  CBSE Class 10 Quadratic Equations MCQ New Pattern

## CBSE Class 10 Quadratic Equations MCQ New Pattern By Clarify Knowledge

CBSE Class 10 Quadratic Equations MCQ New Pattern 2022

## CBSE Class 10 Quadratic Equations MCQ Here

1. Equation of (x+1)2-x2=0 has number of real roots equal to:

(a) 1

(b) 2

(c) 3

(d) 4

Explanation: (x+1)2-x2=0

X2+2x+1-x2 = 0

2x+1=0

x=-½

Hence, there is one real root.

2. The roots of 100x2 – 20x + 1 = 0 is:

(a) 1/20 and 1/20

(b) 1/10 and 1/20

(c) 1/10 and 1/10

(d) None of the above

Explanation: Given, 100x2 – 20x + 1=0

100x2 – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1)2 = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

3. The sum of two numbers is 27 and product is 182. The numbers are:

(a) 12 and 13

(b) 13 and 14

(c) 12 and 15

(d) 13 and 24

Explanation: Let x is one number

Another number = 27 – x

Product of two numbers = 182

x(27 – x) = 182

⇒ x2 – 27x – 182 = 0

⇒ x2 – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

⇒ x = 13 or x = 14

4. If ½ is a root of the quadratic equation x2-mx-5/4=0, then value of m is:

(a) 2

(b) -2

(c) -3

(d) 3

Explanation: Given x=½ as root of equation x2-mx-5/4=0.

(½)2 – m(½) – 5/4 = 0

¼-m/2-5/4=0

m=-2

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:

(a) Base=10cm and Altitude=5cm

(b) Base=12cm and Altitude=5cm

(c) Base=14cm and Altitude=10cm

(d) Base=12cm and Altitude=10cm

Explanation: Let the base be x cm.

Altitude = (x – 7) cm

In a right triangle,

Base2 + Altitude2 = Hypotenuse2 (From Pythagoras theorem)

∴ x2 + (x – 7)2 = 132

By solving the above equation, we get;

⇒ x = 12 or x = – 5

Since the side of the triangle cannot be negative.

Therefore, base = 12cm and altitude = 12-7 = 5cm

6. The roots of quadratic equation 2x2 + x + 4 = 0 are:

(a) Positive and negative

(b) Both Positive

(c) Both Negative

(d) No real roots

Explanation: 2x2 + x + 4 = 0

⇒ 2x2 + x = -4

Dividing the equation by 2, we get

⇒ x2 + 1/2x = -2

⇒ x2 + 2 × x × 1/4 = -2

By adding (1/4)2 to both sides of the equation, we get

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2

⇒ (x + 1/4)2 = 1/16 – 2

⇒ (x + 1/4)2 = -31/16

The square root of negative number is imaginary, therefore, there is no real root for the given equation.

7. The value ofis:

(a) 4

(b) 3

(c) 3.5

(d) -3

Explanation: Let,

Hence, we can write, √(6+x) = x

6+x=x2

x2-x-6=0

x2-3x+2x-6=0

x(x-3)+2(x-3)=0

(x+2) (x-3) = 0

x=-2,3

Since, x cannot be negative, therefore, x=3

8. The sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3. The present age of Rehman is:

(a) 7

(b) 10

(c) 5

(d) 6

Explanation: Let, x is the present age of Rehman

Three years ago his age = x – 3

Five years later his age = x + 5

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x2 + 2x – 15

⇒ x2 – 4x – 21 = 0

⇒ x2 – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

We know age cannot be negative, hence the answer is 7.

9. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

(a) 30 km/hr

(b) 40 km/hr

(c) 50 km/hr

(d) 60 km/hr

Explanation: Let x km/hr be the speed of train.

Time required to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x2 + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

Negative value is not considered for speed hence the answer is 40km/hr.

10. If one root of equation 4x2-2x+k-4=0 is reciprocal of the other. The value of k is:

(a) -8

(b) 8

(c) -4

(d) 4

Explanation: If one root is reciprocal of others, then the product of roots will be:

α x 1/α = (k-4)/4

k-4=4

k=8

11. Which one of the following is not a quadratic equation?

(a) (x + 2)2 = 2(x + 3)

(b) x2 + 3x = (–1) (1 – 3x)2

(c) (x + 2) (x – 1) = x2 – 2x – 3

(d) x3 – x2 + 2x + 1 = (x + 1)3

Answer: (c) (x + 2) (x – 1) = x2 – 2x – 3

Explanation:

We know that the degree of a quadratic equation is 2.

By verifying the options,

(a) (x + 2)2 = 2(x + 3)

x2 + 4x + 4 = 2x + 6

x2 + 2x – 2 = 0

(b) x2 + 3x = (–1) (1 – 3x)2

x2 + 3x = -1(1 + 9x2 – 6x)

x2 + 3x + 1 + 9x2 – 6x = 0

10x2 – 3x + 1 = 0

(c) (x + 2) (x – 1) = x2 – 2x – 3

x2 + x – 2 = x2 – 2x – 3

x2 + x – 2 – x2 + 2x + 3 = 0

3x + 1 = 0

This is not a quadratic equation.

12. Which of the following equations has 2 as a root?

(a) x2 – 4x + 5 = 0

(b) x2 + 3x – 12 = 0

(c) 2x2 – 7x + 6 = 0

(d) 3x2 – 6x – 2 = 0

Answer: (c) 2x2 – 7x + 6 = 0

Explanation:

If 2 is a root then substituting the value 2 in place of x should satisfy the equation.

Let us verify the given options.

(a) x2 – 4x + 5 = 0

(2)2 – 4(2) + 5 = 1 ≠ 0

So, x = 2 is not a root of x2 – 4x + 5 = 0

(b) x2 + 3x – 12 = 0

(2)2 + 3(2) – 12 = -2 ≠ 0

So, x = 2 is not a root of x2 + 3x – 12 = 0

(c) 2x2 – 7x + 6 = 0

2(2)2 – 7(2) + 6 = 0

Here, x = 2 is a root of 2x2 – 7x + 6 = 0

13. A quadratic equation ax2 + bx + c = 0 has no real roots, if

(a) b2 – 4ac > 0

(b) b2 – 4ac = 0

(c) b2 – 4ac < 0

(d) b2 – ac < 0

Answer: (c) b2 – 4ac < 0

A quadratic equation ax+ bx + c = 0 has no real roots, if b2 – 4ac < 0. That means, the quadratic equation contains imaginary roots.

14. The product of two consecutive positive integers is 360. To find the integers, this can be represented in the form of quadratic equation as

(a) x2 + x + 360 = 0

(b) x2 + x – 360 = 0

(c) 2x2 + x – 360

(d) x2 – 2x – 360 = 0

Answer: (b) x2 + x – 360 = 0

Explanation:

Let x and (x + 1) be the two consecutive integers.

According to the given,

x(x + 1) = 360

x2 + x = 360

x2 + x – 360

15. The equation which has the sum of its roots as 3 is

(a) 2x2 – 3x + 6 = 0

(b) –x2 + 3x – 3 = 0

(c) √2x2 – 3/√2x + 1 = 0

(d) 3x2 – 3x + 3 = 0

Answer: (b) –x2 + 3x – 3 = 0

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