{"id":177866,"date":"2025-09-25T17:55:00","date_gmt":"2025-09-25T12:25:00","guid":{"rendered":"https:\/\/www.practicemock.com\/blog\/?p=177866"},"modified":"2025-11-26T15:37:41","modified_gmt":"2025-11-26T10:07:41","slug":"80-most-asked-quadratic-equation-questions-for-lic-aao","status":"publish","type":"post","link":"https:\/\/www.practicemock.com\/blog\/80-most-asked-quadratic-equation-questions-for-lic-aao\/","title":{"rendered":"80+ Most Asked Quadratic Equation Questions for LIC AAO 2025 Exam"},"content":{"rendered":"\n

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Home<\/a><\/span> \u00bb LIC AAO<\/a><\/span> \u00bb 80+ Most Asked Quadratic Equation Questions for LIC AAO 2025 Exam<\/span><\/span><\/div>\n\n\n

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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<\/strong> 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<\/strong>, 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<\/strong>. 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.<\/p>\n\n\n\n

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<\/strong> to ensure that you are fully prepared for the LIC AAO 2025 exam.<\/p>\n\n\n\n

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Take LIC AAO 2025 Free Mock Test to know your Performance<\/strong><\/a><\/div>\n<\/div>\n\n\n\n

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Understand Quadratic Equations<\/h3>\n\n\n\n

A quadratic equation is a second-degree polynomial equation in a single variable, typically expressed as:<\/p>\n\n\n

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where aaa, bbb, and ccc are constants, and a\u22600a \\neq 0a\ue020=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: <\/p>\n\n\n

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In the LIC AAO exam, quadratic equations are often presented in the form of inequalities or comparisons between two equations.<\/p>\n\n\n\n

80+ Most Asked Quadratic Equation Questions PDF Link<\/h2>\n\n\n\n

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.<\/p>\n\n\n\n

Question 1:<\/strong> 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.<\/p>\n\n\n\n

I. x2<\/sup>+ 9x \u201352 = 0<\/p>\n\n\n\n

II. y2<\/sup>+19y \u201366 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 2:<\/strong> 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.<\/p>\n\n\n\n

I. x3<\/sup> = 1331<\/p>\n\n\n\n

II. y2<\/sup>\u201326y + 168 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 3:<\/strong> 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.<\/p>\n\n\n\n

I. 3x \u2013 2y = 31<\/p>\n\n\n\n

II. x + 3y = 69<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 4:<\/strong> 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.<\/p>\n\n\n\n

I. x2<\/sup>+ 21x +90 = 0<\/p>\n\n\n\n

II. y2<\/sup>+25y + 114 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 5:<\/strong> 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.<\/p>\n\n\n\n

I. 2x2<\/sup>= 392<\/p>\n\n\n\n

II. y2<\/sup>\u201329y + 210 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 6:<\/strong> 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.<\/p>\n\n\n\n

I. x2<\/sup> \u2013 7x + 12 = 0<\/p>\n\n\n\n

II. y2<\/sup> \u2013 20y + 75 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 7:<\/strong> 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.<\/p>\n\n\n\n

I. x2<\/sup> + 11x + 30 = 0<\/p>\n\n\n\n

II. y2<\/sup> + 19y + 90 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 8:<\/strong> 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.<\/p>\n\n\n\n

I. x2<\/sup> \u2013 22x + 120 = 0<\/p>\n\n\n\n

II. y2<\/sup> \u2013 26y + 165 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 9:<\/strong> 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.<\/p>\n\n\n\n

I.x2<\/sup> + 23x + 120 = 0<\/p>\n\n\n\n

II.y2<\/sup> \u2013 4y \u2013 96 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

Question 10:<\/strong> 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.<\/p>\n\n\n\n

I. x2<\/sup> \u2013 10x \u2013 119 = 0<\/p>\n\n\n\n

II. y2<\/sup> + 21y + 108 = 0<\/p>\n\n\n\n

A) x > y<\/p>\n\n\n\n

B) x < y<\/p>\n\n\n\n

C) x = y or the relationship cannot be established<\/p>\n\n\n\n

D) x \u2265 y<\/p>\n\n\n\n

E) x \u2264 y<\/p>\n\n\n\n

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ANSWER KEYS and SOLUTIONS:<\/strong><\/p>\n\n\n\n

1) – C)<\/td>2) – B)<\/td>3) – A)<\/td>4) – C)<\/td>5) – E)<\/td>6) – B)<\/td><\/tr>
7) – A)<\/td>8) – C)<\/td>9) – E)<\/td>10) – A)<\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution 1: C)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup>+ 9x \u201352 = 0<\/p>\n\n\n\n

x2<\/sup>+ 13x \u2013 4x\u201352 = 0<\/p>\n\n\n\n

x(x + 13) \u2013 4(x +13) = 0<\/p>\n\n\n\n

(x + 13) (x \u2013 4) = 0<\/p>\n\n\n\n

x = -13, 4<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup>+19y \u2013 66 = 0<\/p>\n\n\n\n

y2<\/sup>+22y \u2013 3y \u2013 66 = 0<\/p>\n\n\n\n

y(y + 22) \u2013 3(y + 22) = 0<\/p>\n\n\n\n

(y + 22) (y \u2013 3) = 0<\/p>\n\n\n\n

y = -22, 3<\/p>\n\n\n\n

So, no relation can be established between x and y.<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 2: B)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x3<\/sup> = 1331<\/p>\n\n\n\n

x = 11<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup>\u201326y + 168 = 0<\/p>\n\n\n\n

y2<\/sup>\u201312y \u2013 14y + 168 = 0<\/p>\n\n\n\n

y(y \u2013 12) \u2013 14(y \u2013 12) = 0<\/p>\n\n\n\n

(y \u2013 12) (y \u2013 14) = 0<\/p>\n\n\n\n

y = 12, 14<\/p>\n\n\n\n

So, x < y.<\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

Solution 3: A)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

3x \u2013 2y = 31<\/p>\n\n\n\n

3x = 31 + 2y<\/p>\n\n\n\n

x = (31 + 2y)\/3 \u2026..(1)<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

x + 3y = 69<\/p>\n\n\n\n

(31 + 2y)\/3 + 3y = 69<\/p>\n\n\n\n

31 + 2y + 9y = 69 \u00d7 3<\/p>\n\n\n\n

31 + 11y = 207<\/p>\n\n\n\n

11y = 207 \u2013 31<\/p>\n\n\n\n

11y = 176<\/p>\n\n\n\n

y = 16<\/p>\n\n\n\n

Put the value of \u2018y\u2019 in equation (1)<\/p>\n\n\n\n

x = (31 + 2 \u00d7 16)\/3<\/p>\n\n\n\n

= (31 + 32)\/3<\/p>\n\n\n\n

= 63\/3 = 21<\/p>\n\n\n\n

So, x > y.<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

Solution 4: C)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup>+ 21x +90 = 0<\/p>\n\n\n\n

x2<\/sup>+ 15x + 6x +90 = 0<\/p>\n\n\n\n

x(x + 15) + 6(x +15) = 0<\/p>\n\n\n\n

(x + 15) (x + 6) = 0<\/p>\n\n\n\n

x = -15, -6<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup>+25y + 114 = 0<\/p>\n\n\n\n

y2<\/sup>+19y + 6y + 114 = 0<\/p>\n\n\n\n

y(y + 19) + 6(y + 19) = 0<\/p>\n\n\n\n

(y + 19) (y + 6) = 0<\/p>\n\n\n\n

y = -19, -6<\/p>\n\n\n\n

So, no relation can be established between x and y.<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 5: E)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

2x2<\/sup>= 392<\/p>\n\n\n\n

x2<\/sup>= 196<\/p>\n\n\n\n

x = \u00b1 14<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup>\u201329y + 210 = 0<\/p>\n\n\n\n

y2<\/sup>\u201314y \u2013 15y + 210 = 0<\/p>\n\n\n\n

y(y \u201314) \u2013 15(y \u2013 14) = 0<\/p>\n\n\n\n

(y \u201314) (y \u2013 15) = 0<\/p>\n\n\n\n

y = 14,15<\/p>\n\n\n\n

So, x \u2264 y.<\/p>\n\n\n\n

Hence, option e.<\/p>\n\n\n\n

Solution 6: B)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup>\u2013 7x + 12 = 0<\/p>\n\n\n\n

x2<\/sup> \u2013 4x \u2013 3x + 12 = 0<\/p>\n\n\n\n

x(x \u2013 4) \u2013 3(x \u2013 4) = 0<\/p>\n\n\n\n

(x \u2013 4)(x \u2013 3) = 0<\/p>\n\n\n\n

x = 4,3<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup> \u2013 20y + 75 = 0<\/p>\n\n\n\n

y2<\/sup> \u2013 15y \u2013 5y + 75 = 0<\/p>\n\n\n\n

y(y \u2013 15) \u2013 5(y \u2013 15) = 0<\/p>\n\n\n\n

(y \u2013 15)(y \u2013 5) = 0<\/p>\n\n\n\n

y = 15, 5<\/p>\n\n\n\n

Hence, option b.<\/p>\n\n\n\n

Solution 7: A)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> + 11x + 30 = 0<\/p>\n\n\n\n

x2<\/sup> + 6x + 5x + 30 = 0<\/p>\n\n\n\n

x(x + 6) + 5(x + 6) = 0<\/p>\n\n\n\n

(x + 6)(x + 5) = 0<\/p>\n\n\n\n

x = -6, -5<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup> + 19y + 90 = 0<\/p>\n\n\n\n

y2<\/sup> + 10y + 9y + 90 = 0<\/p>\n\n\n\n

y(y + 10) + 9(y + 10) = 0<\/p>\n\n\n\n

(y + 10)(y + 9) = 0<\/p>\n\n\n\n

y = -10, – 9<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

Solution 8: C)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup>\u2013 22x + 120 = 0<\/p>\n\n\n\n

x2<\/sup>\u2013 12x \u2013 10x + 120 = 0<\/p>\n\n\n\n

x(x \u2013 12) \u2013 10(x \u2013 12) = 0<\/p>\n\n\n\n

(x \u2013 12)(x \u2013 10) = 0<\/p>\n\n\n\n

x = 12, 10<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup>\u2013 26y + 165 = 0<\/p>\n\n\n\n

y2<\/sup>\u2013 11y \u2013 15y + 165 = 0<\/p>\n\n\n\n

y(y \u2013 11) \u2013 15(y \u2013 11) = 0<\/p>\n\n\n\n

(y \u2013 11)(y \u2013 15) = 0<\/p>\n\n\n\n

y = 11, 15<\/p>\n\n\n\n

Hence, option c.<\/p>\n\n\n\n

Solution 9: E)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup> + 23x + 120 = 0<\/p>\n\n\n\n

x2<\/sup> + 8x + 15x + 120 = 0<\/p>\n\n\n\n

x(x + 8) + 15(x + 8) = 0<\/p>\n\n\n\n

(x + 8)(x + 15) = 0<\/p>\n\n\n\n

x = -8, -15<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup> \u2013 4y \u2013 96 = 0<\/p>\n\n\n\n

y2<\/sup> \u2013 12y + 8y \u2013 96 = 0<\/p>\n\n\n\n

y(y \u2013 12) + 8(y \u2013 12) = 0<\/p>\n\n\n\n

(y \u2013 12)(y + 8) = 0<\/p>\n\n\n\n

y = 12, -8<\/p>\n\n\n\n

Hence, option e.<\/p>\n\n\n\n

Solution 10: A)<\/strong><\/p>\n\n\n\n

From I:<\/p>\n\n\n\n

x2<\/sup>\u2013 10x\u2013 119 = 0<\/p>\n\n\n\n

x2<\/sup> + 7x \u2013 17x \u2013 119 = 0<\/p>\n\n\n\n

x(x + 7) \u2013 17(x + 7) = 0<\/p>\n\n\n\n

(x + 7)(x \u2013 17) = 0<\/p>\n\n\n\n

x = -7, 17<\/p>\n\n\n\n

From II:<\/p>\n\n\n\n

y2<\/sup>+ 21y + 108 = 0<\/p>\n\n\n\n

y2<\/sup>+ 12y + 9y + 108 = 0<\/p>\n\n\n\n

y(y + 12) + 9(y + 12) = 0<\/p>\n\n\n\n

(y +12)(y + 9) = 0<\/p>\n\n\n\n

y = -9, -12<\/p>\n\n\n\n

Hence, option a.<\/p>\n\n\n\n

For More Questions, Download the PDF<\/strong><\/p>\n\n\n\n

80+ Most Asked Quadratic Equation Questions PDF Link<\/a><\/strong><\/h3>\n\n\n\n

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How to Solve Quadratic Equations Easily<\/h2>\n\n\n\n

Mastering quadratic equations is all about understanding the methods and practising smartly. Here\u2019s a step-by-step approach to solving them efficiently for the LIC AAO exam:<\/p>\n\n\n\n

1. Solve by Factoring (Master Factorisation)<\/h3>\n\n\n\n

Factoring is usually the fastest way to solve quadratic equations in exams.<\/p>\n\n\n\n

Example:<\/strong>
<\/strong> x\u00b2 \u2013 5x + 6 = 0<\/p>\n\n\n\n