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 
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)
View AnswerAns:
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 = x^{2} – 100x
⇒ x^{2} – 100x – 560000 = 0
⇒ x^{2} – 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.
⇒
⇒ 21y^{2} + 21 = 58y
⇒ 21y^{2} – 58y + 21 = 0
⇒ 21y^{2} – 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.
Ans: Let α and β be the roots of given quadratic equation 2x^{2}  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 px^{2} – 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
⇒ 6a^{2} = 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 2x^{2} + 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 = k^{2} 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 4x^{2}  5 = 0 and hence comment on the nature of roots of the equation. (2023)
Ans: Given quadratic equation is 4x^{2}  5 = 0
Discriminant, D = b^{2}  4ac = 0^{2}  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
⇒ px^{2}  2xp +6 = 0
On comparing with ax^{2}+ bx + c = 0, we get
a = p, b = 2p and c = 6
Since, the quadratic equations has two equal real roots.
∴ Discriminant D = 0
⇒ b^{2}  4ac = 0
⇒ (2p)^{2 } 4 x p x 6 = 0
⇒ 4p^{2 } 24p = 0
⇒ p^{2}  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?
Can any rational value of x make the new area equal to 220 cm^{2} ? (2023)
View AnswerAns: 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 + x^{2} = 432
⇒ x^{2} + 30x  216 = 0
(iii) x^{2} + 30x  216 = 0
⇒ x^{2}+ 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
According to question (13 + x) (12 + x) = 220
⇒ 216 + 30x + x^{2} = 220
⇒ x^{2 }+ 30x + 216  220 = 0
⇒ x^{2} + 30x  4 = 0
For rational value of x. discriminant (D) must be perfect square.
So, D = b^{2}  4ac
= (30)^{2}  4(1) (4) = 900 + 16 = 916
∴ 916 is not a perfect square.
So, no rational value of x is possible.
Q9: If the sum of the roots of the quadratic equation ky^{2} – 11y + (k – 23) = 0 is 13/21 more than the product of the roots, then find the value of k. (2022)
View AnswerAns: Given, quadratic equation is ky^{2} – 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 = k23/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: x^{2} – 2ax – (4b^{2} – a^{2}) = 0
Ans: x^{2} – 2ax – (4b^{2} – a^{2}) = 0
⇒ x^{2} + (2b – a)x – (2b + a)x – (4b^{2} – a^{2}) = 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 inground 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 +4x^{2}
⇒ 4x^{2}  38x + 48 = 0
⇒ 2x^{2}  19x + 24 = 0, is the required quadratic equation.
Area of the pool given by quadratic equation is2x^{2}  19x + 24 = 0
⇒ 2x^{2}  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 x^{2}  108 + 3x = 260
⇒ x^{2 } 39x +368 = 0
⇒ 4x^{2} 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, b^{2} = c^{2} + a^{2}
⇒ (6 + 2a)^{2} = (3a – 6)^{2} + a^{2} ...[Using (i) and (ii)]
⇒ 36 + 4a^{2} + 24a = 9a^{2} + 36 – 36a + a^{2}
⇒ 60a = 6a^{2}
⇒ 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: x^{2} – 2ax + (a^{2 }– b^{2}) = 0 for x. (2022)
Ans: We have, x^{2} – 2ax + (a^{2} – b^{2}) = 0
⇒ x^{2} – ((a + b) + (a – b))x + (a^{2} – b^{2}) = 0
⇒ x^{2} – (a + b)x – (a – b)x + (a + b)(a – b) = 0 ......[∵ a^{2} – b^{2} = (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) x^{2} + 2 (m  1) x + 1 = 0 has two real and equal roots. (2022)
Ans: We have
(m  1) x^{2} + 2 (m  1) x + 1 = 0 (i)
On comparing the given equation with ax^{2} + bx + c = 0,
we have a = (m  1), b = 2 (m  1), c = 1
Discriminant, D = 0
⇒ b^{2}  4ac = 0 ⇒ 4m^{2} + 4  8m  4m + 4 = 0
⇒ 4m^{2} 12m + 8 = 0
⇒ m^{2}  3m + 2 = 0 ⇒ m^{2}  2m  m + 2= 0
⇒ m(m  2)  1 (m  2) = 0
⇒ (m  1)(m  2) = 0 ⇒ m = 1, 2
Q16: The quadratic equation (1 + a^{2})x^{2} + 2abx + (b^{2}  c^{2}) = 0 has only one root. What is the value of c^{2}(1 + a^{2})? (2022)
Ans: (1 + a^{2})x^{2} + 2abx + (b^{2}  c^{2})= 0
Comparing on Ax^{2} + Bx + C = 0
A = 1 + a^{2}, B = 2ab & C = (b^{2}  c^{2})
Now, B^{2}  4AC = 0
⇒ (2ab)^{2}  4 × (1 + a^{2}) × (b^{2}  c^{2}) = 0
⇒ 4a^{2}b^{2}  4(b^{2}  c^{2} + a^{2}b^{2 } a^{2}c^{2}) = 0
⇒ 4a^{2}b^{2}  4b^{2} + 4c^{2}  4a^{2}b^{2} + 4 a^{2}c^{2} = 0
⇒  b^{2}+ c^{2} + a^{2}c^{2} = 0
⇒ c^{2} + a^{2}c^{2} = b^{2}
∴ c^{2} (1 + a^{2}) = b^{2}
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
x^{2}  (sum of roots)x + Product of roots = 0
⇒ x^{2} + 3x  10 = 0
Ans: Let the sides of the two squares be x m and y m, where ; x > y.
Then, their areas are x^{2 }and y^{2} and their perimeters are  4x and 4y respectively.
By the given condition, x^{2} + y^{2 }= 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} + y^{2} = 544
⇒ y^{2} + 64 + 16y + y^{2 }= 544
⇒ 2y^{2} + 16y  480 = 0
⇒ y^{2} + 8y  240 = 0
⇒ y^{2} + 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 2x^{2} + kx + 2 = 0 has equal roots, is
(a) 4
(b) ± 4
(c)  4
(d) 0 (2020, 1 Mark)
Ans: (b)
Given Quadratic equation is 2x^{2} + kx + 2 = 0
Since, the equation has equal roots.
∴ Discriminant = 0
⇒ k^{2 } 4 x 2 x 2 = 0
⇒ k^{2} 16 = 0
⇒ k^{2} = 16
⇒ k = ±4
Q21: Find the value of k for which x = 2 is a solution of the equation kx^{2} + 2x  3 = 0. (2019)
View AnswerAns: Since x = 2 is a solution of kx^{2} + 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 m^{2}. 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, x^{2} + y^{2}= 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^{2 }+ (17  x)^{2} = 157
⇒ x^{2} + 289 + x^{2}  34x = 157
=> 2x^{2}  34x +132 = 0
⇒ x^{2}  17x + 66 = 0
⇒ x^{2}  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.
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1. What is a quadratic equation? 
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3. What is the quadratic formula and how is it used to solve quadratic equations? 
4. Can a quadratic equation have more than two solutions? 
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