MSBSHSE Solutions For SSC (Class 10) Maths Part 1 Chapter 2- Quadratic Equations

MSBSHSE Solutions For SSC (Class 10) Maths Part 1 Chapter 2 Quadratic Equations are provided here to help students understand the concepts, right from the beginning. To score good marks in Class 10 Mathematics examination, it is advised they solve questions provided at the end of each chapter in the Maharashtra Board Textbooks for SSC Part 1. These Maharashtra Board Solutions for Class 10 Maths help the students in understanding all the concepts in a better way. Chapter 2 Quadratic Equations for Class 10 Maths explains the concepts related to quadratic equations and methods of solving them. Maharashtra State Board Solutions for Chapter 2, students will learn and solve problems based on topics like Quadratic polynomials, standard form, roots, solutions of a quadratic equation by factorisation and completing the square, nature of roots and relation between roots of the quadratic equation and coefficients.

Download the PDF of Maharashtra Solutions For SSC Maths Part 1 Chapter 2 Quadratic Equations

 

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Access answers to Maths MSBSHSE Solutions For SSC Part 1 Chapter 2 – Quadratic Equations

Practice set 2.1 Page no: 34

1. Write any two quadratic equations.

Solution:

Two quadratic equations are

a2 + 16 = 0 and x2 + 2x + 6 = 0

2. Decide which of the following are quadratic equations.

(1) x2 + 5x – 2 = 0

Solution:

x2 + 5x – 2 = 0 is a quadratic equation because it is the form of ax2 + bx + c = 0 and it has degree 2.

(2) y2 = 5y – 10

Solution:

y2 = 5y – 10 is a quadratic equation because it is the form of ax2 + bx + c = 0 and it has degree 2.

(3) y2 + 1/y = 2

Solution:

y2 + 1/y = 2 is a quadratic equation because it is the form of ax2 + bx + c = 0 and it has degree 2.

(4) x + 1/x = -2

Solution:

Given equation can be written as

x2 + 1 = -2x

x2 + 2x + 1 = 0

It is a quadratic equation because it is the form of ax2 + bx + c = 0 and it has degree 2.

(5) (m + 2) (m–5) = 0

Solution:

Given equation can be written as

m (m – 5) + 2 (m – 5)

= m2 – 5m + 2m – 10

= m2 – 3m + 10 = 0

It is a quadratic equation because it is the form of ax2 + bx + c = 0 and it has degree 2.

(6) m3 + 3m2 – 2 = 3 m3

Solution:

Given m3 + 3m2 – 2 = 3 m3

It is not a quadratic equation because it is not in the form of ax2 + bx + c = 0 and it has degree 3.

3. Write the following equations in the form ax2 + bx + c = 0, then write the values of a, b, c for each equation.

(1) 2y = 10 – y2

Given

2y = 10 – y2

2y + y2 – 10 = 0

y2 + 2y – 10 = 0;

a = 1, b = 2, c = -10

(2) (x-1)2 = 2x + 3

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 1

(3) x2 + 5x = – (3-x)

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 2

(4) 3m2 = 2m2 – 9

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 3

(5) P (3 + 6p) = – 5

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 4

(6) x2 – 9 = 13

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 5

4. Determine whether the values given against each of the quadratic equation are the roots of the equation.

(1)  x2 + 4x – 5 = 0, x = 1, -1

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 6

(2) 2m2 – 5m = 0, m = 2, 5/ 2

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 7

∴ m = 2 is not root of the equation and m = 5/2 is a root of the equation.

5. Find k if x = 3 is a root of equation kx2 – 10x + 3 = 0.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 8

6. One of the roots of equation 5m2 + 2m + k = 0 is -7/5. Complete the following activity to find the value of ‘k’.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 9

Practice set 2.2 Page no: 36

1. Solve the following quadratic equations by factorization.

(1) x2 – 15x + 54 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 10

(2) x2 + x – 20 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 11

(3) 2y2 + 27y + 13 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 12

(4) 5m2 = 22m + 15

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 13

(5) 2x2 – 2x + ½ = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 14

(2x – 1) (2x – 1)

2x – 1 = 0

x = ½, ½

Hence x = ½, ½ are the roots of the equation

(6) 6x -2/x = 1

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 15

(7) √2 x2 + 7 x + 5 √2 = 0 to solve this quadratic equation by factorization, complete the following activity.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 16

(8) 3x2 – 2 √6 x + 2 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 17

(9) 2m (m-24) = 50

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 18

(10) 25m2 = 9

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 19

(11) 7m2 = 21m

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 20

(12) m2 – 11 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 21

Practice set 2.3 Page no: 39

1. Solve the following quadratic equations by completing the square method.

(1) x2 + x – 20 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 22

(2) x2 + 2x – 5 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 23

(3) m2 – 5m = -3

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 24

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 25

(4) 9y2 – 12y + 2 = 0

Solution:

Given

9y2 – 12y + 2 = 0

The above equation can be written as

(3y)2 – 2 × 3y × 4 + (4)2 – (4)2 + 2 = 0

(3y)2 – 2 × 3y × 4 + (4)2 – 16 + 2 = 0

(3y – 4)2 – 14 = 0

(3y – 4)2 = 14

3y – 14 = ±√143y = 14 ± √14y = (14 ± √14)/3

(5) 2y2 + 9y + 10 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 26

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 27

(6) 5x2 – 4x + 7 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 28

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 29

Practice set 2.4 Page no: 43

1. Compare the given quadratic equations to the general form and write values of a, b, c.

(1) x2 – 7x + 5 = 0

Solution:

Given

x2 – 7x + 5 = 0

comparing with ax2 + bx + c

we get

a = 1, b = -7, c = 5

(2) 2m2 = 5m – 5

Solution:

Given

2m2 = 5m – 5

comparing with ax2 + bx + c

we get

a = 2, b = -5, c = 5

(3)  y2 = 7y

Solution:

Given

y2 = 7y

comparing with ax2 + bx + c

we get

a = 1, b = -7, c = 0

2. Solve using formula.

(1) x2 + 6x + 5 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 30

(2) x2 – 3x – 2 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 31

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 32

(3) 3m2 + 2m – 7 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 33

(4) 5m2 – 4m – 2 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 34

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 35

(5) y2 + 1/3 y = 2

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 36

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 37

(6) 5x2 + 13x + 8 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 38

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 39

3. With the help of the flow chart given below solve the equation  

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 40 using the formula.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 41

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 42

Practice set 2.5 Page no: 49

1. Fill in the gaps and complete.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 43

Solution:

Roots are distinct and real when b2 – 4ac = 5, not real when b2 – 4ac = -5.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 44

Solution:

x2 + 7x + 5 = 0

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 45

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 46

2. Find the value of discriminant.

(1) x2 + 7x – 1 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 47

(2) 2y2 – 5y + 10 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 48

(3) √2x2 + 4x + 2 √2 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 49

3. Determine the nature of roots of the following quadratic equations.

(1) x2 – 4x + 4 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 50

(2) 2y2 – 7y + 2 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 51

(3) m2 + 2m + 9 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 52

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 53

4. Form the quadratic equation from the roots given below.

(1) 0 and 4

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 54

(2) 3 and -10

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 55

(3) ½, – ½

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 56

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 57

(4) 2 – √5, 2 + √5

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 58

5. Sum of the roots of a quadratic equation is double their product. Find k if equation is x2 – 4kx + k + 3 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 59

6. a, b are roots of y2 – 2y – 7 = 0 find,

(1) α2 + β2

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 60

(2) α3 + β3

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 61

7. The roots of each of the following quadratic equation are real and equal, find k.
(1) 3y2 + ky + 12 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 62

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 63

(2) kx (x-2) + 6 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 64

Practice set 2.6 Page no: 52

1. Product of Pragati’s age 2 years ago and 3 years hence is 84. Find her present age.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 65

2. The sum of squares of two consecutive natural numbers is 244; find the numbers.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 66

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 67

3. In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 68

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 69

4. Vivek is older than Kishor by 5 years. The sum of the reciprocals of their ages is 1/6. Find their present ages.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 70

5. Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 71

Hence, score of first test is 10 as marks are not negative.

6. Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is ₹40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is ` 600, find production cost of one pot and number of pots he makes per day.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 72

7. Pratik takes 8 hours to travel 36 km downstream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 73

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 74

8. Pintu takes 6 days more than those of Nishu to complete certain work. If they work together, they finish it in 4 days. How many days would it take to complete the work if they work alone.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 75

x = -4 is not possible, as no of days can’t be negative.

Nishu will take 6 days alone and Pintu takes 12 days alone

9. If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and divisor.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 76

10. In the adjoining fig. â–¡ABCD is a trapezium AB||CD and its area is 33 cm2. From the information given in the figure find the lengths of all sides of the â–¡ABCD. Fill in the empty boxes to get the solution.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 77

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 78

Problem set 2 Page no: 53

1. Choose the correct answers for the following questions.

(1) Which one is the quadratic equation?

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 79

Solution:

B. x (x + 5) = 2

Explanation:

It is in the form of ax2 + bx + c

(2) Out of the following equations which one is not a quadratic equation?

A. x2 + 4x = 11 + x2
B. x2 = 4x
C. 5x2 = 90
D. 2x – x2 = x2 + 5

Solution:

A. x2 + 4x = 11 + x2

Explanation:

In all other options highest degree of equation is 2, which also the degree of quadratic equation. But in Option A, degree of polynomial is 1

(3) The roots of x2 + kx + k = 0 are real and equal, find k.
A. 0
B. 4
C. 0 or 4
D. 2

Solution:

C. 0 or 4

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 80

4. For √2x2 – 5x + √ 2 = 0 find the value of the discriminant.
A. -5
B. 17
C. 2

D. 2 √2 – 5

Solution:

B. 17

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 81

5. Which of the following quadratic equations has roots 3, 5?
A. 
x2 – 15x + 8 = 0
B. x2 – 8x + 15 = 0
C. x2 + 3x + 5 = 0
D. x2 + 8x – 15 = 0

Solution:

B. x2 – 8x + 15 = 0

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 82

6. Out of the following equations, find the equation having the sum of its roots -5.
A. 3x2 – 15x + 3 = 0
B. x2 – 5x + 3 = 0
C. x2 + 3x – 5 = 0
D. 3x2 + 15x + 3 = 0

Solution:

A. 3x2 – 15x + 3 = 0

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 83

7. √5 m2 – √5m + √5 = 0 which of the following statement is true for this given equation?
A. Real and unequal roots
B. Real and equal roots
C. Roots are not real
D. Three roots.

Solution:

C. Roots are not real

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 84

8. One of the roots of equation x2 + mx – 5 = 0 is 2; find m.
A. -2
B. – ½
C. ½
D. 2

Solution:

C. ½

Explanation:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 85

2. Which of the following equations is quadratic?

(1) x2 + 2x + 11 = 0
(2) x2 – 2x + 5 = x2
(3) (x + 2)2 = 2x2

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 86

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 87

3. Find the value of discriminant for each of the following equation.

(1) 2y2 – y + 2 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 88

(2) 5m2 – m = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 89

(3) √5x2 – x – √5 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 90

4. One of the roots of quadratic equation 2x2 + kx – 2 = 0 is -2, find k.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 91

5. Two roots of quadratic equations are given; frame the equation.

(1) 10 and -10

Solution:

Let α = 10 and β = -10

∴ α + β = 10 – 10 = 0 α β = 10(-10)      = – 100

Quadratic equation is

x2 – (α + β) x + α β = 0

⇒ x2 – 0(x) – 100 = 0

⇒ x2 – 100 = 0

(2) 1–3√5 and 1 + 3√5

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 92

(3) 0 and 7

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 93

6. Determine the nature of roots for each of the quadratic equation.

(1) 3x2 – 5x + 7 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 94

(2) √3x2 + √2x – 2 √ 3 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 95

(3) m2 – 2m + 1 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 96

7. Solve the following quadratic equation.

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 97

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 98

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 99

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 100

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 101

The roots are

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 102

(3) (2x + 3)2 = 25

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 103

(4) m2 + 5m + 5 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 104

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 105

(5) 5m2 + 2m + 1 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 106

(6) x2 – 4x – 3 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 107

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 108

8. Find m if (m – 12) x2 + 2 (m – 12) x + 2 = 0 has real and equal roots.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 109

9. The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 110

10. Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation
2x2 + 2 (p + q) x + p2 + q2 = 0

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 111

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 112

11. Mukund possesses ₹50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 113

12. The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 114

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 115

13. Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 116

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 117

14. Mr. Dinesh owns an agricultural farm at village Talvel. The length of the farm is 10 meters more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is 1/3 of the breadth of the farm. The area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and of the pond.

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 118

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 119

15. A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?

Solution:

Maharashtra Board Solutions for Class 10 Maths Part 1 Chapter 2 - Image 120

Comprehending the textbooks and diligently revising all the concepts is the most reliable method of studying. Students can depend on these Solutions to understand all the topics completely. Stay tuned to learn more about Quadratic Equations, MSBSHSE Exam pattern and other information.

Frequently Asked Questions on Maharashtra State Board Solutions for Class 10 Maths Part 1 Chapter 2 Quadratic Equations

Q1

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No. We have made available these solutions free for download by entering the login details. Or else, the scrollable PDF is also accessible. Those who wish to can also check out the questions and the solutions from our webpage.

Q2

Are these solutions helpful?

Yes. These solutions set the basis for questions that could get asked in the board exams. Students are advised to practise these questions first and then refer back to the solutions to analyse one’s performance and then rectify the mistakes so that they can avoid making any during the board exams.
Q3

Is Maharashtra Board Class 10 Maths Solutions of Chapter 2 difficult?

No, Maharashtra Board Class 10 Maths Solutions of Chapter 2 isn’t half as difficult as it seems. You will be able to understand concepts of the chapter if you refer to this chapterwise solutions . You can also go through your syllabus course structure and unit-wise weightage thoroughly to know how much weightage the chapter will carry for your exam.

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