1.1 PROGRESSIONS :

 

Those sequence whose terms follow certain patterns are called progression. Generally there are three types of progression.

(i) Arithmetic Progression (A.P.)

(ii) Geometric Progression (G.P.) 

(iii) Harmonic Progression (H.P.)

 1.2 ARTHMETIC PROGRESSION :

A sequence is called an A.P., if the difference of a term and the previous term is always same. i.e. 

d = tn+1 – tn

= Constant for all

nN

. The constant difference, generally denoted by ‘d’ is called the common difference.

Ex.1 Find the common difference of the following A.P. : 1,4,7,10,13,16 ......

Sol. 4 - 1 = 7 - 4 = 10 - 7 = 13 - 10 = 16 - 13 = 3 (constant).

 Common difference (d) = 3.

6.3 GENERAL FORM OF AN A.P. :

If we denote the starting number i.e. the 1st number by ‘a’ and a fixed number to the added is ‘d’ then a, a +

d, a + 2d, a + 3d, a + 4d, ...... forms an A.P.

Ex.2 Find the A.P. whose 1st term is 10 & common difference is 5.

Sol. Given : First term (a) = 10 & Common difference (d) = 5.

 A.P. is 10, 15, 20, 25, 30, .....

6.4 nth TERM OF AN A.P. :

Let A.P. be a, a + d, a + 2d, a + 3d, .....

Then, First term (a1

) = a + 0.d

Second term (a2

) = a + 1.d

Third term (a3

) = a + 2.d

. .

. .

. .

nth term (an) = a + (n - 1) d

 an = a + (n - 1) d is called the nth term.

Ex.3 Determine the A.P. whose their term is 16 and the difference of 5th term from 7th term is 12.

Sol. Given : a3

= a + (3 - 1) d = a + 2d = 16 .....(i)

a7

- a5

= 12 ....(ii)

(a + 6d) - (a + 4d) = 12

a + 6d - a - 4d = 12

2d = 12

d = 6

Put d = 6 in equation (i)

a = 16 - 12

a = 4

 A.P. is 4, 10, 16, 22, 28, ......

real number cbse previous question

1. Use Euclid’s division algorithm to find the HCF of :

(i) 56 and 814 (ii) 6265 and 76254

2. Find the HCF and LCM of following using Fundamental Theorem of Arithmetic method.

(i) 426 and 576 (ii) 625, 1125 and 2125

3. Prove that √3 is an irrational number.

4. Prove that √5 is irrational number.

5. Prove that 5 +√2 is irrational.

6. Prove that 2 +√ 3 is irrational.

7. Can we have any n belongs to N , where (7)^n ends with the digit zero.

8. Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non - terminating decimal expansion :

(i)77/210 (ii) 15/1600

9. An army contingent of 616 members is to march behind and 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?

10. 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 the same time, and go in the same direction. After how many minutes will they meet again at the starting point ?

11. Write a rational number between √2 and √3 .

12. Use Euclid’s’ Division Lemma to show that the square of any positive integer is either of the form 3m of 3m + 1 for some integer m. [CBSE - 2008]

Real Number class 10 short notes

 DIVISIBILITY :

A non-zero integer ‘a’ is said to divide an integer ‘b’ if there exists an integer ‘c’ such that b= ac. The intege ‘b’ is called dividend, integer ‘a’ is known as the divisor and integer ‘c’ is known as the quotient.

For example, 5 divides 35 because there is an integer 7 such that 35 = 5 × 7.

If a non-zero integer ‘a’ divides an integer b, then it is written as a | b and read as ‘a a divides b’, a/b is

written to indicate that b is not divisible by a.

EUCLID’S DIVISION LEMMA :

Let ‘a’ and ‘b’ be any two positive integers. Then, there exists unique integers ‘q’ and ‘r’ such that 

                         a = b + r,  

where 0≤ r<b. If b|a, than r = 0.

Ex.1 Show that any positive odd integer is of the form 6q + 1 or, 6q + 3 or, 6q + 5, where q is some integer.

Sol. 

 Let ‘a’ be any positive integer and b = 6. 

Then, by Euclid’s division lemma there exists integers ’a’ and ‘r’ 

such that

 a = 6q + r, where 0≤ r < 6.

     ⇒a = 6q or, a = 6q + 1 or, a = 6q + 2 or, a = 6a + 3 or, a = 6q + 4 or, a = 6q + 5.

[∴ 0 ≤  r < 6  r = 0, 1,2,3,4,5]

 ⇒ a = 6q + 1 or, a = 6q + 3 or, a = 6q + 5.

[∴ a is an odd integer, ∴6q, a ≠  6q + 2, a ≠  6q + 4]

Hence, any odd integer is of the form 6q + 1 or, 6q + 3 or, 6q + 5.

Ex.2 Use Euclid’s Division Lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9 m+ 8, for some integer q.

Sol

do your self

Ex.3 Prove that the square of any positive integer of the form 5q + 1 is of the same form.

EUCLID’S DIVISION ALGORITHM : 

If ‘a’ and ‘b’ are positive integers such that 

a = bq + r, 

then every common divisor of ‘a’ and ‘b’ is a common divisor of ‘b’ and ‘r’ and vice-versa.

Ex.4 Use Euclid’s division algorithm to find the H.C.F. of 196 and 38318. 

Sol. 

Applying Euclid’s division lemma to 196 and 38318.

 38318 = 195 × 196 + 98

 196 = 98 × 2 + 0

 The remainder at the second stage is zero.

 So, the H.C.F. of 38318 and 196 is 98.

Ex.5 If the H.C.F. of 657 and 963 is expressible in the form 657x + 963 × (-15), find x. 

Sol. Applying Euclid’s division lemma on 657 and 963.

 963 = 657 × 1 + 306 

657 = 306 × 2 + 45

 306 = 45 × 6 + 36 

45 = 36 × 1 + 9 

36 = 9 × 4 + 0

 So, the H.C.F. of 657 and 963 is 9. 

Given : 657x + 963 × (-15) = H.C.F. of 657 and 963. 

657 x + 963 × (-15) = 9 

657 x = 9 + 963 × 15

 657 x = 14454 

x=14454 /657 = 22

 Ex.6 What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.

 Sol. Clearly, the required number is the H.C.F. of the number 626 - 1 = 625, 3127 - 2 3125 and 

15628 - 3 = 15625.

 15628 - 3 = 15625. 

Using Euclid’s division lemma to find the H.C.F. of 625 and 3125. 

3125 = 625 × 5 + 0

 Clearly, H.C.F. of 625 and 3125 is 625. 

Now, H.C.F. of 625 and 15625 

15625 = 625 × 25 + 0

 So, the H.C.F. of 625 and 15625 is 625. 

Hence, H.C.F. of 625, 3125 and 15625 is 625. 

Hence, the required number is 625. 

Ex.7 144 cartons of coke cans and 90 cartons of Pepsi cans are to be stacked is a canteen. If each stack is of same height and is to contains cartons of the same drink, what would be the greatest number of cartons each stack would have ?

 Sol. In order to arrange the cartons of the same drink is the same stack, we have to find the greatest number that divides 144 and 90 exactly. Using Euclid’s algorithm, to find the H.C.F. of 144 and 90. 

144 = 90 × 1 + 54

 90 = 54 × 1 + 36 

54 = 36 × 1 + 18 

36 = 18 × 2 + 0 

So, the H.C.F. of 144 and 90 is 18. 

Number of cartons in each stack = 18. 

FUNDAMENTAL THEOREM OF ARITHMETIC :

Every composite number can be expressed as a product of primes, and this factorisation is unique, except for the order in which the prime factors occurs

SOME IMPORTANT RESULTS : 

(i) Let ‘p’ be a prime number and ‘a’ be a positive integer. If ‘p’ divides a 2, then ‘p’ divides ‘a’. 

(ii) Let x be a rational number whose decimal expansion terminates. Then, x can be expressed in the form  p/q ,where p and q are co-primes, and prime factorisation of q is of the form2n× 5ⁿ  where m, n are non-negative integers. 

(iii) Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2m × 5n where m, n are non - negative integers. Then, x has a decimal expansion which is non - terminating repeating.

Ex.8 Determine the prime factors of 45470971. 

Sol. ∴45470971 = 72 × 132 × 172 × 19. 

Ex.9 Check whether 6n can end with the digit 0 for any natural number.

 Sol. Any positive integer ending with the digit zero is divisible by 5 and so its prime factorisations must contain the prime 5.

 6ⁿ  = (2 × 3)ⁿ  = 2ⁿ × 3ⁿ  

⇒ The prime in the factorisation of 6 n is 2 and 3.

⇒ 5 does not occur in the prime factorisation of 6n for any n.

⇒ 6ⁿ does not end with the digit zero for any natural number n. 

Ex.10 Find the LCM and HCF of 84, 90 and 120 by applying the prime factorisation method.

 Sol. 84 = 22 × 3 × 7, 90 = 2 × 32 × and 120 = 23 × 3 × 5.

                                        

                                         Prime factors     Least exponent 

                                                    2                            1            

                                                    3                            1            

                                                    5                            0

                                                    7                            0

               ∴  HCF = (2)^1 × (3 )^1= 6.

                                Common prime factors     Greatest exponent 

                                                2                                         3

                                                3                                         2 

                                                5                                         1

                                                7                                         1

                  ∴ LCM = (2)^3 × (3)^3 × (5)^1 ×( 7)^1 = 8 × 9 × 5 × 7 = 2520. 

Ex.11 In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that they can cover the distance in complete steps ? 

Sol. Required minimum distance each should walk so, that they can cover the distance in complete step is the L.C.M. of 80 cm, 85 cm and 90 cm 

80 = (2)^4 × 5 

85 = 5 *17

 90 = 2 × (3)^2 × 5 

 ∴ LCM = (2)^4 × (3)^2 × (5)^1 × (17)^1

     LCM = 16 × 9 × 5 × 17 

     LCM = 12240 cm, = 122 m 40 cm. 

Ex.12 Prove that √ 2 is an irrational number.

 Sol. Let assume on the contrary that √ 2 is a rational number. 

Then, there exists positive integer a and b such that

 √ 2=a/b where, a and b are co primes i.e. their HCF is 1. 

 Ex.13 Prove that 3  5 is an irrational number. 

Sol. 

Ex.14 Without actually performing the long division, state whether 3125 13 has terminating decimal expansion or not. 

Sol. 

        13/3125  = 13/(2)^0(5)^5

 This, shows that the prime factorisation of the denominator is of the form 2m × 5n. 

Hence, it has terminating decimal expansion.

 Ex.15 What can you say about the prime factorisations of the denominators of the following rationals : (i) 43.123456789 (ii) 43. 123456789 

Sol. (i) Since, 43.123456789 has terminating decimal, so prime factorisations of the denominator is of the form 2m × 5n, where m, n are non - negative integers. 

(ii) Since, 43. 123456789 has non-terminating repeating decimal expansion. So, its denominator has factors other than 2 or 5

problems on probability

 OBJECTIVE 

1. If there coins are tossed simultaneously, then the probability of getting at least two heads, is

(A) 1/4 (B) 3/8 (C) 1/2 (D) 1/4

2. A bag contains three green marbles four blue marbles, and two orange marbles. If marble is picked at random, then the probability that it is not a orange marble is

(A) 1/4  (B) 1/3 (C) 4/9  (D) 7/9

3. A number is selected from number 1 to 27. The probability that it is prime is

(A) 2/3  (B) 1/6 (C) 1/3 (D) 2/9

4. IF (P(E) = 0.05, then P (not E) =

(A) -0.05 (B) 0.5 (C) 0.9 (D) 0.95

5. A bulb is taken out at random from a box of 600 electric bulbs that contains 12 defective bulbs. Then the probability of a non-defective bulb is

(A) 0.02 (B) 0.98 (C) 0.50 (D) None

SUBJECTIVE 

1. To dice are thrown simultaneously. Find the probability of getting :

(i) An even number of the sum

(ii) The sum as a prime number

(iii) A total of at least 10

(iv) A multiple of 2 on one dice and a multiple of 3 on the other.

2. Find the probability that a leap year selected at random will contain 53 Tuesdays.

3. A bag contains 12 balls out of which x are white.(i) If one ball is drawn at random, what is the probability it will be a white ball ? (ii) If 6 more white balls are put in the box. The probability of drawing a white ball will be double than that is (i). Find x.

4. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil a card and puts them into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is: (i) The name of a girl (ii) The name of boy ?

5. The probability of selecting a green marble at random from a jar that contains only green, white and yellow marbles is 1/4. The probability of selecting a white marble from the same jar is 1/3. If this jar contains 10 yellow marbles. What is the total number of marbles in the jar ?

6. A card is drawn at random from a well suffled desk of playing cards. Find the probability that the card drawn is (i) A card of spade or an ace (ii) A red king (iii) Neither a king nor a queen (iv) Either a king or a queen

7. There are 30 cards of same size in a bag on which number 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number of the selected card is not divisible by 3.

8. In figure points A,B,C and D are the centers of four circles that each have a radius of length on unit. If a point is selected at random from the interior of square ABCD. What is the probability that the point will be chosen from the shaded region ?

9. A bag contains 5 white balls, 6 red balls, 6 black balls and 8 green balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is (i) White (ii) Red or black (iii) Note green (iv) Neither white nor black  [CBSE - 2006]

10. A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. Find the probability of getting a black ball. [CBSE - 2008]

11. Cards. marked with number 5 to 50, are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is (i) a prime number less than 10. (ii) a number which is a perfect square.

#jee #math #probability 

Probability

1.1 EXPERIMENT : The word experiment means an operation, which can produce well defined outcomes. 

The are two types of experiment : 

                   (i) Deterministic experiment 

                    (ii) Probabilistic or Random experiment 

(i) Deterministic Experiment : Those experiment which when repeated under identical conditions, produced the same results or outcome are known as deterministic experiment. For example, Physics or Chemistry experiments performed under identical conditions.

 (ii) Probabilistic or Random Experiment :- In an experiment, when repeated under identical conditions donot produce the same outcomes every time. For example, in tossing a coin, one is not sure that if a head or tail will be obtained. So it is a random experiment. 

Sample space : The set of all possible out comes of a random experiment is called a sample space associated with it and is generally denoted by S. 

For example, When a dice is tossed then S = {1, 2, 3, 4, 5, 6}.

 Event : A subset of sample space associated with a random experiment is called an event. For example, In tossing a dive getting an even no is an event. 

Favorable Event : Let S be a sample space associated with a random experiment and A be event associated with the random experiment. The elementary events belonging to A are know as favorable events to the event A. 

For example, in throwing a pair of dive, A is defined by “Getting 8 as the sum”. Then following elementary events are as out comes : (2, 6), (3, 5), (4, 4) (5, 3) (6, 2). So, there are 5 elementary events favorable to event A. 

1.2 PROBABILITY : If there are n elementary events associated with a random experiment and m of them are favorable to an event A, then the probability of happening or occurrence of event A is denoted by P(A)

 Thus,

    

P(A)     =   (Total number of favourable outcomes)/(Total number of possible outcomes)

                   =  m/n 

And 0 £ P(A) £ 1

 If, P(A) = 0, then A is called impossible event

If, P(A) = 1, then A is called sure event

                                P(A) + P (A") = 1

Where P(A) = probability of occurrence of A. 

            P (A') = probability of non - occurrence of A.

 

Ex.1 A box contains 5 red balls, 4 green balls and 7 white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is (i) white (ii) neither red nor white

Sol. Total number of balls in the bag = 5 + 4 + 7 = 16

\ Total number of elementary events =16

(i)                There are 7 white balls in the bag.

\ Favorable number of elementary events = 7

P(Getting a white ball ) = (Total No. of elementaryevents) /(Total No. favourable elementaryevents)

                    = 7 /16

 (ii) There are 4 balls that are neither red nor white 

 Favorable number of elementary events = 4 

Hence, P (Getting neither red not white ball) = 4/16 =1/4

 Ex.2 All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting  [CBSE - 2007] (i) black face card                     (ii) a queen             (iii) a black card.

 Sol. After removing three face cards of spades (king, queen, jack) from a deck of 52 playing cards, there are 49 cards left in the pack. 

Out of these 49 cards one card can be chosen in 49 ways. 

Therefore, Total number of elementary events = 49 

(i) There are 6 black face cards out of which 3 face cards of spades are already removed. 

So, out of remaining 3 black face cards one black face card ban be chosen in 3 ways. 

Therefore, Favorable number of elementary events = 3 

Hence, P (Getting a black face card ) = 3/49

 (ii) There are 3 queens in the remaining 49 cards.

 So, out of these three queens, on queen can be chosen in 3 ways 

Therefore, Favorable number of elementary events = 3

 Hence P (Getting a queen) = 3/49

 (iii) There are 23 black cards in the remaining 49 cards, 

So, out to these 23 black card, one black card can be chosen in 23 ways 

Therefore, Favorable number of elementary events = 23 

Hence, P (Getting a black card) =  23/49

 Ex.3 A die is thrown, Find the probability of (i) prime number (ii) multiple of 2 or 3 (iii) a number greater than 3.

Sol.

In a single throw of die any one of six numbers 1,2,3,4,5,6 can be obtained. 

Therefore, the tome number of elementary events associated with the random experiment of throwing a die is 6.

 (i) Let A denote the event “Getting a prime no”. 

Clearly, event A occurs if any one of 2,3,5 comes as out come. 

Therefore, Favorable number of elementary events = 3 

Hence, P (Getting a prime no.) = 3/6=1/2

(ii) An multiple of 2 or 3 is obtained if we obtain one of the numbers 2,3,4,6 as out comes.

Therefore, Favorable number of elementary events = 4 

Hence, P (Getting multiple of 2 or 3) = 4/6=2/3

 (iii) The event “Getting a number greater than 3” will occur, if we obtain one of number 4,5,6 as an out come. 

Therefore, Favorable number of out comes = 3

 Hence, required probability = 3/6=1/2

 Ex.4 Two unbiased coins are tossed simultaneously. Find the probability of getting (i) two heads (ii) at least one head (iii) at most one head. 

Sol. 

If two unbiased coins are tossed simultaneously, 

we obtain any one of the following as an out come : HH, HT, TH, TT 

Therefore, Total number of elementary events = 4

 (i) Two heads are obtained if elementary event HH occurs. 

Therefore, Favorable number of events = 1 Hence, P (Two heads) =  1/4

 (ii) At least one head is obtained if any one of the following elementary events happen : HH, HT, TH 

Therefore, favorable number of events = 3 

Hence P (At least one head) = 3/4

 (iii) If one of the elementary events HT, TH, TT occurs, than at most one head is obtained 

Therefore, favorable number of events = 3 

Hence, P (At most one head) = 3/3

Ex.5 A box contains 20 balls bearing numbers, 1,2,3,4…...20. A ball is drawn at random from the box. What is the probability that the number of the ball is (i) an odd number (ii) divisible by 2 or 3 (iii) prime number 

Sol. Here, total numbers are 20. 

 Total number of elementary events = 20 

(i) The number selected will be odd number, if it is elected from 1,3,5,7,9,11,13,15,17,19 

 Therefore, Favorable number of elementary events = 10 

Hence, P (An odd number ) = 10/20=1/2

 (ii) Number divisible by 2 or 3 are 2,3,4,6,8,9,10,12,14,15,16,18,20 

Therefore, Favorable number of elementary events = 13

 P (Number divisible by 2 or 3) =  13/20

 (iii) There are 8 prime number from 1 to 20 i.e., 2,3,5,7,11,13,17,19

Therefore, Favorable number of elementary events = 8 

P (prime number ) = 8/20=2/5

 Ex.6 A die is drop at random on the rectangular region as shown in figure. What is the probability that it will land inside the circle with diameter 1m ? 

comment answer of this question

Sunday math class notes 25.07.2021

Click here to download class notes oF AP

Class 10 new syllabus session 2021-2022

             COURSE STRUCTURE

               CLASS –X (2021-22)

       FIRST TERM

 One Paper

 90 Minute

UNIT-NUMBER SYSTEMS

1. REAL NUMBER

Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and 

after illustrating and motivating through examples. Decimal representation of rational 

numbers in terms of terminating/non-terminating recurring decimals.

UNIT-ALGEBRA

2. POLYNOMIALS 

Zeroes of a polynomial. Relationship between zeroes and coefficients of quadratic 

polynomials only.

3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 

Pair of linear equations in two variables and graphical method of their solution, 

consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a 

pair of linear equations in two variables algebraically - by substitution and by elimination. 

Simple situational problems. Simple problems on equations reducible to linear 

equations.

UNIT-COORDINATE GEOMETRY

4. COORDINATE GEOMETRY

 LINES (In two-dimensions) 

 Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. 

Section formula (internal division)

 UNIT-GEOMETRY

 5. TRIANGLES 

Definitions, examples, counter examples of similar triangles. 

1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides 

in distinct points, the other two sides are divided in the same ratio. 

2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.

3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding

sides are proportional and the triangles are similar.

4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding

angles are equal and the two triangles are similar.

5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides

including these angles are proportional, the two triangles are similar.

6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle

to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole

triangle and to each other.

7. (Motivate) The ratio of the areas of two similar triangles is equal to the ratio of the

squares of their corresponding sides.

8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the

squares on the other two sides.

9. (Motivate) In a triangle, if the square on one side is equal to sum of the squares on the

other two sides, the angle opposite to the first side is a right angle.

UNIT- TRIGONOMETRY

6. INTRODUCTION TO TRIGONOMETRY

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well

defined). Values of the trigonometric ratios of 30

, 45 and 60

. Relationships between the

ratios.

TRIGONOMETRIC IDENTITIES

Proof and applications of the identity . Only simple identities to be given

UNIT-MENSURATION

7. AREAS RELATED TO CIRCLES

Motivate the area of a circle; area of sectors and segments of a circle. Problems based on

areas and perimeter / circumference of the above said plane figures. (In calculating area of

segment of a circle, problems should be restricted to central angle of 60° and 90° only.

Plane figures involving triangles, simple quadrilaterals and circle should be taken.)

UNIT- STATISTICS & PROBABILITY

 8. PROBABILITY

Classical definition of probability. Simple problems on finding the probability of an event

UNIT-ALGEBRA

1. QUADRATIC EQUATIONS (10 Periods)

Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic

equations (only real roots) by factorization, and by using quadratic formula. Relationship

between discriminant and nature of roots. Situational problems based on quadratic

equations related to day to day activities (problems on equations reducible to quadratic

equations are excluded)

2. ARITHMETIC PROGRESSIONS

Motivation for studying Arithmetic Progression Derivation of the nth term and sum of

the first n terms of A.P. and their application in solving daily life problems.

(Applications based on sum to n terms of an A.P. are excluded)

UNIT- GEOMETRY

3. CIRCLES

Tangent to a circle at, point of contact

1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the

point of contact.

2. (Prove) The lengths of tangents drawn from an external point to a circle are equal.

4. CONSTRUCTIONS

1. Division of a line segment in a given ratio (internally).

2. Tangents to a circle from a point outside it.

UNIT-TRIGONOMETRY

5. SOME APPLICATIONS OF TRIGONOMETRY

 HEIGHTS AND DISTANCES-Angle of elevation, Angle of Depression.

Simple problems on heights and distances. Problems should not involve more than two

right triangles. Angles of elevation / depression should be only 30°, 45°, 60°.

UNIT-MENSURATION

6. SURFACE AREAS AND VOLUMES

1. Surface areas and volumes of combinations of any two of the following: cubes,

cuboids, spheres, hemispheres and right circular cylinders/cones.

2. Problems involving converting one type of metallic solid into another and other mixed

problems. (Problems with combination of not more than two different solids be taken).

UNIT-STATISTICS & PROBABILITY

 7. STATISTICS

Mean, median and mode of grouped data (bimodal situation to be avoided). Mean by

Direct Method and Assumed Mean Method only

Pair of Linear Equations in Two Variables Linear Equation

* An equation which can be put in the form

 ax + by + c = 0, where a, b and c are real numbers 

and both a and b are nonzero is called a linear equation in two variables.

Solution of an Equation

 

• Each solution (x, y) of a linear equation in two variables. 
• ax + by + c = 0, corresponds to a point on the line representing the equation, and vice-versa

Pair of Linear Equations in Two Variables

• 
  
  
  
  
• 
  
  
  
  
• 
  
  
  
  
  
  
  The solution is (4, 2), the point of intersection.

• To summarize the behavior of lines representing a pair of linear equations in two variables:
• The lines may intersect in a single point. In this case, the pair of equations has a unique solution(consistent pair of equations).
• The lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).
• The lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations]

Substitution Method

      The following are the steps:

• Find the value of one variable, say y in terms of the other variable, i.e., x from either equation,whichever is convenient.
• Substitute this value of y in the other equation, and reduce it to an equation in one variable,i.e., in terms of x, which can be solved. Sometimes, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.
• Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

Elimination Method

 

Steps in the elimination method:

• First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
• Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3. If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions. If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
• Solve the equation in one variable (x or y) so obtained to get its value.
•  Substitute this value of x (or y) in either of the original equations to get the value of the other

Cross Multiplication Method

Linear equation in two variables Previous year question paper class 10.

Q1). Formulate the following problems as a pair of equations, and hence find their solutions:

Anjan can row downstream 40 km in 4 hours, and upstream 2 km in 1 hours. Find her speed of rowing in still water and the speed of the current.

Q2). Solve the following pair of linear equations:

y -4x= 1
6x- 5y= 9

Q3.) Solve using cross multiplication method:

x+y=1

2x – 3y = 11.

Q4). Solve for x and y

x + 2y – 3= 0
3x – 2y + 7 = 0

Q5). Solve for x and y,

2x= 5y + 4;

3x-2y + 16 = 0

Q6). 6(ax + by) = 3a + 2b

6(bx – ay) = 3b -2a

Find the value of x and y.

Q7). If the system of equations

6x + 2y = 3 and kx + y = 2 has a unique solution, find the value of k.(CBSE 2013)

Q8). How many solutions does the pair of equations y = 0 and y = -5 have? (2013)

Q9). For what value of k, the pair of equations 4x – 3y = 9, 2x + ky = 11 has no solution? (2017D)

Q10). Calculate the area bounded by the line x + y = 10 and both the co-ordinate axes. (2012)

Q11). Find whether the following pair of linear equations is consistent or inconsistent: (2015)

3x + 2y = 8 6x – 4y = 9

Q12). Draw the graph of

2y = 4x – 6; 2x = y + 3 and determine whether this system of linear equations has a unique solution or not.

Q13). Draw the graph of

2y = 4x – 6; 2x = y + 3 and determine whether this system of linear equations has a unique solution or not.

Q14). Represent the following pair of equations graphically and write the coordinates of points where the lines intersect y-axis.

Q15). Solve the following pair of linear equations for x and y:

141x + 93y = 189;

93x + 141y = 45 (2013)

Q16).Solve by elimination: (2014)

3x = y + 5

5x – y = 11

Q17). Solve by elimination: 2015

3x – y – 7

2x + 5y + 1 = 0

Solve the following pair of equations: (2014)

49x + 51y = 499

51x + 49 y = 501

Q18). Find the two numbers whose sum is 75 and difference is 15. (2014)

Q19). Solve the following pair of linear equations by the cross multiplication method: x + 2y = 2; x – 3y = 7 (2015)

Q20). Srikant earns ₹600 per month more than his wife. One-tenth of the Srikant's salary and l/6th of the wife’s salary amount to ₹1,500, which is saved every month. Find their incomes. (2014)

Q21). The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number. (2015)

Q22). The age of the father is twice the sum of the ages of his 2 children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father. (2012)

Real numbers NCERT CHAPTER 1 previous question paper class 10 MATHEMATICS

Q 1. Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively

Q2. Express 98 as a product of its primes?

Q3.If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.

Q4. Find HCF and LCM of 13 and 17 by prime factorisation method. (2013)

Q5.  HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number. (2012)

Q6.  Find the LCM of 96 and 360 by using fundamental theorem of arithmetic. (2012)

Q7. Find the HCF (865, 255) using Euclid’s division lemma. (2013)

Q8. Find the largest number which divides 70 and 125 leaving remainder 5 and 8 respectively. (2015)

Q9. Prove that 2 + 3√5 is an irrational number. (2014)

Q10. Show that 3√7 is an irrational number. (2016)

Q11.  Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number? (2015)

Q12.  Check whether 4n can end with the digit 0 for any natural number n. (2015)

Q13.   Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons. (2017 OD)

Q14. Prove that √5 is irrational and hence show that 3 + √5 is also irrational. (2012) 3 marks

Q16.  Prove that 3 + 2√3 is an irrational number. (2014)

Q15. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together? (2013)

Q17. Two tankers contain 850 liters and 680 liters of petrol. Find the maximum capacity of a container which can measure the petrol of each tanker in the exact number of times. (2012)

Q18.   The length, breadth, and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly. (2015)

Q19. Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they start ringing together, after how much time will they next ring together? (2015)

Q20.  By using Euclid’s algorithm, find the largest number which divides 650 and 1170.

2. POLYNOMIAL NCERT EXERCISE

                                       EXERCISE 2.1

EXERCISE 2.2

EXERCISE 2.3

 

Polynomial Previous year question paper class 10. NCERT CHAPTER 2

1. For what value of k, (–4) is a zero of the polynomial x2 – x – (2k + 2)?

(CBSE 2009)

2.  For what value of p, (–4) is a zero of the polynomial x2 – 2x – (7p + 3)?

(CBSE 2009)

3.  If 1 is a zero of the polynomial p(x) = ax2 – 3(a – 1) x – 1, then find the value of a.

(Al CBSE 2009)

4.  If (x + a) is a factor of 2x2 + 2ax + 5x + 10 find a.

(Al CBSE 2008 F)

5.  Write the zeroes of the polynomial x2 + 2x + 1.

(CBSE 2008)

6.  Write the zeroes of the polynomial x2 – x – 6.

(CBSE 2008)

7.  Write a quadratic polynomial, the sum and product of whose zeroes are 3 and –2 respectively.

12. Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficient of the polynomial.

(CBSE 2008)

13. Find the zeroes of the quadratic polynomial 5x2 – 4 – 8x and verify the relationship between the zeroes and the coefficient of the polynomial.

(AI CBSE 2008)

14. Find the quadratic polynomial, the sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.

(CBSE 2008)

15. If one zero of the polynomial (a2 – 9) x2 + 13x + 6a is reciprocal of the other, find the value of ‘a’.

(AI CBSE 2008)

16. If the product of zeroes of the polynomial ax2 – 6x – 6 is 4, find the value of ‘a’.

(AI CBSE 2008)

17. Find all the zeros of the polynomial x4 + x3 – 34x2 – 4x + 120, if two of its zeroes are 2 and – 2.

(AI CBSE 2008)

18. Find all the zeroes of the polynomial 2x4 + 7x – 19x2 – 14x + 30, if two of its zeroes are 

(AI CBSE 2008)

19. Find the quadratic polynomial whose zeroes are 1 and –3. Verify the relation between the coefficients and the zeroes of the polynomial.

(CBSE 2008 C)

20. Find the zeroes of the quadratic polynomial 4x2 – 4x – 3 and verify the relation between the zeroes and its coefficients.

(CBSE 2008 C)

21. Obtain all other zeroes of the polynomial 2x3 – 4x – x2 + 2, if two of its zeroes are 

(CBSE 2008 C)

22. Find all the zeroes of x4 – 3x3 + 6x – 4, if two of its zeroes are 

(AI CBSE 2008 C)

23. Find a quadratic polynomial whose zeroes are –4 and 3 and verify the relationship between the zeroes and the coefficients.

(AI CBSE 2008 C)

24. Using division algorithm, find the quotient and remainder on dividing f(x) by g(x), where f(x) = 6x3 + 13x2 + x – 2 and g(x) = 2x + 1

(AI CBSE 2008 C)

25. If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1 then the remainder comes out to be ax + b, find ‘a’ and ‘b’

(CBSE 2009)

26. If the polynomial x4 + 2x3 + 8x2 + 12x + 18 is divided by another polynomial x2 + 5, the remainder comes out to be px + q. Find the value of p and q.

(CBSE 2009)

27. Find all the zeroes of the polynomial x3 + 3x2 – 2x – 6, if two of its zeroes are – 

(AI CBSE 2009)

28. Find all the zeroes of the polynomial 2x3 + x2 – 6x – 3, if two of its zeroes are – 

(AI CBSE 2009)

29. If α and β are zeroes of the quadratic polynomial x2 – 6x + a; find the value of ‘a’ if 3α + 2β = 20.