# Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

 Table of contents Previous Year Questions 2024 Previous Year Questions 2023 Previous Year Questions 2022 Previous Year Questions 2021 Previous Year Questions 2020 Previous Year Questions 2019

## Previous Year Questions 2024

Q1: In flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by100km/h and by doing so, the time of flight is increased by 30 minutes. Find the original duration of the flight.     (2024)

Ans:
Let the speed of aircraft be x km/hr.
Time taken to cover 2800 km by speed of x km/hr = 2800/x  hrs.
New speed is (x – 100) km/hr
so time taken to cover 2800 km at the speed of

⇒ 560000 = x2 – 100x
⇒ x2 – 100x – 560000 = 0
⇒ x2 – 800x + 700x – 560000 = 0
⇒ x(x – 800) + 700(x – 800) = 0
⇒ (x – 800) (x + 700) = 0
⇒ x = 800, – 700 (Neglect)
⇒ x = 800
Speed = 800 km/hr
Time = 2800/800
= 3 hr 30 min.

Q2: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is  find the fraction.      (2024)

Ans: Let the numerator be x.
Denominator = 2x + 1

Let,
Then, the equation will be.

⇒ 21y2 + 21 = 58y
⇒ 21y2 – 58y + 21 = 0
⇒ 21y2 – 49y – 9y + 21 = 0
⇒ 7y(3y – 7) – 3(3y – 7) = 0
⇒ (3y – 7) (7y – 3) = 0

∴ Required fraction will be 7/3 and 3/7.

## Previous Year Questions 2023

Q3: Find the sum and product of the roots of the quadratic equation 2x2 - 9x + 4 = 0.  (2023)  View Answer

Ans: Let α and β be the roots of given quadratic equation  2x2 - 9x + 4 = 0.
Sum of roots = α + β = -b/a = (-9)/2 = 9/2
and Product of roots, αβ = c/a = 4/2 = 2

Q4: Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.  (2023)

Ans: Let the first root be α, then the second root will be 6a
Sum of roots  = -b/a
⇒ a + 6a = 14/p
⇒ 7a = 14/p
⇒ a = 2/p
Product of roots = c/a
⇒ a x 6a  = 8/p
⇒ 6a2 = 8/p

⇒ p = 6 x 4/8
⇒ p = 3
Hence, the value of p is 3.

Q5: The least positive value of k for which the quadratic equation 2x2 + kx + 4 = 0 has rational roots, is  (2023)
(a) ±2√2
(b) 4√2
(c) ±2
(d) √2

Ans: (b)
Sol: For rational roots, D is a perfect square
D = k2- 32
If k = 4√2  then,
D = (4√2)2 - 32 = 0.

Which is a perfect square

Q6: Find the discriminant of the quadratic equation 4x2 - 5 = 0 and hence comment on the nature of roots of the equation.  (2023)

Ans: Given quadratic equation is 4x2 - 5 = 0
Discriminant, D = b2 - 4ac = 02 - 4(4)(-5) = 80 > 0
Hence, the roots of the given quadratic equation are real and distinct.

Q7: Find the value of 'p' for which the quadratic equation px(x - 2) + 6 = 0 has two equal real roots.  (2023)

Ans: The given quadratic equation is px(x - 2) + 6 = 0
⇒ px2 - 2xp +6 = 0
On comparing with ax2+ bx + c = 0, we get
a = p, b = -2p and c = 6
Since, the quadratic equations has two equal real roots.

∴ Discriminant D = 0
⇒ b2 - 4ac = 0
⇒ (-2p)- 4  x p x 6 = 0
⇒  4p- 24p = 0
⇒ p2 - 6p = 0
⇒  p(p - 6) = 0
⇒  p = 0 or p = 6
But p ≠ 0 as it does not satisfy equation
Hence, the value of p is 6.

Q8: Case Study : While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by n units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide.
Based o n the above information. answer the following Questions:
(i) Write an algebraic equation depicting the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii) What should be the new dimensions of the enlarged photo?

OR

Can any rational value of x make the new area equal to 220 cm2 ?      (2023)

Ans: Area = 18 x 12 cm
Length (l) is increased by x cm
So,  new length =(18 + x ) cm
New width = (12 + x) cm

(i) Area of photo after increasing the length and width
= (18 + x)(12 + x) = 2 x 18 x 12
i.e., (18 + x) (12 + x) = 432 is the required algebric equation.
(ii) From part (i) we get, (18 + x) (12 + x) = 432
216 + 18x + 12x + x2 = 432
x2 + 30x - 216 = 0
(iii)  x2 + 30x - 216 = 0
x2+ 36x - 6x - 216 = 0
x(x+ 36) - 6 (x+ 36) = 0 ⇒  x = 6, -36
-36 is not possible.
So, new length = (18 + 6) cm = 24 cm
New width = (12 + 6) cm = 18cm
So. new dimension = 24cm x 18 cm

OR

According to question (13 + x) (12 + x) = 220
216 + 30x + x2 = 220
x+ 30x + 216 - 220 = 0
x2 + 30x - 4 = 0
For rational value of x. discriminant (D) must be perfect square.
So, D = b2 - 4ac
= (30)2 - 4(1) (-4) = 900 + 16 = 916
916 is not a perfect square.
So, no rational value of x is possible.

## Previous Year Questions 2022

Q9: If the sum of the roots of the quadratic equation ky2 – 11y + (k – 23) = 0 is 13/21 more than the product of the roots, then find the value of k.    (2022)

Ans: Given, quadratic equation is ky2 – 11y + (k – 23) = 0
Let the roots of the above quadratic equation be α and β.
Now, Sum of roots, α + β = -(-11)/k = 11/k ...(i)
and Product of roots, αβ = k-23/k  ...(ii)
According to the question,
α + β = αβ + 13/21

⇒ 21(34 – k) = 13k
⇒ 714 – 21k = 13k
⇒ 714 = 13k + 21k
⇒ 34k = 714
⇒ k = 714/34
⇒ k = 21

Q10: Solve the following quadratic equation for x: x2 – 2ax – (4b2 – a2) = 0

Ans: x2 – 2ax – (4b2 – a2) = 0
⇒ x2 + (2b – a)x – (2b + a)x – (4b2 – a2) = 0
⇒ x(x + 2b – a) – (2b + a)(x + 2b – a) = 0
⇒ (x + 2b – a)(x – 2b – a) = 0
⇒ (x + 2b – a) = 0, (x – 2b – a) = 0
∴ x = a − 2b, a + 2b

Q11: In the picture given below, one can see a rectangular in-ground swimming pool installed by a family In their backyard. There is a concrete sidewalk around the pool of width x m. The outside edges of the sidewalk measure 7 m and 12 m. The area of the pool is 36 sq. m.

Based on the information given above, form a quadratic equation in terms of x
Find the width of the sidewalk around the pool.
(2022)

Ans: Given, width of the sidewalk = xm.
Area of the pool = 36 sq.m
∴ Inner length of the pool
= (12 - 2x)m
Inner width of the pool = (7 - 2x)m
∴ Area of the pool. A = l x b
⇒ 36 = (12 - 2x) x  (7 - 2x)
⇒ 36 = 84 - 24x - 14x +4x2
⇒  4x2 - 38x + 48 = 0
⇒ 2x2 - 19x + 24 = 0, is the required quadratic equation.

Area of the pool given by quadratic equation is2x2 - 19x + 24 = 0
⇒ 2x2 - 16x  - 3x + 24 = 0
⇒  2x(x - 8) - 3(x - 8) = 0
⇒ (x - 8)(2x - 3) = 0
⇒ x = 8 (not possible) or x = 3/2 = 1.5
Width of the sidewalk =1.5m

Q12: The sum of two numbers is 34. If 3 Is subtracted from one number and 2 is added to another. the product of these two numbers becomes 260. Find the numbers.      (2022)

Ans: Let one number be x and another number be y.
Since, x + 4y = 34 ⇒  y = 34 - x      (i)
Now. according to the question. (x - 3) (y + 2) = 260       (ii)
Putting the value or y from (i) in (ii), we get
⇒ (x - 3)(34 - x + 2) = 260
⇒ (x - 3)(36 - x) = 260
⇒ 36x -x2 - 108 + 3x = 260
⇒ x- 39x +368 = 0
⇒ 4x2- 23x - 16x + 368 = 0
⇒ x(x - 23) - 16(x - 23) = 0
⇒ (x - 23)(x - 16) =0
⇒ x = 23 or 16
Hence; when x  = 23 from (i),  y = 3+ - 23 = 11
When x = 16. then y = 34 - 16 = 18
Hence the required numbers are 23 and 11 or 16 and 18.

Q13: The hypotenuse (in cm) of a right angled triangle is 6 cm more than twice the length of the shortest side. If the length of third side is 6 cm less than thrice the length of shortest side, then find the dimensions of the triangle.     (2022)

Ans:

Let ΔABC be the right angle triangle, right angled at B, as shown in the figure.
Also, let AB = c cm, BC = a cm and AC = b cm
Then, according to the given information, we have
b = 6 + 2a  .....(i) (Let a be the shortest side)
and c = 3a – 6  ...(ii)
We know  that, b2 = c2 + a2
⇒ (6 + 2a)2 = (3a – 6)2 + a2  ...[Using (i) and (ii)]
⇒ 36 + 4a2 + 24a = 9a2 + 36 – 36a + a2
⇒ 60a = 6a2
⇒ 6a = 60  ...[∵ a cannot be zero]
⇒ a = 10 cm
Now, from equation (i),
b = 6 + 2 × 10 = 26
and from equation (ii),
c = 3 × 10 – 6 = 24
Thus, the dimensions of the triangle are 10 cm, 24 cm and 26 cm.

Q14: Solve the quadratic equation: x2 – 2ax + (a– b2) = 0 for x.     (2022)

Ans: We have, x2 – 2ax + (a2 – b2) = 0
⇒ x2 – ((a + b) + (a – b))x + (a2 – b2) = 0
⇒ x2 – (a + b)x – (a – b)x + (a + b)(a – b) = 0  ......[∵ a2 – b2 = (a + b)(a – b)]
⇒ x(x – (a + b)) – (a – b)(x – (a + b)) = 0
⇒ (x – (a + b))(x – (a – b) = 0
⇒ x = a + b, a – b

Q15: Find the value of m for which the quadratic equation (m - 1) x2 + 2 (m - 1) x + 1 = 0 has two real and equal roots.   (2022)

Ans:  We have
(m - 1) x2 + 2 (m - 1) x + 1 = 0   ----(i)
On comparing the given equation with ax2 + bx + c = 0,
we have a = (m - 1), b = 2 (m - 1), c = 1
Discriminant, D = 0
⇒ b2 - 4ac = 0 ⇒ 4m2 + 4 - 8m - 4m + 4 = 0
⇒ 4m2 -12m + 8 = 0
⇒ m2 - 3m + 2 = 0 ⇒ m2 - 2m - m + 2= 0
⇒ m(m - 2) - 1 (m - 2) = 0
⇒ (m - 1)(m - 2) = 0 ⇒ m = 1, 2

Q16: The quadratic equation (1 + a2)x2 + 2abx + (b2 - c2) = 0 has only one root. What is the value of c2(1 + a2)?    (2022)

Ans: (1 + a2)x2 + 2abx + (b2 - c2)= 0
Comparing on Ax2 + Bx + C = 0
A = 1 + a2, B = 2ab & C = (b2 - c2)
Now, B2 - 4AC = 0
⇒ (2ab)2 - 4 × (1 + a2) × (b2 - c2) = 0
⇒ 4a2b2 - 4(b2 - c2 + a2b- a2c2) = 0
⇒ 4a2b2 - 4b2 + 4c2 - 4a2b2 + 4 a2c2 = 0
⇒ - b2+ c2 + a2c2 = 0
⇒ c2 + a2c2 = b2
∴ c2 (1 + a2) = b2

## Previous Year Questions 2021

Q17: Write the quadratic equation in x whose roots are 2 and-5.     (2021)

Ans: Roots of quadratic equation are given as 2 and - 5.
Sum of roots = 2 + (-5) = -3
Product of roots = 2 (-5) = -10
Quadratic equation can he written as
x2 - (sum of roots)x + Product of roots = 0
⇒ x2 + 3x - 10 = 0

## Previous Year Questions 2020

Q18: Sum of the areas of two squares is 544 m2. If the difference of their perimeters is 32 m, find the sides of the two squares.     (2020)

Ans: Let the sides of the two squares be x m and y m, where ; x > y.
Then, their areas are xand y2 and their perimeters are - 4x and 4y respectively.
By the given condition, x2 + y= 544    --------(i)
and 4x - 4y = 32
⇒ x - y = 8
⇒ x = y + 8 ------------ (ii)
Substituting the value of x from (ii) in (i) we get
⇒ (y + 8)2 + y2 = 544
⇒ y2 + 64 + 16y + y= 544
⇒ 2y2 + 16y - 480 = 0
⇒ y2 + 8y - 240  = 0
⇒ y2 + 20y - 12y - 240 = 0
⇒ y(y + 20) - 12(y + 20) = 0
⇒ (y - 12) (y + 20) = 0
⇒ y = 12    (∵ y ≠ 20 as length cannot be negative)
From (ii), x = 12 + 8 = 20 Thus, the sides of the two squares are 20 m and 12 m.

Q19: A motorboat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.     (2020)

Ans: Let the speed of the stream be x km/hr.
Speed of the boat upstream = (18 - x) km/hr
Speed of the boat downstream = (18 + x) km/hr
According to question,

(∵ x ≠ -54 as speed can’t be negative)
Hence, the speed of the stream is 6 km/hr.

Q20: The value(s) of k for which the quadratic equation 2x2 + kx + 2 = 0 has equal roots, is
(a) 4
(b) ± 4
(c) - 4
(d) 0   (2020, 1 Mark)

Ans: (b)
Given Quadratic equation is 2x2 + kx + 2 = 0
Since, the equation has equal roots.
∴ Discriminant = 0
⇒ k- 4 x 2 x 2 = 0
⇒ k2- 16 = 0
⇒  k2 = 16
⇒ k = ±4

## Previous Year Questions 2019

Q21: Find the value of k for which x = 2 is a solution of the equation kx2 + 2x - 3 = 0.     (2019)

Ans: Since x = 2 is a solution of kx2 + 2x - 3 = 0
k(2)2 + 2(2) - 3 = 0
= 4k + 4 - 3 = 0
⇒ k = -1/4

Q22: Sum of the areas of two squares is 157 m2. If the sum of their perimeters is 68 m, find the sides of the two squares.    (2020)

Ans: Let the length of the side of one square be x m and the length of the side of another square be y m.
Given, x2 + y2= 157    _(i)
and 4x + 4y=68    _(ii)
x + y = 17
y = 17 - x    _(iii)
On putting the value of y in (i), we get
x+ (17 - x)2 = 157
⇒ x2 + 289 + x2 - 34x = 157
=> 2x2 - 34x +132 = 0
⇒ x2 - 17x + 66 = 0
⇒ x2 - 11x - 6x + 66 =0
⇒ x(x - 11)-6(x - 11) = 0
⇒ (x - 11) (x - 6) = 0
⇒ x = 6 or x = 11
On putting the value of x in (iii), we get
y = 17 - 6 = 11 or y = 17 - 11 = 6
Hence, the sides of the squares be 11 m and 6 m.

The document Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

 1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants, and a is not equal to 0.
 2. How do you solve a quadratic equation by factoring?
Ans. To solve a quadratic equation by factoring, you need to factor the quadratic expression into two binomial expressions and set each factor equal to zero to find the solutions.
 3. What is the quadratic formula and how is it used to solve quadratic equations?
Ans. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a, which is used to find the roots of a quadratic equation ax^2 + bx + c = 0 by substituting the values of a, b, and c into the formula.
 4. Can a quadratic equation have more than two solutions?
Ans. No, a quadratic equation can have at most two real solutions, which can be the same if the discriminant (b^2 - 4ac) is equal to zero.
 5. How are quadratic equations used in real-life applications?
Ans. Quadratic equations are used in various real-life applications such as calculating projectile motion, determining profit and loss in business, optimizing shapes in architecture, and predicting the trajectory of objects.

## Mathematics (Maths) Class 10

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