NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths in PDF & Video form, हिंदी मीडियम as well as English Medium session 2020-21 for CBSE, UP, Gujrat Board, MP Board, Bihar, Uttarakhand board and all other boards following new CBSE Curriculum 2020-2021.

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NCERT Solutions for Class 10 Maths

Class:10
Subject:Maths – गणित
Medium:English – हिंदी

CBSE NCERT Solutions for Class 10 Maths

NCERT – CBSE Solutions for class 10 Maths all chapters are given below updated for new academic session 2020-21. You may download to use it offline or use online as it is. Download 10 Maths App in Hindi or 10 Maths in English for offline free.




Class 10 Maths Important Points related to all Chapters.

Chapter 1: Real Numbers

Real Numbers is one of the scoring chapters for Board exams. It follows the Euclid’s Division Algorithms and Lemma. The Fundamental Theorem of Arithmetic helps to split the composite numbers into primes. It is suitable for solving the questions of exercise 1.2 of chapter 1, class 10 Maths. It also helps to evaluate HCF or LCM of 2 or 3 numbers. Lemma helps split the composites into primes. First and third exercises contain the related sums of lemma. It requires step by step algorithm. One can check if the number is terminating or not.

Chapter 2: Polynomials

This chapter is related to Class 9 Maths Chapter 2 Number systems. Here, we will learn how to find zeros of a quadratic polynomial and its uses. Set relations between zeroes and coefficient of variables. We will also learn to derive a quadratic polynomial using the sum and product of zeroes. Long division method for a polynomial helps us in 10th Maths Exercise 2.4. The last exercise 2.4 is an optional exercise not for CBSE Board Exam March 2021.

Chapter 3: Pair of Linear Equations in Two Variables

Class 10 Maths Chapter 3 is similar to Class 9 Maths Chapter 4 Linear Equations in Two Variables. However, the pair of linear equations of class 10 help to find out their solutions. Use any of methods like Graphical method, elimination, substitution and cross multiplication to solve. Exercise 3.1 of Class 10 Maths tells us how to form an equation for the word problem, whereas Exercise 3.2 provides a graphical way for the sums. Similarly, exercises 3.3, 3.4 & 3.5 follow Elimination, Substitution and Cross Multiplication Method to get the result. Exercise 3.6 contains a variety of problems which are non-linear format. So, convert first into a linear format to solve them. Exercise 3.7 is an optional exercise, not for CBSE Board Exams.

Chapter 4: Quadratic Equations

An equation is whether quadratic or not a quadratic; this is the concept of Exercise 4.1 of Class 10 Maths. Exercise 4.2 includes the solving of quadratic equations by factor method. Just like Chapter 3, it also contains the word problems based on the quadratic forms in Exercise 4.3. Questions based on Speed, Distance and time are asked in CBSE and UP Board exams most of the time. Before the finding roots of a quadratic equation, we have to check if roots exist or not. Exercise 4.4 covers the fact that roots are real or imaginary.

Chapter 5: Arithmetic Progression

Class 10 Maths Chapter 5 AP explains the basics of sequences in increasing or decreasing order. Exercise 5.1 includes the questions, which are just for the formation of AP. The last Exercise 5.4 is optional, but its first question frequently comes in-class tests as well as in CBSE Board exams. Other exercises contain the questions to find the missing term of an AP or sum of the series up to a specified term. Word problems given in Exercise 5.3 are examples of daily life-based.

Chapter 6: Triangles

There are many theorems in chapter 6 of Class 10 Maths, but only a few theorems are there which comes in CBSE Board Exams. As per the CBSE Syllabus 2020-21, only four theorems named, Thales Theorem, Area Theorem, Pythagoras and Converse of Pythagoras may ask for proving in final exams. Sums given in Exercise 6.5 based on Pythagoras theorem need more practice for proving questions. Exercise 6.6 contains optional sums for the practice of similar triangles. We have explained all exercises with reasons to make it easy for all.

Chapter 7: Coordinate Geometry

Coordinate geometry has a vast use in the field of Maths and Science. Using this, one can calculate the distance between two points using the coordinates of the points. Class 10 Maths Exercise 7.1 provides the way to do so. Similarly, the next sections tell how to dividing a particular distance in the given ratio. Exercise 7.2 of Math illustrate such examples. We can find the area of triangles by measuring its sides. Coordinate geometry helps to find the area of a triangle without measuring its sides. Exercise 7.3 illustrates the same fact. Optional exercise 7.4 is also essential for boards.

Chapter 8: Introduction to Trigonometry

Class 10 Maths Chapter 8 is a little bit difficult as compared to others. This concept is new for class 10 students. Students should do exercise 8.1, sincerely understanding the concepts. Don’t try to learn the answers or methods. Try to grasp the derivation of each trigonometric value. Exercise 8.4 is a difficult one as compared to others. It consists of the questions of proving. Apply identities to solve it. Derive & learn trigonometric identities before doing the questions.

Chapter 9: Some Applications of Trigonometry

If one has gone through Chapter 8 entirely, it will be a more straightforward chapter for him. Applications of Trigonometry help in engineering works. Here, we have to apply trigonometry to find the height and distance between two objects. The last three questions are an important one for exams. This chapter is more comfortable to score good marks. Draw the layout of the questions first and then go for solving.

Chapter 10: Circles

Class 10 Maths Chapter 10 elaborates the multiple cases of tangents. Two important theorems, given in this chapter, may ask in boards. Exercise 10.2 is essential. It contains various examples based on tangents to a circle. Various questions are on the concepts of tangents. Using the concepts of tangents, schools ask a variety of questions in the tests. If one has done class 9 Maths Chapter 10 with flying colours, this chapter will be good for him.




Chapter 11: Constructions

Class 10 Maths Chapter 11 have two exercises. It demonstrates how to draw a similar triangle and tangents. The first exercise provides an idea about to draw a similar triangle to a given one. The second one illustrates the method of drawing tangents to a circle. The chapter helps to develop creative skills. Students get perfection in construction just by practice only. Try to draw neet and clean diagrams. Use only a ruler and compass to draw.

Chapter 12: Area Related to Circles

Area of segments and sectors are the basics of this chapter. Use always the direct formula to find the area of sectors. In the case of a segment, never use the same way. Subtract the area of the triangle from the sector to get the segment. Exercise 12.1 consists of the sum on circumference and area of a circle. The word segment refers to a minor segment of a circle.

Chapter 13: Surface Areas and Volumes

This chapter involves the questions based on the surface area of 3-D figures. The sums are a combination of two figures. Volumes are a bit easier than others. From the concepts of volume convert one shape to another. The frustum is the new concept for 10th standard. That is why the last exercise is exhausting. Do additional sums for more practice. Practice more to go through this chapter as it needs enough practice for perfection.

Chapter 14: Statistics

In real-life situations, statistics make a meaningful study of vital data. Class 10 Statistics help us to find mean, mode and median of grouped data. First three exercises demonstrate to find out mean, mode and median respectively. In the end, Exercise 14.4 describes how to plot a cumulative frequency curve. It is a good scoring chapter as an examination point of view. Students can score more in fewer efforts in chapter 14 of class 10 Maths.

Chapter 15: Probability

Probability is a topic to score good marks in CBSE Boards. In Class 10, Chapter 15, we learn about the equally likely trials. Here, we take only events. We will assume that all the experiments have Equally Likely results. Questions of dice and Cards are occurring in CBSE Board exams frequently. Students should use the theoretical or classical probability approach to get the answers. Exercise 15.2 is an optional one, not for the exams point of view.



10th Maths in Hindi & English Medium

All solutions are available in Hindi medium as well as English Medium format updated for new academic session 2020-21. Solutions are prepared by the expert and experience teachers based on latest NCERT Books 2020-2021. All the steps of solutions are mention properly so that student can understand easily. After completing NCERT Books students should do NCERT Exemplar books for better practice. Download 10 Maths Apps in Hindi for offline use or 10 Maths in English for offline free.

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Sample Papers and study material in Hindi medium for class 10 Maths will be uploaded very soon for academic year 2020-21. We are working on Hindi medium at large scale for UP Board (High School), Delhi Board as well as Bihar Board.



Important Questions on Class 10 Maths

3825 को अभाज्य गुणनखंडो के गुणनफल के रूप में व्यक्त कीजिए।
3825
= 3×3×5×5×17
What is meant by degree of a polynomial?
If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x).
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let the larger angle = x
Let the smaller angle = y
According to question,
x = y + 18 … (1)
Both angles are supplementary, therefore
x + y = 180 … (2)
Putting the value of x in equation (2), we get
y + 18 + y = 180
⇒ 2y = 162
⇒ y = 81
Putting the value of y in equation (1), we get
x = 81 + 18 = 99
Hence, one angle is 81° and the other one is 99°.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Let, Shefali’s marks in Mathematics = x
Therefore, Shefali’s marks in English = 30 – x
If she got 2 marks more in Mathematics and 3 marks less in English,
Marks in Mathematics = x + 2
Marks in English = 30 – x – 3
According to questions, Product
= (x + 2)(27 – x) = 210
⇒ 27x – x^2 + 54 – 2x = 210
⇒〖-x〗^2 + 25x – 156 = 0
⇒ x^2 – 25x + 156 = 0
⇒ x^2 – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x – 12) = 0
⇒ (x – 12)(x – 13) = 0
⇒ (x – 12) = 0 or (x – 13) = 0
Either x = 12 or x = 13
If x = 12
then, marks in Maths = 12 and marks in English = 30 – 12 = 18
If x=13
then, marks in Maths = 13 and marks in English = 30 – 13 = 17
Write first four terms of the AP, when the first term a and the common difference d are given as follows: a = 10, d = 10.
a = 10, d = 10
First term a_1 = a = 10
Second term a_2 = a_1 + d = 10 + 10 = 20
Third term a_3 = a_2 + d = 20 + 10 = 30
Fourth term a_4 = a_3 + d = 30 + 10 = 40
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Let OA is wall and AB is ladder.
In ∆AOB, by Pythagoras theorem
AB^2 = OA^2 + OB^2
⇒〖10〗^2 =8^2 + BO^2
⇒ 100 = 64 + BO^2
⇒ BO^2 = 36
⇒ BO = 6 m
Hence, the distance of the foot of the ladder from the base of the wall is 6 m.
In∆PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin⁡ (P), cos⁡〖P〗 and tan⁡〖P〗
Given that: in ∆PQR, angle Q is right angled.
Let QR = x, therefore, PR = 25 – x
In ∆PQR, by Pythagoras theorem, we have
〖PR〗^2= 〖PQ〗^2 + 〖OQ〗^2
⇒ (25 – x)^2 = (5)^2 + 〖(x)〗^2
⇒ 625 + x^2 – 50x = 25 + x^2
⇒ 625 – 50x = 25
⇒ 50x = 600
⇒ x = 12
⇒ QR = 12
Therefore,
PR = 25 – 12 = 13
Now,
sin⁡(P) = QR/PR = 12/13,
cos(P) = PQ/PR = 5/13
tan⁡(P) = QR/PQ = 12/5
How can we find the height of an object without measuring?
The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let AB is diameter, PQ and RS are tangents drawn at ends of diameter.
We know that the radius is perpendicular to tangent. Therefore, OA ⊥ RS and OB ⊥ PQ.
∠OAR = 90° and ∠OAS = 90°
∠OBP = 90° and ∠OBQ = 90°
From the above, we have
∠OAR = ∠OBQ [Alternate angles]
∠OAS = ∠OBP [Alternate angles]
Since, alternate angles are equal.
Hence, PQ is parallel to PS.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is (A) 2 units (B) π units (C) 4 units (D) 7 units
Let the radius of circle = r, Circumference of circle (perimeter) = 2πr, Area of circle = πr2
Given that the perimeter and area are equal in magnitude. Therefore
2πr = πr2
⇒ 2 = r
Therefore, the radius of circle is 2 units.
Hence, the option (A) is correct.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Width of canal = 6 m,
Depth of canal = 1.5 m
Speed of water = 10 km/h = 1000/60 m/min
Volume of water in 1 minute = 6 × 1.5 ×1000/60 = 1500 m^3
Therefore, the volume of water in 30 minutes
= 30 × 1500 = 45000 m^3
Let the area of irrigated field = A m^2
Therefore,
Volume of water from canal in 30 minutes = volume of water in the field
⇒ 45000 = (A × 8)/100
⇒ A = (45000 × 100)/8 = 562500 m^2
Hence, canal irrigate 562500 m^2 area in 30 minutes.
How is the mean an important tool in getting information about a data?
The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, i.e., the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions. For example, by comparing the average (mean) results of students of different schools of a particular examination, we can conclude which school has a better performance.
यदि P(E)=0.05 है, तो ‘E नहीं’ की प्रायिकता क्या है?
हम जानते हैं कि P(E)+P(नहीं E)=1
इसलिए 0.05+P(नहीं E)=1
⇒ P(नहीं E)=1-0.05=0.95
एक थैले में 5 लाल गेंद और कुछ नीली गेंदे हैं यदि इस थैले में से नीली गेंद निकालने की प्रायिकता लाल गेंद निकालने की प्रायिकता की दुगुनी है, तो थैले में नीली गेंदों की संख्या ज्ञात कीजिए।
माना नीली गेंदों की कुल संख्या = x
लाल गेंदों की कुल संख्या = 5
कुल गेंदे = x + 5
P (लाल गेंद) = 5/(5 + x)
P (नीली गेंद) = x/(5 + x)
दिया है,
2(5/(5 + x)) = x/(5 + x)
⇒ 10(x + 5) = x^2 + 5x
⇒ x^2-5x-50=0
⇒ x^2 – 10x + 5x – 50 = 0
⇒ x(x – 10) + 5 (x – 10) = 0
⇒ (x – 10)(x + 5) = 0
⇒ x – 10 = 0 या x + 5 = 0
⇒ x = 10 या x = – 5
क्योंकि गेंदों की संख्या ऋणात्मक नहीं हो सकती, अतः नीली गेंदों की कुल संख्या 10 है ।