2}$

OB = $√2$

Answer: (c)

65x – 33y = 97 .... (i)

(given)

33x – 65y = 1 ....(ii)

(given)

From eq (i) + eq (ii)

98x – 98y = 98

98(x – y) = 98 ∴ x – y = 1 ....(iii)

Now

From eq (i) × 1 – eq (iii) × 33

65x - 33y = 97

- 33x + 33y = - 33

32x = 64

∴ $x = 64/32$ = 2

From eq (iii)

x – y = 1

2 – y = 1

∴ y = 2 – 1 = 1

Hence, xy = 2 × 1 = 2

Answer: (a)

Let x be the first digit and y be the second digit of two digit number.

According to question,

x + y = 8 ... (i)

(10x + y) – (10y + x) = 18

⇒9x – 9y = 18

⇒x – y = 2 ... (ii)

Adding (i) and (ii),

x + y = 8

x - y = 2

We get

2x = 10

⇒ x = 5

⇒ y = 8 – 5 = 3

⇒ x = 5 and y = 3

∴ Required difference of digits, x – y

= 5 – 3 = 2.

Answer: (d)

∴ x = ${135}/5$ = 27