Class 10 CBSE Maths Solutions of Important Questions Chapter 4 ‘Quadratic equations’
Class 10 CBSE Maths Solutions of Important Questions Chapter 4 ‘Quadratic equations’ is one of the important parts of algebra. The quadratic equations are used to evaluate different kinds of physical quantities like speed, length, height, profit, loss, and many other scientific calculations. The standard form of a quadratic equation is ax² + bx + c =0, where a ,b and c are numbers such that a≠0, x is variable with power 2. Here you can study 10 most important questions which we have selected from chapter number 4 keeping in view your CBSE board exams.You can study in future study point NCERT solutions, sample papers, assignments of maths and science, solutions of previous question papers, articles related to science and maths, most important questions with solutions, and posts related to your carrier.
Class X Maths Solutions of Important Questions Chapter 4 ‘Quadratic equations’
Q1. Find the roots of the following quadratic equations.
Answer.(a) The given equation is as following
√2x ² + 7x + 5√2 = 0
The product of √2 × 5√2 =10=5×2
√2x ² + 5x +2x+ 5√2 = 0
x(√2x + 5) + √2(√2x + 5)=0
(√2x + 5)(x + √2) =0
(b) The given equation is as following
16x² –8x + 1 = 0
The product of 16 and 1 is 16 =4 × 4
16x² –4x –4x + 1 = 0
4x(4x –1) – 1( 4x –1) =0
(4x –1)(4x –1) = 0
Q2.A cottage industry produces certain number of toys in a day .The cost of production of each toys (in rupees) was found to be 55 minus the number of toys produced in a day.On a particular day,the total cost of production was Rs 750 .Find out the number of toys produced on that day.
Answer. Let the number of toys produced in a day = x
The cost of each toy according to the question is = 55 –x
On a day the cost of production is = Rs 750
The cost of production in a day = number of toys produced in a day × cost of a toy
According to question
x(55 –x) = 750
–x² + 55x –750 = 0
x² – 55x +750 = 0
750 = 5 × 3 × 5 ×5×2= 25 × 30
x² – 25x –30x +750 = 0
x(x– 25) – 30(x – 25) = 0
(x– 25)(x – 30) = 0
x = 25, 30
Hence number of the toys produced on that day 25 or 30.
If the number of toys produced on that day = 25 then the cost of a toy is 55 – 25 =30(rupees) and if the number of toys produced on that day=30 then the cost of a toy is 55 – 30=25(rupees) in both cases the cost production is Rs 750.
What is second law of of motion ?
Q3. Find two consecutive positive integers, the sum of whose squares is 365.
Answers. Let one of the positive integers is x
Next positive integer is = x+ 1
According to question the sum of squares of these number = 365
(x +1)² + x² = 365
x² + 1 + 2x + x²= 365
2x² + 2x –364 = 0
x² + x –182 = 0
182 = 2 ×13×7 =14 × 13
x² + 14x –13x –182 = 0
x(x + 14) – 13(x – 14) = 0
(x + 14)(x –13) = 0
x =–14, 13
Neglecting negative integers,therefore two consecutive positive integers are 13 and 14 .
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
Q4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm , find the other two sides.
Answer. Let the base of given right Δ is x then according to question its altitude is x – 7
The hypotenuse of given Δ is =13 cm
Applying Pythagoras theorem
(x – 7)² + x² = 13²
x² + 49 – 14x + x²= 169
2x² – 14x – 120 = 0
x² – 7x – 60 = 0
60= 2 ×3×2×5 = 12×5
x² – 12x + 5x – 60 = 0
x(x – 12) + 5(x – 12) = 0
(x – 12)(x +5) = 0
x = 12, – 5
Side can not be negative so neglecting negative sign
Therefore altitude of given Δ is 12 cm and base is 12 – 7 = 5 cm.
Q5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particle day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total number of production on that day Rs90.Find the number of articles produced and the cost of each article.
Answer. Let the number of articles produced on that day = x
Since the cost of each article was observed on that day 3 more than twice the number of articles produced
So, the cost of each article is = 2x + 3
The total cost of production as per the question=x(2x + 3)
It is given in question the total cost of production on that day= Rs90
According to question
x(2x + 3) = 90
2x² + 3x –90 = 0
2 × 90 = 2×3×3×5 ×2 =15×12
2x² + 15x –12x –90 = 0
x(2x + 15) –6(2x +15) =0
(2x + 15)(x –6) = 0
The number of articles produced can’t be negative so number of articles produced on that day equal to = 6 and the cost of each article is = Rs (2x +3 )=Rs(2×6 + 3) = Rs15
Myopia, Hypermetropia and Presbyopia
Q6. Find the roots of the following equation by the method of complete square method .
2x² –7x + 3 = 0
Answer. The identity of complete square is (a +b)² = a² + 2ab+ b²
Let’s make the given equation 2x² –7x + 3 = 0 a complete square
⇒(√2x)² –7x + 3
⇒2ab = 7x
⇒2×√2x×b=7x
Adding and subtracting to the given equation
Q7. The sum of reciprocal of Rehman’s ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.
Answer. Let Rehman’s present age is = x
His age 3 years ago = x –3
Reciprocal of his age 3 years ago =
His age 5 years from now = x +5
Reciprocal of his age 5 years from now =
According to the question we have
x² +5x –3x – 15 = 6x +6
x² – 4x –21 = 0
21 = 7 × 3
x² – 7x +3x –21 = 0
x(x –7) + 3(x –7) =0
(x –7)(x + 3) = 0
x = 7, –3
The age can’t be negative so Rehman’s present age is 7 years.
electrical circuits resistance and conductance
Q8. The diagonal of a rectangular field is 60 m more than its shorter side. If the longer side is 30 m more than the shorter side. Find the sides of the field.
Answer. Let the shorter side of the rectangle ABCD is = x
The diagonal of rectangle = x + 60
Longer side of rectangle = x +30
ΔADC is a right triangle so applying Pythagoras theorem
(x +60)² = x² + (x +30)²
x² + 3600 + 120x = x² + x² +900 + 60x
x²–60x –2700= 0
a = 1, b = –60, c = –2700
Applying quadratic formula
x =90,–30
Length can’t be negative so length of shorter side of the rectangle=90 m
Length of longer side = 90 + 30 = 120 m
Q9. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number.Find the two numbers.
Answer. Let the larger number is = x
Its square is = x²
Since we are given the difference of squares of both numbers
Square of larger number – Square of smaller number =180
x² = Square of smaller number +180
Since it is given that the square of the smaller number is 8 times the larger number
So,
Square of smaller number = 8x
x²= 8x + 180=0
x² – 8x – 180= 0
180 = 2 ×3×3×5×2= 18× 10
x² –18x +10x – 180= 0
x(x –18) + 10(x –18) = 0
(x –18)(x + 10) = 0
x = 18, –10
If we suppose x =–10 since we are given that square of the smaller number is 8 times of larger number,it gives the square of smaller number =8 ×–10=–80, the square of any number can’t be negative so the required larger number is = 18
square of smaller number =8x= 8 ×18 = 144
smaller number =
Therefore the larger number and smaller numbers are 18 and 12 respectively.
Q10.A train travels 360 km at a uniform speed.If the speed had been 5 km/h more , it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer. Let the speed of the train = x km/h
Distance covered by the train = 360 km
Time is taken by the train with x km/h =
The modified speed of the train = (x +5) km/h
Time taken by the train with (x +5) km/h =
According to question
x² + 5x –1800 = 0
1800 = 2 × 3 ×3×5×2×2×5= 45× 40
x² + 45x –40x –1800 = 0
x(x + 45) – 40(x + 45) = 0
(x + 45)(x – 40) = 0
x = – 45, 40
The speed of the train can’t be negative,so the speed of the train is 40 km/h.
Study our Important posts for Class 10 Maths CBSE Board
Important notes on science and maths for achieving 100 % marks
Class X maths frequently asked questions with solutions
How to solve questions of mensuration
How to solve linear and quadratic equations
How to write a linear equation
Class 10 maths questions 3-4 marks asked last years with solutions
Technics of achieving 100% marks in maths
CBSE solutions of the most important maths question of 3-4 marks
CBSE Class X science most important questions with answers part-III
CBSE class X science most important questions with answers part-1
NCERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions for class 9 maths
NCERT Solutions for class 9 science
NCERT Solutions for class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
NCERT Solutions for Class 10 Science
NCERT Solutions for class 11 maths
Chapter 1-Sets | Chapter 9-Sequences and Series |
Chapter 2- Relations and functions | Chapter 10- Straight Lines |
Chapter 3- Trigonometry | Chapter 11-Conic Sections |
Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
Chapter 8- Binomial Theorem | Chapter 16- Probability |
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
NCERT Solutions for Class 11 Physics
chapter 3-Motion in a Straight Line
NCERT Solutions for Class 11 Chemistry
Chapter 1-Some basic concepts of chemistry
NCERT Solutions for Class 11 Biology
NCERT solutions for class 12 maths
Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
Chapter 4-Determinants | Chapter 12-Linear Programming |
Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
Chapter 7- Integrals | |
Chapter 8-Application of Integrals |
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution