Class 10

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Coordinate Geometry Class 10 Exercise 7.1 Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Solution:
(i) Here x₁ = 2, y₁ = 3 and x₂ = 4, y₂ = 1
∴ The required distance

(ii) Here x₁ = -5, y₁ = 7 and x₂ = -1, y₂ = 3
∴ The required distance

(iii) Here x₁ = a, y₁ = b and x₂ = -a, y₂ = -b
∴ The required distance

Ncert Class 10 Maths 7.1 Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns /I and B discussed below as following: ‘A town 6 is located 36 km east and 15 km north of the town A’.
Solution:
Part-I
Let the points be A(0, 0) and B(36, 15)

Part-II
We have A(0, 0) and B(36,15) as the positions of two towns.

Class 10 Maths Chapter 7 Exercise 7.1 Solutions Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let the points be A( 1, 5), B(2, 3) and C(-2, -11). A, B and C are collinear, if
AB + BC = AC or AC + CB = AB or BA + AC = BC

Here, AB + BC ≠ AC, AC + CB ≠ AB, BA + AC ≠ BC
∴ A, B and C are not collinear.

Chapter 7 Maths Class 10 Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let the points be A(5, -2), 6(6, 4) and C(7, -2).

We have AB = BC ≠ AC.
∴ ∆ABC is an isosceles triangle.

Class 10th Maths Exercise 7.1 Question 5.
In a classroom, 4 friends are seated at the points A B, Cand D as shown in the figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution:
Let the horizontal columns represent the x-coordinates whereas the vertical rows represent the y-coordinates.
∴ The points are A(3, 4), B(6, 7), C(9, 4) and D(6, 1)


i.e., BD = AC ⇒ Both the diagonals are also equal.
∴ ABCD is a square.
Thus, Champa is correct.

Class 10th Maths Chapter 7.1 Question 6.
Name the type of quadrilateral formed if any, by the following points, and give reasons for your answer:
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let the points be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) of a quadrilateral ABCD.

⇒ AB = BC = CD = AD i.e., all the sides are equal.
Also, AC and BD (the diagonals) are equal.
∴ ABCD is a square.

(ii) Let the points be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).


We see that \(\sqrt{13}+\sqrt{13}=2 \sqrt{13}\)
i. e., AC + BC = AB
⇒ A, B and C are collinear. Thus, ABCD is not a quadrilateral.

(iii) Let the points be A(4, 5), B(7, 6), C(4, 3) and D(1, 2).

Since, AB = CD, BC = DA i.e., opposite sides of the quadrilateral are equal.
And AC ≠ BD ⇒ Diagonals are unequal.
∴ ABCD is a parallelogram.

Exercise 7.1 Class 10 Question 7 Question 7.
Find the point on x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
We know that any point on x-axis has its ordinate = 0
Let the required point be P(x, 0)
Let the given points be A(2, -5) and B(-2, 9)


∴ The required point is (-7, 0)

Maths Class 10 Chapter 7 Question 8.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Solution:
The given points are P(2, -3) and Q(10, y).

Squaring both sides, y² + 6y + 73 = 100
⇒ y² + 6y – 27 = 0
⇒ y² – 3y + 9y – 27 = 0
⇒ (y – 3)(y + 9) = 0
Either y – 3 = 0 ⇒ y = 3
or y + 9 = 0 ⇒ y = -9
∴ The required value of y is 3 or -9.

Class 10 Ka Math Chapter 7.1 Question 9.
If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution:

Squaring both sides, we have x² + 25 = 41
⇒ x² + 25 – 41 = 0
⇒ x² – 16 = 0 ² x = \(\pm \sqrt{16}\) = ±4
Thus, the point R is (4, 6) or (-4, 6)
Now,

Class 10th Coordinate Geometry Exercise 7.1 Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Solution:
Let the points be A(x, y), B(3, 6) and C(-3, 4).
∴ AB = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
And AC = \(\sqrt{[(-3)-x]^{2}+(4-y)^{2}}\)
Since, the point (x, y) is equidistant from (3, 6) and (-3, 4).
∴ AB = AC
⇒ \(\sqrt{(3-x)^{2}+(6-y)^{2}}=\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
Squaring both sides,
(3 – x)² + (6 – y)² = (-3 – x)² + (4 – y)²
⇒ (9 + x² – 6x) + (36 + y² – 12y)
⇒ (9 + x² + 6x) + (16 + y² – 8y)
⇒ 9 + x² – 6x + 36 + y² – 12y – 9 – x² – 6x – 16 – y² + 8y = 0
⇒ – 6x – 6x + 36 – 12y – 16 + 8y = 0
⇒ – 12x – 4y + 20 = 0
⇒ -3x – y + 5 = 0
⇒ 3x + y – 5 = 0 which is the required relation between x and y.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Pdf, Class 10 Maths Chapter 4, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Class 10th Maths Chapter 4 Exercise 4.1 Question 1.
Check whether the following are quadratic equations:
(i) (x + 1 )² = 2(x – 3)
(ii) x² – 2x = (-2)(3 – x)
(iii) (x – 2)(x + 1) = (x – 1 )(x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1 )(x – 3) = (x + 5)(x – 1)
(vi) x² + 3x + 1 = (x – 2)²
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ – 4x² – x + 1 = (x – 2)³
Solution:
(i) We have, (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x² + 2x + 1 – 2x + 6 = 0
⇒ x² + 7 = 0
Since, x² + 7 is a quadratic polynomial.
∴ (x + 1)² = 2(x – 3) is a quadratic equation.

(ii) We have, x² – 2x = (-2) (3 – x)
⇒ x² – 2x = – 6 + 2x
⇒ x² – 2x – 2x + 6 = 0
⇒ x² – 4x + 6 = 0
Since, x² – 4x + 6 is a quadratic polynomial
∴ x² – 2x = (-2)(3 – x) is a quadratic equation.

(iii) We have, (x – 2)(x + 1) = (x – 1) (x + 3)
⇒ x² – x² = x² + 2x³
⇒ x² – x – 2 – x² – 2x + 3 = 0
⇒ – 3x +1 = 0
Since, – 3x + 1 is a linear polynomial.
∴ (x – 2)(x + 1) = (x – 1)(x + 3) is not a quadratic equation.

(iv) We have, (x – 3) (2x + 1) = x(x + 5)
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² – 5x – 3 – x² – 5x = 0
⇒ x² – 10x – 3 = 0
Since, x² – 10x – 3 is a quadratic polynomial.
∴ (x – 3)(2x + 1) = x(x + 5) is a quadratic equation.

(v) We have, (2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² – 7x + 3 = x² + 4x – 5
⇒ 2x² – 7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
Since, x² – 11x + 8 is a quadratic polynomial
∴ (2x – 1)(x – 3) = (x + 5)(x -1) is a quadratic equation.

(vi) We have, x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ x² + 3x + 1 – x² + 4x – 4 = 0
⇒ 7x – 3 = 0
Since, 7x – 3 is a linear polynomial.
x² + 3x + 1 = (x – 2)² is not a quadratic equation.

(vii) We have, (x + 2)³ = 2x(x² – 1)
⇒ x³ + 3x²(2) + 3x(2)² + (2)³ = 2x³ – 2x
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
⇒ x³ + 6x² + 12x + 8 – 2x³ + 2x = 0
⇒ -x³ + 6x² + 14x + 8 = 0
Since, -x³ + 6x² + 14x + 8 is a polynomial of degree 3.
∴ (x + 2)³= 2x(x² – 1) is not a quadratic equation.

(viii) We have, x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ + 3x²(-2) + 3x(-2)² + (-2)³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ x³ – 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0 ⇒ 2x² – 13x + 9 = 0
Since, 2x² – 13x + 9 is a quadratic polynomial.
∴ x³ – 4x² – x + 1 = (x – 2)³ is a quadratic equation.

Class 10 Chapter 4 Maths Exercise 4.1 Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
(i) Let the breadth = x metres
∵ Length = 2(Breadth) + 1
∴ Length = (2x + 1)metres
Since, Length × Breadth = Area
∴ (2x + 1) × x = 528 ⇒ 2x² + x = 528
⇒ 2x² + x – 528 = 0
Thus, the required quadratic equation is 2x² + x – 528 = 0

(ii) Let the two consecutive positive integers be x and (x + 1).
∵ Product of two consecutive positive integers = 306
∴ x(x + 1) = 306 ⇒ x² + x = 306
⇒ x² + x – 306 = 0
Thus, the required quadratic equation is x² + x – 306 = 0

(iii) Let the present age of Rohan be x years
∴ Mother’s age = (x + 26) years
After 3 years,
Rohan’s age = (x + 3) years
Mother’s age = [(x + 26) + 3] years
= (x + 29) years
According to the condition,
(x + 3) × (x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 29x + 3x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Thus, the required quadratic equation is x² + 32x – 273 = 0

(iv) Let the speed of the train = u km/hr Distance covered = 480 km Distance

In second case,
Speed = (u – 8) km/hour

⇒ 480u – 480(u – 8) = 3u(u – 8)
⇒ 480u – 480u + 3840 = 3u² – 24u
⇒ 3840 – 3u² + 24u = 0
⇒ u² – 8u – 1280 = 0
Thus, the required quadratic equation is u² – 8u – 1280 = 0

In this article, we will share MP Board Class 10th English Solutions General Unseen Passages Pdf, Passage For Class 10, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th General English Unseen Passages

In this part of your syllabus you will be given three unseen passages of 150 words each. You are required to read the passage carefully and answer the questions given below it. Each passage will carry 5 marks with 1 mark for vocabulary included in it.

 

Discursive Passages

1. Read the following passage and answer the questions given below it.

Man’sourney of life from childhood to old age is very charming and colorful. Youth is the most exciting period of man’s life when it is time to grow and dream. A young man is full of hope, energy and zeal. Nothing is difficult or impossible or dangerous for him. The old people say that youth is not daring but thoughtless. A young man bums the candle at both ends. He commits mistakes and leams only after burning his finger. Sometimes the young men misuse their freedom and thus invite difficulties by their foolish actions. They are full of strength, energy and enthusiasm. They become rebels and are no longer afraid of facing the forces of realities. A young man accepts the challenge of evil difficulties and hardships, to win or lose the game of life is the mission of his career. He loves to lead an adventurous life and has a keen desire to build up a new world of his dream. But the period of youth does not last long. Soon it is followed by old age when he regrets his past mistakes and failures. The weak old man feels helpless, depressed and disappointed. He becomes unfit for any adventure. But some fortunate old people never grow old and continue to feel young and active and make the most of even the last years of their lives. It will not be wrong to say that youth brings honor and old age commands respect.

Unseen Passage For Class 10 MP Board Questions:
(a) Man’sourney is :
(i) Charming and colorful (ii) exciting (iii) dangerous
(b) What is the opinion of the old people about the young men?
(c) What is the mission of a young man’s career?
(d) Why does the old man feel helpless and disappointed?
(e) Find the antonym of ‘clever’ from the passage.
Answers:
(a) (i) charming and colorful.
(b) The old people say that youth is not daring but thoughtless.
(c) To win or lose the game of life is the mission of his, career.
(d) He feels helpless and disappointed because he becomes unfit for any adventure.
(e) Foolish.

2. Read the following passage and answer the questions given below it.

Whatever may be the cause of their suffering, we have to treat the handicapped with sympathy and understanding. In many instances, physically handicapped children are neglected and left to themselves in their homes. This makes them extremely sad and lonely. Our first duty is to make these children happier and less lonely. Secondly, we have to educate these children and help them to live useful lives. We should secure for them the benefits of education in schools specially intended for them. We have to make them useful citizens by creating suitable opportunities in them to be employed. They will then have a sense of achievement and we can be happy that we have done our duty towards them.

Unseen Passage For Class 10 Questions:
(a) The physically handicapped children suffer from:
(i) fever (ii) neglect (iii) cold
(b) Which are the two important duties towards the handicapped children that we should perform?
(c) How can we make them useful citizens?
(d) What will be the outcome of these efforts?
(e) Give the synonym of ‘disabled’ from the passage?
Answers:
(a) (ii) neglect.
(b) Our first duty is to make handicapped children happier and less lonely. Secondly, we should educate them and help them to live useful lives.
(c) We can make them useful citizens by creating suitable opportunities for them to be employed.
(d) The outcome of these efforts will be that they will have a sense of achievement and we can be happy that we have done our duty towards them.
(e) Physically handicapped.

3. Read the following passage and answer the questions given below it.

Reading has a variety of meanings. To some people it means little more than the ability to pronounce aloud the printed word; to others it means it an ability to gain merely a general impression of what they read. Even students daily engaged in the study of books develop a superficial ability to read rapidly, and with apparent understanding, what they it subsequently prove to have understood imperfectly. Ability to read properly, to understand not only the general sense of a given passage but its particular implications, to appreciate so to speak, the light and shade of the passage, the precise meaning of the parts as well as of the whole, what it hints at as well as what it states, to distinguish between what is clearly proved and established and what is merely suggested or put forward as a supposition is still a comparatively rare quality.

Unseen Passage For Class 10th MP Board Questions:
(a) has a variety of meanings.
(i) Reading (ii) Studying (iii) Understanding.
(b) What does it mean to other people?
(c) How do students develop?
(d) What is a comparatively rare quality?
(e) precise’ means
(i) exact (ii) short (iii) incorrect.
Answers:
(a) (i) Reading.
(b) Others think that it is an ability to gain merely a general impression of what they read.
(c) They develop a superficial ability to read quickly and with apparent understanding, what they subsequently prove to have understood imperfectly.
(d) To distinguish between what is clearly proved and established and what is merely suggested or put forward as a supposition is still a comparatively rare quality.
(e) (i) exact.

4. Read the following passage aid answer the questions given below it.

A cheerful person is always more disposed to be happy than miserable. He tends to look at the bright side of things and thus often derives pleasure from circumstances which would ordinarily sadden a person. “^ cheerful beggar is happier than a low-spirited millionaire. As a source of happiness neither wealth nor fame nor beauty nor power nor even health can be compared even for a moment with cheerful temperament. A great advantage of cheerfulness is that it enables man to do his work more efficiently and prevents him from being easily exhausted. The laborer who whistles over his work goes homeless tired and can work harder than another who deeply thinks over real or imaginary troubles. ’

Which Word In The Passage Means Newly Married Man Questions:
(a) A cheerful person is always :
(i) miserable (ii) happy (iii) beautiful
(b) What is the tendency of a cheerful person?
(c) What cannot be compared with cheerful moment?
(d) What is the real advantage of cheerfulness?
(e) ‘Exhausted’ means :
(i) tired (ii) vigor (iii) low-spirited.
Answers:
(a) (ii) happy.
(b) A cheerful person tends to look at the bright side of things.
(c) Neither wealth nor fame nor beauty nor power nor even health can be compared with a cheerful moment.
(d) The real advantage of cheerfulness is that it enables man to do his work more efficiently and prevents him from being easily exhausted.
(e) (i) tired.

5. Read the following passage and answer the questions given below it.

Discipline means obedience to the established rules of conduct. Certain rules have been laid in every society to control and regulate the life and activities of its members so that the society as a whole may progress in harmony and peace. If any of these rules is broken, there is trouble and society suffers. In fact, discipline is the very basis of progress in every sphere, public or private. A man without discipline is like an engine without a brake. A society that has no rules or whose members do not conform to its rules soon falls into pieces. In game too, discipline is necessary. Every player has to obey his captain and carry out his commands whether he likes them or not. In army, discipline is more necessary. An army without discipline is no better than a lawless mob. In the same way a school or a college cannot run if the boys do not observe the rules and regulations of the institution. Teaching is impossible if the boys do not keep discipline. Discipline cultivates a spirit of respect for elders and superiors, teaches gentlemanly behavior in society and meek submission to any punishment that may be inflicted due to indiscipline. It is the duty of every student to observe them if they want to build their character and prosper in life.

Unseen Passage For Class 10 With Answers 2023 Questions:
(a) When rules are broken suffers :
(i) life (ii) society (iii) man
(b) Why have certain rules of conduct been laid down by the society?
(c) What is a man without discipline?
(d) What does discipline cultivate among the students?
(e) The synonym of’ crowd’ is :
(i) society (ii) mob (iii) members.
Answers:
(a) (ii) society.
(b) Certain rules of conduct have been laid down in every society to control and regulate the life and activities of its members so that society as a whole may progress in harmony and peace.
(c) A man without discipline is an engine without a brake.
(d) Discipline cultivates a spirit of respect for elders and superiors, teaches gentlemanly behavior in society and meek submission to any punishment that may be inflicted due to indiscipline.
(e) (ii) mob.

6. Read the following passage and answer the questions given below it.

Home is the first and the most important school of character. It is here that every human being receives his moral training, or his worst, for it is here! that he imbibes those principles of conduct which endure throughout manhood, and cease only with life.

It is a common saying that, “Manners make the man” and there is a second that, “Mind makes the man”, but truer than either is a third that “Home makes the man”. For the home, training includes not only manners and mind, but also character. It is mainly in the home that the heart is opened, the habits are formed, the intellect is awakened, and the character is molded for good or for evil.

From that source be it pure or impure, issue the principles and maxims of society. Law itself is but the reflection of homes. The finest bits of opinions sown in the minds of children in private life afterward issue forth to the world and become its public opinion, for nations are grown out of nurseries. Those who hold the leading-strings of children may even exercise greater power than those who wield the reins of Government.

Unseen Passage For Class 10 MP Board Special English Questions:
(a) Which is the most important school of character:
(i) Office (ii) Home (iii) Factory
(b) What does hole training include?
(c) How is public opinion formed at home?
(d) What is done at home with the man?
(e) Find a word from the passage which means ‘to bear’.
(i) intellect (ii) endure (iii) imbibe.
Answers:
(a) (ii) Home.
(b) Home training includes not only the manners and mind, but also character.
(c) The small opinions sown in the minds of children in private life afterward issue forth to the world and become its public opinion.
(d) At home heart is opened, habits are formed, the intellect is awakened and the character of man is molded for good or for evil.
(e) (ii) endure.

literary Passages

1. Read the following poem and answer the questions given below it.

Life ! I know not what thou art,
But know that, thou and I must part.
And when or how, or where we met
I own to me’s secret yet.
Life ! We’ve been long together
Through pleasant and through cloudy weather
Tis hard to part when friends are dear
Perhaps it will cost a sigh, a tear;
Then steal away, give little warning,
Choose thine own time :
Say not good night; but in some better clime,
Bid me good morning.

Unseen Passage Class 10 Questions:
(a) Whom is the poet addressing to :
(i) Friends (ii) Weather (iii) Life
(b) When is it hard to part?
(c) What is a secret to the poet?
(d) How were life and poet associated together?
(e) A word from the poem that means the same as ‘ to leave’ is :
(i) part (ii) steal (iii) clime (iv) none of these.
Answers:
(a) (iii) Life.
(b) It is hard to part from life when the friends and dear ones are close.
(c) It is a secret to the poet that when, how and where he met with life.
(d) Poet and life were associated together in all good and bad times.
(e) (ii) part.

2. Read the following poem and answer the questions given below it.

We have no wings, we cannot soar
But we have feet to scale and climb,
By slow degrees, by more and more,
The cloudy summits of our time.
The heights by great men reached and kept,
Were not attained by sudden flight
But they, while their companions slept,
Were toiling upward in the night.

Discursive Passage For Class 10 Questions:
(a) What do we not have :
(i) feet (ii) hands (iii) wings
(b) How did great men reach heights and kept them?
(c) What do we have if not wings?
(d) What lesson do you get from the poem?
(e) Which word in the poem means ‘to fly’?
(i) climb (ii) flighty (iii) soar (iv) scale
Answers:
(a) (iii) wings.
(b) Great men reached heights by working hard even when their friends were sleeping.
(c) We have feet to climb upward if not wings to fly.
(d) We learn from the poem that if one works hard continuously then he reaches great heights.
(e) (iii) soar.

3. Read the following passage and answer the questions given below it.

A poor, villager once saved the life of a wealthy goldsmith by attacking a robber who was about to kill him. When the villager had knocked the robber down with his lathi and bound his hands and feet, the goldsmith said to the villager, “I have no money with me. So I shall give you my watch.” He did so and went on his way. The man was greatly pleased with his watch and spent hours in listening to its ticking and watching the second-hand go round. Next day the watch stopped as the man did not know how to wind it. He was very sad and said, “Alas! It is dead.” Thinking its dead body might be of value, he took the watch to a Mahajan who gave him fifty rupees for it, as it was well worth for two hundred. As he was leaving the room, the villager who was at heart an honest man, turned back and said, “Here take your money. It is dead and I have cheated you.” But the Mahajan only laughed and told him to keep the money and go.

Unseen Passage For Class 10 With Questions And Answers Questions:
(a) The goldsmith presented to the villager:
(i) a watch (ii) money (iii) a pen
(b) Why did the watch stop?
(c) What did the villager do with the dead watch?
(d) What did the villager admit in front of Mahajan?
(e) The synonym of ‘dead’ in the passage is :
(i) stopped (ii) lifeless (iii) immovable
Answers:
(a) (i) a watch.
(b) The watch stopped because the villager did I not know how to wind it.
(c) The villager sold the dead watch to a Mahajan for fifty rupees.
(d) The villager admitted in front of the Mahajan that the watch was dead and he had cheated him.
(e) (i) stopped.

4. Read the following passage and answer the questions given below it.

A deer who was very thirsty went to a pool to s quench his thirst. At the time of drinking water, he saw himself in the clear water “How handsome lam! i thought he. The horns on my head are branching like trees. My coat is smooth and glossy. My eyes sparkle like stars. Only my legs are so long and thin that I am ashamed of them.” just then he heard the sound; of the hunter’s foot steps. He dashed away through the forest. His long, thin legs bearing him swiftly on. ; The forest grew thicker and at last he could not run on account of his branching horns. So he was caught by the hunter. :

“How foolish I have been! ” cried the dying deer.
“Oh ! my splendid horns are the cause of my death.”

Unseen Passage For Class 10th Question :
(a) The cause of deer’s death was :
(i) his legs (ii) his coat (iii) his horns
(b) What was he ashamed of?
(c) Why could he not escape?
(d) What did he think at the time of his death?
(e) Which word in the passage means ‘bright’?
(i) splendid (ii) glossy (iii) sparkle
Answers:
(a) (iii) his horns.
(b) He was ashamed of his long and thin legs.
(c) He could not escape because of his horns which got stuck in the dense forest.
(d) At the time of his death, he thought of his foolishness that the horns for whose beauty he was proud were the cause of his death.
(e) (ii) glossy.

5. Read the following passage and answer the questions given below it.

How delightful to Sita, Ram and Lakshman were the years of their forest exile. Wherever they went, they were welcomed by the companies of hermits and admitted to the forest ways of life. Thus they were quickly established in huts made of leaves and carpeted with the sacred grass, like other ascetics. Quickly they had also arranged their articles of worship, and gathered together their small stores of necessities and without any loss of time, Sita fell into the habit of cooking for her husband and brother like any peasant-woman and serving them with her own fair hands. Now and‘then it would happen, during their first years in the forest, that they came across some great saint, who would recognize Ram at the glance as the Lord himself.

Unseen Passage For Class 10 Maharashtra State Board Questions.:
(a) The years of forest exile of Ram, Laxman and Sita were :
(i) delightful (ii) gloomy (iii) frustrating
(b) What were the huts made of?
(c) How did Sita help Ram and Laxman?
(d) What did the saint recognize?
(e) Which word in the passage means ‘articles of need’?
(i) necessities (ii) ascetics (iii) none of these.
Answers :
(a) (i) delightful.
(b) The huts were made of leaves and carpeted with the sacred grass.
(c) Sita helped Ram and Laxman by cooking and serving them food.
(d) The saint recognized Ram as the Lord himself.
(e) (i) necessaries.

6. Read the following passage and answer the questions given below it.

Mahmud of Ghazni had conquered so many countries that he could not rule over them properly. In one of these countries, robbers attacked a caravan of merchants and killed many of them and stole then- goods. The mother of one of the merchants walked a long way to Ghazni and made a complaint to the sultan. “My good woman,” said Mahmud, “how can I keep order in that distant land? It is hundreds of miles from Ghazni. I cannot put down robbers nor keep the road safe so far away.” “Why then, ” replied the old woman, “do you take countries which you cannot rule? For the bad rule of every country of which you are the king, God will call you to account, when you die.”

Class 10 Unseen Passage English 2023 Questions:
(a) Where did the woman go to :
(i) Ghazni (ii) Agra (iii) Delhi
(b) What had the robbers done?
(c) Whiat did the old woman question him?
(d) Why did she say that he would have to give an account to God?
(e) Find the antonym of ‘near’ from the passage.
Answers:
(a) (i) Ghazni.
(b) The robbers had attacked a caravan of merchants and killed many of them and stolen their goods.
(c) The old woman questioned him that why had he taken over countries which he could not rule.
(d) She said so because of his bad rule of every country of which he was the king.
(e) distant.

Factual Passages

1. Read the following passage and answer the questions given below it.

Kashmir is economically a backward state. It has been ruled despotically for several centuries. Progress therefore has been extremely slow. Arts and crafts of Kashmir could not flourish as our handicrafts did not find any market due to difficulties of transport. The peasant in Kashmir still clings to the age-old methods of farming.

As agriculture occupies the most important position in the plan, special attention was given to promote it. More and more facilities were offered to farmers to improve the conditions of agriculture. The launching of the National Extension Service Scheme is the most important step in this direction. This aims at the social, economic and educational uplift of the poor people of Kashmir. The people engaged in the service are enthusiastic workers. The Government is determined to see the prompt accomplishment of responsibilities of reconstructing the countryside through the basic principle of Extension.

Class 10 Unseen Passage Questions:
(a) Kashmir is an economical state :
(i) forward (ii) developed (iii) backward
(b) Why could art and craft not flourish in Kashmir?
(c) What was done to improve agriculture?
(d) What does the National Extension Service Scheme aim at?
(e) Find out the synonyms of ‘sticks’ from the passage.
Answers:
(a) (iii) backward.
(b) Art and craft did not flourish in Kashmir because the handicrafts did not find any market due to difficulties of transport.
(c) More and more facilities were given to the farmers to improve agriculture.
(d) National Extension Service Scheme aims at the social, economic and educational uplift of the poor people in Kashmir.
(e) clings.

2. Read the following passage and answer the questions given below it.

Indira Gandhi was the first woman Prime Minister of India. She was born in Allahabad on November 19, 1917. She was the only child of Pt.awahar Lai Nehru and Kamla Nehru. Indira was a lovely child so her parents and grandparents called her Indira Priyadarshini. They lived in a big house. Its name was Anand Bhawan.

Little Indira had many dolls. Some of them were foreign. She loved to play with them. She dressed them like brides and bridegrooms or Raja and Rani or Satyagrahis and Policemen. At the age of four, Indira went to Gandhiji’s I Ashram at Sabarmati. There she slept on the floor, and ate simple food.

CBSE Class 9 English Beehive Book Poem 6 No Men are Foreign MCQ Questions with Answers from Beehive Book.

Class 10 English Unseen Passage Questions:
(a) Gandhiji’s ashram was at:
(i) Allahabad (ii) Lucknow (iii) Sabarmati
(b) When and where was she born?
(c) What is Anand Bhawan?
(d) How did she dress her dolls?
(e) Which word in the passage means ‘newly married man’?
Answers:
(a) (iii) Sabarmati.
(b) She was born at Allahabad on November 19, 1917.
(c) Anand Bhawan is the name of the house I where Indira Gandhi and her parents lived,
(d) She dressed her dolls as brides and bridegrooms or Raja and Rani or Satyagrahis and policemen.
(e) bridegroom.

3. Read the following passage and answer the questions given below it.

A hockey team from Delhi went to Sri Lanka last month. They came to Chennai by plane. They left Delhi at 7 o’clock in the morning and reached Chennai at 12 o’clock. They stayed in a hotel for the day. They visited the museum and went in for shopping.

The next morning they went by bus to the railway station. They went in two buses. They traveled from Chennai to Dhanushkoti by ship. They went by train to Talaimannar in Sri Lanka. From Talaimannar they went to Colombo by taxi The team stayed in Sri Lanka for ten days.

Class 10th Unseen Passage Questions:
(a) The team stayed in Sri Lanka for:
(i) ten days (ii) one month (iii) three months.
(b) When and where did the hockey team go?
(c) What did the team do in Chennai?
(d) How did the team travel from Chennai to Colombo?
(e) Find a word from the passage which means a place where antique things are kept’.
Answers:
(a) (i) ten days.
(b) Hockey team went from Delhi to Sri Lanka last month.
(c) The team visited the museum and went in for shopping in Chennai.
(d) The team traveled from Chennai to Dhanushkoti by ship. They went by train to Talaimannar in Sri Lanka. From Talaimannar they went to Colombo by taxi.
(e) museum.

4. Read the following passage and answer the questions given below it.

A peacock is a beautiful bird. It is the national bird of India. It feeds on plants and animals. Seeds, fruits, bulbs, roots, grass and leaves are its staple food. It also eats white ants, insects, spiders and worms. It devours lizards and frogs. It is an enemy of all kinds of snakes. It drinks water an hour after sunrise and returns to roost after a heavy meal at dusk. In the monsoon when there is rainfall, it is ready to dance. Its breeding season starts with the rainy season. It dances to see the clouds in the sky. It is considered as an auspicious bird.

Class 10 English Unseen Passage 2023 Questions:
(a) Which is our National Bird :
(i) Sparrow (ii) Crow (iii) Peacock
(b) What does it feed on?
(c) When does it dance?
(d) When is its breeding season?
(e) Which word in the passage means ‘swallows’?
Answers:
(a) (iii) Peacock.
(b) It feeds on plants and animals. Seeds, fruits, bulbs, roots, grass and leaves are its staple food. It also eats white ants, insects, spiders and worms.
(c) It dances in the monsoon when there is rainfall.
(d) Its breeding season starts with the rainy season.
(e) devours.

5. Read the following passage and answer the questions given below it.

Pandit Jawaharlal Nehru was our first Prime Minister. His father was Pandit Motilal Nehru. He was an advocate and he was a rich man. He lived , in Allahabad. The name of his house is ‘Anand Bhawan’. It is a big house and there is beautiful . garden around it.

Pandit Jawaharlal Nehru studied in India and England. He played cricket too. He was a good writer and a good speaker. He was brave and kind.
Nehru loved children and roses very much. Children lovingly called him ‘Chacha Nehru’. His birthday, 14th November, is celebrated as Children’s
Day. He was a great freedom-fighter. He will always be remembered.

Read The Following Passage And Answer The Questions Given Below Questions:
(a) Pt. Motilal Nehru was :
(i) a businessman (ii) an advocate (iii) an industrialist
(b) Mention the qualities of Jawaharlal Nehru.
(c) Where did Jawaharlal Nehru study?
(d) When is Children’s Day celebrated?
(e) Find the synonym of ‘author’ from the passage.
Answers:
(a) (ii) an advocate.
(b) Jawaharlal Nehru could play cricket. He was a good writer and speaker. He was brave and kind.
(c) He studied in India and England.
(d) Children’s day is celebrated on 14th November on Jawaharlal Nehru’s birthday.
(e) Writer.

6. Read the following passage and answer the questions given below it.

Kalpana Chawla, the astronaut who died on board the space shuttle Columbia, had been sponsoring two students from her school (Tagore Bal Niketan in Kamal) each year since 1997 for. International Space School Camp in Houston.

Manpreet Kaur and Namita Along visited NASA in August 2002. “We spent an excellent day with KC she even cooked for us and made us feel totally at home”, said Namita. These two and other youngsters who benefitted from this program, say, KC, as she was popularly known, might have been the first Indian-born woman in space but was entirely unaffected by her success.

Gaurav Goel of the 1999 batch and now an engineering student at Ambala said, “Not only was KC down to earth, she still retained the Indian in herself despite living in US for so many years”.

Even as a student, Kalpana had looked out for other students. She used to pay the fees for two of her college mates who could not afford it. “It is possible that to this day they are not aware that Kalpana used to pay their tuition fees,” says Sovina Sood, Kalpana’s sunior in the Punjab Engineering College.

Sovina who now teaches Civil Engineering said that sometimes Kalpana would hand over the money to her “but ask me not to disclose it to anybody”.

Unseen Passage For Grade 10 Questions:
(a) Kalpana Chawla’s school was at:
(i) Ambala (ii) Kamal (iii) Houston
(b) Who visited NASA in August 2002?
(c) What did Kalpana Chawla do for her college mates when she was a student?
(d) Give a word from a passage that means the same as ‘getting advantage’.
(e) Sponsoring means :
(i) to provide support (ii) contributing (iii) participating.
Answers:
(a) (ii) Kamal.
(b) Manpreet Kaur and Namita Along visited NASA in August 2002.
(c) Kalpana Chawla used to pay the fees for two of her college mates when she was a student.
(d) benefitted.
(e) (ii) contributing.

In this article, we will share MP Board Class 10th Social Science Book Solutions Chapter 1 Resources of India: Soil, Water, Forest and Wild Life Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Social Science Solutions Chapter 1 Resources of India: Soil, Water, Forest and Wild Life

MP Board Class 10th Social Science Chapter 1 Text book Exercise

We have provided Class 10 SST MCQ Questions and Answers Chapter Wise PDF Download to help students understand the concept very well.

Objective Type Questions

Question 1.
Multiple Choice Questions
(Choose the correct answer from the following)

MP Board Class 10 Geography Chapter 1 Question 1.
Which factor does not help in the formation of the soil?
(a) Wind and water
(b) Decomposed plants and animals
(c) Rocks and temperature
(d) Water accumulation.
Answer:
(a) Wind and water

MP Board 10th Maths Solution

MP Board Class 10th Social Science Chapter 1 Question 2.
Which soil is generally found in the delta region of Andhra Pradesh and Orissa and the plains of Ganges?
(a) Red soil
(b) Alluvial soil
(c) Black soil
(d) Laterite soil.
Answer:
(b) Alluvial soil

Social Science Class 10 MP Board English Medium Question 3.
In which region, the method of making contour bunds is used for soil conservation?
(a) Delta region
(b) Plateau region
(c) Hills
(d) Plains
Answer:
(c) Hills

MP Board Class 10 History Chapter 1 Question 4.
Man uses the most?
(a) underground water
(b) Oceanic Water
(c) Surface Water
(d) Atmospheric Water
Answer:
(a) underground water

MP Board Solutions Class 10 Social Science Question 5.
Which of the following states is known as Tiger state?
(a) Rajasthan
(b) Madhya Pradesh
(c) Uttarakhand
(d) Assam.
Answer:
(b) Madhya Pradesh

Class 10 Social Science Lesson 1 Question Answer Question 6.
The founder of Vanmahostava was?
(a) Mahatma Gandhi
(b) Pt.Jawaharlal Nehru
(c) K.M. Munshi
(d) Acharya Vinobha Bhave
Answer:
(c) K.M. Munshi

Class 10 Social Science Chapter 1 Question 7.
Most forested state is –
(a) Madhya Pradesh
(b) Uttar Pradesh
(c) Assam
(d) Tamil Nady.
Answer:
(c) Assam

MP Board Class 10 Economics Chapter 1 Question 8.
Ghana Bird Santuary is located in –
(a) Kerala
(b) Rajasthan
(c) West Bengal
(d) Madhya Pradesh
Answer:
(b) Rajasthan

Question 2.
Fill in the blanks:

1.  ………………. has an important place in Joint Forest Management System.
2.  Social Forestry Scheme is getting financial assistance from ……………….
3.  Forest Fire Control Project is working in association with ……………….
4.  ……………….. and are ……………… established to protect and conserve wild life.

Answer:

1. 1.  Forest protection
   2.  World Bank
   3.  W.W.F.

1.  Sancturay, Naitonal Parks.

Question 3.

Answer:

1. (a)
2. (c)
3. (b)
4. (e)
5. (d)

MP Board Class 10th Social Science Chapter 1 Very Short Answer Type Questions

MP Board Class 10th History Chapter 1 Question 1.
What is meant by soil erosion?
Answer:
Removal of soil at a large scale from one place to another by some natural agent is known as soil erosion.

Pariksha Adhyayan Class 10th Social Science Question 2.
What do you mean by soil conservation?
Answer:
The ever increasing population resulted in the destruction of natural resources. Therefore, to prevent destruction soil conservation is necessary. There are various methods of soil conservation.

Class 10th Sst Chapter 1 Question 3.
What are the sources of underground water?
Answer:
Rain water is a main source of underground water. Some part of rain water is soaked by the land. Rest of the water percolates and is collected below the surface as underground water.

Class 10 MP Board Question 4.
What is the basis of modified forest policy of 1988?
Answer:
The main basis of modified forest policy is to maintain environmental stability, to conserve the natural heritage and to check on soil erosion.

Resources And Development Chapter Question 5.
What is the basis of the success of social forestry?
Answer:
This programme of plantation has been started in association with World Bank. It aims to plant useful trees in waste lands, road side and canal embankments near villages. One tree for every child’ this slogan is developed in schools and colleges. People’s participation is increased by publishing Vanmahotsava and by farm forestry, by planting trees road side, railway side and canal embankments.

MP Board Social Science Class 10 Question 6.
Why Indian Institute of Forest Management had been established?
Answer:
This institute had been established in Ahemdabad in 1978 in collaboration with a Swidish Company for the development of the forest. Central Government has also established Indian Institute of Forest Management in Bhopal for training, research and consultancy purposes.

MP Board Class 10th Social Science Chapter 1 Short Answer Type Questions

Question 1.
What is meant by soil profile? Explain?
Answer:
Soil profile is the sequence, colour, texture and nature of the horizons (layers) superimposed one above the other and exposed in a pit – section dug through the soil mantle,

1.  Upper most layer is top soil
2.  Second layer is sub soil
3.  Third layer is weathered parent rock material
4.  Fourth layer consists of parent rocks.

Top soil of the upper most layer is the real soil. Its important characteristic is the presence of humans and organic matter. Second layer is sub soil which consists of rocks, sand particles and clay. Third layer consists of weathered parent rock material and the fourth layer is made of parent rocks.

What Is Bangar Class 10th Question 2.
What is importance of soil in human life? Explain?
Answer:
Soil is very important for human life, especially for farmers. Human life depends on soil. All living organisms get their food directly or indirectly from soil. We get cotton, silk, jute and wool for making
clothes from soil, either directly or indirectly, e.g. sheep eats grass and gives us wool, silk worms survive on vegetation and vegetation grows in soil. Our industries like animal rearing, agriculture and forest – based industries all depend on soil. So soil is the basis of our life. According to Wil Cox, ‘The history of civilization is the history of the soil and the education of the individual begins from the soil.’

Soil Map Of India Class 10 Question 3.
Differentiate between Red Soil and Laterite Soil ?
Answer:
Red Soil:

1. Red soil is formed due to weathering of igneous and metamorphic rocks.
2. It is highly porous and less fertile but where it is deep it is fertile.
3. It is less crystalline.
4. It is red in colour due to the presence of iron in it.
5. It occurs in part of Tamilnadu, Karnataka, Orissa, Jharkhand and Andhra Pradesh.

Laterite Soil:

1. Laterite soil is formed by the leaching process in the heavy rainfall areas of tropical India.
2. It is less fertile, only grass grows on it in abundance.
3. It is crystalline.
4. It is red in colour due to little clay and much gravel of red sandstones.
5. It is found is hills of the Deccan, Karnataka, Orissa, Assam, Megha­laya and Kerala.

Major Soil Types In India Map Class 10 Question 4.
What are the measures of water conservation?
Answer:
The judicious utilization, conservation and management of water resources is necessary. In the view of the limited water available, increasing demand and its uneven availability it has become imperative to conserve the water resources. Following three steps are essential in this direction:

1.  To collect the rainwater and stop it from draining off.
2.  Scientific management of the water resources of all the river watersheds minor to major.
3.  Prevention of water resources from pollution.

Water Resources Map Class 10 Question 5.
Rain water harvesting is important. Why?
Answer:
Natural water is precious but abundantly available during rainy season. But due to carelessness of the people it goes wasted. We know that the crisis of fresh and pure water has became a world wide problem. So it is call of time to collect such a huge quantity of water by constructing reservoir on the roof or nearby the house or roadways.

Social Science Class 10 Chapter 1 Question Answer Question 6.
Conservation of forest is necessary, why?
Answer:
Conservation of forests is necessary, because of the following:

1. Plants provide food for men and animals.
2. They help in the maintenance of ecosystem.
3. They give oxygen necessary for the survival of men and animals.
4. They provide us timber for building, doors, houses, etc.
5. They give us herbal medicines.
6.  They help in soil erosion.

Water Resources Class 10 Map Question 7.
Explain forest based industries?
Answer:
Forests provide a large number of minor produce which are essential for industries such as lac, tanning materials, gum, honey, katha, wax, resins, bamboo, medicinal herbs, horn and hides of animals etc. Forests provide materials for basic industries i.e., wood is useful raw material for several industries like paper, match, lac, leather, oil and herbal medicines. Small scale industries developed from the minor forest provide like tendupatta, cane, honey, wax, etc.

Major Rivers And Dams In India Map Class 10 Question 8.
How does forest control the climate?
Answer:
It is necessary for a country to have a proper echological balance. A country should therefore have at least one third of its land area covered with the forests. A larger area under forest is must for absorption of carbon – di – oxide, the accumulation of which is likely to accentuate green house effect. This ‘effect’ may further give rise to general increase in temperature globally and ultimately melt the icecaped areas of the world. This would cause great loss to the life and property of the people living in low lying areas of the world. Thus, the forests would vanish by these natural calamities brought by the man.

Class 10 Soil Map 9.
Write down the chief characteristics of Forests Policy of December 1988?
Answer:
The following are the chief characteristics of the Forests Policy of Decemeber 1988:

1. Substantial increase in forest tree cover through massive forestation and social forestry programmes.
2. Steps to meet requirements of fuel wood, fodder and minor forest produce and timber for tribal and rural populations.
3. Increase in productivity of forest to meet the national needs.
4. Encouragement of efficient utilisation of forest produce and optimum substitution of wood.
5. Steps to massive people’s movement with involvement of women to achieve the objective and minimise pressure on existing forests.

Soil Map Class 10 Question 10.
What is Social Forestry Scheme?
Answer:
Social Forestry Scheme means the scheme for an awareness of tree plantation with the help of government and non – governmental institution. There is a known slogan ‘one tree for every child’ geared up to the students of schools and colleges is a serious measure to implement this variety of ecological scheme. This scheme of plantation is started in association with World Bank.

MP Board Class 10th Social Science Chapter 1 Long Answer Type Questions

Question 1.
What is soil? Describe different types of soil, their characteristics and distribution?
Answer:
The – uppermost layer of the earth’s crust, which is useful for cultivations and the basic resource of agriculture is called soil. The soils of India are:

1. Alluvial soil
2. Black soil
3. Red soil, and
4. Laterite soil.

1. Alluvial Soil:
Alluvial soil is considered to be a most fertile soil which forms the largest and the most important soil group of India. It contributes the largest share to the country’s agricultural production.

(a) Area:
Alluvial soil covers about 43.7% of the total land area under cultivation. The entire northern plain of India is made up of alluvial soil.

(b) Composition:
These soils are made up of new alluvium and old alluvium. These soils contain fine particles of soil called alluvium. The soil is called new and old depending upon their period of deposit.

(c) Fertility:
Alluvium soils are very fertile soils as they contain adequate amount of potash, phosphoric acid and lime. All the river basin generally have alluvial soils. It supports over half the Indian population.

2. Black Soil:
These soils are black in colour and are very suitable for the cultivation of cotton.

(a) Area:
These soils are spread all over the Deccan trap and are made up of lava flows. They cover the plateaus of Maharashtra, Saurastra, Malwa and southern Madhya Pradesh and extend eastwards in the south along the Godavari and Krishan Valley.

(b) Composition:
The Regur soils or Black soils contain calcium carbonate, magnesium carbonate, potash and lime. They are generally poor in phosphoric content. The soils consisting of extremely fine clay material known for their sticky characteristics are also called Black soils.

(c) Fertility:
Regur soils are very important and suitable for the cultivation of cotton, that is why they are sometimes also called cotton soil.

3. Red Soil:
These soils are derived from crystalline and metamorphic rocks rich in minerals.

(a) Area:
The southern half of peninsular block is covered by red soils of different shades of red and yellow. This type of soil can be seen in Chotanagpur plateau, Orissa, east Madhya Pradesh, Telangana, Nigiries and Tamilnadu plateau. These soil areas are also found northwards in the west along the Konkan coast of Maharashtra.

(b) Composition:
Red soils are loamy in deep depressions and in uplands. They contain loose grave (a highly coarse material and are deficient in phosphoric acid, organic matter and nitrogenous material.

(c) Fertility:
Since these soils are loamy and are made up of coarse material, they are not fertile. They are deficient in the organic matter and nitrogenous material that makes it less fertile.

4. Laterite Soil:
The laterite soil is a result of intense leaching owing to heavy tropical rains.

(a) Area:
They are found in western coastal regions receiving very heavy rainfall. They are also found in patches along the edge of the plateau in the east covering small parts of Tamilnadu and Orissa, a small part of Chotanagpur and Meghalaya in the north east.

(b) Composition:
These soil’s have resulted due to intense leaching owing to heavy tropical rainfall.

(c) Fertility:
These soils are invariably poor and support only pastures and scrub forests.

Question 2.
What is soil erosion? Explain the causes of soil erosion and methods of conservation of soil?
Answer:
The term erosion means the loosening and removal of soil from its previous resting place by the action of water and other agents. In India, soil erosion is in many places a serious menace. The extent to which erosion is liable to occur will vary with the condition but at any point its incidence is determined by the following factors:

1. The configuration, and particularly the slope of the land.
2. The credibility of the soil.
3. The amount distribution and intensity of the rainfall.
4. The vegetable cover.
5. The system of husbandary and soil management practiced.

Causes of soil erosion are:

1. Deforestation
2. Overgrazing
3. Shifting agriculture
4. Wind erosion
5. Agriculture by non-scientific methods.

Soil conservation:
The ever increasing population resulted in the destruction of natural resources. Therefore, to prevent destruction, soil conservation is necessary. There are various methods of soil conservation.

1. Contour farming
2. To prevent gully erosion by making field ridge.
3. Prevention of soil erosion by planting trees as wind breaks in deserts which check the velocity of wind. By doing plantation on the follow land and mountain slopes and by controlled grazing.
4. By collecting the run off water in mountain slopes and uneven areas.
5. By developing grazing land in the rural areas.

Question 3.
Draw a labelled diagram of soil profile?
Answer:
Soil profile:
Soil profile is the sequence, colour, texture, nature of the horizons (layers) superimposed one above the other and exposed in a pit – section dug through the soil mantle. Top soil of the upper most layer is the real soil. Its important characteristic is the presence of humans and organic matter.

Second layer is sub soil which consists of rocks, sand particles and clay. Third layer consists of weathered parent rock material and the fourth layer is made of parent rocks.

Question 4.
What are the main sources of water resource? What is importance of water resources in human life?
Answer:
There are four major sources of water.

1. Surface water
2. Ground water
3. Atmospheric water, and
4. Oceanic Water.

1. Surface water:
The surface water is available in rivers, ponds and lakes. Rivers are the main source of surface water in India. Rivers and its tributaries are found in each and every part of India. Three main rivers are the Indus, the Ganga and the Brahmaputra and they carry nearly 60 percent of the total surface water in India. Among the major rivers of the world, the Brahmaputra and the Ganga are at eighth and tenth places respectively.

2. Ground water:
Some part of rain water is soaked by the land. Only 60 percent reaches the upper layer of the soil, which is very useful for agriculture and vegetation to grow. Rest of the water percolates and is collected below the surface as ground water. It is obtained on the surface through wells and tubewells and is used by human beings for irrigation purposes, gardening and industrial purposes.

3. Atmospheric water:
This is in the form of water vapour therefore, it is not used.

4. Ocean water:
This type of water is mainly used for transport and fishing industry. Arabian Sea, Bay of Bengal and Indian Ocean are in the West, East and South of India respectively. Importance of water in human life is as follows:

1. To provide irrigational facilities.
2. To conserve the soil fertility as the flood.
3. To generate hydro – electric power.
4. To promote navigation through canals and rivers.
5. To promote tourism as beach tourism.
6. To promote fish culture, and
7. To store water, which can be used when it is in greater demand.

Question 5.
Describe the methods of water conservation. Why is it necessary?
Or
Why is water conservation necessary? Describe its main methods? (MP Board 2009,)
Answer:
The following are the crucial methods for the water conservation:

1. To collect rain water and stop it from draining off.
2. Scientific management of water resources.
3. Prevention of water resources from industrial and domestic pollution.

The availability of water for agricultural and other purpose is inadequate and irregular in our country. Being the monsoon, the bulk of rainfall is confined to a brief period of three to four months. Even the places of high rainfall like Cherrapunji and Konkan having heavy rainfall face scarcity of water during the dry season.

Secondly the distribution of rainfall is unequal for example our ground water resources are abundant only in the northern and coastal plains and in the other parts of the country its supply is inadequate. The river water of the Country is also not well connected by canals. In short, the supply of water in India depends on Monsoons and also the topography of land.

Therefore we feel that it is necessary to conserve this precious natural resource. Running water of rivers may be used for irrigation by constructing a canal. Similarly, dams may to constructed to produce electricity.

Question 6.
Describe direct and indirect advantages of forests? (MP Board .2009, 2013)
Or
Mention the direct and indirect advantages of forests? (MP Board 2009)
Answer:
Direct advantages of the forests are:
1. Forest provides wood:
Wood from forests is an important fuel. Wood and cow dung produces 34.6 per cent of total power resources. They provide us Teak, Sal, Shisham, Pine, Abnoos, Sandle wood and Deodar. Wood is also used for making furniture.

2. Forests provide minor forest produce:
Forests produce a large number of minor product which are essential for industries such as lac, tanning materials, gum, honey, katha, wax, resins, bamboo, medicinal herbs, horn and hides of animals etc.

Forests provide materials for basic industries i.e., wood is useful raw material for several industries like paper, match, lac, leather, oil and herbal medicines. Small scale industries developed from the minor forest provide like tendupatta, cane, honey, wax, etc.

3. Grazing land for animals:
Forests provide natural pastures for grazing animals.

4. Employment:
About 7.8 crore people depend on forests for their livelihood. Many industries are based on raw materials from forests giving employment to crores of people.

5. Revenue generation:
Government receive crores of rupees from the forests as revenue and royalty. Presently this revenue is 670 crore rupees per annum.

Indirect advantages of the forests are:

1. Control soil erosion:
Trees firmly enclose and considerably reduce soil erosion. Trees hold the fertile top layer of the soil.

2. Control the climate:
Forest act as wind breaks which check the velocity of wind. The climate of forest area remains temperate.

3. Check floods:
Speed of water is reduced by the existence of forests. Water is reduced by the trees. The force of water is reduced by the extensive forest cover.

4. Control expansion of desert areas:
Sardar Patel said, “If expansion of deserts are to be controlled and human civilization is to be prevented then the destruction of forest wealth is to be prevented.”

Question 7.
Describe the efforts of government in forest conservation?
Answer:
In 1950 after independence, Central Forest Board was established. New forest policy was made. Its four main points were:

1. Forest area should be increased to 33.3 per cent.
2. Forestation
3. Protection of forests
4. Forestry research.

The policy was revised on 7th December 1988. The main aim of the forest policy of 1988 is protection, conservation and development of forests. Policy holds the following objectives:

1. Maintenance of environmental stability through preservation and restoration of ecological balance.
2. Conservation of natural heritage.
3. Check on soil erosion and denudation in catchment areas of river, lakes and reservoirs.

In 1990, a 20 – year National Forestry Action programme was launched to make National Forest Policy of 1988 functional. For the development of the forests following activities are taking place:

1. Establishment of Central Forest Commission.
2. Indian Forest Survey Organisation.
3. Council of Forestry Research and Education.
4. Establishment of Wood Craft Training Centre.
5. State Forest Development Corporations.
6. Indian Institute of Forest Management.

Question 8.
Why wildlife conservation is necessary? What are the measures of wildlife conservation? (MP Board, 2011)
Answer:
Depletion of forest has endangered plant and wildlife. Several species have already become extinct. In order to preserve natural habitat and protect them from becoming extinct the government has set up many programmes. There are also programmes for conservation of wetland mangroves and coral reefs under the preservation of special ecosystem.

Coral reefs are characterized by high biomass production and rich floral and faunal diversity four coral reefs have been identified for conservation and management. Periodic census of wild animals are undertaken to check the number of certain species getting reduced. The hunting of wild animals and. birds has been banned and hunters are penalised. Conservation of wildlife: Following efforts can be made to protect the wildlife:

1. Safeguarding the national habitat of the wild animals.
2. Poaching should be restricted.
3. Establishing biosphere reserves in forest areas.
4. Educating public for environmental protection at levels of education.
5. Implementation of wildlife management programmes.

Project Work

Question 1.
Prepare a plan for a geographical tour of your area and collect the following information:

1. Type of soil in your region and its characteristics.
2. Causes of soil erosion in the area.
3. What could be the measures to prevent soil erosion in that area.
4. Characteristics of soil on the basis of crops grown in that area.

Answer:
Do yourself with the help of your subject teacher.

Question 2.
Prepare a map of India and show the following:
Kaziranga, Gir, Gim, Keoladeo, Simlipal National Park, Sunderban, Ranthambore, Savisca, Manas, Corbett Tiger Project, Nilgiri, Nandadevi, Great Nicobar, Pachmarhi Biosphere Reserve.
Answer:
Do yourself with the help of map under the guidance of your subject teacher.

Question 3.
Visit a National Park and prepare a report on the following points:

1. Wild life
2. Habitat of wild animals.
3. Their food and methods of hunting.
4. Forest produce collected from that park.

Answer:
Do yourself with the help of your subject teacher.

MP Board Class 10th Social Science Chapter 1 Additional Important Questions

Objective Type Questions

Question 1.
Multiple Choice Questions:
(Choose the correct answer from the following)

Question (a)
Which soil variety in known a ‘regur’?
(a) Laterite soil
(b) Black soil
(c) Red soil
(d) Alluvial soil.
Answer:
(b) Black soil

Question (b)
Which of the following river is the longest one?
(a) Indus
(b) Narmada
(c) Godavari
(d) Ghagra.
Answer:
(b) Narmada

Question (c)
Indian Institute of Forest Management is in?
(a) Mumbai
(b) Kanpur
(c) Dehradoon
(d) Ahamedabad.
Answer:
(d) Ahamedabad.

Question (d)
Dachigam National Park is located in the state?
(a) Madhya Pradesh
(b) Jammu & Kashmir
(c) Kerala
(d) Uttar Pradesh
Answer:
(c) Kerala

Question (e)
Project Tiger was started in?
(a) 1971
(b) 1972
(c) 1973
(d) 1974
Answer:
(c) 1973

Question (f)
Which type of soil is found in western Ghats?
(a) Alluvial soil
(b) Black soil
(c) Laterite soil
(d) Red soil
Answer:
(a) Alluvial soil

Question 2.
Fill in the blanks:

1. Petrol and coal are ………………… resources.
2. Clay soil is called ………………….. soil.
3. Deforestation is the cause of soil …………………
4. The National Forest Policy was formulated in …………………….
5. Jim Corbett National Park is located in …………………
6. Central Forest Commission is established in …………………… (MP Board 2009, Set D)
7. Ghana Bird Sanctuary is located in …………………….. (MP Board 2009, Set C)
8. All the elements which are capable of fulfilling human wants, are called ……………………..
9. Soil erosion ocwrs due to ………………… , ……………… and ………………. activities.

Answer:

1. non – renewable
2. alluvial
3. erosion
4. 1988
5. Uttarakhand
6. 1965
7. Rajasthan
8. Resources
9. water, wind, human.

Question 3.
True and False type questions:

1. Sunderlal Bahuguna was the founder of Vanmahotsava in 1950.
2. 33 per cent is the minimum criteria of the forestation as the biological balance.
3. About 47,000 species are found in India according to Zoological Survey of India.
4. In 1965, K.M. Munshi started Vanmahostsava. (MP Board 2009, Set C)
5. Gir National Park is located in Maharashtra.
6. There are 14 boisphere reserves which have been established in India.

Answer:

1. False
2. True
3. False
4. False
5. False
6. True.

Question 5.
Match the Columns:

Answer:

1. (c)
2. (e)
3. (d)
4. (a)
5. (b)

Question 1.
What per cent of our land area is plain?
Answer:
About 43%.

Question 2.
What per cent of land is mountainous?
Answer:
About 30%.

Question 3.
What is pasture?
Answer:
Land covered with natural grasses is knows as pasture.

Question 4.
Name the most endangered species in India?
Answer:
The tiger, the rhino, the bustard and the lion.

Question 5.
Which is the first biological reserve in India?
Answer:
The first biological reserve was set up in the Nilgiries.

Question 6.
Where is the Gulf of Mannar?
Answer:
Tamil Nadu.

Question 7.
What percentage of India is covered with forests?
Answer:
About 23.3%.

Question 8.
What are the sources of surface water?
Answer:
Ponds, tanks, rivers and reservoirs, etc.

Question 9.
What do you mean by rain water harvesting?
Answer:
It simply means capturing rains where it falls.

Question 10.
What are the major sources of irrigation in India?
Answer:
Canals, Tanks, Wells and Tubewells.

Question 11.
Which elements enhance the fertility of soil?
Answer:
Fine vegetal and animal remains add to the fertility of the soil.

Question 12.
What is Bangar?
Answer:
This is the old alluvium soil spread at the outer areas of khadar.

Question 13.
Where are the elephants found in India?
Answer:
Elephants in India are found in the jungles of Assam and those of Kerala and Karnataka.

Question 14.
How many species of animals, birds and fishes are found in India?
Answer:
India has approximately 89000 species of animals, 1200 of birds and 2500 species of fishes.

Question 15.
What is the primary source of water on the earth?
Answer:
The primary source of water on earth is precipitation that comes in the forms of rain and snowfall.

MP Board Class 10th Social Science Chapter 1 Very Short Answer Type Questions

Question 1.
Name the various elements of nature which contribute to evolution of soil?
Answer:
Changing temperature, running water and wind etc., contribute in the evolution of soil.

Question 2.
Which soil supports over half the Indian population?
Answer:
The alluvial soil. The new alluvium soil near the river banks or the catchment areas is called khadar.

Question 3.
What is National Park?
Answer:
It is a place where wildlife is in their natural setting. It is a place where animals mpve freely. Food for the animals is provided by animals themselves.

Question 4.
What do you mean by hydro – electricity?
Answer:
The electricity derived from the running or falling water. It is one of the neatest, cheapest and pollution free form of energy.

Question 5.
What are the types of forests according to administration?
Answer:
On the basis of administration, forests are classified into three types:

1. Reserved Forests
2. Protected Forests
3. Unclassified Forests.

Question 6.
What are the main causes of growing scarcity of water?
Answer:
Main causes of growing scarcity of water are:

1. Rapidly growing population.
2. Rising demand for. food and cash crops.
3. Increasing urbanisaffon and rising standard of living.

Question 7.
What is Humus?
Answer:
Deposited organic matter of plants and animals is called humus. It is found in top soil and helps in making soil fertile.

Question 8.
Define weathering?
Answer:
Disintegration of rocks on the earth surface due to weathering.

MP Board Class 10th Social Science Chapter 1 Short Answer Type Questions

Question 1.
Write any four characteristics of black soil?
Answer:
Black soil is also known as Regur soil. It is:

1. most suitable for the cultivation of cotton.
2. made up of extremely fine material.
3. known for its capacity to hold moisture.
4. rich in soil nutrients such as calcium carbonate, magnesium carbonate, potash and lime.

Question 2.
Why should we increase the land under forest?
Answer:
India has only 19 per cent of forest cover out its total land area. It is far below the scientific norms which prescribes minimum availability of forest to be 33 per cent. The forest area in India must therefore be increased to maintain ecological balance and absorption of carbon dioxide. The forests help us to preserve the wild ‘life. They cause precipitation and help us in decreasing the possibility of droughts. Forests regulate the flow of river water.

Question 3.
Write the main differences between Khadar soil and Bangar soil?
Answer:
Khadar Soil:

1. This is the new alluvial soil deposited by the rivers in the northern plains. The new alluvial soil deposited at the bank of rivers is locally kndwn as khadar.
2. This is very fertile land and the soil contains adequate amount of potash, phosphoric acid and lime. This has fine particles.

Bangar Soil:

1. This is the old alluvial soil ‘ deposited by the rivers in the northern plains. The old alluvial soil which is deposited beyond the new alluvium is locally called bangar in the norhtern plains.
2. This is also fertile soil but not as fertile as the Khadar is. This old alluvium often contains kankar nodules with- calcium carbonate in sub soil.

Question 4.
What is bio – reserve?
Answer:
Efforts are being made on the basis of specific regions to protect and preserve every plant and animal species found in India so that this natural heritage can be transmitted to the future. This concept is known as bio – reserve. Till now there are 14 bio – reserves have been established in India. Nilgiri of Kerala is one of them.

Question 5.
Differentiate between surface water and ground water resources?
Answer:
Surface Water:

1. Surface water is the volume of water present over the surface of the earth whether in the stationary state or the running state.
2. The volume of water on the surface changes with the change in the seasons and weather conditions.
3. The surface water can be used by constructing dams and canals etc.

Ground Water:

1. The underground water is the volume of water present below the earth surface whether in the stationary state or the running state.
2. The volume of underground water also depends on the amount of water received on the surface and at the same time, the amount of underground water utilized.
3. The undergournd water is widely used for drinking and irrigation through well and tube- wells.

Question 6.
Explain the agents of soil formation?
Answer:
Soil is a renewable resource. It takes hundreds of years in the formation of one centimetre thick layer of soil. Plain surface is best for the formation of soil because least problems are created here during the formation. There are different factors which help in formation of soil e.g., parent rocks and topography, climatic conditions which helps in weathering of rocks, plants, animals and there remains.

Question 7.
Write five steps for conervation of wild life?
Answer:

1. Safeguarding the national habital of the wild animals.
2. Hunting of wild animals should be restricted.
3. Establishing Biosphere Reserves in forest areas.
4. Educating public for environmental protection at levels of education.
5. Implementation of wildlife management programmes.

MP Board Class 10th Social Science Chapter 1 Long Answer Type Questions

Question 1.
Describe the concept of soil conservation. Outline the measures of soil conservation?
Answer:
Conservation of Soil. Soil is the most important resource to support human life. The soil on which we depend so much for our survival has evolved over thousands and thousands of years. It is therefore necessary to conserve this resource to enable people to produce a variety of crops.

What is needed to day is the scientific management of soil, their proper conservation, avoidance of their erosion and maintainence of their fertility through organic and bio manures.

Soil erosion is one single factor which cause great harm to our agricultural productivity. Secondly, salinity and alkalinity has also spoiled the fertility of the soil in India. Running water and wind have been causing regular soil erosion. So, conservation is necessary to ensure sustained productivity of land.
The soil can be conserved by adopting the following measures:

1. Aforestation and a ban on blind deforestation.

2. Contour Ploughing:
It is the method by which field are ploughed, harrowed and sown along the contours instead of up and down the slopes.

3. Terracing:
This means making a series of winding steps on mountain slope.

4. Strip Cropping:
Cover crop such as grasses and small grains are planted alternately with cultivated crops. Cover crops absorb moisture and hold the soil together.

5. Stopping clearance of forests on a large scale.

6. Preventing overgrazing.

7. Construction of bunds.

Question 2.
How are forests useful to man?
Answer:
The economical uses of forests are as follows:

1. Forests maintain environmental stability, ecological balance for the existence of all life forms.
2. They provide timber and fuel wood. Soft wood is used in furniture, packages, and also other building materials.
3. Pulp is made from soft wood and is used in paper making.
4. Forest provide many things to meet our food requirements like wild fruits, nuts, berries, etc. Many tribes are dependent on gathering of these products in forests.
5. They modify climatic conditions and bring better rainfall in the area.
6. They reduce wind force and reduce air temperature during summer which have a positive effect on the overall environment.
7. They add to the forest floor large quantities of leaves, twigs and branches, which after decomposition form humus. This increases the fertility of soil.

Question 1.
Show distribution of soil on a outline map of India?
Answer:

Question 2.
Show Wildlife and Biosphere Reserves on a outline map of India?
Answer:

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2

MP Board Class 10 Maths Chapter 5 Exercise 5.2 Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P.
Solution:

Solution:
(i) a_n = a + (n- 1)d
a₈ = 7 + (8 – 1)3 = 7 + 7 × 3 = 7 + 21
⇒ a₈ = 28

(ii) a_n = a + (n – 1)d
⇒ a₁₀ = -18 + (10 – 1)7 ⇒ 0 = -18 + 9d
⇒ 9d = 18 ⇒ d = \(\frac{18}{9}=2\)
∴ d = 2

(iii) a_n = a + (n – 1)d
⇒ -5 = a + (18 – 1) × (-3)
⇒ -5 = a + 17 × (-3)
⇒ -5 = a – 51 ⇒ a = -5 + 51 = 46
Thus, a = 46

(iv) a_n = a + (n – 1)d
⇒ 3.6 = -18.9 + (n – 1) × 2.5
⇒ (n – 1) × 2.5 = 3.6 + 18.9

⇒ n = 9 + 1 = 10
Thus, n = 10

(v) a_n = a + (n- 1)d
⇒ a_n = 3.5 + (105 – 1) × 0
⇒ a_n = 3.5 + 104 × 0 ⇒ a_n = 3.5 + 0 = 3.5
Thus, a_n = 3.5

5.2 Class 10 Question 2.
Choose the correct choice in the following and justify:
(i) 30^th term of the AP: 10,7,4, , is, ….,
(A) 97
(B) 77
(C) -77
(D) -87

(ii) 11^th term of the AP: -3, \(-\frac{1}{2}\), 2, …. ,is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
(i) (C): Here, a = 10, n = 30
∵ T₁₀ = a + (n – 1)d and d = 7 – 10 = -3
∴ T₃₀ = 10 + (30 – 1) × (-3)
⇒ T₃₀ = 10 + 29 × (-3)
⇒ T₃₀ = 10 – 87 = -77

(ii) (B): Here, a = -3, n = 11 and

Common difference calculator is the difference between two successive terms of an arithmetic progression.

Ex 5.2 Class 10 Question 3.
In the following APs, find the missing terms in the boxes:

Solution:
(i) Here, a = 2, T₃ = 26
Let common difference = d
∴ T_n = a + (n- 1 )d
⇒ T₃ = 2 + (3 – 1)d
⇒ 26 = 2 + 2 d
⇒ 2d = 26 – 2 = 24

(ii) Let the first term = a
and common difference = d
Here, T₂ = 13 and T₄ = 3
T₂ = a + d = 13, T₄ = a + 3d = 3
T₁ – T₂ = (a + 3d) – (a + d) = 3 – 13

(iii)

(iv) Here, a = – 4, T₆ = 6
∵ T_n = a + (n -1 )d
T₆ = – 4 + (6 – 1)d ⇒ 6 = -4 + 5d ⇒ 5d = 6 + 4 = d = 10 – 5 = 2
T₂ = a + d = -4 + 2 =-2
T₃ = a + 2d = -4 + 2(2) = 0
T₄ = a + 3d = -4 + 3(2) = 2
T₅ = a + 4d = -4 + 4(2) = 4

(v) Here, T₂ = 38 and T₆ = -22
∴ T₂ = a + d = 38, T₆ = a + 5d = -22
⇒ T₆ – T₂ = a + 5d – (a + d) = -22 – 38 -60
⇒ 4d = -60 ⇒ d = \(\frac{-60}{4}\) = -15
a + d = 38 ⇒ a + (-15) = 38
⇒ a = 38 + 15 = 53
Now,
T₃ = a + 2d = 53 + 2(-15) = 53 – 30 = 23
T₄ = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T₅ = a + 4d = 53 + 4(-15) = 53 – 60 = -7
Thus missing terms are

5.2 Maths Class 10 Question 4.
Which term of the AP: 3, 8, 13, 18, is 78?
Solution:
Let the n^th term be 78
Here, a = 3 ⇒ T₁ = 3 and T₂ = 8
∴ d = T₂ – T₁ = 8 – 3 = 5
And, T_n = a + (n- 1 )d
⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5
⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15
⇒ n = 15 + 1 = 16
Thus, 78 is the 16^th term of the given AP.

Maths Class 10 Ex 5.2 Question 5.
Find the number of terms in each of the following APs:
(i) 7,13,19, …….. ,205
(ii) 18, \(15 \frac{1}{2}\), 13, …… ,-47
Solution:
(i) Here, a = 7,d = 13 – 7 = 6
Let total number of terms be n.
∴ T_n = 205
Now, T_n = a + (n – 1) ×d
= 7 + (n – 1) × 6 = 205
⇒ (n – 1) × 6 = 205 – 7 = 198
⇒ n – 1 = \(\frac{198}{6}=33\)
∴ n = 33 + 1 = 34
Thus, the required number of terms is 34.

(ii)

Thus, the required number of terms is 27.

Class 10 5.2 Question 6.
Check whether -150 is a term of the AP:
11, 8, 5, 2…
Solution:
For the given AP,
we have a = 11, d = 8 -11 = -3
Let -150 be the n^th term of the given AP
∴ T_n = a + (n – 1 )d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 – 11 = (n – 1) × (-3)
⇒ -161 = (n – 1) ⇒ (-3)

But n should be a positive integer.
Thus, -150 is not a term of the given AP

Class 10th Math 5.2 Question 7.
Find the 31^st term of an AP whose 11^th term is 38 and the 16^th term is 73.
Solution:
Here, T₁₁ = 38 and T₁₆ = 73
Let the first term = a and the common difference = d.
T_n = a + (n – 1 )d
Then, T_n = a + (11 – 1)d = 38
⇒ a + 10d = 38 …(1)
and T₁₆ = a + (16 – 1)d = 73
⇒ a + 15d = 73 …(2)
Subtracting (1) from (2), we get
(a + 15d) – (a + 10d) = 73 – 38

Class 10 Maths Exercise 5.2 Question 8.
An AP consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term.
Solution:
Here, n = 50, T₃ = 12, T_n = 106
⇒ T₅₀ = 106
Let the first term = a and the common difference = d
T_n = a + (n – 1 )d
T₃ = a + 2d = 12 …(1)
T₅₀ = a + 49d = 106 …(2)
Subtracting (1) from (2), we get

Maths Ch 5 Ex 5.2 Class 10 Question 9.
If the 3^rd and the 9^th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Here, T₃ = 4 and T₉ = -8
T_n = a + (n – 1)d
T₃ = a + 2d = 4 …. (1)
T₉ = a + 8d = – 8 …. (2)
Subtracting (1) from (2), we get

10th Standard Maths Exercise 5.2 Question 10.
The 17^th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let a be the first term and d the common difference of the given AP
Now, using _n = a + (n – 1 )d, we have

Thus, the common difference is 1.

Arithmetic Progression Exercise 5.2 Question 11.
Which term of the AP : 3, 15, 27, 39,… will be 132 more than its 54^th term?
Solution:
Here, a = 3, d = 15 – 3 = 12
Using T_n = a + (n – 1 )d, we get

Thus, 132 more than 54^th term is the 65^th term.

10th Class Chapter 5 Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let for the 1^st AP, the first term = a and common difference = d

And for the 2^nd AP, the first term = a and common difference = d

According to the condition,

Ncert Solutions For Class 10th Maths Chapter 5 Exercise 5.2 Question 13.
How many three-digit numbers are divisible by 7?
Solution:
The first three-digit number divisible by 7 is 105.
The last such three-digit number divisible by 7 is 994.
∴ The AP is 105,112,119, ,994
Let n be the required number of terms Here, a = 105, d = 7 and _n = 994

Thus, 128 numbers of 3-digits are divisible by 7.

Class 10 Maths Solutions MP Board Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 beyond 10 is 12.
The multiple of 4 just below 250 is 248.

Thus, the required number of terms is 60.

Class 10 Math Ap Ex 5.2 Question 15.
For what value of n, are the n^th terms of two APs: 63,65,67 … and 3,10,17, …. equal?
Solution:

Thus, the 13^th terms of the two given AP’s are equal.

Class 10 Maths Chapter 5 Exercise 5.2 Question 16.
Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Solution:
Let the first term = a and the common difference = d

Class 10 Maths Ch 5 Ex 5.2 Question 17.
Find the 20^th term from the last term of the AP : 3, 8, 13, …, 253.
Solution:
We have, the last term (l) = 253
Here, d = 8 – 3 = 5
Since, the n^th term before the last term is given by l – (n – 1 )d
We have 20^th term from the last term = l – (20 – 1) × 5 = 253 – 19 × 5 = 253 – 95 = 158

Class 10 Maths 5.2 Question 18.
The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
Let the first term = a and the common difference = d

Class 10 Maths Ex 5.2 Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ? 200 each year. In which year did his income reach ₹ 7000?
Solution:
Here, a = ₹ 5000 and d = ₹ 200
Let, in the nth year Subba Rao gets ₹ 7000.

Exercise 5.2 Class 10 Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the n^th week, her weekly savings become ₹ 20.75, find n.
Solution:
Here, a = ₹ 5 and d = ₹ 1.75

Thus, the required number of weeks is 10.

In this article, we will share MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Science Solutions Chapter 2 Acids, Bases and Salts

MP Board Class 10th Science Chapter 2 Intext Questions

Intext Questions Page No. 18

Mp Board Class 10 Science Chapter 2 Question 1.
You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Answer:

• Put the red litmus paper turn by turn in each of the three test tubes. The solution which turns the red litmus paper to blue will be a basic solution here, the blue litmus paper formed can now be used to test the acidic solution.
• Put this blue litmus paper in the remaining two test tubes one by one. The solution which turns the blue litmus paper to red will be the acidic solution.
• The solution which has no effect on any litmus paper will be neutral and hence it will be distilled water.

Intext Questions Page No. 22

Question 1.
Why should curd and sour substances not be kept in brass and copper vessels?
Answer:
Curd and other sour substances are acidic in nature. So, when they are kept in brass and copper vessels, harmful products along with hydrogen gas are produced which spoil the food.

Question 2.
Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?
Answer:
Hydrogen gas is usually liberated when an acid reacts with a metal. Let us illustrate it with the following examples:

1. Add some pieces of zinc granules into 5ml of a dilute solution of sulphuric acid (H₂SO₄).
2. Shake it well.
3. Pass the produced gas into a soap solution.
4. Now, soap bubbles are formed in the soap solution and these soap bubbles contain hydrogen.
5. Bring a burning candle near a gas-filled bubble. A candle burns with a pop sound. So, the following reaction takes place:
   H₂SO₄(aq) + Zn(s) → ZnSO₄(aq) + H₂↑
   
6. We can test the evolved hydrogen gas by its burning with a pop sound when a candle is brought near the soap bubbles.

pH Calculator is a free online tool that displays the pH value for the given chemical solution.

Mp Board Class 10th Science Chapter 2 Question 3.
Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:

Intext Questions Page No. 25

Question 1.
Why do HCl, HNO₃ etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
HCl or HNO₃ dissolve in water to form H⁺ or H₃O⁺ ions in aqueous solutions which show their acidic character. The following reactions take place when HCl or HNO₃ are mixed with water:
HCl(aq) → H⁺ + Cl^–
H⁺ + H₂O → H₃O⁺
On the other hand, when alcohol and glucose are mixed with water they do not dissolve to form ions due to the presence of hydrogen bonds and basic character. Hence, they do not show acidic character.

Table 2.2 Class 10 Science Solution Question 2.
Why does an aqueous solution of acid conduct electricity?
Answer:
In the aqueous solution, acid forms ions and these ions are conductor of electricity.

Question 3.
Why does dry HCl gas not change the colour of the dry litmus paper?
Answer:
Dry HCl gas does not change the colour of the dry litmus paper because it has no hydrogen ions (H⁺) or hydronium (H₃O⁺) ions in it.

Question 4.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
Answer:
Mixing water to acid is an exothermic reaction. Hence while diluting an acid it is recommended that the acid should be added to water and not water to acid. If we mix water to acid explosion occurs and burning take place.

Table 2.4 Class 10 Science Answers Question 5.
How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?
Answer:
Concentration of hydronium ions (H₃O⁺) decreases and becomes weak. In this way concentration of hydronium ion affects when a solution of acid is diluted.

Question 6.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
When excess base is dissolved in a solution of sodium hydroxide concentration of OH^– Hydroxide ion is more.

Intext Questions Page No. 28

Activity 2.11 Class 10 Science Question 1.
You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:
Solution ‘A’ is acidic because pH of the solution A is 6 which is less than 7 while solution ‘B’ is basic because pH of the solution ‘B’ is 8 which is greater than 7. Solution ‘A’ has more hydrogen ion concentration in comparison to solution ‘B’ because solution ‘A’ is acidic.

If the pH value is less than 7, it represents an acidic solution.
If the pH value is more than 7, it represents a base.
It the pH – 6 is acidic it has more concentration of ions than pH-8 which is a base.

Question 2.
What effect does the concentration of H⁺(aq) ions have on the nature of the solution?
Answer:
If the concentration of H⁺(aq) ions is increased then the solution becomes acidic and if the concentration of H⁺(aq) ions is decreased then the solution becomes basic in nature.

Difference Between Acid And Base Class 10 Question 3.
Do basic solutions also have H⁺(aq) ions? If yes, then why are these basic?
Answer:
Yes. H⁺ ions are always present in basic solution. Concentration of Basic is more than OH^– ions.

Question 4.
Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?
Answer:
If the soil is acidic and improper for cultivation, then to increase the basicity of soil, the farmer would treat the soil with quick lime or slaked lime or chalk.

Intext Questions Page No. 33

Question 1.
What is the common name of the compound CaOCl₂?
Answer:
Bleaching powder.

Table 2.2 Class 10 Science Question 2.
Name the substance which on treatment with chlorine yields bleaching powder.
Answer:
Dry slaked lime or calcium hydroxide.

Question 3.
Name the sodium compound which is used for softening hard water.
Answer:
Sodium carbonate.
Na₂CO₃, 10 H₂O is the compound of sodium to soften hard water.

Class 10 Science Chapter 2 Question 4.
What will happen if a solution of sodium hydrocarbon is heated? Give the equation of the reaction involved.
Answer:
When sodium hydrocarbon is heated then sodium carbonate and water are formed along with the evolution of carbon dioxide gas.
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

Question 5.
Write an equation to show the reaction between Plaster of Paris and water.
Answer:
Plaster of Paris reacts with water to form gypsum.

MP Board Class 10th Science Chapter 2 Ncert Textbook Exercises

Ch 2 Science Class 10 Question 1.
A solution turns red litmus blue, its pH is likely to be
(a) 1
(b) 4
(c) 5
(d) 10
Answer:
(d) 10
Bases turn red litmus to blue. pH value of 7 is greater than 7. Hence this solution changes red litmus to blue.

Question 2.
A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains
(a) NaCl
(b) HCl
(c) LiCl
(d) KCl
Answer:
b) the solution contains HCl.

Question 3.
10 ml of a solution of NaOH is found to be completely neutralised by 8 ml of a given solution of HCl. If we take 20 ml of the same solution of NaOH, the amount of HCl solution (the same solution as before) required to neutralise it will be
(a) 4 ml
(b) 8 ml
(c) 12 ml
(d) 16 ml
Answer:
d) 16 mL HCl solution is required.

Ph Value Table Class 10 Question 4.
Which one of the following types of medicines are used for treating indigestion?
(a) Antibiotic
(b) Analgesic
(c) Antacid
(d) Antiseptic
Answer:
c) Antacid is used to treat indigestion.

Question 5.
Write word equations and then balanced equations for the reaction taking place when:
(a) dilute sulphuric acid reacts with zinc granules.
(b) dilute hydrochloric acid reacts with magnesium ribbon.
(c) dilute sulphuric acid reacts with aluminium powder.
(d) dilute hydrochloric acid reacts with iron filings.
Answer:
a) Sulphuric acid + Zinc ➝ zinc Sulphate + Hydrogen
H₂SO(aq) + Zn(s) ➝ ZnSO₄(aq) + H₂(g)
b) hydro Chloric acid + magnesium ➝ Magnisium Chloride + Hydrogen
2HCl(aq) + Mg(s) ➝ MgCl₂(aq) + H₂(g)
(c) Sulphuric Hydrigen Sulphate + Aluminium ➝ Aluminium + Hydrogen Chloride
3H₂SO₂(aq) + Mg(s) ➝ MgCl₂(aq) + H₂(g)
d) Hydrochloric acid  + Iron ➝ Ferric + Hydrogen
6HCl(aq) + 2Fe(s) ➝ 2FeCl₂(aq) + 3H₂(g)

Indicator Table Class 10 Question 6.
Compounds such as alcohols and glucose also contain hydrogen but are not categorised as acids. Describe an activity to prove it.
Answer:
Experiment: Fix two nails on the cork and keep this in 100 ml beaker. Two nails are fixed to 6 volt battery, bulb and switch. Then pour some dilute HCl in the beaker and switch on the current. Repeat the experiment separately with glucose and alcohol solutions.
Observation: Bulb glows in HCl solution but do not glows in glucose solution.
Result: HCl ➝ H+ and Cl- ions.
These ions conduct electricity and bulb glows.
By this experiment we conclude that All acids contain Hydrogen.

Question 7.
Why does distilled water not conduct electricity, whereas rainwater does?
Answer:
Distilled water cannot conduct electricity because it does not contain ions while rainwater conducts electricity as it contains ions due to the presence of dissolved salts in it.

Table 2.2 Class 10 Science Byju’s Question 8.
Why do acids not show acidic behavior in the absence of water?
Answer:
Acids do not show acidic property in the absence of water. Because Hydrogen ions dissociates in presence of water. Hydrogen ions are responsible for acidic nature.

Question 9.
Five solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9, respectively. Which solution is
(a) Neutral?
(b) Strongly alkaline?
(c) Strongly acidic?
(d) Weakly acidic?
(e) Weakly alkaline?
Arrange the pH in increasing order of hydrogen-ion concentration.
Answer:
a) Neutral ➝ solution D ➝ pH value of pH is 7.
b) strongly alkaline ➝ solution C ➝ pH is 11
c) strongly acidic ➝ solution B ➝ pH is 1
d) weakly acidic ➝ solution A ➝ pH is 4
e) weakly alkaline ➝ solution E —> pH is 9
We can arrange the pH in increasing order of hydrogen ion concentration as 11 < 9 <7 <4<1

Table 2.2 Science Class 10th Question 10.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A, while acetic acid (CH₃COOH) is added to test tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing
occur more vigorously and why?
Answer:
The fizzing will occur strongly in test tube A, in which hydrochloric acid (HCl) is added. This is because HCl is a stronger acid than CH₃COOH and therefore, produces hydrogen gas at a faster speed due to which fizzing occurs.

Question 11.
Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain your answer.
Answer:
pH value of fresh milk is 6, but when it converts into curd value of pH decreases because curd is acidic. Hence this value is becoming less.

Table 2.4 Class 10 Science Question 12.
A milkman adds a very small amount of baking soda to fresh milk.

1. Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
2. Why does this milk take a long time to set as curd?

Answer:

1. The milkman shifts the pH of the fresh milk from 6 to slightly alkaline because, in alkaline condition, milk does not set as curd easily. Hence, it does not get spoiled for longer period of time, in which he can sell it to make a profit.
2. Since this milk is slightly basic than usual milk, acids produced to set the curd are neutralised by the base. Therefore, it takes a longer time for the curd to set which is usually acidic.

Question 13.
Plaster of Paris should be stored in a moisture-proof container. Explain why?
Answer:
The Plaster of Paris should be stored in a moisture-proof container as it absorbs water from moisture and turns into a hard substance (Gypsum) as shown in the following chemical equation:

10th Science 2 Chapter 2 Question Answer Question 14.
What is a neutralisation reaction? Give two examples.
Answer:
The reaction between an acid a base to give salt and water is known as a neutralisation reaction.

Question 15.
Give two important uses of washing soda and baking soda.
Answer:
1. Washing Soda:
(a) This is used in glass, soap and paper industries,
(b) It is used for removing permanent hardness of water.

2. Baking Soda:
(a) This is used cooking mixture of Baking soda and acid (tartaric acid – weak acid) is called Baking powder. When it is heated or combined with water. CO₂ is evolved and soften the bread.
(b) It is also used in soda-acid fire extinguishers.

MP Board Class 10th Science Chapter 2 Additional Important Questions

MP Board Class 10th Science Chapter 2 Multiple Choice Questions

Question 1.
The range of a pH scale is:
(a) 1 – 10
(b) 1 – 100
(c) 0 – 14
(d) 1 – 14
Answer:
(c) 0 – 14

Science Class 10 Chapter 2 Question 2.
pH is defined as:
(a) The logarithm of hydrogen ion concentration
(b) The negative logarithm of hydrogen ion concentration
(c) Hydrogen ion concentration
(d) None of the above
Answer:
(a) The logarithm of hydrogen ion concentration

Question 3.
Which of the following solution will have pH = 7?
(a) Tea
(b) The salt solution in distilled water
(c) Hydrochloric acid solution
(d) Water distilled with chlorine gas.
Answer:
(b) The salt solution in distilled water

Difference Between Acid And Base Class 10th Question 4.
Which colour indicate neutral solution on a pH paper?
(a) Brown
(b) Green
(c) Purple
(d) White or transparent
Answer:
(b) Green

Question 5.
A solution is acidic if:
(a) it releases H⁺ ions in the solution.
(b) it has a pH of less than 7.
(c) it has dark red, orange or greenish-yellow colour on a pH paper.
(d) all of the above.
Answer:
(d) all of the above.

Question 6.
Which of the following solutions will have pH < 7?

Choose correct combination:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) All of the above.
Answer:
(b) (ii) and (iii)

Class 10th Science Table 2.2 Question 7.
“p” in pH stands for:
(a) Phosphorus
(b) Potenz
(c) Potential
(d) Polarity
Answer:
(b) Potenz

Question 8.
Four different jars A, B, C, D contains hydrochloric acid, black coffee, ammonia and soap solution. Choose the order of decreasing acidic strength.

(a) A > B > C > D
(b) B > C > A > D
(c) D > C > B > A
(d) D = C > B > A
Answer:
(a) A > B > C > D

Activity 2.4 Class 10 Science Question 9.
How can we find the pH of a solution?
(a) By dipping pH paper in it.
(b) By dropping some solution over pH paper.
(c) By heating pH paper in vapours of solution.
(d) By pouring all the solution over pH paper.
Answer:
(a) By dipping pH paper in it.

Question 10.
In a class, while doing practical on different solutions, four students give their observations. Which student/s has/ve given a correct explanation if their observation is as follows?

Choose the correct combination of students with wrong observations:
(a) A and B
(b) C and D
(c) A, B, C
(d) All students
Answer:
(b) C and D

Question 11.
Lemon juice gives orange colour over pH paper. What is its nature?
(a) Strong Acid
(b) Basic
(c) Neutral
(d) Moderate acid
Answer:
(a) Strong Acid

Activity 2.10 Class 10 Science Question 12.
Pure water has a pH = 7, while distilled water has pH = 8 – 10, it represents that distilled water is:
(a) Slightly basic
(b) Strong base
(c) Mild acid
(d) Strong acid
Answer:
(a) Slightly basic

Question 13.
On a pH paper, pH values of 1 and 8 are represented by colours:
(a) Yellow and Orange
(b) Purple and Greenish
(c) Red and Bluish Green
(d) Green and Red
Answer:
(c) Red and Bluish Green

Acids Bases And Salts Class 10 Solutions Question 14.
Which of the following acids gives a dark red colour?
(a) Lemon juice
(b) Hydrochloric acid
(c) Acetic acid
(d) Nitric acid
Answer:
(b) Hydrochloric acid

Question 15.
How pOH can by represented?
(a) -Log [H⁺] = pOH
(b) -Log [H^–] = pOH
(c) -Log [pH] = pOH
(d) -Log [OH^–] = pOH
Answer:
(d) -Log [OH^–] = pOH

Question 16.
If [H⁺] is 1.0 × 10^-9 mole, what will be pH of solution?
(a) 10^-9
(b) 1
(c) 9
(d) -9
Answer:
(c) 9

Chapter 2 Class 10 Science Question 17.
What is the nature of citric acid?
(a) Basic
(b) Acidic
(c) Neutral
(d) Both a and b
Answer:
(b) Acidic

Question 18.
What is the nature of Sodium Hydroxide?
(a) Basic
(b) Acidic
(c) Neutral
(d) Unpredictable
Answer:
(a) Basic

Question 19.
pH > 7 represents?
(a) Basic solutions
(b) Acidic solutions
(c) Neutral solution
(d) All
Answer:
(a) Basic solutions

Question 20.
pH < 7 represents?
(a) Basic solutions
(b) Acidic solutions
(c) Neutral solution
(d) All
Answer:
(b) Acidic solutions

Ch 2 Science Class 10 Activities Question 21.
Hydrogen ion concentration for pure water is:
(a) 7
(b) 10^-7
(c) 10^-7 mole/litre
(d) 10⁷ moles/litre
Answer:
(c) 10^-7 mole/litre

MP Board Class 10th Science Chapter 2 Very Short Answer Type Questions

Question 1.
How can pH be represented using log?
Answer:
pH = – log [H⁺].

Question 2.
What is the range of pH on pH paper?
Answer:
0 – 14.

Question 3.
What is the pH of a strong acid?
Answer:
0 – 2.

Question 4.
Why is water neutral?
Answer:
On dissociation, water has equal numbers of H⁺ and OH^– ions. So, it does not go through any change and remains neutral.

Question 5.
What is the pH of concentric HCl?
Answer:
1 – 2.

Question 6.
What will be the colour of pH paper when coffee is poured over it?
Answer:
Reddish as it is slightly acidic in nature.

Question 7.
Which is more acidic-lemon juice or baking powder?
Answer:
Lemon juice.

Question 8.
What is the best medium to check any chemical’s nature?
Answer:
Water.

Question 9.
What is the universal solvent?
Answer:
Aqua regia.

Question 10.
What kind of reactions are neutralisation reactions?
Answer:
Any reaction between an acid and a base to form salt and water is called neutralisation reaction.

Question 11.
Which solution is considered to be neutral?
Answer:
Solutions with no acidity or alkalinity are neutral. Acids and bases are present in equal amounts.

Question 12.
Write the formula of brine and bleaching powder.
Answer:
NaCl and CaOCl₂.

Question 13.
Name two products which we can be obtained by chemical processing of common salt.
Answer:
Baking soda and bleaching powder.

Question 14.
What is the common name of sodium hydrogen carbonate?
Answer:
Baking soda.

Question 15.
Write the formula for washing soda.
Answer:
Na₂CO₃.10H₂O.

Question 16.
What is the source of naturally occurring acid lactic acid?
Answer:
Curd.

Question 17.
Which acid is present in tomato?
Answer:
Oxalic acid.

Question 18.
Name two olfactory indicators.
Answer:
Vanilla and clove.

Question 19.
What is the colour of methyl orange in acidic solution?
Answer:
Red.

Question 20.
What is the colour of phenolphthalein in basic medium?
Answer:
Pink.

MP Board Class 10th Science Chapter 2 Short Answer Type Questions

Question 1.
How acids are different from bases when dissolved in water?
Answer:
Acids on dissolving in water produce H⁺ ions while the base produces OH^– ions.

Question 2.
Name two indicators and write their colour in different mediums.

Name of indicator	Colour in acidic medium	Colour in basic medium
Methyl Orange	Red	Yellow Pink
Phenolphthalein	Colourless	Pink

Question 3.
Name any three hydrated salts.
Answer:

1. Barium chloride, BaCl₂. 2H₂O.
2. Copper sulphate, CuSO₄. 5H₂O.
3. Ferrous sulphate, FeSO₄. 6H₂O.

Question 4.
Give an equation of neutralisation reaction.
Answer:

Question 5.
What causes acidity in our body? How can it be cured?
Answer:
Our stomach produces hydrochloric acid which helps in digestion of food. During indigestion, the stomach produces too much acid and this causes pain and irritation. This can be cured by using bases called antacids.

Question 6.
What is the result of the reaction between an acid and a metal?
Answer:
Corresponding salt is formed with the evolution of hydrogen gas when a metal reacts with acid.

Question 7.
Write two important uses of pH in everyday life. Also, give an example.
Answer:
pH balance and its particular range of maintenance are very important in nature because it affects animal and plant life very much.

For example:

1. Curd formation: Atmospheric bacteria change the pH of milk which causes the curd formation.
2. Aids in digestion: Slight acidic conditions in the stomach due to the presence of hydrochloric acid aids in the digestion of food.

Question 8.
Give examples of two acids and bases present in nature.
Answer:

1. Acids: Citric acid, acetic acid.
2. Bases: Calcium carbonate, sodium hydroxide.

Question 9.
Discuss the various types of salts.
Answer:
There are three types of salts:

1. Neutral salts: Salts formed by the mixing of strong acid and strong base, e.g., NaCl, K₂SO₄ etc.
2. Acidic salts: Salts formed by the mixing of a strong acid and weak base e.g., NH₄Cl, CaSO₄.
3. Basic salts: Salts formed by the mixing of a strong base and weak acid e.g., Na₂CO₃, CH₃COONa etc.

Question 10.
Common salt acts as raw material for many important daily use chemicals. Name some of them and also write their chemical formula.
Answer:
Sodium hydroxide (NaOH), Baking Soda (NaHCO₃), Washing soda (Na₂CO₃. 10H₂O) etc.

Question 11.
What are the products of Chlor-alkali process?
Answer:
The products of Chlor-alkali process are chlorine and sodium hydroxide.

Question 12.
Name two uses of each of the given salts:

1. Bleaching powder
2. Baking soda

Answer:
Use of given salts are:
1. Bleaching powder:

• It is used as an oxidising agent in chemical industries.
• It is used for disinfecting water to make it free of germs.

2. Baking soda:

• It is used in soda acid fire extinguishers.
• It is an ingredient in antacids.

Question 13.
What is the water of crystallisation?
Answer:
The water of crystallisation is the fixed number of water molecules present in one formula unit of salt.

Question 14.
Give reaction to show the formation of sodium zincate?
Answer:

Question 15.
Name the products of electrolysis of brine and also give on use of each.
Answer:
Chlorine gas, H₂ gas, and sodium hydroxide are the products of electrolysis of brine:

1. Use of chlorine gas: It is used as a disinfectant.
2. Use of H₂ gas: It is used in the manufacture of ammonia.
3. Use of sodium hydroxide: It is used for the manufacture of soaps and detergents.

MP Board Class 10th Science Chapter 2 Long Answer Type Questions

Question 1.
Discuss the nature of the solution and its type in brief and also explain the strength of a solution.
Answer:
Nature of the solution:
When a solute is dissolved solvent (generally water), it shows different kind of nature with regard to its reactivity and solubility. On the basis of removal of H⁺ ion or OH^– the solution is formed. The types of solutions are divided as follows:

Strength of solution is determined by the,

1. Speed of reactivity i.e., how fast the ions are found dissociated.
2. Amount of ions (H⁺ or OH^– ) released or their ion concentration.

Question 2.
Give one example in each case:
(a) a weak mineral acid.
(b) a base which is not an alkali.
(c) a hydrogen-containing compound which is not an acid.
(d) a basic oxide soluble in water.
(e) a basic oxide insoluble in water.
Answer:

MP Board Class 10th Science Chapter 2 Textbook Activities

Class 10 Science Activity 2.1 Page No. 18

1. Collect the following solutions from the science laboratory – hydrochloric acid (HCl), sulphuric acid (H₂SO₄), nitric acid (HNO₃), acetic acid (CH₃COOH), sodium hydroxide (NaOH), calcium hydroxide (Ca(OH)₂], potassium hydroxide (KOH), magnesium hydroxide [Mg(OH)₂] and ammonium hydroxide (NH₄OH).
2. Put a drop of each of the above solutions on a watch-glass one by one and test with a drop of the indicators shown in the table.
3. What change in colour did you observe with red litmus, blue litmus, phenolphthalein and methyl orange solutions for each of the solutions taken?
4. Tabulate your observations in the table.

Result:

Class 10 Science Activity 2.2 Page No. 18,19

1. Take some finely chopped onions in a plastic bag along with some strips of clean cloth. Tie up the bag tightly and leave overnight in the fridge. The cloth strips can now be used to test for acids and bases.
2. Take two of these cloth strips and check their odour.
3. Keep them on a clean surface and put a few drops of dilute HCl solution on one strip and a few drops of dilute NaOH solution on the other.
4. Rinse both cloth strips with water and again check their odour.
5. Note your observations.
6. Now take some dilute vanilla essence and clove oil and check their odour.
7. Take some dilute HCl solution in one test tube and dilute NaOH solution in another. Add a few drops of dilute vanilla essence to both test tubes and shake well. Check the odour once again and record changes in odour, if any.
8. Similarly, test the change in the odour of clove oil with dilute HCl and dilute NaOH solution and record your observations.

Observations:

1. On putting the cloth strip into dilute.HCl solution, the red colour of cloth strip changes to pale red.
2. On putting the cloth strip into NaOH solution, the red colour of strip changes to green in colour.

Odour test:

1. Add vanilla essence in dilute NaOH.
2. Add vanilla essence in dilute HCl.
3. Add clove oil in to dilute HCl.
4. Add clove oil in dilute NaOH.

Result:

1. No smell found.
2. The smell of vanilla exists.
3. The smell of clove present.
4. The smell of clove exists.

Class 10 Science Activity 2.3 Page No. 19,20

Caution:

1. This activity needs the teacher’s assistance.
2. Set the apparatus as shown in Figure.
3. Take about 5 ml of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.
4. What do you observe on the surface of zinc granules?
5. Pass the gas being evolved through the soap solution.
6. Why are bubbles formed in the soap solution?
7. Take a burning candle near a gas-filled bubble.
8. What do you observe?
9. Repeat this Activity with some more acids like HCl, HNO₃ and CH₃COOH.
10. Are the observations in all the cases the same or different?

Observations:

1. 5 ml of dilute sulphuric acid + zinc granules → On the surface of zinc granules + bubbles of hydrogen gas are formed.
2. On passing the gas evolved through a soap solution → bubbles are formed due to low surface tension of soap solution. On taking a burning candle near a gas-filled bubbles → gas burns with a pop sound.
3. All other acids HCl, HNO₃ and CH₃COOH show the same observation.

Class 10 Science Activity 2.4 Page No. 20

1. Place a few pieces of granulated zinc metal in a test tube.
2. Add 2 ml of sodium hydroxide solution and warm the contents of the test tube.
3. This activity needs the teacher’s assistance.
4. Set the apparatus as shown in Figure.
5. Take about 5 ml of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.
6. What do you observe on the surface of zinc granules?
7. Pass the gas being evolved through the soap solution.
8. Why are bubbles formed in the soap solution?
9. Take a burning candle near a gas-filled bubble.
10. What do you observe?
11. Repeat this Activity with some more acids like HCl, HNO₃ and CH₃COOH.
12. Are the observations in all the cases the same or different?

Observations:
Reaction:
2NaOH + Zn → Na₂ZnO₃ + H₂ ↑

Conclusion:
Bubbles of hydrogen gas are formed.

Class 10 Science Activity 2.5 Page No. 20

1. Take two test tubes, label them as A and B.
2. Take about 0.5 g of sodium carbonate (Na₂CO₃) in test tube A and about 0.5 g of sodium hydrogen carbonate (NaHCO₃) in test tube B.
3. Add about 2 ml of dilute HCl to both the test tubes.
4. What do you observe?
5. Pass the gas produced in each case through lime water (calcium hydroxide solution) as shown below and record your observations.

Observations:
Test Tube A:
Reaction:
Na₂CO₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) +CO₂(g)

Test Tube B:
Reaction:
NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g)
On passing CO₂ gas through lime water it turns milky because insoluble white precipitate of CaCO₃ is formed as shown below:

On passing excess gas through lime water, it becomes colourless.

Class 10 Science Activity 2.6 Page No. 21

1. Take about 2 ml of dilute NaOH solution in a test tube and add two drops of phenolphthalein solution.
2. What is the colour of the solution?
3. Add dilute HCl solution to the above solution drop by drop.
4. Is there any colour change for the reaction mixture?
5. Why did the colour of phenolphthalein change after the addition of an acid?
6. Now add a few drops of NaOH to the above mixture.
7. Does the pink colour of phenolphthalein reappear?
8. Why do you think this has happened?

Observations:

1. Add 2 drops of phenolphthalein solution into NaOH solution → colour of the solution is pink.
2. Now add dilute HCl solution drop by drop.
3. Now, reaction mixture changes to colourless.
4. This is because neutralisation of HCl and NaOH takes place.
5. Now, add NaOH (few drops) → Pink colour reappears.

Class 10 Science Activity 2.7 Page No. 21

1. Take a small amount of copper oxide in a beaker and add dilute hydrochloric acid slowly while stirring.
2. Note the colour of the solution. What has happened to the copper oxide?

Observations:

1. Add dilute HCl to copper oxide solution:
2. Colour of the solution turns into green and CuCl₂ dissolves.
3. The blue-green colour formed due to formation of copper (II) chloride,
   CuO + 2HCl → CuCl + H₂O

Class 10 Science Activity 2.8 Page No. 22

1. Take solutions of glucose, alcohol, hydrochloric acid, sulphuric acid etc.
2. Fix two nails on a cork and place the cork in a 100 ml beaker.
3. Connect the nails to the two terminals of a 6-volt battery through a bulb and a switch, as shown in Figure.

1. Now pour some dilute HCl in the beaker and switch on the current. Repeat with dilute sulphuric acid.
2. What do you observe?
3. Repeat the experiment separately with glucose and alcohol solutions.
4. What do you observe now?
5. Does the bulb glow in all cases?

Class 10 Science Activity 2.9 Page No. 23

1. Take about 1g solid NaCl in a clean and dry test tube and set up the apparatus as shown in Figure.
2. Add some concentrated sulphuric acid to the test tube.
3. What do you observe? Is there a gas coming out of the delivery tube?
4. Test the gas evolved successively with dry and wet blue litmus paper.
5. In which case does the litmus paper change colour?
6. On the basis of the above Activity, what do you infer about the acidic character of:
   • Dry HCl gas?
   • HCl solution?

Note to teachers:
If the climate is very humid, you will have to pass the gas produced through a guard tube (drying tube) containing calcium chloride to dry the gas.

Observations:

1. Adding concentrated H₂SO₄ to test tube containing NaCl leads to the production of HCl gas. Now, testing this gas with litmus paper following were recorded,
2. HCl gas passes → Colour change
3. Dry litmus paper → No change
4. Wet litmus paper → Blue litmus turns red

Thus, only HCl solution release H⁺ ions and acidic property exist due to H⁺ ions.

Class 10 Science Activity 2.10 Page No. 24

1. Take 10 ml water in a beaker.
2. Add a few drops of concentrated H₂SO₄ to it and swirl the beaker slowly.
3. Touch the base of the beaker.
4. Is there a change in temperature?
5. Is this an exothermic or endothermic process?
6. Repeat the above Activity with sodium hydroxide pellets and record your observations.

Observations:

1. Add a few drops of concentrated H₂SO₄ to the water in a beaker → it becomes hot as the reaction is highly exothermic.
2. Now, add NaOH pellets to water → the beaker becomes hot, the reaction is exothermic.

Class 10 Science Activity 2.11 Page No. 26

1. Test the pH values of solutions given in the table.
2. Record your observations.
3. What is the nature of each substance on the basis of your observations?

Class 10 Science Activity 2.12 Page No. 27

1. Put about 2 g soil in a test tube and add 5 mL water to it.
2. Shake the contents of the test tube.
3. Filter the contents and collect the filtrate in a test tube.
4. Check the pH of this filtrate with the help of universal indicator paper.
5. What can you conclude about the ideal soil pH for the growth of plants in your region?

Observations:

Class 10 Science Activity 2.13 Page No. 28,29

1. Write the chemical formulae of the salts given below.
   Potassium sulphate, sodium sulphate, calcium sulphate. magnesium sulphate, copper sulphate, sodium chloride, sodium nitrate, sodium carbonate and ammonium chloride.
2. Identify the acids and bases from which the above salts may be obtained.
3. Salts having the same positive or negative radicals are said to belong to a family. For example. NaCl and Na₂SO₄ belong to the family of sodium salts. Similarly, NaCl and KCl belong to the family of chloride salts. How many families can you identify among the salts given in this Activity?

Class 10 Science Activity 2.14 Page No. 28

1. Collect the following salt samples – sodium chloride, potassium nitrate, aluminium chloride, zinc sulphate, copper sulphate, sodium acetate, sodium carbonate and sodium hydrogen carbonate (some other salts available can also be taken).
2. Check their solubility in water (use distilled water only).
3. Check the action of these solutions on litmus and find the pH using a pH paper.
4. Which of the salts are acidic, basic or neutral?
5. Identify the acid or base used to form the salt.
6. Report your observations in the table.

Class 10 Science Activity 2.15 Page No. 32

1. Heat a few crystals of copper sulphate in a dry boiling tube.
2. What is the colour of the copper sulphate after heating?
3. Do you notice water droplets in the boiling tube? Where have these come from?
4. Add 2-3 drops of water on the sample of copper sulphate obtained after heating
5. What do you observe? Is the blue colour of copper sulphate restored?

Observations:

1. On heating blue crystals of copper sulphate, it becomes colourless or white and few drops of water are seen on test tube due to condensation of water of crystallisation.
2. On adding few drops of water to heated anhydrous copper sulphate, the blue colour of copper sulphate reappear.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 1 Real Numbers Ex 1.1 PDF, MP Board Class 10 Maths Solutions, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1

Class 10 Maths Chapter 1 MP Board Question 1.
Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) HCF of 135 and 225
Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get
225 = (135 × 1) + 90, since 90 ≠ 0, therefore, applying the Euclid’s lemma to 135 and
90, we get 135 = (90 × 1) + 45
But 45 ≠ 0
∴ Applying Euclid’s lemma to 90 and 45, we get 90 = (45 × 2) + 0
Here, r = 0, so our procedure stops. Since, the divisor at the last step is 45,
∴ HCF of 225 and 135 is 45.

(ii) HCF of 196 and 38220
We start dividing the larger number 38220 by 196, we get
38220 = (196 × 195) + 0
Here, r = 0
∴ HCF of 38220 and 196 is 196.

(iii) HCF of 867 and 255 Here, 867 > 255
∴ Applying Euclid’s Lemma to 867 and 255, we get
867 = (255 × 3) + 102, 102 ≠ 0
∴ Applying Euclid’s Lemma to 255 and 102, we get
255 = (102 × 2) + 51, 51 ≠ 0
∴ Applying Euclid’s Lemma to 102 and 51, we get
102 = (51 × 2) + 0, r = 0
∴ HCF of 867 and 255 is 51.

MP Board Class 10 Maths Chapter 1 Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let us consider a positive odd integer as ‘a’.
On dividing ‘a’ by 6, let q be the quotient and ‘r’ be the remainder.
∴ Using Euclid’s lemma, we get a = 6q + r
where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4 or 5 i.e.,
a = 6q + 0 = 6q or a = 6q + 1
or a = 6q + 2 or a = 6q + 3
or a = 6q + 4 or a = 6q + 5
But, a = 6q, a = 6q + 2, a = 6q + 4 are even values of ‘a’.
[∵ 6q = 2(3q) = 2m₁ 6q + 2 = 2(3q + 1) = 2m₂,
6q + 4 = 2(3 q + 2) = 2m₃]
But ‘a’ being an odd integer, we have :
a = 6q + 1, or a = 6q + 3, or a = 6q + 5

Class 10 Maths Chapter 1 MP Board In English Medium Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
Total number of members = 616
∴ The total number of members are to march behind an army band of 32 members is HCF of 616 and 32.
i. e., HCF of 616 and 32 is equal to the maximum number of columns such that the two groups can march in the same number of columns.
∴ Applying Euclid’s lemma to 616 and 32, we get
616 = (32 × 19) + 8, since, 8 ≠ 0
Again, applying Euclid’s lemma to 32 and 8, we get
32 = (8 × 4) + 0, r = 0
∴ HCF of 616 and 32 is 8
Hence, the required number of maximum columns = 8.

MP Board Solutions Class 10 Maths Chapter 1 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3g, 3q + 1 or 3g + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]
Solution:
Let us consider an arbitrary positive integer as ‘x’ such that it is of the form
3q, (3q + 1) or (3q + 2)
For x = 3q, we have x² = (3q)²
⇒ x² = 9q² = 3(3q²) = 3m ………. (1)
Putting 3q² = m, where m is an integer.
For x = 3q + 1,
x² = (3q + 1)² = 9q² + 6q + 1
= 3(3q² + 2q) + 1 = 3m + 1 ………… (2)
Putting 3q² + 2q = m, where m is an integer.
For x = 3q + 2,
x² = (3q + 2)²
= 9q² + 12q + 4 = (9q² + 12q + 3) + 1
= 3(3q² + 4q + 1) + 1 = 3m + 1 ……….. (3)
Putting 3q² + 4q +1 = m, where m is an integer.
From (1), (2) and (3),
x² = 3m or 3m + 1
Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Class 10 Maths Chapter 1 Exercise 1.1 Solutions MP Board Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q +1) or (3q + 2).
For x = 3q
x³ = (3q)³ = 27q³ = 9(3q³) = 9m ……… (1)
Putting 3q³ = m, where m is an integer.
For x = 3q + 1
x³ = (3 q + 1)³ = 27q³ + 27q² + 9q + 1
= 9(3q³ + 3q² + q) + 1 = 9m + 1 ………… (2)
Putting 3q³ + 3q² + q = m, where m is an integer.
For x = 3q + 2,
x³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8
= 9(3q³ + 6q² + 4q) + 8 = 9m + 8 ……………. (3)
Putting 3q³ + 6q² + 4q = m, where m is an integer.
From (1), (2) and (3), we have
x³ = 9m, (9m + 1) or (9m + 8)
Thus, cube of any positive integer can be in the form 9m, (9m + 1) or (9m + 8) for some integer m.

We hope the given MP Board Class 10th Maths Solutions Chapter 1 Real Numbers Ex 1.1 will help you. If you have any queries regarding NCERT Madhya Pradesh Syllabus MP Board Class 10 Maths Solutions Chapter 1 Real Numbers Ex 1.1, drop a comment below and we will get back to you at the earliest.

MP Board Class 10th Sanskrit Solutions Durva Chapter 4 सुभाषितानि (पद्यम्) (सङ्कलितम्)

MP Board Class 10th Sanskrit Chapter 4 पाठ्यपुस्तक के प्रश्न

Students can also download MP Board 10th Model Papers to help you to revise the complete Syllabus and score more marks in your examinations.

Sanskrit Shlok Class 10 MP Board प्रश्न 1.
एकान उत्तरं लिखत-(एक पद में उत्तर लिखिए)।
(क) गङ्गा किं हन्ति? (गङ्गा क्या नष्ट करती है?)
उत्तर:
पापम् (पाप को)

(ख) केतकीगन्धम् आघ्राय के स्वयम् आयान्ति? (केवड़े की गन्ध को सूंघकर कौन स्वयं आ जाते हैं?)
उत्तर:
षट्पदाः (भौरे)

(ग) सर्वोत्तमं भूषणं किम् अस्ति? (सबसे उत्तम आभूषण क्या है?)
उत्तर:
वाग्भूषणम् (वाणी)

(घ) प्रीतिरसायनं किम् अस्ति? (प्रेमरूपी रस का आश्रय क्या है?)
उत्तर:
मित्रम् (मित्र)

(ङ) सर्वस्य के परीक्ष्यन्ते? (सबकी क्या परीक्षा की जाती है?)
उत्तर:
स्वभावः (स्वभाव)

You can download MP Board 10th sanskrit solution to help you to revise complete syllabus and score more marks in your examinations.

MP Board Class 10 Sanskrit Chapter 4 प्रश्न 2.
एकवाक्येन उत्तरं लिखत-(एक वाक्य में उत्तर लिखिए-)
(क) केषां कृते महोदधिः अपारो नास्ति? किसके लिए महान् समुद्र अगम्य नहीं है?)
उत्तर:
व्यवसायद्वितीयानां कृते पहोदधिः अपारो नास्ति। (उद्यमशील लोगों के लिए महान् समुद्र अगम्य नहीं है।)

(ख) अर्थेः समायुक्तोऽपि कः परिभवपदं याति? (धन से युक्त होते हुए भी कौन पराजयता को प्राप्त होता है?)
उत्तर:
अर्थैः समायुक्तोऽपि कृपणः परिभवपदं याति। (धन से युक्त होते हुए भी कंजूस व्यक्ति पराजय को प्राप्त होते हैं।)

(ग) सन्तः कानि घ्नन्ति? (सज्जन क्या नष्ट करते हैं?)
उत्तर:
सन्तः पापं, तापं दैन्यं च ध्नन्ति। (सज्जन लोग पाप, ताप और दीनता का नाश करते हैं।)

(घ) कान् अतीत्य कः मूर्ध्नि वर्तते? (किसको छोड़कर क्या सर्वोच्च है?)
उत्तर:
सर्वान् गुणान् अतीत्य स्वभावः मूर्ध्नि वर्तते। (सब गुणों को छोड़कर स्वभाव सर्वोच्च है।)

(ङ)
कीदृशो वह्रि स्वयम् उपशाम्योते? (कैसी अग्नि स्वयं शान्त हो जाती है?)
उत्तर:
अतृणे पतितः वह्निः स्वयम् उपशाम्यति। (तिनके से रहित गिरी हुई अग्नि स्वयं शान्त हो जाती है।)

MP Board Class 10th Sanskrit Chapter 4 प्रश्न 3.
अधोलिखितप्रश्नानाम् उत्तराणि लिखत-(नीचे लिखे प्रश्नों के उत्तर लिखिए)
(क) दानेन कानि-कानि भवन्ति? (दान से क्या-क्या होता है?)
उत्तर:
दानेन भूतानि वशीभवन्ति, वैराणि नाशं यान्ति, परः अपि बन्धुत्वम् उपैति, सर्वव्यसनानि च हन्ति। (दान से प्राणी वश में होते हैं, शत्रुभाव नष्ट हो जाता है, पराये भी अपने हो जाते हैं और सब बुरी आदतों का नाश होता है।)

(ख) पुरुष के न विभूषयन्ति? (पुरुष को क्या शोभा नहीं देते?)
उत्तर:
पुरुषः केयूराः न, चन्द्रोज्ज्वलाः हाराः न, स्नानं न, विलेपन न, कुसुमं न, अलङ्कताः मूर्धजा न विभूषयन्ति।

(पुरुष न बाजूबन्द, न चन्द्रमा जैसे उचल हार, न स्नान, न चन्दन आदि के लेप, न फूल और न सुसज्जित केशां से सुशोभित होता है।)

(ग) बुधजनसकाशात् मे किमभवत्? (विद्वानो की सङ्गति से मेरा क्या हुआ?)
उत्तर:
बुधजनसकाशात् मे मदः ज्वर इव व्यपगतः।
(विद्वानों की सङ्गति में मेरा अभिमान ज्वर के समान दूर हो गया।)

Subhasitani Sloka Class 10 प्रश्न 4.
रिक्तस्थानानि पूरयत
(दिए हुए शब्दों से रिक्त स्थान भरिए-.)
(क) दानेन …………… वशी भवन्ति। (भूतानि/प्रेताः)
(ख) द्युतिं सैंही …………… किं धृतकनकमालोऽपि लभते। (श्वा/अश्वाः )
(ग) केतकीगन्धमानाय स्वयमायान्ति ……………। (आपदाः/षट्पदाः)
(घ) क्षमाशस्त्रं करे यस्य …………… किं करिष्यति। (सज्जनः/दुर्जनः)
(ङ) अतीत्य हि गुणान्सर्वान् …………… मूर्ध्नि वर्तते। (स्वभावो/दुर्भावो)
उत्तर:
(क) भूतानि
(ख) श्वा
(ग) पट्पदाः
(घ) दुर्जनः
(ङ) स्वभावो

Class 10th Sanskrit Shlok प्रश्न 5.
यथायोग्यं योजयत-(उचित क्रम से मिलाइए-)

उत्तर:
(क) 2
(ख) 1
(ग) 4
(घ) 5
(ङ) 3

Sanskrit Shlok Class 10th प्रश्न 6.
शुद्धवाक्यानां समक्षम् ‘आम्’ अशुद्धवाक्यानां समक्षं ‘न’ इति लिखत- (शुद्ध वाक्यों के सामने ‘आम’ तथा अशुद्ध वाक्यों के सामने ‘न’ लिखिए-)
(क) अलङ्कताः मूर्धजाः पुरुषं विभूषयन्ति।
(ख) संस्कृता वाणी पुरुष समलङ्करोति।
(ग) तृणे पतितो वह्निः स्वयमेवोपशाम्यति।
(घ) दानेन परोऽपि बन्धुत्वमुपैति।
(ङ) धृतकनकमालो श्वा सैंहीं द्युतिं लभते।
उत्तर:
(क) न
(ख) आम्
(ग) न
(घ) आम्
(ङ) न।

10th Class Sanskrit Chapter 4 Question Answer प्रश्न 7.
अधोलिखितक्रियापदानां लकारं पुरुषं वचनं च लिखत-(नीचे लिखे क्रियापदों के लकार, पुरुष और वचन लिखिए।)

उत्तर:

Subhashitani Class 10 प्रश्न 8.
उदाहरणानुसारं पर्यायशब्दं लिखत-(उदाहरण के अनुसार पर्यायवाची शब्द लिखिए)
यथा- गङ्गा – देवनदी
(क) सन्तः
(ख) शिखरः
(ग) वीरः
(घ) षट्पदाः
(ङ) नयनयोः
उत्तर:
(क) सन्तः – सज्जनाः
(ख) शिखरः – तुंगः
(ग) वीरः – शूरः
(घ) षट्पदाः – भ्रमराः
(ङ) नयनयोः – नेत्रयोः

Class 10th Sanskrit Chapter 4 प्रश्न 9.
रेखात्तिपदान्याधृत्य प्रश्ननिर्माणं कुरुत-(खांकित्त पदों के आधार पर प्रश्न बनाइए।)
(क) ते सर्वत्र मिलन्ति। (वे सब जगह मिलते हैं।)
उत्तर:
के सर्वत्र मिलन्ति? (कौन सब जगह मिलते हैं?)

(ख) दानेन वैराण्यपि नाशं यान्ति। (दान से शत्रुभाव नष्ट होता है।
उत्तर”
केन वैराण्यपि नाशं यान्ति? (किससे शत्रुभाव नष्ट होता है?)

(ग) कृपणः परिभवपदं याति। (कंजूस पराजयता को प्राप्त होता है।)
उत्तर:
कः परिभवपदं याति? (कौन पराजयता, को प्राप्त होता है ?)

(घ) मे मदः ज्वर इव व्यपगतः। (मेरा घमण्ड ज्वर के समान दूर हो गया।)
उत्तर:
कस्य मदः ज्वर इव व्यपगतः? (किसका घमण्ड ज्वर के समान दूर हो गया?)

(ङ) केयूराः पुरुषं न विभूषयन्ति। (बाजूबन्द पुरुष को शोभा नहीं देते।)
उत्तर:
केयूराः कं न विभूषयन्ति? (बाजूबन्द किसको शोभा नहीं देते?)

Subhashitani Class 10th प्रश्न 10.
श्लोकपूर्ति कुरुत। (श्लोक पूरा कीजिए।)
उत्तर:
(क) गङ्गा पापं शशी तापं दैन्यं कल्पतरुस्तथा।
पापं तापं च दैन्यं च घ्नन्ति सन्तोमहाशयाः॥

(ख) नात्युच्चशिखरो मेरु नातिनीचं रसातलम्।
व्यवसायद्वितीयानां नाप्यपारो महोदधिः॥

(ग) गुणाः कुर्वन्ति दूतत्वं दूरेऽपि वसतां सताम् ।
केतकी गन्धमाघ्राय स्वयमायान्ति षट्पदाः॥

(घ) क्षमाशस्त्रं करे यस्य दुर्जनः किं करिष्यति।
अतृणे पतितो वह्निः स्वयमेवोपशाम्यति॥

(ङ) सर्वस्य हि परीक्ष्यन्ते स्वभावाः नेतरेः गुणाः।
अतीत्य हि गुणान्सर्वान् स्वभावो मूर्धिन वर्तते॥

योग्यताविस्तार –

पाठे समागतान् श्लोकान् कण्ठस्थं कुरुत।
(पाठ में आए हुए लोकों को कण्ठस्थ करो।)

पाठ्यपुस्तकेतराणि सुभाषितानि पठत।।
(पाठ्यपुस्तक से अलग सुभाषित श्लोक पढ़ो।)

पाठ्यपुस्तकेतरान् दशसुभाषितश्लोकान् लिखत।
(पाठ्यपुस्तक से अलग दस सुभाषित श्लोक लिखो।)

सुभाषितानि पाठ का सार

प्रस्तुत पाठ में ऐसे श्लोकों का संग्रह है, जिनसे मनुष्य को जीवन में अच्छे कार्य करने की प्रेरणा मिलती है। इनमें सज्जन व दान की, परिश्रमी व वीर पुरुषों की, गुण की, वाणी की, क्षमा की प्रशंसा, मित्र का स्वरूप, स्वभाव की महिमा तथा अल्पज्ञानी के स्वभाव की चर्चा की गई है।

सुभाषितानि पाठ का अनुवाद

1. गङ्गा पापं शशी तापं दैन्यं कल्पतरुस्तया।
पापं तापं च दैन्यं च ध्नन्ति सन्तोमहाशयाः॥1॥

अन्वय :
गङ्गा पापं, शशी तापं कल्पतरुः च दैन्यं (घ्नन्ति) तथा सन्तो महाशयाः पापं तापं दैन्य च (त्रीणि अपि) घ्नन्ति।

शब्दार्था:
शशी-चन्द्रमा-Moon; दैन्यम्-दीनता/गरीबी-humbleness, poverty, indigence, घ्नन्ति-नाश करते हैं-destroy, uproot.

अनुवाद :
गंगा पाप का, चन्द्रमा ताप (गर्मी) का और कल्पवृक्ष गरीबी का नाश करते हैं। वैसे ही महान (संत) लोग पाप, ताप व दीनता तीनों का ही नाश करते हैं।

English :
The saints uproot sin, heat and poverty. They have the qualities of the ganges, the moon and the kalpa tree.

2. दानेन भूतानि वशीभवन्ति दानेन वैराण्यपि यान्ति नाशम्।
परोऽपि बन्धुत्वमुपैति दानैर्दानं हि सर्वव्यसनानि हन्ति॥2॥

अन्वय :
दानेन भूतानि वशीभवन्ति, दानेन वैराणि अपि नाशं यान्ति, दानैः परः अपि बन्धुत्वम् उपैति, हि दानं सर्वव्यसनानि हन्ति ।।

शब्दार्था :
भूतानि-प्राण-animate creatures, वैराणि-शत्रुभाव-enmity; उपैति-बन जाता है-turns into, becomes.

अनुवाद :
दान के द्वारा सभी प्राणी वश में किए जाते हैं, दान के द्वारा शत्रुभाव का भी नाश किया जाता है, दान के द्वारा पराये भी अपने बन जाते हैं, क्योंकि दान सब बुरी आदतों को हर लेता है।

English :
Generosity controls human beings (creatures), removes enmity, befriends enemies and uproots vices.

3. नात्युच्चशिखरो मेरुनातिनीचं रसातलम् ।
व्यवसायद्वितीयानां नाप्यपारो महोदधिः॥3॥

अन्वय :
व्यवसायद्वितीयानां कृते मेरुः शिखरो अत्युच्चः न, रसातलम् अतिनीचं न, महोदधिः अपि अपारो न।

शब्दार्था :
व्यवसायद्वितीयानाम्-उद्यमशील लोगों के लिए-for the adventurous people; अत्युच्चः-अत्यन्त ऊँचा-lofty, very high, रसातलम्-पृथ्वी के नीचे का रसातल नामक छठा लोक-nether region, अपारः-अगम्य-inaccessible.

अनुवाद :
उद्यमशील (परिश्रमी) लोगों के लिए मेरू शिखर अत्यन्त ऊंचा नहीं है, रसातल (पृथ्वी के नीचे का लोक) बहुत नीचा नहीं है, बहुत बड़ा समुद्र भी अगम्य (न पार करने योग्य) नहीं है। अर्थात् परिश्रमी व्यक्तियों के लिए कोई भी कार्य असम्भव नहीं है।)

English :
Adventurous people have access to all the places under the earth, on the earth and above the earth.

4. विनाप्यर्थैर्वीरः स्पृशति बहुमानोन्नतिपदं,
समायुक्तोप्यर्थैः परिभवपदं याति कृपणः।
स्वभावादुद्भूतां गुणसमुदयावाप्तिविषयां,
युतिं सैंही श्वा किं धृतकनकमालोऽपि लभते॥4॥

अन्वय P\वीरः अर्थेः विना अपि बहुमानोन्नतिपदं स्पृशति, कृपणः अर्थः समायुक्तः अपि परिभवपदं याति, गुणसमुदयावाप्तिविषयां स्वभावाद् उद्भूतां सैंही द्युतिं धृतकनकमालो श्वा अपि किं लभते?

शब्दार्था :
अथैः विना-धन से रहित-devoidof riches (money), बहुमानोन्नतिपदम्अत्यन्त सम्मान एवं उन्नति के स्थान.को-Position of respect and progress, परिभवपदम्-पराजयता को-defeat, गुणसमुदयावाप्तिविषयाम्-गुणों के समुदाय को प्राप्त कराने वाली को-whichcauses theattainmentofheapofvirtues, सैंही द्युतिम्-सिंह की कान्ति को-The grace of the lion, धृतकनकमालः-स्वर्णमाला को धारण करने वाला-Wearing golden necklace, श्वा-कुत्ता-a dog.

अनुवाद :
वीर पुरुष धन के बिना भी अत्यधिक सम्मान एवं उन्नति के स्थान को छू लेता है, कंजूस व्यक्ति, धन से युक्त होते भी पराजयंता को प्राप्त होता है। स्वभाव से गुणों के समुदाय को प्राप्त कराने वाली सिंह की कान्ति को स्वर्णमाला को धारण करने वाला कुत्ता भी क्या प्राप्त कर सकता है?

English :
Money is no criterion for greatness. A dog with ornaments can not match a lion in grace.

5. गुणाः कुर्वन्ति दूतत्वं दूरेऽपि वसतां सताम्।
केतकीगन्धमाघ्राय स्वयमायान्ति षट्पदाः॥5॥

अन्वय :
दूरे अपि वसतां सतां गुणाः दूतत्वं कुर्वन्ति, षट्पदाः केतकीगन्धम् आघ्राय स्वयम् आयान्ति। – शब्दार्थाः-दूतत्वम्-दूत के कार्य-duty of messenger, सताम्-सज्जनों के-of the noble persons, षट्पदाः-भौरे-Black bee केतकीगन्धम्-केवड़े के गन्ध को-smell of ‘ketaki’, आघ्राय-सूंघकर-smelling.

अनुवाद :
दूर रहते हुए भी सज्जनों के गुण दूत के कार्य करते हैं, भौरे केवड़े की गन्ध को सूंघकर स्वयं ही आ जाते हैं।

English :
Virtues reveal themselves from a distance. The smell of flowers attracts black bees.

6. केयूरा न विभूषयन्ति पुरुषं हारा न चन्द्रोज्वलाः.
न स्नानं न विलेपनं न कुसुमं नालङ्कता मूर्धजाः।
वाण्येका समलङ्करोति पुरुष या संस्कृता धार्यते,
क्षीयन्ते खलु भूषणानि सततं वाग्भूषणं भूषणम्।।6।।

अन्वय :
पुरुषं केयूराः न, चन्द्रोज्ज्वलाः हाराः न, स्नानं न, विलेपनं न, कुसुमं न, अलङ्कताः मूर्धजाः न विभूषयन्ति, एका वाणी (एव) पुरुषं समलङ्करोति या संस्कृता धार्यते, भूषणानि खलु सततं क्षीयन्ते, वाग्भूषणं भूषणम् (अस्ति)।

शब्दार्था:
विलेपनम्-चन्दनादि सुगन्धित पदार्थों के लेप-bemearing with fragrant pastes, मूर्धजाः-केश-hair, समलङ्करोति-सुशोभित करती है-adorn, संस्कृतासंस्कारमयी-cultured, polished, क्षीयन्ते-नष्ट हो जाते हैं-decay.

अनुवाद :
मनुष्य न बाजूबन्द से, न चन्द्रमा के समान उज्ज्वल हार से, न स्नान से, न चन्दनादि सुगन्धित पदार्थों के लेप से, न फूलों से, न सुसज्जित केश से सुशोभित होता है। एक वाणी ही मनुष्य को सुशोभित करती है, जो संस्कारित रूप से धारण की जाती है। आभूषण तो निरन्तर नष्ट हो जाते हैं, वाणी रूपी आभूषण ही सच्चा आभूषण है।

English :
Ornaments and pastes fail to adorn a person. Cultured speech beautifies one’s personality. Speech is the real ornament which never loses its lustre (fades).

7. क्षमाशस्त्रं करे यस्य दुर्जनः कि करिष्यति।
अतणे पतितोः वद्धिः स्वयमेवोपशाम्यति॥7॥

अन्वय :
यस्य करे क्षमाशस्त्रं (विद्यते), (तस्य) दुर्जनः किं करिष्यति? (यथा) अतृणे पतितः वह्निः स्वयम् एव उपशाम्यति।

शब्दार्था :
अतृणे-तृण (घास) से रहित-withoutstraw (grass), उपशाम्यति-उपशमित हो जाता है-extinguishes..

अनुवाद :
जिसके हाथ में क्षमा रूपी शस्त्र होता है, उसका दुष्ट व्यक्ति क्या कर सकता है? जैसे तिनके से रहित गिरी हुई अग्नि स्वयं ही शान्त हो जाती है।

English :
An enemy cannot harm a forgiving person. The fire lying on bare ground gets extinguished.

8. मित्रं प्रीतिरसायनं नयनयोरानन्दनं चेतसः
पात्रं यत्सुखदुःखयोः सह भवेन्मित्रेण तद् दुर्लभम्।
ये चान्ये सुहृदः समृद्धिसमये द्रव्याभिलाषाकुला
स्ते सर्वत्र मिलन्ति तत्त्वनिकषग्रावा तु तेषां विपत्॥8॥

अन्वय :
मित्रं प्रीतिरसायनं, नयनयोः चेतसः (च) आनन्दनं, यत् मित्रेण सह सुखदुःखयोः पात्रं भवति तदुर्लभम्, ये च अन्ये सुहृदः समृद्धिसमये द्रव्याभिलाषाकुलाः (भवन्ति) ते सर्वत्र मिलन्ति, तेषां (कृते) विपत् तु तत्त्वनिकषग्रावा (इव भवति)।

शब्दार्था :
प्रीतिरसायनम्-प्रेमरूपी रस का आश्रय-uf the heart, चेतसः-चिन्न का-of the heart, आनन्दनम्-आनन्दित करने वाला-which delights, अन्ये-दूसरे-others; द्रव्याभिलाषाकुलाः-धन प्राप्ति की कामना करने वाले-those who crave for attainment of wealth; तत्त्वनिकषग्रावा-तत्त्व रूपी कसौटी का पत्थर-touchstone.

अनुवाद :
प्रेमरूपी रस का आश्रय मित्र नेत्रों व मन को आनन्दित करने वाला होता है। जो मित्र के साथ सुख-दुख का भागी होता है, वह (मित्र) मुश्किल से मिलता है और जो, दूसरे, मित्र की खुशहाली के समय पर धन प्राप्ति की कामना करने वाले होते हैं, वे सब जगह मिलते हैं, उनके लिए विपत्ति तो तत्त्व रूपी कसौटी का पत्थर समान है।

English :
A loving friend delights one’s eyes and heart. Shares weal and woe of a friend. He is rare. A false friend sticks during prosperity but leaves in adversity.

9. सर्वस्य हि परीक्ष्यन्ते स्वभावा नेतरे गुणाः।
अतीत्य हि गुणान्सर्वान्स्वभावो मूर्ध्नि वर्तते॥9॥

अन्वय:
सर्वस्य स्वभाषाः हि परीक्ष्यन्ते, इतरे गुणाः न, हि सर्वान् गुणान् अतीत्य स्वभावो मनि वर्तते।

शब्दार्था :
परीक्ष्यन्ते-परीक्षा की जाती है-is tested, अतीत्य-छोड़कर-surpassing, leaving behind, मूर्जि-मस्तक पर (सर्वोच्च)-on the head, above all.

अनुवाद :
सबके स्वभाव की ही परीक्षा की जाती है, दूसरे गुणों की नहीं, क्योंकि सब गुणों को छोड़कर स्वभाव सर्वोच्च है।

English :
Nature is above other virtues. It is tested everywhere leaving aside other qualities.

10. यदा किञ्चिज्ज्ञोऽहं गज इय मदान्धः समभवं,
तदा सर्वज्ञोऽस्मीत्यभवदवलिप्तं मम मनः।
यदा किञ्चित् किञ्चिद् बुधजनसकाशादवगतं
तदा मूखोऽस्मीति ज्वर इव मदो मे व्यपगतः॥10॥

अन्वय :
यदा अहं किञ्चिद् ज्ञः (आसम्) तदा (अह) गज इव मदान्धः समभवम्, सर्वज्ञः अस्मि इति मम मनः अवलिप्तम् अभवत्। यदा किञ्चित् किञ्चिद् बुधजनसकाशात् अवगतं तदा मूर्खः अस्मि इति मे मदः ज्वर इव व्यपगतः।

शब्दार्था :
किञ्चिज्ज्ञः-थोड़ा जानने वाला-having lesser knowledge, मदान्धः-घमण्ड में चूर-vain, full of pride, अवलिप्तम्-अभिमानी-proud, बुधलनराकाशात्-विद्वानों की सङ्गति से-Incompanyof the wise,अवगतम्-जाना-knew, व्यपगतः-दूर हो गया-removed, abated.

अनुवाद :
जब मैं थोड़ा जानने वाला होता, तब मैं हाथी के समान घमण्ड में चूर हो जाता, मैं सब जानता हूँ, ऐसा मेरा मन अभिमानी हुआ। जब थोड़ा-थोड़ा विद्वानों की सङ्गति से जाना, तब ‘मैं मूर्ख हूँ’ ऐसा मेरा अभिमान (घमण्ड) बुखार के समान दूर हो गया।

English :
Aman with lesser knowledge is proud like an elephantthinks himself knowledgeable-Pride shattered when came in association with learned persons.

MP Board Class 10th Sanskrit Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.3 Pdf, Class 10 Subject Maths Chapter 3 Exercise 3.3, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Find the value of x in each of the given triangles.

MP Board Class 10th Maths Chapter 3 Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14; x – y = 4
(ii) s – f = 3; \(\frac{s}{3}+\frac{t}{2}\) = 6
(iii) 3x – y = 3; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
(v) \(\sqrt{2} x+\sqrt{3} y\) = 0; \(\sqrt{3} x-\sqrt{8} y\) = 0
(vi) \(\frac{3 x}{2}-\frac{5 y}{3}\) = -2, \(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
(i) x + y = 14 … (1),
x – y = 4 …. (2)
From (1) , we get x = (14 – y) …. (3)
Substituting value of x in (2) , we get
(14 – y) – y = 4 ⇒ 14 – 2y = 4 ⇒ -2y = -10 ⇒ y = 5
Substituting y = 5 in (3), we have
x = 14 – 5 ⇒ x = 9
Hence, x = 9, y = 5

(ii) s – t = 3 … (1)

From (1), we have s = (3 + t) … (2)
Substituting this value of s in (2), we get

Substituting, t = 6 in (3) we get,
S = 3 + 6 = 9
Thus, S = 9, f = 6

(iii) 3x – y = 3 … (1),
9x – 3y = 9 … (2)
From (1) , y = (3x – 3)
Substituting this value of y in (2),
9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 9 ⇒ 9 = 9 which is true,
Eq. (1) and eq. (2) have infinitely many solutions.

(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 …. (2)
From the equation (1),

Substituting the value of y in (2), we have


Substituting the value of x in (1), we have

Solve using Substitution Calculator is a free online tool that displays the solution of the pair of linear equations using the substitution method.

MP Board Solutions Class 10 Maths Chapter 3 Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + c
Solution:

Enter the equation A and B in the substitution calculator for solving the linear equations.

Class 10 Maths Chapter 3 MP Board Question 3.
Form the pair of linear equations for the following problems and find their solution by
substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for 13800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) Let the two numbers be X and y such that x > y
It is given that
Difference between two numbers = 26
∴ x – y = 26 … (1)
Also one number = 3 [the other number]
⇒ x = 3y … (2)
Substituting x = 3y in (1) , we get 3y – y = 26 ⇒ 2y = 26
Now, substituting y = 13 in (2) , we have
x = 3(13) ⇒ x = 39
Thus, two numbers are 39 and 13.

(ii) Let the two angles be x and y such that x > y
∵ The larger angle exceeds the smaller by 18° (Given)
∴ x = y + 18°…. (1)
Also, sum of two supplementary angles = 180°
∴ x + y = 180° … (2)
Substituting the value of x from (1) in (2) , we get,
(18° + y) + y = 180°

Substituting, y = 81° in (1) , we get
x = 18° + 81° = 99°
Thus, x = 99° and y = 81°

(iii) Let the cost of a bat = ₹ x
And the cost of a ball = ₹ y
∵ [cost of 7 bats] + [cost of 6 balls] = ₹ 3800
⇒ 7x + 6y = 3800 … (1)
Also, [cost of 3 bats] + [cost of 5 balls] = ₹ 1750
3x + 5y = 1750 …. (2)

Substituting this value of y in (1) , we have

(iv) Let fixed charges = ₹ x
and charges per km = ₹ y
∵ Charges for the journey of 10 km = ₹ 105 (Given)
∴ x + 10y = 105 … (1)
and charges for the journey of 15 km = ₹ 155
∴ x + 15y = 155 … (2)
From (1) , we have, x = 105 – 10y …. (3)
Putting the value of x in (2) , we get
(105 – 10y) + 15y = 155
⇒ 5y = 155 – 105 = 50 ⇒ y = 10
Substituting y = 10 in (3) , we get
x = 105 – 10(10) ⇒ x = 105 – 100 = 5
Thus, x = 5 and y = 10
⇒ Fixed charges = ₹ 5
and charges per km = ₹ 10
Now, charges for 25 km = x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255
∴ The charges for 25 km journey = ₹ 255

(v) Let the numerator = x
and the denominator = y
∴ Fraction = \(\frac{x}{y}\)
According to the given condition,

⇒ 11(x + 2) = 9(y + 2)
⇒ 11x + 22 = 9y + 18
⇒ 11x – 9y + 4 = 0 … (1)

Substituting this value of x in (1) , we have

(vi) Let the present age of Jacob = x years
and the present age of his son = y years
∴ 5 years hence: Age of Jacob = (x + 5) years
Age of his son = (y + 5) years
According to given condition,
[Age of Jacob] = 3[Age of his son]
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15
⇒ x – 3y -10 = 0 … (1)
5years ago : Age of Jacob = (x – 5) years,
Age of his son = (y – 5) years
According to given condition,
[Age of Jacob] = 7[Age of his son]
∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35
⇒ x – 7y + 30 = 0 … (2)
From (1) , x = [10 + 3y] … (3)
Substituting this value of x in (2) , we get
(10 + 3y) – 7y + 30 = 0
⇒ -4y = -40 ⇒ y = 10
Now, substituting y = 10 in (3) ,
we get x = 10 + 3(10)
⇒ x = 10 + 30 = 40
Thus, x = 40 and y = 10
⇒ Present age of Jacob = 40 years and present age of his son = 10 years