Boost your LIC AAO 2025 preparation with the most asked quadratic equation questions and solved examples
Quadratic Equations are one of the most important topics in the Quantitative Aptitude section of the LIC AAO 2025 exam. Typically, the Prelims exam contains 5 to 6 questions from this topic, making it crucial for aspirants to have a clear understanding and strategy. In this article, we will provide detailed knowledge of quadratic equations, including the basic concepts, methods of solving, types of questions asked in previous LIC AAO exams, and the most commonly asked questions with step-by-step solutions. By going through this article, candidates can strengthen their conceptual understanding, learn shortcut methods, and practice effectively to attempt all quadratic equation questions with confidence and accuracy in the exam.
Quadratic equations are not just about solving standard problems; they also appear in the form of word problems, inequalities, and comparisons of roots. This article aims to cover everything from basics to advanced techniques to ensure that you are fully prepared for the LIC AAO 2025 exam.
A quadratic equation is a second-degree polynomial equation in a single variable, typically expressed as:
where aaa, bbb, and ccc are constants, and a≠0a \neq 0a=0. The solutions to this equation are known as the roots, which can be found using methods such as factoring, completing the square, or the quadratic formula:
In the LIC AAO exam, quadratic equations are often presented in the form of inequalities or comparisons between two equations.
For aspirants aiming to ace the LIC AAO 2025 exam, practising a wide variety of questions is key. To help you prepare effectively, we have compiled 80+ Most Asked Quadratic Equation Questions with detailed solutions in a handy PDF. This resource covers all important question types, including factorisation the quadratic formula. Download the PDF and start practising to improve your speed and accuracy in the exam.
Question 1: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2+ 9x –52 = 0
II. y2+19y –66 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 2: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x3 = 1331
II. y2–26y + 168 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 3: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. 3x – 2y = 31
II. x + 3y = 69
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 4: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2+ 21x +90 = 0
II. y2+25y + 114 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 5: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. 2x2= 392
II. y2–29y + 210 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 6: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 – 7x + 12 = 0
II. y2 – 20y + 75 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 7: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 + 11x + 30 = 0
II. y2 + 19y + 90 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 8: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 – 22x + 120 = 0
II. y2 – 26y + 165 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 9: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I.x2 + 23x + 120 = 0
II.y2 – 4y – 96 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Question 10: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 – 10x – 119 = 0
II. y2 + 21y + 108 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
ANSWER KEYS and SOLUTIONS:
| 1) – C) | 2) – B) | 3) – A) | 4) – C) | 5) – E) | 6) – B) |
| 7) – A) | 8) – C) | 9) – E) | 10) – A) |
Solution 1: C)
From I:
x2+ 9x –52 = 0
x2+ 13x – 4x–52 = 0
x(x + 13) – 4(x +13) = 0
(x + 13) (x – 4) = 0
x = -13, 4
From II:
y2+19y – 66 = 0
y2+22y – 3y – 66 = 0
y(y + 22) – 3(y + 22) = 0
(y + 22) (y – 3) = 0
y = -22, 3
So, no relation can be established between x and y.
Hence, option c.
Solution 2: B)
From I:
x3 = 1331
x = 11
From II:
y2–26y + 168 = 0
y2–12y – 14y + 168 = 0
y(y – 12) – 14(y – 12) = 0
(y – 12) (y – 14) = 0
y = 12, 14
So, x < y.
Hence, option b.
Solution 3: A)
From I:
3x – 2y = 31
3x = 31 + 2y
x = (31 + 2y)/3 …..(1)
From II:
x + 3y = 69
(31 + 2y)/3 + 3y = 69
31 + 2y + 9y = 69 × 3
31 + 11y = 207
11y = 207 – 31
11y = 176
y = 16
Put the value of ‘y’ in equation (1)
x = (31 + 2 × 16)/3
= (31 + 32)/3
= 63/3 = 21
So, x > y.
Hence, option a.
Solution 4: C)
From I:
x2+ 21x +90 = 0
x2+ 15x + 6x +90 = 0
x(x + 15) + 6(x +15) = 0
(x + 15) (x + 6) = 0
x = -15, -6
From II:
y2+25y + 114 = 0
y2+19y + 6y + 114 = 0
y(y + 19) + 6(y + 19) = 0
(y + 19) (y + 6) = 0
y = -19, -6
So, no relation can be established between x and y.
Hence, option c.
Solution 5: E)
From I:
2x2= 392
x2= 196
x = ± 14
From II:
y2–29y + 210 = 0
y2–14y – 15y + 210 = 0
y(y –14) – 15(y – 14) = 0
(y –14) (y – 15) = 0
y = 14,15
So, x ≤ y.
Hence, option e.
Solution 6: B)
From I:
x2– 7x + 12 = 0
x2 – 4x – 3x + 12 = 0
x(x – 4) – 3(x – 4) = 0
(x – 4)(x – 3) = 0
x = 4,3
From II:
y2 – 20y + 75 = 0
y2 – 15y – 5y + 75 = 0
y(y – 15) – 5(y – 15) = 0
(y – 15)(y – 5) = 0
y = 15, 5
Hence, option b.
Solution 7: A)
From I:
x2 + 11x + 30 = 0
x2 + 6x + 5x + 30 = 0
x(x + 6) + 5(x + 6) = 0
(x + 6)(x + 5) = 0
x = -6, -5
From II:
y2 + 19y + 90 = 0
y2 + 10y + 9y + 90 = 0
y(y + 10) + 9(y + 10) = 0
(y + 10)(y + 9) = 0
y = -10, – 9
Hence, option a.
Solution 8: C)
From I:
x2– 22x + 120 = 0
x2– 12x – 10x + 120 = 0
x(x – 12) – 10(x – 12) = 0
(x – 12)(x – 10) = 0
x = 12, 10
From II:
y2– 26y + 165 = 0
y2– 11y – 15y + 165 = 0
y(y – 11) – 15(y – 11) = 0
(y – 11)(y – 15) = 0
y = 11, 15
Hence, option c.
Solution 9: E)
From I:
x2 + 23x + 120 = 0
x2 + 8x + 15x + 120 = 0
x(x + 8) + 15(x + 8) = 0
(x + 8)(x + 15) = 0
x = -8, -15
From II:
y2 – 4y – 96 = 0
y2 – 12y + 8y – 96 = 0
y(y – 12) + 8(y – 12) = 0
(y – 12)(y + 8) = 0
y = 12, -8
Hence, option e.
Solution 10: A)
From I:
x2– 10x– 119 = 0
x2 + 7x – 17x – 119 = 0
x(x + 7) – 17(x + 7) = 0
(x + 7)(x – 17) = 0
x = -7, 17
From II:
y2+ 21y + 108 = 0
y2+ 12y + 9y + 108 = 0
y(y + 12) + 9(y + 12) = 0
(y +12)(y + 9) = 0
y = -9, -12
Hence, option a.
For More Questions, Download the PDF
Mastering quadratic equations is all about understanding the methods and practising smartly. Here’s a step-by-step approach to solving them efficiently for the LIC AAO exam:
Factoring is usually the fastest way to solve quadratic equations in exams.
Example:
x² – 5x + 6 = 0
Split the middle term:
x² – 2x – 3x + 6 = 0
Factorise:
(x – 2)(x – 3) = 0
Roots: x = 2, 3
Tip: Memorise common factor pairs for numbers up to 100. This will save time in exams.
For equations that don’t factorise easily, use:
x = [-b ± √(b² – 4ac)] / (2a)
When comparing roots of two equations, a simple table method helps you organise your calculations quickly.
Example:
Equation: 40x² – 47x + 12 = 0
Repeat the same steps for the second equation and then compare the roots. This makes comparison problems much easier.
Going through the last 5 years of LIC AAO or RRB PO papers is extremely helpful.
Daily practice with topic-focused tests can drastically improve your speed and accuracy.
If you feel weak in the Quant section, do mini mock tests or sectional tests.
Full-length mocks help you understand your overall performance.
Disclaimer: The quadratic equation questions, solutions, and strategies provided are compiled from expert references and exam trends. They are intended for practice and guidance only, not official LIC AAO exam content. Actual exam questions may differ. Candidates should always verify details through official LIC notices before relying on them.
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