Showing posts with label class10-maths. Show all posts
Showing posts with label class10-maths. Show all posts

Friday 7 October 2016

CBSE Class 10 - Maths - Quadratic Equations - Summary

Quadratic Equations - Summary

CBSE Class 10 - Maths - Quadratic Equations - Summary

A polynomial of degree 2 is called quadrilateral polynomial. The general form of a quadrilateral polynomial is ax2 + bx+ c, where a, b, c are real numbers such that a ≠ 0 and x is a real variable.


If p(x) = ax2  + bx + c, a ≠ 0 is a quadratic polynomial and α is a real number, then p(α) = aα2 + bα + c is known as the value of the quadratic polynomial p(x).

A real number α is said to be a zero of the quadratic polynomial p(x) = ax2  + bx + c, if p(α) = 0.

If p(x) = ax2  + bx + c is a quadratic polynomial,then p(x) = 0 i.e. ax2  + bx + c = 0, a ≠ 0 is called a quadratic equation.

A real number α is said to be root of the quadratic equation ax2  + bx + c = 0, if aα2 + bα + c = 0.
In other words, α is a root of ax2  + bx + c = 0 if and only if α is zero pf the polynomial p(x) = ax2  + bx + c.

Tuesday 20 September 2016

CBSE Class 10 - Real Numbers Problems on Euclid's Division Algorithm(2016)

Real Numbers Problems on
Euclid's Division Algorithm(2016)

CBSE Class 10 - Real Numbers   Problems on Euclid's Division Algorithm(2016)

Q1(CBSE 2012): Using Euclid’s division algorithm, find the HCF of 240 and 228.

Answer: By Euclid’s division algorithm,
⇒ 240 = 228 × 1 + 12
⇒ 228 = 12 × 19 + 0
∴ HCF (240, 228) = 12


Q2(CBSE 2014): The length, breadth and height of a room are 8m 25 cm, 6m 75 cm and 4 m 50 cm respectively. Find the length of the longest rod that can measure the three dimensions of the room exactly.

Answer:
∵ 1m = 100 cm
∴ 8 m 25 cm = 825 cm
6 m 75 cm = 675 cm
4 m 50 cm = 450 cm
The length of the longest rod = HCF(825, 675, 450)
⇒ 825 = 675 × 1 + 150
675 = 150 × 4 + 75
150 = 75 × 2 + 0
∴ HCF(825, 675) = 75
450 = 75 × 6 + 0
∴ HCF(450, 75) = 75
∴ HCF (825, 675, 450) = 75×
Thus, the length of the longest rod is 75 cm.

Q3(NCERT Exemplar): Write whether every positive integer can be of the form 4q + 2, where q is an integer.Justify your answer.

Answer: No, every positive integer cannot be expressed as only of the form 4q + 2.

Wednesday 14 September 2016

CBSE Class 10 - Maths - Polynomials (SA1) - Few Questions from NCERT Exemplar

POLYNOMIALS 

(Questions and Answers from NCERT Exemplar)

CBSE Class 10 - Maths - Polynomials (SA1) - Few Questions from NCERT Exemplar

 

Here follows few questions that have asked in previous examination papers.






Sunday 28 August 2016

Thursday 25 February 2016

CBSE Class 10 - Maths - Probability - 5 Questions That Your Teacher Can Ask To Confuse You

Probability

5 Questions Your Maths Teacher Can Ask To Confuse You


Q1: A fair coin is tossed twice, find the probability of getting different outcomes of this experiment. What will be the sum of all these probabilities?


Answer: A coin is tossed twice, the sample space (possible outcomes) will be:

S = {HH, HT, TH, TT}

Let A be the event getting two heads then

Number of Favourable Outcomes for event A 1
P(A) =   divided by  = divided by
Total Number of Outcomes 4


Let B be the event getting first head and then tail (HT) then


Number of Favourable Outcomes for event B 1
P(B) =   divided by  = divided by
Total Number of Outcomes 4

Let C be the event getting first tail then head (TH) then

Number of Favourable Outcomes for event C 1
P(C) =   divided by  = divided by
Total Number of Outcomes 4

Let D be the event getting two tails (TT) then

Number of Favourable Outcomes for event D 1
P(D) =   divided by  = divided by
Total Number of Outcomes 4

Sum of all probabilities P(A) + P(B) + P(C) + P(D) = 4/4 = 1


Q2: Find the probability that a leap year selected randomly will have 53 Sundays? 
(Note: If first two days lands on a Sunday in a leap year it would have 53 Sundays. How to prove?). 

Sunday 21 February 2016

CBSE Class 10 - Maths - Circles (Quiz 2016)

Circles 

CBSE Class 10 - Maths - Circles (Quiz 2016)

MCQs


Q1: A tangent to a circle intersects it at how many points?

(a) 1
(b) 2
(c) 3
(d) none of these


Q2: Both tangents and radii

(a) orignate from the center of a circle.
(b) meet a circle at exactly one point.
(c) are half a circle’s length.
(d) are straight angles.



Q3: A tangent to a circle (center O) drawn from an exterior point to the circle and touches the circle at a point P on the circumference. Which of the following statement is TRUE?

(a) ∠TPO > 90°
(b) ∠TPO = 90°
(c) ∠TPO < 90°
(d) ∠TPO = 180°


Q4: Two concentric circles are of radii 13 cm and 12 cm. What is the length of the chord of the larger circle which touches the smaller circle?

(a) 8 cm
(b) 6 cm
(c) 10 cm
(d) 4 cm


Q5: A parallelogram circumscribing a circle a ____

(a) square
(b) rhombus
(c) rectangle
(d) incircle

Friday 19 February 2016

CBSE Class 10 - Maths - 5 Minutes Revision of Probability

Probability


5 Minutes Revision



1. The sample space of an experiment is the set of all possible outcomes of the experiment.


2. The sample space when a coin is tossed three times is

     S  = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}


3. Basic formula to find Probability of an event (E) is

Number of Favourable Outcomes
P(E) =  divided by
Total Number of Outcomes


4. An event that is not very likely has a probability close to 0.


5. An event that is very likely has a probability close to 1 (100%).


6. If probability of happening an event is x then probability of not happening that event is (1-x).


7. If probability of winning a game is 0.6 then probability of loosing it is (1-0.6) = 0.4

Wednesday 16 September 2015

Sunday 2 August 2015

CBSE Class 10 - Introduction to Trigonometry - NCERT Ex 8.3

Introduction to Trigonometry 


CBSE Class 10 - Introduction to Trigonometry - NCERT Ex 8.3


NCERT Exercise 8.3 Chapter Answers 


NCERT Chapter 8.3 solutions are available at www.eduvictors.com site.
Please share your feedback and suggestions here.

Thursday 16 July 2015

CBSE Class 10 Maths - CH8 - Introduction to Trigonometry (NCERT Ex 8.2)

Introduction to Trigonometry 

 NCERT Exercise 8.2


NCERT Exercise 8.2 Chapter Answers 


NCERT Chapter 8.2 solutions are available at www.eduvictors.com site.
Please share your feedback and suggestions here.

Sunday 12 July 2015

CBSE Class 10 - CH 8: Introduction to Trigonometry - NCERT Ex 8.1

Introduction to Trigonometry 

NCERT Exercise 8.1 Chapter Answers 

Introduction to Trigonometry Ex 8.1
NCERT Chapter 8.1 solutions are available at www.eduvictors.com site.
Please share your feedback and suggestions here.

Saturday 27 June 2015

CBSE Class 10 Maths CH6 Triangles (Important Points You must Know)

Triangles: Important Points You must Know

CBSE Class 10 Maths CH6 Triangles (Important Points You must Know)

1. Two figures are said to be similar if and only if they have same shape but not necessarily of same size.

2. All the congruent figures are similar but the converse is not true.

3. Two polygons of the same number of sides are similar if and only if
   (i) their corresponding angles are equal
   (ii) their corresponding sides are in proportion (same ratio).
 
4Basic Proportionality Theorem (Thales Theorem):
   If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.(see figure above)
 

5. Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.

6. Similar Triangles: Two triangles are similar if and only if
   i.  their corresponding angles are equal i.e. they are equi-angular and
   ii. their corresponding sides are in the same ratio.
 
7. Criteria for establishing similarity of two triangles:
   i.   AA (or AAA) criterion of similarity
   ii.  SSS criterion
   iii. SAS criterion
   iv.  Altitude of right angle triangles

Wednesday 20 May 2015

CBSE Class 10: Maths - CH2 - Polynomials (NCERT EX 2.2)

Polynomials

CBSE Class 10 - Maths - Polynomials - NCERT Ex 2.2NCERT Solutions Ex 2.2

Q1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i)  x2-2x-8
(ii) 4s2-4s+1
(iii) 6x2-3-7x
(iv) 4u2+8u
(v) t2-15
(vi) 3x2 - x -4

Answer: (i)  x2-2x-8 = (x - 4)(x +2)
⇒x2-2x-8 = 0 when x - 4= 0 or x + 2 = 0
∴ Zeroes of x2- 2x - 8 are 4 and -2.
Sum of zeroes =
Product of zeroes =
Hence relationship between zeroes and coefficients are verified.

(ii) 4s2-4s+1
= (2s -1 )2
⇒ 4s2-4s+1 = 0 when 2s -1 = 0
⇒ s = 1/2
∴ Zeroes of 4s2-4s+1 are ½ and ½.
Sum of zeroes =
Product of zeroes =
Hence relationship between zeroes and coefficients are verified.

(iii) 6x2-3-7x
= (3x + 1)(2x - 3)
⇒ 6x2-3-7x = 0 when 3x + 1 = 0 OR 2x - 3 = 0
⇒ x = -1/3 or x = 3/2 ∴ Zeroes of 6x2-3-7x are -1/3 and 3/2
Sum of zeroes =
Product of zeroes =
Hence relationship between zeroes and coefficients are verified.

(iv) 4u2+8u
= 4u(u + 2)
⇒ 4u2+8u = 0 when 4u = 0 OR u + 2 = 0
⇒ u = 0 OR u = -2
∴ Zeroes of 4u2+8u are 0 OR -2
Sum of zeroes =
Product of zeroes =

Hence relationship between zeroes and coefficients are verified.

(v) t2-15
= t2 - 0t -15 = (t - √15)(t + √15)
⇒ The value of t2-15 is 0 when t - √15 = 0 OR t + √15 = 0
⇒ t = √15 or t = -√15.
Sum of zeroes = √15 -√15 = 0
Product of zeores = (√15) ☓ (-√15) = -15
Hence relationship between zeroes and coefficients are verified.

Saturday 25 April 2015

CBSE Class 10 - Maths - Polynomials - NCERT Ex 2.1

POLYNOMIALS
NCERT Exercise 2.1 and other Q & A

Q1: The graphs of y = p(x) are given in below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

CBSE Class 10 - Maths - Polynomials - NCERT Ex 2.1

Answer
(i) As shown in the graph, it does not intersect x-axis. Hence it has no real zeroes.
(ii) From the graph, it intersects x-axis at one point. ∴ it has one real zero.
(iii) As the graph shows, it intersects x-axis at three points. ∴ the graph polynomial has three real zeroes.
(iv) From the graph, it is clear it intersects x-axis at two points. Hence it has two zeroes.
(v) As the graph shows, it meets x-axis at four points. ∴ it has four real zeroes.

Saturday 18 April 2015

CBSE Class 10 - Maths - CH2 - Polynomials - Study Points

Polynomials

CBSE Class 10 - Maths - CH2 - Polynomials - Study Points


Study Points

1. An expression of the form a0 + a1x + a2x2 + ----- + anxn where an is called a polynomial in variable x of degree n. where; a0 ,a1, ----- an are real numbers and are called terms/co-efficients of the polynomial and each power of x is a non negative integer.

2. Polynomials in the variable x are denoted by f(x), g(x), h(x) etc.
    e.g. f(x) = a0 + a1x + a2x2 + ----- + anxn

3. A polynomial p(x) = a (where a is constant) is of degree 0 and is called a constant polynomial.

4. A polynomial p(x) = ax + b is of degree 1 and is called a linear polynomial. e.g. 4x - 3, 5x...

5. A polynomial p(x) = ax2 + bx + c of degree 2 and is called a quadratic polynomial.
    e.g. 3x2 -x + 5, 1- x2...

6. A polynomial p(x) = ax3 + bx2 + cx + d of degree 3 and is called a cubic polynomial.
    e.g. √5x3 - 2x2 + 5x -1...

Sunday 29 March 2015

CBSE Class 10 - Maths - CH1 - Real Numbers (Euclid's Division Lemma)

Euclid's Division Lemma

Important Questions asked in Examination...
Euclid (4th BC)
credits: wikipedia

Q1: State Euclid's Division Lemma.

Answer: Given positive integers a and b ( b ≠ 0 ), there exists unique integer numbers q and r satisfying a = bq + r, 0 ≤ r < |b|. where
    a is called dividend
    b is called divisor
    q is called quotient
    r is called remainder.

    e.g. 17 = 5 × 3 + 2

Q2: Prove Euclid's Division Lemma.

Answer: According to Euclid's Division lemma, for a positive pair of integers there exists unique integers q and r, such that
                      a = bq + r, where 0 ≤ r < b

Let us assume q and r are not unique i.e. let there exist another pair q0 and r0 i.e. a = bq0 + r0, where 0 ≤ r0 < b

⇒ bq + r = bq0 + r0
⇒ b(q - q0) = r - r0                                          ................ (I)

Since 0 ≤ r < b and 0 ≤ r0 < b, thus 0 ≤ r - r0 < b ......... (II)

The above eq (I) tells that b divides (r - r0) and (r - r0) is an integer less than b. This means (r - r0) must be 0.
⇒ r - r0 = 0 ⇒ r = r0

Eq (I) will be, b(q - q0) = 0
Since b ≠ 0, ⇒ (q - q0) = 0 ⇒ q = q0

Since r = r0 and q = q0, ∴ q and r are unique.

Q3: Prove that the product of two consecutive positive integers is divisible by 2?