CBSE Class 10 - Maths- SA2- Sample Question Paper (2016)

CBSE Class 10 - Maths- SA2- Sample Question Paper (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) =		=
	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) =		=
	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) =		=
	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) =		=
	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?). 

CBSE Class 10 - Maths - Circles (Quiz 2016)

Circles 

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

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) =
	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

CBSE Class 10 - Maths- SA2- Sample Question Paper and Marking Scheme (2016)

CBSE Class 10 - Maths- SA2- Sample Question Paper and Marking Scheme (2016) 

CBSE Class 10 - Maths- SA1- Sample Question Paper (2015-16) Second Set

CBSE Class 10 - Maths- SA1- Sample Question Paper (2015-16) Second Set

CBSE Class 10 - Maths - SA1- Sample Question Paper (2015-16)

CBSE Class 10 - Maths - SA1- Sample Question Paper (2015-16)
Paper sent by one of the reader. Tough One.

CBSE Class 10 - Introduction to Trigonometry - NCERT Ex 8.3

Introduction to Trigonometry 

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.

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

Introduction to Trigonometry 

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.

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

Introduction to Trigonometry 

NCERT Exercise 8.1 Chapter Answers 

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

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

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).
 
4. Basic 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

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

Polynomials

NCERT 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.

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.

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.

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

Polynomials

Study Points

1. An expression of the form a₀ + a₁x + a₂x² + ----- + a_nxⁿ where a_n is called a polynomial in variable x of degree n. where; a₀ ,a₁, ----- a_n 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 + ----- + a_nxⁿ

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...

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?

CBSE Class 10 - Maths - Design of Sample Question Paper (SA2)

CBSE Class 10 - Maths - Modified Syllabus, Question Paper Design

CBSE Class 10 - Maths - CH 14 - Statistics (Ex 14.1)

Statistics

(NCERT Ex 14.1)

Q1: A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.


Answer:

We use the following relation to find the class mark (x_i) for each interval:

Class mark (x_i) =

Calculating x_i and f_ix_i as follows:

Number of plants	Number of houses(fi)	xi	fixi
0 -2	1	1	1 ☓ 1 = 1
2 -4	2	3	2 ☓ 3 = 6
4 -6	1	5	1 ☓ 5 = 5
6 -8	5	7	5 ☓ 7 = 35
8 - 10	6	9	6 ☓ 9 = 54
10 -12	2	11	2 ☓ 11 = 22
12 - 14	3	13	3 ☓ 13 = 39
Total	20		162

From the table, it is observed that

Σ f_i = 20

Σ f_ix_i = 162

Mean , x =

= = 8.1

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (x_i) and f_i are small.


Q2: Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages(in Rs)	100 - 120	120 - 140	140 - 160	160 - 180	180 - 200
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer:

CBSE Class 10 - Maths HOTS Questions by CBSE

CBSE Class 10 - Maths HOTS Questions by CBSE 

CBSE Class 10 - Maths - Real Numbers (NCERT Ex 1.1)

Real Numbers

(NCERT Ex 1.1)

1. Use Euclid’s division algorithm to find the HCF of :
 
(i) 135 and 225 
(ii) 196 and 38220 
(iii) 867 and 255

Answer:
(i) 135 and 225
Step 1 : Since 225 > 135, apply Euclid’s division lemma to 225 and 135 to get
                                225 = 135 × 1 + 90

Step 2: Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to get
                                135 = 90 × 1 + 45

Step 3: Now the remainder r = 45 ≠ 0 so we apply Euclid’s division lemma to 90 and 45 to have
                                 90 = 2 × 45 + 0

Step 4: The remainder has now become zero. Since the divisor at this stage is 45,
            ∴ HCF(135,225) =45

(ii) 196 and 38220
Step 1: Since 38220 > 196, apply the division lemma to 38220 and 196 to get
                            38220 = 196 × 195 + 0

Step 2: Since remainder is zero, the process stops. Since the divisor at this stage is 196,
             ∴ HCF(38220, 196) = 196

(iii) 867 and 255

Step 1: Since 867 > 255, we apply the division lemma to 867 and 255 to get
                   867 = 255 × 3 + 102
Step 2: Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to have
                   255 = 102 × 2 + 51
Step 3: Now the new divisor  is 102 and new remainder is 51. Apply the division lemma to get
                   102 = 51 × 2 + 0
Step 4: Since the remainder is zero, the process stops. Since the divisor at this stage is 51,
              ∴ HCF(867, 255) = 51

CBSE Class 10 - Maths - Real Numbers (Study Points)

Real Numbers

Study Points

1. A lemma is a proven statement used for proving another statement.

2. An Algorithm is a series of well defined steps which gives a procedure for solving a type of problem.

3. Muḥammad ibn Mūsā al-Khwārizmī, a persian mathematician coined the term 'algorithm'.

4. Euclid's division lemma: Given positive integers a and b there exist whole numbers q and r satisfying a = bq +r, 0 ≤ r < b.

5. Euclid's division algorithm: In order to compute the HCF of two positive integers say a and b, with a > b by using Euclid's algorithm we follow the given below steps:

STEP-❶:	Apply Euclid's division lemma to a and b and obtain whole numbers q1 and r1 such that a = bq1 + r1, 0 ≤ r < b
STEP-❷:	If r1 = 0, b is the HCF of a and b.
STEP-❸:	If r1 ≠ 0, apply Euclid's division lemma to b and r1 and obtain two whole numbers q1 and r2 such that b = q1r1 + r2
STEP-❹:	If r2 = 0, r1 is the HCF of a and b
STEP-❺:	If r2 ≠ 0, apply Euclid's division lemma to r1 and r2 and continue the above process till the remainder rn is zero. The divisor at this stage i.e. rn-1or the non-zero remainder at the previous stage is the HCF of a and b.

6. Euclid’s Division Algorithm is stated for only positive integers but it can be extended for all integers except zero, i.e, b ≠ 0.

7. The Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorized) as product of primes, and this factorization is unique except for the order in which the prime factors occur.