Written as per the new syllabus prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. STD.

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1 Written as per the new syllabus prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Mathematics Part I STD. X Salient Features Written as per the new textbook. Exhaustive coverage of entire syllabus. Topic-wise distribution of textual questions and practice problems at the beginning of every chapter. Covers solutions to all Practice Sets and Problem Sets. Includes additional problems and MCQs for practice. Chapter-wise assessment for every chapter. Model Question Paper in accordance with the latest paper pattern. Printed at: India Printing Works, Mumbai Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher. TEID: 12530_JUP P.O. No

2 The division of marks in question papers as per objectives will be as follows. Mathematics - Part I Mathematics - Part II Objectives Percentage of marks Objectives Percentage of marks Previous knowledge 20 Previous knowledge 20 Knowledge and understanding 30 Knowledge and understanding 30 Application 40 Application 30 Skill 10 Skill 20 [Reference: ceneje<ì^ jep³e HeeþîeHegmlekeÀ efveefce&leer Je DeY³eeme eàce mebmeesoeve ceb[u, HegCes efveefce&le cetu³eceeheve DeejeKe[e] [P.S. Scan this Q.R. Code to get a better understanding of the New Syllabus as well as Paper Pattern] Contents No. Topic Name Page No. 1 Linear Equations in Two Variables 1 2 Quadratic Equations 47 3 Arithmetic Progression 82 4 Financial Planning Probability Statistics 153 Challenging Questions (4 Marks) Mark and 2 Marks Questions (from Std. IX) 226 Model Question Paper Part I 232 Solved examples from textbook are indicated by +.

3 1 ÒekeÀjCe 2: JeemleJe mebk³ee Linear Equations in Two Variables Type of Problems Practice Set Q. Nos. 1.1 Q.1, 2 Solving Simultaneous linear equations using substitution and elimination method Q.1 to Q.10 (Based on Practice Set 1.1) 1.2 Q. 1 Graph of linear equations (Based on Practice Set 1.2) Q. 1 Problem set 1 Q Q. 2 Solving Simultaneous equations using Graphical method (Based on Practice Set 1.2) Q. 2 Problem Set 1 Q Q.1, 2 Value of the determinant (Based on Practice Set 1.3) Q.1 Problem Set 1 Q Q. 3 Solving Simultaneous equations using Cramer s rule (Based on Practice Set 1.3) Q. 2, 3 Problem Set 1 Q Q.1 Equations reducible to a pair of linear Q.1 equations in two variables (Based on Practice Set 1.4) Problem Set 1 Q Q.1 to Q.6 Application of Simultaneous equations (Based on Practice Set 1.5) Q.1 to Q.12 Problem Set 1 Q.7 1

4 Std. X: Maths (Part - I) An equation of the form ax + by + c = 0 is called a linear equation in two variables x and y, where a, b, c R and a 0, b 0. The equation ax + by + c = 0 is also the general form of a linear equation. Examples: i. 3x + 2y 13 = 0 ii. 5x y 14 = 0 Note: Linear equation contains two variables and the degree of each of them is one. Complete the following table. (Textbook pg. no. 1) Is the equation a No. Equation linear equation in 2 variables 1 4m + 3n = 12 Yes 2 3x 2 7y = 13 No 3 2x 5y = 16 Yes 4 0x + 6y 3 = 0 No 5 0.3x + 0y 36 = 0 No = 4 x y No 7 4xy 5y 8 = 0 No Methods for solving simultaneous linear equations: Elimination method: Examples: 1. Solve: 3x + 2y = 29; 5x y = 18 (Textbook pg. no. 3) 3x + 2y = 29 (i) and 5x y = 18 Let s solve the equations by eliminating y. Fill suitably the boxes below. Multiplying equation (ii) by 2, we get 5x 2 y 2 = x 2y = 36 (iii) Add equations (i) and (iii). 3x + 2y = 29 10x 2 y = 36 13x = 65 x = = 5 Substituting x = 5 in equation (i). 2 2 Let s Recall Linear equation in two variables Simultaneous linear equations 3x + 2y = y = y = 29 2y = y = 14 y = 14 2 = 7 (x, y) = 5, 7 is the required solution. 2. Solve: 4x 5y = 172 ; 5x 4y = 251 4x 5y = 172 (i) 5x 4y = 251 In the above equations, the coefficients of x and y are interchanged. By adding and subtracting these equations, we get two simple equations. After solving these two simple equations, we get the solution. i. Adding equations (i) and (ii), we get 4x 5y = 172 5x 4y = 251 9x 9y = 423 x y = [Dividing both sides by 9] x y = 47 (iii) ii. 4x 5y = 172 5x 4y = x y = 79 x + y = 79 (iv) [Multiplying both sides by 1] iii. Adding equations (iii) and (iv), we get x y = 47 x + y = 79 2x = 126 x = = 63 iv. Substituting x = 63 in equation (iv), we get 63 + y = 79 y = = 16 (x, y) = (63, 16) is the solution of the given Substitution method: Example: 1. Solve: 3x + y = 14, 2x 3y = 2 i. 3x + y = 14 y = 14 3x (i) 2x 3y = 2

5 Chapter 1: Linear Equations in Two Variables ii. Substituting y = 14 3x in equation (ii), we get 2x 3 (14 3x) = 2 2x x = 2 2x + 9x = x = 44 x = x = 4 iii. Substituting x = 4 in equation (i),we get y = 14 3(4) y = y = 2 (x, y) = (4, 2) is the solution of the given 1. Complete the following activity to solve the 5x + 3y = 9 (i) 2x 3y = 12 Add equations (i) and (ii). 5x + 3y = 9 2x 3y = 12 7 x = 21 x = 21 7 x = 3 Put x = 3 in equation (i) y = 9 3y = y = 6 y = Practice Set y = 2 Solution is (x, y) = 3, 2 2. Solve the following i. 3a + 5b = 26; a + 5b = 22 ii. x + 7y = 10; 3x 2y = 7 iii. 2x 3y = 9; 2x + y = 13 iv. 5m 3n = 19; m 6n = 7 v. 5x + 2y = 3; x + 5y = 4 1 vi. 3 x + y = 10 3 ; 2x y = 11 4 vii. 99x + 101y = 499; 101x + 99y = 501 viii. 49x 57y = 172; 57x 49y = 252 i. 3a + 5b = 26 (i) a + 5b = 22 3a + 5b = 26 a + 5b = 22 2a = 4 a = 4 2 = 2 Substituting a = 2 in equation (ii), we get 2 + 5b = 22 5b = b = 20 b = 20 5 = 4 (a, b) = (2, 4) is the solution of the given ii. x + 7y = 10 x = 10 7y (i) 3x 2y = 7 Substituting x = 10 7y in equation (ii), we get 3(10 7y) 2y = y 2y = 7 23y = y = y = 23 = 1 Substituting y = 1 in equation (i), we get x = 10 7(1) = 10 7 = 3 (x, y) = (3, 1) is the solution of the given iii. 2x 3y = 9 (i) 2x + y = 13 2x 3y = 9 2x + y = 13 4y = 4 4 y = 4 = 1 Substituting y = 1 in equation (ii), we get 2x + 1 = 13 2x = 12 x = 12 2 = 6 (x, y) = (6, 1) is the solution of the given iv. 5m 3n = 19 (i) m 6n = 7 m = 6n 7 Substituting m = 6n 7 in equation (i), we get 5(6n 7) 3n = 19 3

6 Std. X: Maths (Part - I) 30n 35 3n = 19 27n = n = 54 n = = 2 Substituting n = 2 in equation (ii), we get m = 6(2) 7 = 12 7 = 5 (m, n) = (5, 2) is the solution of the given v. 5x + 2y = 3 (i) x + 5y = 4 x = 4 5y Substituting x = 4 5y in equation (i), we get 5(4 5y) + 2y = y + 2y = 3 23y = y = y = 23 = 1 Substituting y = 1 in equation (ii), we get x = 4 5(1) = 4 5 = 1 (x, y) = (1, 1) is the solution of the given 1 vi. 3 x + y = 10 3 x + 3y = 10 [ Multiplying both sides by 3] x = 10 3y (i) 2x y = x + y = 11 [Multiplying both sides by 4] Substituting x = 10 3y in equation (ii), we get 8(10 3y) + y = y + y = 11 23y = y = y = 23 = 3 Substituting y = 3 in equation (i), we get x = 10 3(3) = 10 9 = 1 (x, y) = (1, 3) is the solution of the given vii. 99x + 101y = 499 (i) 101x + 99y = 501 Adding equations (i) and (ii), we get 99x + 101y = x + 99y = x + 200y = x + y = 1000 [Dividing both sides by 200] 200 x + y = 5 (iii) 99x + 101y = x + 99y = 501 2x + 2y = 2 2 x y = 2 [Dividing both sides by 2] x y = 1 (iv) Adding equations (iii) and (iv), we get x + y = 5 x y = 1 2x = 6 x = 6 2 = 3 Substituting x = 3 in equation (iii), we get 3 + y = 5 y = 5 3 = 2 (x, y) = (3, 2) is the solution of the given viii. 49x 57y = 172 (i) 57x 49y = 252 Adding equations (i) and (ii), we get 49x 57y = x 49y = x 106y = 424 x y = [Dividing both sides by 106] x y = 4 (iii) 49x 57y = x 49y = x 8y = x + y = 8 [Dividing both sides by 8] x + y = 10 (iv) Adding equations (iii) and (iv), we get x y = 4 x + y = 10 2x = 14 x = 14 2 = 7 Substituting x = 7 in equation (iv), we get 7 + y = 10 y = 10 7 = 3 (x, y) = (7, 3) is the solution of the given