 NCERT Exemplar Solution for CBSE Class 10 Mathematics: Quadratic Equations (Part-IVA)

In this article you will get CBSE Class 10 Mathematics chapter 4, Quadratic Equations: NCERT Exemplar Problems and Solutions (Part-IVA). Every question has been provided with a detailed solution. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018. Here you get the CBSE Class 10 Mathematics chapter 4, Quadratic Equations: NCERT Exemplar Problems and Solutions (Part-IVA). This part of the chapter includes solutions of Question Number 1 to 3 from Exercise 4.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Quadratic Equations. This exercise comprises only of the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

NCERT Exemplar Solution for CBSE Class 10 Mathematics: Quadratic Equations (Part-I)

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

NCERT Exemplar Solution for CBSE Class 10 Mathematics: Quadratic Equations (Part-II)

NCERT Exemplar Solution for CBSE Class 10 Mathematics: Quadratic Equations (Part-III)

Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Quadratic Equations:

Exercise 4.4

Long Answer Type Questions

Question1. Find whether the following equations have real roots. If real roots exist, find them Solution:

Concept used: For a quadratic equation of the form, ax2 + bx + c = 0 to have real roots, its discriminant, D = b2 – 4ac > 0.     Question2. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Solution:

Let the required natural number be x.

According to the question,

x2 – 84 = 3 (x + 8)

⟹       x2 – 84 = 3x + 24

⟹       x2 – 3x – 108 = 0

⟹       x2 – 12x + 9x – 108 = 0  [By splitting the middle trem]

⟹       x (x – 12) + 9 (x – 12) = 0

⟹       (x – 12) (x + 9) = 0

⟹       x = 12, ‒ 9

Reject x = ‒ 9 value as a natural number can’t be negative.

Hence, the required natural number is 12.

Question3. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Solution:

Let the natural number be x. Reject x = – 20 value as a natural number can’t be negative.

Hence, the required natural number is 8.

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CBSE Class 10 Mathematics Syllabus 2017-2018

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