CBSE MATH STUDY: December 2012

Wednesday, 26 December 2012

CBSE I NCERT Arithmetic Progression-X Solved Problems

1. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top .If the top and the bottom rungs are two and a half meter apart, what is the length of the wood required for the rungs?

2. The sum of first n terms of an AP is given by Sn = 3n2 + 5n find the nth term of the AP.

3. How many terms of the ap -6,-11/2 , -5,...... are needed to give the sum -25

4. Find a, b such that 27, a, b - 6 are in A.P.

5. Find the sum of all the odd numbers between 50 and 150 divisible by 7

6. The sum of third and seventh term of an AP is 6 and their product is 8.

7. If Sn = n2p and Sm = m2p, (m not equal ton), is an A.P. prove that Sp = p3.

8.Show that the sum of (m+n)th term and (m-n)thterm of an A.Pis equal to twice the mth term.

9. If the ratio of the sums of n terms of 2 APs is n+1:3n+1, then find the ratio of the 7th terms of the AP.

10. Determine the sum of the first 30 terms of the sequence whose nth term is given by tn=2n+9/3

11. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

12.The sum of the first and the last terms of an AP is 60,the sum of n terms of the AP is720.what is n?

13. If pth , qth and rth term of an AP are a,b,c respectively , then show that (a-b)r +(b-c)p + (c-a)q = 0

14. if (b+c)/a, (c+a)/b, (a+b)/c are in A.P. show that bc,ca,ab re in A.P.

15. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms

Class X ARITHMETIC PROGRESSIONS (8) Periods

Motivation for studying AP. Derivation of standard results of finding the nth term and sum of first n terms and their application in solving daily life problems.

www.jsuniltutorial.weebly.com/ (To get solution in pdf)

Wednesday, 19 December 2012

11th Maths Chapter Set Theory

Class XI Maths Chapter Set Sets

1.1 Sets and their Representations

1.2 The Empty Set

1.3 Finite and Infinite Sets

1.4 Equal Sets

1.5 Subsets

1.6 Power Set

1.7 Universal Set

1.8 Venn Diagrams

1.9 Operations on Sets

1.10 Complement of a Set

The theory of sets was developed by German mathematician Georg Cantor (1845-1918). He first encountered sets while working on “problems on trigonometric series”.

A set is a well define collection of object.

Greek symbol ∈ (epsilon) is used to denote the phrase ‘belongs to’.

If ‘a’ is an element of a set A, we write a ∈ A.

If ‘b’ is not an element of a set A, we write b ∉ A and read “b does not belong to A”.

There are two methods we used to represents a set

a. Roaster or Tabular form: We use commas to separate elements of sets within braces { } without repainting elements. The order of listing elements has no relevance.

For example, the set of all even positive integers less than 5 is described in roster form as {2, 4,}.

b. Set builder form: We use this form if all the elements of a set possess a single common property which is not possessed by any element outside the set.

Example: V = {x : x is a vowel in English alphabet}

V is read as “the set of all x such that x is a vowel of the English alphabet”. In this description the braces stand for “the set of all”, the colon stands for “such that”.

The set of all natural numbers which divide 42 is written as A= {x : x is a natural number which divides 42}

Q. Write the set {x: x is a positive integer and x² < 40} in the roster form.

Solution: x² < 40 here, x = 1, 2, 3, 4, 5, 6.(Whose square is less than 40 )

x = {1, 2, 3, 4, 5, 6)

Q. Write the set A = {1, 4, 9, 16, 25…} in set-builder form

Ans: 1, 4, 9, 16, 25… are square of 1, 2, 3, 4, 5… respectively. Or Square of natural number

A = {x: x is the square of a natural number}

Or,

A = { x : x = n² , Where n ∈ N}

Q. Write the set {1/2 , 2/3, ¾, 4/5, 5/6, 6/7 } in the set-builder form

Solution We see that each member in the given set has the denominator one more than the numerator

Also, the numerators begin from 1 and do not exceed 6.

{x : x = n/n+1 where is a natural number and 1£ n£6}

Finite and Infinite set

Let us see new set: B = { x : x is a student presently studying in both Classes X and XI }

We observe that a student cannot study simultaneously in both Classes X and XI. Thus, the set B contains no element at all. Such type of set is called Empty Set.

A set which does not contain any element is called the empty set or the null set or the void set. The empty set is denoted by the symbol φ or { }

Let us see another new set

A = {men living presently in your town}

here we do not know the numbers of element of this set

B= A = {1, 2, 3, 4, 5}

This set has a definite number of elements

Such type of sets ate called Infinite and Finite respectively.

Hence, A set which is empty or consists of a definite number of elements is called finite otherwise, the set is called infinite.

Note : All infinite sets cannot be described in the roster form. For example, the set of real numbers cannot be described in this form, because the elements of this set do not follow any particular pattern.

Q. State which of the following sets are finite or infinite:

(i) {x : x ∈ N and (x – 1) (x –2) = 0}

(ii) {x : x ∈ N and x2 = 4}

(iii) {x : x ∈ N and 2x –1 = 0}

(iv) {x : x ∈ N and x is prime}

(v) {x : x ∈ N and x is odd}

Solution (i) Given set = {1, 2}. Hence, it is finite.

(ii) Given set = {2}. Hence, it is finite.

(iii) Given set = φ. Hence, it is finite.

(iv) The given set is the set of all prime numbers and since set of prime numbers is infinite. Hence the given set is infinite

(v) Since there are infinite numbers of odd numbers, hence, the given set is infinite.

Equal Sets

Two set A and B are said to be equal if they have exactly the same elements and we

Q. write A = B. Otherwise, the sets are said to be unequal

Q. Find the pairs of equal sets, if any, give reasons:

A = {0}, B = {x : x > 15 and x < 5},

C = {x : x – 5 = 0 }, D = {x: x2 = 25},

E = {x : x is an integral positive root of the equation x2 – 2x –15 = 0}.

Solution: Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E, it follows that, A ≠ B, A ≠ C, A ≠ D, A ≠ E.

Since B = φ but none of the other sets are empty. Therefore B ≠ C, B ≠ D and B ≠ E. Also C = {5} but –5 ∈ D, hence C ≠ D.

Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E.

Thus, the only pair of equal sets is C and E.

Subsets

Consider the sets : X = set of all students in your school, Y = set of all students in your

class.

We note that every element of Y is also an element of X; we say that Y is a subset of X. The fact that Y is subset of X is expressed in symbols as Y ⊂ X. The symbol ⊂ stands for ‘is a subset of’ or ‘is contained in’.

Definition 4 A set A is said to be a subset of a set B if every element of A is also an element of B

Class 11^th and 12^th Study materials

11th Study Materials Maths and Science

12th Study Materials Maths and Science

11th Study Materials Accountancy

11^th and 12^th Sample papers

Tuesday, 18 December 2012

8th Construction of Quadrilaterals CBSE Guess Paper

Chapter: Practical Geometry

Q.1 Construct a quadrilateral ABCD in which AB = 4.4 cm, BC = 4 cm, CD = 6.4 cm, DA = 3.8 cm and BD = 6.6 cm

Q.2 Construct a parallelogram PQRS such that PQ = 5.2 cm, PR = 6.8 cm and QS = 8.2 cm

Q.3 Construct a rhombus with side 6 cm and one diagonal 8 cm. Measure the other diagonal.

Q.4 Construct if possible a quadrilateral ABCD give AB = 6cm, BC = 3.7 cm, CD =5.7cm, AD = 5.5 cm and BD = 6.1 cm. If not possible then give reasons for not able to construct it.

Q.5 Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3cm, AD = 2.3 cm, AC = 4.5 cm and BD = 3.8 cm

Q.6 Construct a quadrilateral PQRS in which QR = 4cm, RP = 5.6 cm, PS = 4.5 cm, RS = 5cm and QS = 6.5 cm.

Q.7 Construct a quadrilateral ABCD in which AB = 6 cm, BC = 4cm, CD = 4cm, <B= 95=, and <C = 90⁰

Q.8 Construct a quadrilateral STUV where ST = 4.2 cm, TU = 3.6 cm, UV = 4.8 cm,

<T = 30= and <U = 150⁰

Q.9 Construct a quadrilateral ABCD in which AB = BC = 3cm, AD = 5cm , <A = 90⁰

and < B = 105⁰

Q.10 Construct a quadrilateral BDEF where DE = 4.5 cm, EF = 3.5cm, FB = 6.5cm,

< F = 50= and < E = 100⁰

9th Construction of Triangle CBSE Guess Paper

Section - A

Q.1 With a ruler and compass which of the following angles cannot be constructed?

(a) 60⁰ (b) 80⁰ (c) 900 105⁰

Q.2 With a ruler and compass which of the following angles can be constructed?

(a) 80⁰ (b) 90⁰ (c) 100⁰ 110⁰

Section - B

Q.3 Construct an angle of 45⁰ at the initial point of a given ray and justify the construction.

Q.4 Construct the following angles and verify by measuring them by a protractor.

(i) 75⁰ (ii) 135⁰

Section - C

Q.5 Construct Triangle PQR with base QR = 3.8 cm and <Q=75⁰ and PQ + PR = 7.9cm

Q.6 Construct a with base Triangle PQ R a with base QR = 3.4 cm and <R=75⁰ and PQ - PR = 1.2cm

Q.7 Construct an equilateral triangle with sides 4cm.

Section -D

Q.8 Construct a triangle ABC in which and < B = 60⁰ and <C = 45⁰ and AB+BC+CA = 13 cm.

Q.9 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18cm.

Q.10 Construct a with its perimeter = 11cm and the base angles of 75⁰ and 30⁰.

Answers:

Q.1 b Q.2 b

For more help Visit : JSUNIL TUTORIAL

Saturday, 15 December 2012

8th(VIII)CBSE Chapter: Practical Geometry

Chapter: Practical Geometry:Constrtucton of Quadrilaterals