Class 10 Maths NCERT Solutions and NCERT Exampler Solutions — Free.
Class 10 maths NCERT and NCERT Exemplar section of mathsdiscussion.com include all exercises questions given in class 10 NCERT textbook and NCERT Exemplar book . We team mathsdiscussion.com provide free Solutions for all exercises and questions in class 10 maths NCERT and NCERT Exemplar for students help. Further share the website link and address with your class mates , friends and relatives so that they can also get benefit of Solutions.
In addition to the text solution of word problems where it is difficult for a student to understand that concept , In addition we have given detailed explanatory video solutions for those questions from Class 10 maths NCERT textbook. Now one can do each and every question of class 10 maths NCERT and NCERT Exemplar with complete understanding not just memorizing the solution.
Still if any student face any difficulty in any solution , on any question from any book the BIGGEST ADVANTAGE OF mathsdiscussion.com is its Discussion forum , where one can get the solutions for all challenging questions not only from school text book , and someone from forum will give the solution. To avail the discussion forum you need to only register on our website for free.
NCERT TextBook and NCERT Exemplar for Class 10 Maths —- Free Exercise Solution
Being in class 10 , it will be the first time when you will appear for external exam i.e Board Exams , where in particular mathematics is a subject in which students fear the most. For class 10 maths , through knowledge and visualization or feel of concept can be experienced by our visual lecture i.e video lectures and solution to class 10 maths NCERT word problems with video solution will help in boasting the confidence.
Most important thing to score perfect 100 marks in class 10 maths board exam is , understanding of chapter and extensive practice of real board level questions , also not to mention previous years board question papers.
In mathsdiscussion.com we have taken at most care for understanding of maths , concepts and solutions by free video lecture supported by discussion forum to clarify all your droughts , before appearing for class 10 maths board exam in math in Febarury-march 2020. For extensive practice not only the class 10 NCERT textbook solutions we also provide the solution to class 10 maths NCERT Exemplar , and there are number of practice sample paper for class 10 maths board exam is available on mathsdiscussion.com , you can also practice previous years question papers from our site absolutely free.
CHAPTER 1 ——– REAL NUMBERS
Concepts of Chapter
- How to prove number like $\sqrt2 $ , $ \sqrt 3 $ are irrational.
- Euclid’s division algorithm .
- LCM and HCF of numbers.
- Number of the form p/q , whether they have terminating or non terminating decimals.
NCERT Chapter 1 Exercise Solution
Online Solution Chapter 1 Real Numbers
CHAPTER 1.
EXERCISE 1.1
QUESTION 3
An army contingent of 616 members is to march behind an army band of 32 members in a parade. Two groups are to march in the same number of columns . What is the maximum number of column in which they can march ?
CHAPTER 1 .
EXERCISE 1.2
QUESTION 7
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field , while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at same time , and go in the same direction. After how many minuets will they meet again at starting point ?
NCERT Exemplar Chapter 1 Questions Solutions.
CHAPTER 2 ——- POLYNOMIALS
Concepts of chapter
- What are zeros of polynomials.
- Relation between zeros of quadratic polynomial and coefficients of quadratic polynomial.
- Formation of quadratic polynomial if zeros of polynomials are known,
- Relation between zeros of cubic polynomial and coefficients of cubic polynomial equation.
NCERT Chapter 2 Exercise Solutions.
NCERT Exemplar Chapter 2 Solutions.
CHAPTER 3 —– PAIR OF LINEAR EQUATIONS IN TWO VARIABLE
Concepts of Chapter
- Formation of pair of linear equation in two variable.
- Conditions for two linear equations represent pair of parallel,Intersecting or Coincident lines.
- Methods to find the solutions of linear equations . Graphical method , Substitution Method , Elimination method and Cross Multiplication method.
- Solutions of equation reducible to linear equation in two variable form.
$\\$ CHAPTER 3
CONCEPT
Questions related to If 3 women and 2 men can do a work in 4 days and 2 women and 3 men can do the same work 5 days then find the time 1 men and 1 women will take to finish the same work.
NCERT Exemplar Chapter 3 Solutions
CHAPTER 4 —– Quadratic Equations
Concept of Chapter
1. A polynomial of degree 2 is called a quadratic polynomial. Its general form is $ax^2 + bx + c$, where $a, b, c \in R$ such that $a\neq 0$ and x is a variable.
2. If $p(x) = ax^2 + bx + c$ is a quadratic polynomial and let $\alpha$ be a real number. Then, $a\alpha^2 + b\alpha + c$ is known as the value of the quadratic polynomial $p(x)$ and it is denoted by $p(\alpha)$. i.e., $p(\alpha) = a\alpha^2 + b\alpha + c$.
3. A real number $’ \alpha ‘$ is called a zero of a quadratic polynomial $p(x)=ax^2+bx+c$ if $ p(\alpha)=0$
4. If $p(x)$ is a quadratic polynomial, then $p(x) = 0$ is called a quadratic equation. Its general form is $ax^2 + bx + c$ , where $a, b, c \in R$ and $a\neq 0$.
5. If $p(x) = 0$ is a quadratic equation, then the zeros of the polynomial $p(x)$ are called the roots of the equation $p(x) = 0$.
6. A quadratic equation has two roots, which may be :
(i) real and distinct or
(ii) real and coincident or
(iii) imaginary (not to be taken is this class)
7. Solution of quadratic equation by factorization. The following steps should be taken :
(i) Write all terms on one side by making R.H.S. zero.
(ii) Resolve in linear factors of the type (ax + b)(cx + d) of the terms on the L.H.S.
(iii) Put each factor equal to zero.
(iv) Get the required solution.
Remarks : If the product of two polynomials is zero, then each of them is zero.
i.e., if (ax + b)(cx + d) = 0, then either ax + b = 0 or cx + d = 0.
8. A quadratic equation can be solved by completing the square.
9. If $ax^2+bx+c=0$ , $a,b,c \in R$ and $a \neq 0 $ , then this is known as quadratic formula.
10. Nature of the roots of quadratic equation : In the equation $ax^2 + bx + c = 0$ , $a, b, c \in R$ , $a\neq 0$ , $b^2 – 4ac$ determine whether the quadratic equation $ax^2 + bx + c = 0$ has real roots or not, $b^2 – 4ac$ is called the discriminant of this quadratic equation.
So, a quadratic equation $ax^2 + bx + c = 0$.
(i) has no real roots if $b^2 – 4ac \lt 0$.
(ii) has two equal real roots if $b^2 – 4ac = 0$.
(iii) has two distinct real roots if $b^2 – 4ac \gt 0$.
NCERT Chapter 4 Exercise Solutions.
NCERT Exemplar Chapter 4 Solutions
CHAPTER 5 —– ARITHMETIC PROGRESSION
Concept of Chapter
1. A sequence is an arrangement of numbers in a definite order according to some rule.
2. An arithmetic progression is a sequence in which terms increase or decrease regularly by the same constant. This constant is called the common difference of the progression (series).
In other words, an arithmetic progression is that list of numbers in which the first term is given and each term is obtained by adding a fixed number d to the preceding term.
This fixed number d is called the common difference. This can be positive, negative or zero.
3. We usually denote the first term of an AP by a, the common difference by d and the last term by l.
4. If the first term is a and the common difference is d, then a, a + d, a + 2d, a + 3d, ….. represents an arithmetic progression for different values of a and d. This is the general form of an AP.
5. In an AP,d = a2 – a1 = a3 – a2 = a4 – a3 = ….. = an – an – 1, where an denotes the nth term.
6. In general, if an + 1 – an, is same for different values of n or in other words, an + 1 – an is independent of n, then the given list of numbers form an AP.
7. The nth term of an AP with first term a and common difference d is an = a + (n – 1)d, an is also called the general term of the AP. If there are n terms in the AP, then an represents the last term l.
8. The sum of first n terms of an AP is given by.
(i)$\,S_n\,=\,(\frac{n}{2})(\,a\,+\,l\,)$ , where a is the first term and l is the last term.
(ii)$\,S_n\,=\,(\frac{n}{2})(\,2a\,+\,(n-1)d\,)$ , where a is the first term and d is the common difference.
NCERT Chapter 5 Exercise Solutions.
NCERT Exemplar Chapter 5 Solutions
CHAPTER 6 —– TRIANGLES
1. Two geometric figure having the same shape and size are known as congruent figure.
2. Geometric figures having the same shape but different sizes are known as similar figures.
3. Two congruent figures are always similar but similar figures need not be congruent.
4. Two polygons of the same number of sides are said to similar to each other, if :
(i) their corresponding angles are equal, and
(ii) the lengths of their corresponding sides are proportional.
5. Two triangles are said to be similar to each other, if
(i) their corresponding angles are equal, and,
(ii) their corresponding sides are proportional.
i.e., Two $\Delta ABC$ and $\Delta DEF$ are similar, if $$(i) \angle A = \angle D, \angle B = \angle E, \angle C = \angle F$$ $$(ii) \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$$
Note : If two $\Delta ABC$ and $\Delta DEF$ are similar, we write $\Delta ABC \sim \Delta DEF$.
6. Some Basic Results on Proportionality
(i) Basic proportionality theorem of Thale’s theorem : If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
(ii) Converse of basic proportionality theorem : If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
7. Criteria for Similarity of Triangles :
(i) AAA-Criterion of similarity of triangles : If the corresponding angles of two triangles are equal i.e., if the two triangles are equiangular, then they are similar.
(ii) SSS-Criterion of similarity of triangles : If the corresponding sides of two triangles are proportional, then they are similar.
(iii) SAS-Criterion of similarity of triangles : If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.
8. Areas of two Similar Triangles : The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
9. Pythagoras Theorem : In a right-angled triangles, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
10. Converse of Phythagoras Theorem : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.
11. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles of both sides of the perpendicular are similar to the whole triangle and to each other.
NCERT Chapter 6 Exercise Solutions
NCERT Exemplar Chapter 6 Solutions
CHAPTER 7 —– COORDINATE GEOMETRY
1. The abscissa of a point is its perpendicular distance from y-axis.
2. The ordinate of a point is its perpendicular distance from x-axis.
3. The abscissa of every point situated on the right side of y-axis is positive and the abscissa of every point situated on the left side of y-axis is negative.
4. The ordinate of every point situated above x-axis is positive and that of every point below x-axis is negative.
5. The abscissa of every point on y-axis is zero.
6. The ordinate of every point on x-axis is zero.
7. Coordinate of the origin are (0, 0).
8. The distance between two points P(x1, y1) and Q(x2, y2) is given by $PQ=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
Note : If O is the origin and P(x, y) is any point, then $OP=\sqrt{x^2+y^2}$
9. Some useful results
(a) In order to prove that a given figure is a
(i) square, prove that the four sides are equal and the diagonals are also equal.
(ii) rhombus, prove that the four sides are equal.
(iii) rectangle, prove that opposite sides are equal and the diagonals are also equal.
(iv) parallelogram, prove that the opposite sides are equal.
(v) parallelogram but not a rectangle, prove that its opposite sides are equal but diagonals are not equal.
(vi) rhombus but not a square, prove that its all sides are equal but the diagonals are not equal.
(b) For three points to be col linear, prove that the sum of the distance between two pairs of points of points is equal to the third pairs of points.
10. Section formula (Internal division) : The coordinate of the point R(x, y) which divides internally the straight line joining points P(x1, y1) and Q(x2, y2) in the ratio m : n is given by .
$$\Bigl( \frac{nx_1+mx_2}{m+n}\,,\,\frac{ny_1+my_2}{m+n} \Bigr) $$
(i) The coordinates of the mid-point of the line joining the points P(x1, y1) and Q(x2, y2) is given by $$\Bigl(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} \Bigr )$$
(ii) If a point R divides the line joining the points P(x1, y1) and Q(x2, y2) in the ratio k : 1, then the coordinates of R are given by $$\Bigl(\frac{x_1+kx_2}{k+1},\frac{y_1+ky_2}{k+1}\Bigr)$$
11. Area of triangle : The area of the triangle having vertices as (x1, y1), (x2, y2) and (x3, y3) is given by $$Area=\frac{1}{2} \vert (x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)) \vert $$
12. Condition of collinearity of three points : Three points A(x1, y1), B(x2, y2) and C(x3, y3) will be collinear if and only if the area of $\Delta ABC$ is zero. That is,
(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y1 – x1y3) = 0
or x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
13. The coordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) are given by $$\Bigl(\frac{x_1+x_2+x_3}{3} , \frac{y_1+y_2+y_3}{3} \Bigr)$$
NCERT Chapter 7 Exercise Solutions.
NCERT Exemplar Chapter 7 Solutions
CHAPTER 8 —– INTRODUCTION TO TRIGONOMETRY
1. Trigonometric Ratio : We shall now define certain ratios involving the sides of a right triangle, and cell them trigonometric ratios.
The trigonometric ratios of the angle A in right triangle ABC are defined as $$SinA=\frac{Perpandicular\,opposite\,to\,angle\,A}{Hypotenuse}=\frac{BC}{AC}$$ $$CosA=\frac{Side\,adjusent\,to\,angle\,A}{Hypotenuse}=\frac{AB}{AC}$$ $$tanA=\frac{Perpandicular\,opposite\,to\,angle\,A} {Side\,adjusent\,to\,angle\,A} =\frac{BC}{AB}$$ $$CosecA=\frac{Hypotenuse}{Perpandicular\,opposite\,to\,angle\,A}=\frac{AC}{BC}$$ $$SecA=\frac{Hypotenuse}{Side\,adjusent\,to\,angle\,A}=\frac{AC}{AB}$$ $$CotA=\frac{Side\,adjusent\,to\,angle\,A} {Perpandicular\,opposite\,to\,angle\,A} =\frac{AB}{BC}$$
2. Relationship between Trigonometrical Ratios
From the definition of t-ratios are
$$CosecA=\frac{1}{sinA}$$
$$SecA=\frac{1}{CosA}$$
$$CotA=\frac{1}{tanA}$$
3. Thevalue of the trigonometric ratios of an angle does not depend on the size of the triangle. It only depends on the angle.
4. Since hypotenuse is the longest side in a right triangle, therefore, the value of sin A or cos A is always less than or equal to 1.
5. Trigonometric Ratios of Complementary Angles : Two angles are said to be complementary if their sum equals 90°. In $\Delta ABC$ , right angled at B, can you see that $\angle BAC$ (or $\angle A)$ and $\angle ACB$ (or $\angle C$) are complementary angles, i.e., $\angle A + \angle C$ = 90° ? (Because $\angle A + \angle C $ = 180° – $\angle B$ = 180° – 90° = 90°).
For all values of angle A lying between 0° and 90°, we have sin (90° – A) = cos A; cos (90° – A) = sin A; tan (90° – A) = cot A ; cot (90° – A) = tan A ; sec (90° – A) = cosec A ; cosec (90° – A) = sec A
Note : (i) Add co, if that is not there. Remove co, if that is there.
(ii) tan 0° = 0 = cot 90°, sec 0° = 1 = cosec 90° and sec 90°, cosec 0°, tan 90° and cot 0° are not defined.
6. Trigonometric Identities : We may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angle.
For any acute angle A, we have
(i) sin2 A + cos2 A = 1, sin2 A = 1 – cos2 A, cos2 A = 1 – sin2 A
(ii) 1 + tan2 A = sec2 A, tan2 A = sec2 A – 1, sec2 A – tan2 A = 1
(iii) 1 + cot2 A = cosec2 A, cot2 A = cosec2 A – 1, cosec2 A – cot2 A = 1
NCERT Chapter 8 Exercise Solutions.
NCERT Exemplar Chapter 8 Solutions
CHAPTER 9 —– SOME APPLICATIONS OF TRIGONOMETRY
1. Trigonometry enables us to find the heights of the objects and the distances between the points, the actual measurement of these heights and distances being difficult or impossible.
2. Angles of Elevation and Depression
Let P be the position of an object above the horizontal OX, where O is the eye of the observer looking at the object. Then \angle XOP$ is called the angle of elevation.
Let P be the position of an object below the horizontal OX, where O is the eye of the observer. Then, $\angle XOP$ is called the angle of depression.
Notes :
1. Numerically, angle of elevation is equal to the angle of depression.
2. The angle of elevation and the angle of depression both are measured along the horizontal.
3. To determine one side (unknown side) of a right angled triangle, other side (known side) and acute angle are given, we combine known (given) with unknown (required) and use the suitable t-ratio of the given angle.
$i.e\;\frac{Known\,side}{Unknown\,side}=\frac{Given\,side}{Required\,side}$= Suitable t ration of given angle.
NCERT Chapter 9 Exercise Solutions.
NCERT Exemplar Chapter 9 Solutions
CHAPTER 10 —– CIRCLES
1. A line which intersects a circle at two distinct points is called a secant of the circle.
2. A tangent to a circle is a line that intersects the circle at exactly one point.
3. A point where a tangent touches the circle is called the point of contact.
4. A tangent to a circle is perpendicular to the radius through the point of contact.
5. A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.
6. The lengths of two tangents drawn from an external point to a circle are equal.
7. If two tangents are drawn to a circle from an external point, then
(i) they subtend equal angles at the centre.
(ii) they are equally inclined to the segment, joining the centre to the that point.
NCERT Chapter 10 Exercise Solutions.
NCERT Exemplar Chapter 10 Solutions
CHAPTER 11 —– CONSTRUCTIONS
In this chapter, we shall do the following constructions:
1. To divide a line segment in a given ratio.
2. To construct a triangle similar to a given triangle as per given scale factor, which may be less than 1 or greater than 1.
Note: The number of points should be greater than m and n in the scale factor $\frac{m}{n}$ .
3. To construct a pair of tangents from an external point to a circle.
NCERT Chapter 11 Exercise Solutions.
NCERT Exemplar Chapter 11 Solutions
CHAPTER 12 —– AREAS RELATED TO CIRCLES
1. A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point always remains the same.
The fixed point is called the centre and the constant distance is known as the radius of the circle.
2. Area and Circumference of a Circle: Ifr is the radius of a circle, then
(i) Circumference = $2\pi d$ or $\pi d$ , where d = 2r.
(ii) Area = $\pi r^2$ or $\pi \frac{d^2}{2}$ , where d = diameter
(iii) Area of a semicircle $=\frac{1}{2} \pi r^2$
(iv) Area of a quadrant of a circle $=\frac{1}{4} \pi r^2$
3. Area Enclosed by Two Concentric Circles:
If R and r are radii of two concentric circles, then Area enclosed by the two circles $=\pi (R^2-r^2)$ = $\pi (R + r)(R – r) $.
4. Some Important Results
(i) The distance moved by a rotating wheel in one revolution is equal to the circumference of the wheel.
(ii) The number of revolutions completed by a rotating wheel in one minute $=\Bigl( \frac{Distance\,moved\,in\,one\,minute}{Circumference} \Bigr) $
5. If an arc subtends an angle $\theta $ at the centre, then its arc length $=\Bigl(\frac{\theta}{180} \pi r \Bigr)$
If an arc subtends an angle $\theta $ , then the area of the corresponding sector $=\Bigl( \frac{\theta}{360} \pi r^2 \Bigr)$
7. Area of a Segment: If r is the radius and $\theta$ is the angle which chord AB makes at the centre, then
area of a segment $=r^2 \Bigl( \frac{\pi \theta}{360}-\frac{1}{2}Sin \theta \Bigr)$.
NCERT Chapter 12 Exercise Solutions.
NCERT Exemplar Chapter 12 Solutions
CHAPTER 13 —– SURFACE AREAS AND VOLUME
1. Cylinder: A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides.
i.e., by one complete revolution of a rectangle ABCD about one of its sides AB, a right circular cylinder is generated.
Radius of the cylinder: AB or CD is the radius (r) of the cylinder, as shown in the figure.
Height of the cylinder: AD or BC is the height (h) of the cylinder, as shown in figure.
Surface area of a right circular cylinder
Surface area of a right circular cylinder: If r is the radius of the cylinder and h is the height of the cylinder, then surface area of cylinder $= 2\pi r$.
(i) Curved surface of the cylinder = Area of the rectangle ( Solid yellow rectangle ) $=2\pi rh$.
(ii) Area of each end (area of base/top) $=\pi r^2 $.
(iii) Total surface area = Curved surface area + Area of two end $$ = 2\pi rh + 2\pi r2 = 2\pi r(h + r) $$
(iv) Volume of a cylinder: The space enclosed by a cylinder has volume and this volume is known as the volume of the cylinder.
Volume of the cylinder = Area of the base x Height $= \pi r^2 h $
Surface area and volume of a hollow cylinder.
Hollow cylinder is a solid bounded by two co-axial cylinders of the same height and radii.
(i) Area of each end $= \pi R^2– \pi r^2= \pi (R^2– r^2)$
(ii) Curved surface area of he hollow cylinder = External surface + Internal surface $$= 2\pi Rh + 2\pi rh = 2\pi h(R + r) $$
(iii) Total surface area of the hollow cylinder $$ $$= Curved surface area + 2 x Area of the base ring . $$= 2\pi Rh + 2\pi rh + 2\pi (R^2– r^2) $$ $$= 2\pi h(R + r) + 2\pi (R – r)(R + r) $$
NCERT Chapter 13 Exercise Solutions.
NCERT Exemplar Chapter 13 Solutions
CHAPTER 14 —– STATISTICS
1. MEAN :: We have the following techniques to evaluate the mean of a given data.
Case I: Simple distribution or Individual observation. The mean of a set of numbers $x_1 , x_2 , x_3 ,………x_n $ is denoted by $ \bar x $ and is defined as $$\bar x =\frac{x_1+x_2+x_3…….+x_n}{n} =\frac {\sum_{r=1}^{n} x_i}{n}$$
Case II: Mean of a Discrete, i.e., Ungrouped Frequency Distribution.
Formula 1. If $x_1,x_2,x_3,……,x_n$ are the values of a variable with frequencies $f_1,f_2,f_3,……..,f_n$ , then $$\bar x =\frac{x_1f_1+x_2f_2+x_3f_3+…..+x_nf_n}{f_1+f_2+f_3+……..+f_n}=\frac{\sum_{r=1}^{n} x_if_i}{\sum_{r=1}{n} f_i} $$
Formula 2. Assumed mean method . If the frequencies and the values of the variable are quite large numbers, the product $f_ix_i$ will also be large. Then, a short cut method has been devised as $$\bar x = a+\Bigl( \frac{\sum_{r=1}^{n} f_id_i}{\sum_{r=1}^{n} f_i}\Bigr) \; where\,d_i=(x_i-a)$$ And ‘a’ is the assumed mean.
Formula 3. Step deviation mathod $d_i’s$ are found to be divisible by a common number h (say), then we can make the calculation easier by putting $$u_i=\frac{x_i-a}{h}\; \forall i=1,2,3…..n $$
where a is the assumed mean, h is the length of the class interval the mean $$ \bar x = a + h\Bigl( \frac{\sum_{r=1}^{n} u_if_i}{\sum_{r=1}^{n} f_i} \Bigr)$$ where ‘a’ is the assumed mean, h is the length of the class interval and $u_i=\frac{x_i-a}{h}$
Case III: Mean of Continuous or Grouped Frequency Distribution. In this case also, the mean is computed by applying any of the formulae given for discrete frequency distribution. The values $x_1 , x_2 , x_3 ,……,x_n$ are taken as the mid-values or class marks of the various class intervals. If the given frequency distribution is inclusive, then it should first be converted to exclusive distribution.
2. Median :: Case I: For an individual series . Arrange the raw data in ascending or descending order. Then the following rule gives the median.
When n, the number of observations, is odd, then median is $\Bigl( \frac{n+1}{2} \Bigr) $ observation.
When n, the number of observations, is even , then median is the mean of $\Bigl( \frac{n}{2} \Bigr)^{th} $ and $\Bigl( \frac{n}{2}+1 \Bigr)^{th} $ observation.
Case II: For a frequency distribution (i) When the series is discrete.
Arrange the data in ascending or descending order of magnitude and then prepare the cumulative frequency table. Let $n$ be the total frequency. The if $n$ is odd, median is the value of $\Bigl( \frac{n+1}{2} \Bigr)^{th} $ item, and if $n$ is even, median is the mean of the values of $\Bigl( \frac{n}{2} \Bigr)^{th} $ and $\Bigl( \frac{n}{2}+1 \Bigr)^{th} $ items.
(ii) When the series is continuous. Median of the grouped data is obtained using the formula $$ Median =l+\Bigl( \frac{\frac{n}{2}-cf}{f} \Bigr) h $$
Where , $l$=lower limit of median class.
$n$=number of observations
$cf$=Cumulative frequency of class preceding the median class.
$f$ = frequency of median class
$h$ = class size.
3. Mode :: The value of the variable which occurs most frequently in a distribution is called mode.
The mode for grouped data is found by formula :$$ Mode = l +\Bigl( \frac{f_1-f_2}{2f_1-f_0-f_2} \Bigr) h$$
Where $l$=Lower limit of the modal class.
$h$ = Size of the class interval.
$f_1$ = frequency of the modal class.
$f_0$ = frequency of the class preceding the modal class.
$f_2$ = frequency of the class succeeding the modal class.
4. RELATIONSHIP AMONG MEAN, MEDIAN AND MODE
The mode is approximately = 3 median – 2 mean
5. CERTAIN POINTS TO BE NOTED
(i) Arithmetic mean is based on all the observations. It is not useful if there are some extreme values in the data.
(ii) Mode is more useful to business persons who may not be interested in the magnitude but only if the most common value of fashionable value. For example, while ordering for shows of various sizes for resale, a modal size will be more appropriate.
(iii) Median is useful when there are open and extreme classes, because median is not affected by extreme open values.
6. Representing a cumulative frequency distribution graphically as a cumulative frequency curve, or an ogive of the less than type and of the more than type.
7. The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for this data.
NCERT Chapter 14 Exercise Solutions.
NCERT Exemplar Chapter 14 Solutions
CHAPTER 15 —– PROBABILITY
1. A random experiment is an experiment in which;
(i) all the outcomes of the experiment are known in advance, and
(ii) the exact outcome of any specific performance of the experiment is unpredictable, i.e., not known in advance.
2. The set of all possible outcomes of a random experiment is known as a sample space associated with the random experiment.
3. An event is something that happens.
4. Ordinarily speaking, that probability of an event denotes the likelihood of its happening.
5. Definition of probability of an event: If there are n events associated with a random experiment and m of them are favorable to an event A, then the probability of happening of A is denoted by P(A) and is defined as he ratio $\frac{m}{n}$.
Thus , $P(A) =\Bigl( \frac{Number\,of\,favorable\,outcome}{Total\,Number\,of\,possible\,outcomes} \Bigr)$ i.e $P(A)=\frac{m}{n}$
8. An event which is certain (sure) to happen has a probability of 1 (or 100%). For example, the probability that the sun will rise in the east is 1.
9. An event which is impossible has a probability of 0 (or 0%). For example, the probability that the sun will never set again is 0.
NCERT Chapter 15 Exercise Solutions.