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

*CBSE Class 10 Quadratic Equations MCQ New Pattern 2022*

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## CBSE Class 10 Quadratic Equations MCQ Table

## CBSE Class 10 Quadratic Equations MCQ Here

**1. Equation of (x+1) ^{2}-x^{2}=0 has number of real roots equal to:**

(a) 1

(b) 2

(c) 3

(d) 4

Answer: **(a) 1**

Explanation: (x+1)^{2}-x^{2}=0

X^{2}+2x+1-x^{2} = 0

2x+1=0

x=-½

Hence, there is one real root.

**2. The roots of 100x ^{2} – 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

Answer:** (c) 1/10 and 1/10**

Explanation: Given, 100x^{2} – 20x + 1=0

100x^{2} – 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

Answer: **(b) 13 and 14**

Explanation: Let x is one number

Another number = 27 – x

Product of two numbers = 182

x(27 – x) = 182

⇒ x^{2} – 27x – 182 = 0

⇒ x^{2} – 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 x ^{2}-mx-5/4=0, then value of m is:**

(a) 2

(b) -2

(c) -3

(d) 3

Answer: **(b) -2**

Explanation: Given x=½ as root of equation x^{2}-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

Answer: **(b) Base=12cm and Altitude=5cm**

Explanation: Let the base be x cm.

Altitude = (x – 7) cm

In a right triangle,

Base^{2} + Altitude^{2} = Hypotenuse^{2} (From Pythagoras theorem)

∴ x^{2} + (x – 7)^{2} = 13^{2}

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 2x ^{2} + x + 4 = 0 are:**

(a) Positive and negative

(b) Both Positive

(c) Both Negative

(d) No real roots

Answer:** (d) No real roots**

Explanation: 2x^{2} + x + 4 = 0

⇒ 2x^{2} + x = -4

Dividing the equation by 2, we get

⇒ x^{2} + 1/2x = -2

⇒ x^{2} + 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 of****is:**

(a) 4

(b) 3

(c) 3.5

(d) -3

Answer: **(b) 3**

Explanation: Let,

Hence, we can write, √(6+x) = x

6+x=x^{2}

x^{2}-x-6=0

x^{2}-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

Answer: **(a) 7**

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 = x^{2} + 2x – 15

⇒ x^{2} – 4x – 21 = 0

⇒ x^{2} – 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

Answer:** (b) 40 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

⇒ x^{2} + 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 4x ^{2}-2x+k-4=0 is reciprocal of the other. The value of k is:**

(a) -8

(b) 8

(c) -4

(d) 4

Answer: **(b) 8**

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) x^{2} + 3x = (–1) (1 – 3x)^{2}

(c) (x + 2) (x – 1) = x^{2} – 2x – 3

(d) x^{3} – x^{2} + 2x + 1 = (x + 1)^{3}

Answer: **(c) (x + 2) (x – 1) = x ^{2}**

**– 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)

x^{2} + 4x + 4 = 2x + 6

x^{2} + 2x – 2 = 0

This is a quadratic equation.

(b) x^{2} + 3x = (–1) (1 – 3x)^{2}

x^{2} + 3x = -1(1 + 9x^{2} – 6x)

x^{2} + 3x + 1 + 9x^{2} – 6x = 0

10x^{2} – 3x + 1 = 0

This is a quadratic equation.

(c) (x + 2) (x – 1) = x^{2} – 2x – 3

x^{2} + x – 2 = x^{2} – 2x – 3

x^{2} + x – 2 – x^{2} + 2x + 3 = 0

3x + 1 = 0

This is not a quadratic equation.

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

(a) x^{2} – 4x + 5 = 0

(b) x^{2} + 3x – 12 = 0

(c) 2x^{2} – 7x + 6 = 0

(d) 3x^{2} – 6x – 2 = 0

Answer: **(c) 2x ^{2}**

**– 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) x^{2} – 4x + 5 = 0

(2)^{2} – 4(2) + 5 = 1 ≠ 0

So, x = 2 is not a root of x^{2} – 4x + 5 = 0

(b) x^{2} + 3x – 12 = 0

(2)^{2} + 3(2) – 12 = -2 ≠ 0

So, x = 2 is not a root of x^{2} + 3x – 12 = 0

(c) 2x^{2} – 7x + 6 = 0

2(2)^{2} – 7(2) + 6 = 0

Here, x = 2 is a root of 2x^{2} – 7x + 6 = 0

**13. A quadratic equation ax ^{2}**

**+ bx + c = 0 has no real roots, if**

(a) b^{2} – 4ac > 0

(b) b^{2} – 4ac = 0

(c) b^{2} – 4ac < 0

(d) b^{2} – ac < 0

Answer: **(c) b ^{2}**

**– 4ac < 0**

A quadratic equation ax^{2 }+ bx + c = 0 has no real roots, if b^{2} – 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) x^{2} + x + 360 = 0

(b) x^{2} + x – 360 = 0

(c) 2x^{2} + x – 360

(d) x^{2} – 2x – 360 = 0

Answer: **(b) x ^{2}**

**+ x – 360 = 0**

Explanation:

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

According to the given,

x(x + 1) = 360

x^{2} + x = 360

x^{2} + x – 360

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

(a) 2x^{2} – 3x + 6 = 0

(b) –x^{2} + 3x – 3 = 0

(c) √2x^{2} – 3/√2x + 1 = 0

(d) 3x^{2} – 3x + 3 = 0

Answer: **(b) –x ^{2}**

**+ 3x – 3 = 0**