Q.1. Based on Euclid’s algorithm: a = bq + r ; Using Euclid’s algorithm: Find the HCF of 825 and 175.
Ans:Since 825>175, we apply division lemma to 825 and 175 to get825 = 175 x 4 + 125. Since r ≠ 0, we apply division lemma to 175 and 125 to get 175 = 125 x 1 + 50 Again applying division lemma to 125 and 50 we get, 125 = 50 x 2 + 25. Once again applying division lemma to 50 and 25 we get. 50 = 25 x 2 + 0. Since remainder has now become 0, this implies that HCF of 825 and 125 is 25. Q.2. Based on Showing that every positive integer is either of the given forms: Solved example: Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q. Ans: Let a be any odd positive integer (first line of problem) and let b = 4. Using division Lemma we can write a = bq + r, for some integer q, where 0≤r<4. So a can be 4q, 4q + 1, 4q + 2 or 4q + 3. But since a is odd, a cannot be 4q or 4q + 2. Therefore any odd integer is of the form 4q + 1 or 4q + 3. Q.Find H C F (26,91) if LCM(26,91) is 182 Sol: We know that LCM x HCF = Product of numbers. or 182 x HCF = 26 x 91 or HCF = 26 x 91 = 13 Hence HCF (26, 91) = 13. Q. prove that √5 is irrational. Solution: let us assume on the contrary that √5 is rational. That is we can find coprimes a and b b (≠0) such that √5 = a/b. Or √5b = a. Squaring both sides we get 5b2 = a2. This means 5 divides a2. Hence it follows that 5 divides a. So we can write a = 5c for some integer c. Putting this value of a we get 5b2 = (5c)2 Or 5b2 = 25c2 Or b2 = 5b2. It follows that 5 divides b2. Hence 5 divides b. Now a and b have at least 5 as a common factor. But this contradicts the fact that a and b are coprimes. This contradiction has arisen because of our incorrect assumption that √5 is rational. Hence it follows that √5 is irrational. Q. prove that product of three consecutive positive integers is divisible by 6? Ans: Let three consecutive positive integers be, n, n + 1 and n + 2. Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2. ∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3. ⇒ n (n + 1) (n + 2) is divisible by 3. Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1. ∴ n = 2q or 2q + 1, where q is some integer. If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2. If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2. ⇒ n (n + 1) (n + 2) is divisible by 2. Since, n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6. Q. Express HCF of F 65 and 117 in the form of 65m +117n Ans: Among 65 and 117; 117 > 65 Since 117 > 65, we apply the division lemma to 117 and 65 to obtain 117 = 65 x 1 + 52 … Step 1 Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain 65 = 52 x 1 + 13 … Step 2 Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain 52 = 4 x 13 + 0 … Step 3 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 From Step 2: 13 = 65 – 52 x 1 … Step 4 From Step 1: 52 = 117 – 65 x 1 Thus, from Step 4, it is obtained 13 = 65 – (117 – 65 x 1) x 1 ⇒13 = 65 x 2 – 117 ⇒13 = 65 x 2 + 117 x (–1) In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1 1oth maths term1
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CLASS X SCIENCE CHEMICAL REACTIONS AND EQUATIONS (solved questions) 1 – Mark questions answer 1. What happens when magnesium ribbon burns in air? Ans. When magnesium ribbon burns in air, it combines with the oxygen to form magnesium oxide. 2Mg(s) + O2(g) > 2MgO(s) 2. Name the gas evolved when zinc reacts with dil. HCl. Ans. Hydrogen gas is evolved. 3. What is a chemical equation? Ans. A chemical equation is a symbolic notation that uses formulae instead of words to represent a chemical equation. 4. On what chemical law, balancing of chemical equation is based? Ans. Balancing of a chemical equation is based on the law of conservation of mass. 5. Represent decomposition of ferrous sulphate with the help of balanced chemical equation. Ans. 2FeSO4(s) > Fe2O3(s) + SO2(g) + SO3(g) 6. When carbon dioxide is passed through lime water, it turns milky, why? For full Question and solution X chemical reactions equations_solved questions Download File Related posts and downloadable file X Chemical reactions and equations key consents Download X Chemical Reaction and Equation [Solved Questions] Download X Questions for Formative assessments Download 10th Chemical and chemical Equations (13 pages visit blog Click ) X Chemical reactions and Chemical Equation:FA Evaluation Assignments Solved X Chem. React.and Eq. NCERT solution Download File Chemical reactions and equations Activity Based Question Read and Practice Electric current and its effects links for study Related search and topics 10th science physics : Terminology : Electric current and its effects Electric current and its effects Formative Assessments Guess Paper SA1 2011  X Electricity and its effects 10th science Electric current and its effects notes 10th science physics Electric current and its effects Solved questions (12 marks) 10th science physics Electric current and its effects Test Paper NUMERICAL  Electricity & its effects For (Class X unsolved) Solved Electricity numerical for class 10 Current Electricity Numerical solved MCQ physics 10th Chapter Electric current and its effects Electric current and its effects: An Introduction To Electrical Power And Energy Electric current : An Introduction To Electric Potential and Potential Difference QUESTION BANK Class X Physics (Electricity) CBSE PHYSICS : 10th Electric Current & and Its Effect Multiple choice questions ELECTRIC CURRENT AND CIRCUIT+ Revision Notes 10th Solved Science Sample paper Module : 9P (01 & 02) Chapter: Motion Click Here Contents: 1. Motion in Living and NonLiving things. 2. Uniform and NonUniform Motion. 3. Distance Traversed and Displacement 4. Scalar and Vector Quantities. 5. Speed and velocity Learning Objectives The students will: 1. Understand and distinguish between Rest and Motion 2. Understand Scalar and Vector quantities. 3. Differentiate between Uniform and NonUniform motion. 4. Understand and explain the meaning of Average Speed. Key Term Scalars and Vectors, Distance and Displacement, Average Speed, Velocity, Average Velocity Module: 03 & 04 Chapter: Motion Contents: 1. Velocity and Acceleration 2. Graphs: Distance – Time Displacement – Time Speed – Time Velocity – Time Learning Objectives The students will: 1. Understand the concept of Acceleration. 2. Understand the importance of graphs. 3. Learn how to plot graphs. 4. Compute Speed and Acceleration from graphs. Key Term 1. Acceleration 2. Retardation 3. Area under the graph 4. Slope of graph Module : 05 Chapter : Motion Contents : 1. Derivation of three Equation of Motion (By numerical as well as by graphical method) 2. Numerical based on Equations of Motion 3. Circular Motion (qualitative idea) Learning Objectives The students will: 1. Understand the relation between Acceleration, Velocity and Time. 2. Derive the three Equations of Motion. 3. Solve numericals. 4. Explain Circular Motion and identify it as an accelerated motion. Key Terms 1. Angular displacement 2. Angular velocity IX Physics Motion and Rest Notes
Download File IX Physics Motion and Rest numerical Unsolved Download File IX Physics Motion and Rest solved Numerical Download File IX Physics Motion and Rest Test paper1 Download File Assignment_20122013 science 9th chemistry Download File Assignment_20122013 science 9th Physics Download File The National Council of Educational Research and Training (NCERT) was established by the Government of India in the year 1961 with a view to bringing about qualitative improvement in school education in the country. No sooner the Council was set up than it mounted a number of programmes in this direction. One such programme was to identify and nurture the talented students. This programme took up the shape of a scheme called National Science Talent Search Scheme (NSTSS) in the year 1963 which provided for the identification of talented students and awarding them with scholarships. During the first year of the implementation of the scheme, it was confined to the Union Territory of Delhi wherein only 10 scholarships were awarded to the Class XI students.
Final Result of NTSE 2011 Download Syllabi for classes I  XII Download NCERT BOOK ONLINE Coordinate Geometry [ JSUNIL Tutorial] 1. Name the mathematician who developed Coordinate geometry. 2. How many points are required to locate a line segment? 3. The axis divide the plane into four parts. What these four parts called? 4. What are the coordinates of origin? 5. What is the point of intersection of axis called? 6. What is the distance of a point form y axis called? 7. What is the distance of a point from xaxis called? 9th Co Ordinate GeometryTest Paper Solved1
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Section – A 1. In fig. AP, AQ and BC are tangents to the circle . If AB = 5 cm , AC = 6 cm and BC = 4cm then length of AP is ( in cm) (A) 7.5 (B) 15 (C) 10 (D) 9 Ans: (A) 7.5 2. A solid right circular cone is cut into two part at the middle of its height by plane parallel to its base. The ratio of volume of the smaller cone to the whole cone is (A) 1:2 (B)1:4 (C) 1: 6 (D) 1: 8 Ans: (D) 1: 8 3. A kite is flying at height of 30m from the ground. The length of string from the kite at the ground is (A) 450 (B)300 (C)600 (D)900 Ans: (B) 4. In fig. P(5,3),q(3,y) are the points of trisections of the line segments joining A(7,2) and B(1,5)Then y equals (A) 2 (B)4 (C)4 (D)5/2 Ans: (C)4 5. Card bearing numbers 2,3,4,11 are kept in a bag . A card is drawn at random from the bag . The probability of getting a card with prime number is (A) 1/ 2 (B) 2/5 (C) 3/10 (D) 5/9 Ans: (A) 1/ 2 6. The distance of point (3,4) from the x axis is (A) 3 (B) 3 (C) 4 (D) 5 Ans: (C) 4 7. The circumference of circle is 22cm .The area of its quadrant is (A) 77/2 (B) 77/4 (C) 77/8 (D) 77/16 Ans: (C) 77/8 8. From a point Q , 13 cm away from the centre of circle ,the length of tangent PQ to the circle is 12 cm . The radius of circle (in cm ) (A) 25 (B) 313 (C) 5 (D) 1 Ans: (C) 5 9. If nth term of AP is (2n+1), then the sum of its first three term (A) 6n + 3 (B) 15 (C) 12 (D) 21 Ans: (B) 15 10. The root of the quadratic equation 2x2  x – 6 = 0 (A) 2 , 3/2 (B) 2, 3/2 (C) 2 ,3/2 (D) 2, 3/2 Ans: (B) 2, 3/2 Download full Paper (Allahabad Region)X BOARD SOLVED QUESTION BY http://www.meritnation.com DELI REGION 2012_class_10_set1_sectionA Download File 2012_class_10_set1_sectionB Download File 2012_class_10_set1_sectionc DownloadFile 2012_class_10_set1_sectionD Download File 
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