Written s per the revised syllbus prescribed by the Mhrshtr Stte Bord of Secondry nd Higher Secondry Eduction, Pune. Slient Fetures Written s per the new textbook. Exhustive coverge of entire syllbus. Fcilittes complete nd thorough preprtion of HOTS section. Mrk-wise segregtion of ech lesson. Constructions drwn with ccurte mesurements. Bord questions of HOTS with solutions updted till the ltest yer (July 017). Self evlutive in nture. Printed t: Dinik Smn, Nvi Mumbi Trget Publictions Pvt. Ltd. No prt of this book my be reproduced or trnsmitted in ny form or by ny mens, C.D. ROM/Audio Video Cssettes or electronic, mechnicl including photocopying; recording or by ny informtion storge nd retrievl system without permission in writing from the Publisher. 10195_11990_JUP P.O. No. 77368
Subject No. Topic Nme Pge No. Algebr Geometry Science & Technology 1 Arithmetic Progression 1 Qudrtic Equtions 10 3 Liner Equtions in Two Vribles 4 4 Probbility 4 5 Sttistics I 50 6 Sttistics II 6 1 Similrity 7 Circle 90 3 Geometric Constructions 105 4 Trigonometry 117 5 Co-ordinte Geometry 141 6 Mensurtion 158 1 School of Elements 180 The Mgic of Chemicl Rections 185 3 The Acid Bse Chemistry 189 4 The Electric Sprk 194 5 All bout Electromgnetism 0 6 Wonders of Light Prt I 08 7 Wonders of Light Prt II 14 8 Understnding Metls nd Non-metls 18 9 Amzing World of Crbon Compounds 4 10 Life s Internl Secrets 9 11 The Regultors of Life 36 1 The Life Cycle 41 13 Mpping our Genes 46 14 Striving for better Environment Prt I 51 15 Striving for better Environment Prt II 54
1 Arithmetic Progression 1. If S n = np + 1 n(n 1)Q, where S n denotes the sum of first n terms of n A.P., find the common difference of the A.P. Sum of first n terms of n A.P. is given by S n = n [ + (n 1)d] = n + 1 n(n 1)d Compring with S n = np + 1 n(n 1)Q, we get d = Q. Find t n of the following A.P. 5 6, 1, 1 1 6,.,t n The given A.P. is 5 6, 1, 1 1 6. Here, = 5 6, d = 1 6 t n = + (n 1)d 1 Mrk Questions ----[ d = t t 1 ] = 5 6 + (n 1) 1 6 = 5 6 + n 1 6 t n = n+4 6 3. Find the first term of the following sequence S n = 4n 3 3 n + 3 S n = 4n 3 3 n 3 When n = 1, S 1 = 1 413 3 3 = 43 3 3 133 = = 6 S 1 = 3 The first term of the given sequence is 3. 4. Find the vlue of d for n A.P., if t 5 = 11 nd t 6 = 13. d = t 6 t 5 = 13 11 = d = Mrks Questions 5. Find the sum of ll three digit numbers which leve the reminder 3 when divided by 5. Here, the sequence is 103, 108, 113,., 998 This sequence is n A.P. with = 103, d = 5, t n = 998 Since, t n = + (n 1)d 998 = 103 + (n 1)5 998 = 103 + 5n 5 5n = 900 n = 180 S n = n [ + (n 1)d] = 180 [ 103 + (180 1)5] = 90[06 + 179 5] = 90 1101 S n = 99090 6. If for n A.P., S n = 0.0 ( n 1), find t n. S n = 0.0 ( n 1) We know tht, t n = S n S n 1 t n = 0.0( n 1) [0.0 ( n 1 1)] = 0.0 ( n 1) [0.0 ( n 1 1)] = 0.0 n n.0 0.0 + 0.0 = 0.0 n 0.01 n t n = 0.01 n 3 Mrks Questions 7. Find the sum of ll nturl numbers lying between 50 nd 500, which re multiples of 5. Required numbers re 55, 60, 65, 70,..., 495 This is n A.P with = 55, d = 60 55 = 5 1
Std. X: HOTS (Algebr) Now, t n = + (n 1)d = 495 55 + (n 1) 5 = 495 5(n 1) = 495 55 n 1 = 440 5 n 1 = 88 n = 89 Here, t 1 = 55, t n = 495, n = 89 S n = n (t 1 + t n ) = 89 (55 + 495) = 89 550 S n = 4475 8. The first, second nd lst terms of n A.P. re, b nd. Find the number of terms in the A.P., b,..., re in A.P. Since, t n = + (n 1)d = + (n 1)(b ) = (n 1)(b ) = (n 1)(b ) n 1 = b n = + 1 = b b b b n = b 11 9. How mny terms of n A.P. 6,, 5, re needed to give the sum 5? Explin the reson for getting two nswers. Here, t 1 = = 6, d = 1 ----[ d = t t 1 ] Let 5 be the sum of n terms of this A.P. where, n N. S n = 5 S n = n [ + (n 1)d] n 5 = 1 6 n 1 5 = n n 5 n 5n 5 = 4 5 4 = n 5n 100 = n 5n n 5n + 100 = 0 (n 5)(n 0) = 0 n 5 = 0 or n 0 = 0 n = 5 or n = 0 Both the vlues of n re nturl numbers, hence two nswers re obtined. 10. Divide 45 into three terms which re in A.P. in such wy tht the product of the lst two terms is 55. Let the three terms in A.P. be d,, + d. d + + + d = 45... (given) 3 = 45 = 15 The three terms in A.P. re 15 d, 15, 15 + d It is given tht the product of lst two terms is 55. ( + d) = 55 + d = 55 15 + d = 55 ----[ = 15] 15 15 + d = 17 d = Three terms in A.P. re 13, 15 nd 17. 11. If the p th term of n A.P. is q nd the q th term is p, prove tht n th term is (p + q n). Let be the first term nd d be the common difference of the given A.P. t p = q ----[Given] + (p 1)d = q ----(i) [ t n = + (n 1)d] nd t q = p ----[Given] + (q 1)d = p ----(ii) [ t n = + (n 1)d] Subtrcting eqution (ii) from (i), we get ( ) + (p 1)d d(q 1) = q p d(p 1 q + 1) = q p d(p q) = (p q) d = 1 Substituting vlue of d in eqution (i), we get + (p 1) (1) = q p + 1 = q = p + q 1 Now, t n = + (n 1)d = (p + q 1) + (n 1) (1) = p + q 1 n + 1 t n = p + q n
Chpter 1: Arithmetic Progression 1. Find 31 st term of A.P. whose 11 th term is 38 nd 16 th term is 73. Let be the first term nd d be the common difference. t 11 = + 10d = 38 ----(i) nd t 16 = + 15d = 73 ----(ii) Subtrcting eqution (ii) from (i), we get + 10d = 38 + 15d = 73 5d = 35 d = 7 Substituting d = 7 in eqution (i), we get + 10 7 = 38 = 38 70 = 3 Now, t 31 = + 30d Substituting the vlue of nd d, we get t 31 = 3 + 30 7 t 31 = 3 + 10 t 31 = 178 31 st term of the A.P. = 178 13. Which term of A.P. 3, 15, 7, 39, will be 13 more thn its 54 th term? Given series is 3, 15, 7, 39,. Here, = 3, d = 1 [ d = t t 1 = 15 3 = 1] t n = + (n 1)d t 54 = 3 + (54 1)1 = 3 + (53 1) t 54 = 639 Let us consider the required term s t x. t x = 13 + t 54 ---- [Given] t x = 13+ 639 t x = 771 t x = + (x 1) d 771 = 3 + (x 1)1 771 = 3 + 1x 1 771 = 1x 9 771 + 9 = 1x 780 = 1x x = 65 The 65 th term of A.P. 3, 15, 7 will be 13 more thn its 54 th term. 14. There re 0 rows of sets in concert hll with 0 sets in first row, 1 sets in second row, sets in third row nd so on. Clculte the totl number of sets in tht concert hll. As per the given condition, the sequence of number of sets in the hll is n A.P. with t 1 = 0, n = 0 nd d = 1 n S n = (n 1)d 0 = 10 [40 + 19] = 10 59 S n = 0 (0 1) 1 S n = 590 There re 590 sets in the concert hll. 15. The sum of the first ten terms of n A.P. is three times the sum of the first five terms, then find rtio of the first term to the common difference. We know tht, S n = n [ + (n 1)d] S 10 = 10 [ + (10 1)d] S 10 = 5( + 9d) Similrly, S 5 = 5 [ + (5 1)d] S 5 = 5 ( + 4d) According to the given condition, S 10 = 3S 5 5( + 9d) = 3 5 ( + 4d) 10( + 9d) = 15( + 4d) ( + 9d) = 3( + 4d) 4 + 18d = 6 + 1d = 6d d = 3 1 16. Obtin the sum of the first 56 terms of n A.P. whose 18 th nd 39 th terms re 5 nd 148 respectively. [July 15] Given, t 18 = 5 nd t 39 = 148 Now, t n = + (n 1)d t 18 = + (18 1)d 5 = + 17d + 17d = 5... (i) 3
Std. X: HOTS (Algebr) Also, t 39 = + (39 1)d 148 = + 38d + 38d = 148... (ii) Adding (i) nd (ii), we get + 17d = 5 + 38d = 148 4 + 55d = 00 Also, S n = n [ + (n 1)d] (iii) S 56 = 56 [ + (56 1)d] = 8 [ + 55d] = 8 (00) [From (iii)] S 56 = 5600 Sum of the first 56 terms of n A.P. is 5600. 17. There is n uditorium with 35 rows of sets. There re 0 sets in the first row, sets in the second row, 4 sets in the third row nd so on. Find the number of sets in the twenty second row. [Mr 15] The number of sets rrnged row wise re s follows : 0,, 4,... This sequence is n A.P. with = 0, d = 0 =, n = Now, t n = + (n 1)d t = 0 + ( 1) = 0 + 1 = 6 The number of sets in the twenty second row re 6. 4 Mrks Questions 18. A mn hs to repy lon of ` 350 by pying ` 305 in the first month nd then decreses the pyment by ` 15 every month. How long will it tke to cler his lon? Here, = 305, d = 15, S n = 350 Let the time required to cler the lon be n months. S n = n [ + (n 1)d] 350 = n [ 305 + (n 1)(15)] 6500 = n(610 15n + 15) 6500 = n(65 15n) 6500 = 65n 15n 15n 65n + 6500 = 0 3n 15n + 1300 = 0 3n 60n 65n + 1300 = 0 3n(n 0) 65(n 0) = 0 n 0 = 0 or 3n 65 = 0 n = 0 or n = 65 3 Since n is nturl number, n 65 3 n = 0 The time required to cler the lon is 0 months. 19. If in n A.P. the sum of m terms is equl to n nd the sum of n terms is equl to m, then show tht sum of (m + n) terms is (m + n). Let be the first term nd d be the common difference of A.P. S n = n [ +(n 1) d] According to the given condition, S m = n n = m [ + (m 1)d] n = m[ + md d] n = m + m d md ----(i) Also, S n = m m = n [ + (n 1)d] m = n[ + nd d] m = n + n d nd ----(ii) Subtrcting (ii) from (i), we get m n + m d n d md + nd = n m (m n) + d(m n ) d(m n) = (n m) (m n)[ + (m + n)d d] = (m n) (m n)[ + (m + n 1)d] = (m n) [ + (m + n 1)d] = ----(iii) S m + n = m n [ + (m + n 1)d] = m n () ----[From (iii)] S m + n = (m + n)
Chpter 1: Arithmetic Progression 0. A contrct on construction job specifies penlty for dely of completion beyond certin limit s follows: ` 00 for first dy, ` 50 for second dy, ` 300 for third dy, etc. If the contrctor pys ` 7,750 s penlty, find the numbers of dys for which the construction work is delyed. Here, = 00, d = 50, S n = 7750 S n = n [ + (n 1)d] 7750 = n [ 00 + (n 1)50] 7750 = n [350 + 50n] 7750 = 175n + 5n 5n + 175n 7750 = 0 n + 7n 1110 = 0 n + 37 30n 1110 = 0 n (n + 37) 30 (n + 37) = 0 (n + 37) (n 30) = 0 n = 37 or n = 30 n 37 s no. of dys cnnot be negtive. n = 30 The construction work is delyed by 30 dys. 1. The interior ngles of polygon re in rithmetic progression. The smllest ngle is 5 nd the common difference is 8. Find the number of sides of the polygon. Let n be the number of sides of the polygon. Sum of ll interior ngles of polygon = (n ) 180 Here, = 5, d = 8 S n = n [ + (n 1)d] (n ) 180 = n [ 5 + (n 1)8] 180n 360 = n [104 + 8n 8] 180n 360 = n [96 + 8n] 360n 70 = 96n + 8n 8n + 96n 360n + 70 = 0 8n 64n + 70 = 0 n 33n + 90 = 0 (n 30)(n 3) = 0 n 30 = 0 or n 3 = 0 n = 30 or n = 3 But, when n = 30, the lst ngle is t n = + (n 1)d 5 + (30 1)8 = 84, which is not possible s interior ngle of polygon cnnot be more thn 180. Number of sides of the given polygon re 3.. A mn set out on cycle ride of 50 km. He covers 5 km in the first hour nd during ech successive hour his speed flls by 1 km/hr. How mny hours will he tke to 4 finish his ride? 1 Here, = 5, S n = 50, d = 4 Let number of hours required to finish the ride be n. S n = n [ + (n 1)d] 50 = n 1 5 (n 1) 4 50 = n 1 n 10 4 4 100 = n 41 n 4 4 100 = n 41 n 4 400 = 41n n n 41n + 400 = 0 (n 5)(n 16) = 0 n 5 = 0 or n 16 = 0 n = 5 or n = 16 If n = 5, speed would be negtive. n = 16 16 hours re required to finish the ride. 3. The 11 th term nd the 1 st term of n A.P. re 16 nd 9 respectively then find: i. The first term nd common difference. ii. The 34 th term. iii. n such tht t n = 55. Given, t 11 = 16, t 1 = 9 i. Since, t n = + (n 1)d t 11 = + (11 1)d 16 = + 10d [Mr 16] 5
Std. X: HOTS (Algebr) + 10d = 16 ----(i) Also, t 1 = + (1 1)d 9 = + 0d + 0d = 9 ----(ii) Subtrcting (i) from (ii), we get + 0d = 9 + 10d = 16 () () () 10d = 13 d = 13 10 Substituting d = 13 in (i), we get 10 + 10 13 10 = 16 + 13 = 16 = 16 13 = 3 = 3 nd d = 13 10 = 1.3 6 The 1 st term is 3 nd the common difference is 1.3 ii. t n = + (n 1)d t 34 = 3 + (34 1)1.3 = 3 + 33 1.3 = 3 + 4.9 = 45.9 t 34 = 45.9 iii. t n = + (n 1)d 55 = 3 + (n 1)1.3 55 = 3 + 1.3 n 1.3 55 3 + 1.3 = 1.3 n 53.3 = 1.3 n n = 53.3 1.3 n = 41 4. If the second term nd the fourth terms of n A.P. re 1 nd 0 respectively, then find the sum of first 5 terms. [July 17] Given, t = 1, t 4 = 0 Now, t n = + (n 1)d t = + ( 1)d 1 = + d + d = 1 ----(i) Also, t 4 = + (4 1)d 0 = + 3d + 3d = 0 ----(ii) Subtrcting (i) from (ii), we get + 3d = 0 + d = 1 () () () d = 8 d = 8 = 4 Substituting d = 4 in (i), we get + 4 = 1 = 1 4 = 8 Now, S n = n S 5 = 5 = 5 = 5 [ + (n 1)d] [8 + (5 1)4] [16 + 4 4] [16 + 96] = 5 11 = 5 56 S 5 = 1400 The sum of first 5 terms is 1400. 5 Mrks Questions 5. Prove tht the sequence S n = n + 5n is in A.P. Hence, find t n. S n = n + 5n For n = 1, S 1 = (1) + 5(1) = + 5 = 7 S 1 = 7 = t 1 For n =, S = () + 5() = 4 + 10 = 18 S = 18 t = S S 1 = 18 7 = 11 For n = 3, S 3 = (3) + 5(3) = 9 + 15 = 33 S 3 = 33 t 3 = S 3 S = 33 18 = 15 Now, t t 1 = 11 7 = 4 t 3 t = 15 11 = 4 The sequence is n A.P. Here, = t 1 = S 1 = 7 nd d = 4 t n = + (n 1)d = 7 + (n 1)4 = 7 + 4n 4 t n = 4n + 3