Excel in Mathematics Book 10 Final_CTP (2080)_compressed (1)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 151 Vedanta Excel in Mathematics - Book 10 Sequence and Series or, r2 16 = 1 4 or, r2 = 4 ∴ r = ± 2 Also, r = 1 b n+1 a or, ± 2 = 1 80 n+1 5 or, ± 2 = (16) 1 n+1 or, (± 2)1 = (±2)4 × 1 n+1 or, 1 = 4 n + 1 or, n + 1 = 4 or, n= 3 (iii) Again, g1 = ar = 5 × 2 = 10 g2 = ar2 = 5 × 22 = 20 g3 = ar3 = 5 × 23 = 40 Hence 10, 20 and 40 are the geometric means between 5 and 80. Example 7: There are 5 geometric means between a and b. If the second mean is 4 and the last mean is –32, find: (i) the first term and common ratio (ii) the last term (iii) the remaining means Solution: Let g1 , g2 , g3 , …, g5 be the 5 geometric means between a and b respectively. Then a, g1 , 4, g3 , g4 , –32, b are in a G.P. Here, first term = a, last term = b and number of geometric means (n) = 5 (i) Now, second mean (g2 ) = 4 ∴ar2 = 4 … (i) And, last mean (g5 ) = –32 ∴ar5 = –32 … (ii) Dividing equation (ii) by equation (i), we get ar5 ar2 = –32 4 or, r3 = – 8 or, r3 = (–2)3 ∴ r = –2 Putting the value of r in equation (i), we get a (–2)2 = 4 ∴ a = 1 Hence, the first term is 1 and common ratio is –2. Alternatively, r = 1 b n+1 a = 1 80 n+1 5 = (16) 1 n+1 = (2) 4 n+1 Also, First mean (g1 ) = ar Last mean (gn) = arn Now, g1 gn = 1 4 or, ar arn = 1 4 or, r1 – n = 2 –2 or, (2) 4 n+1 × 1 – n = 2 –2 or, 4 – 4n n + 1 = – 2 or, 4 – 4n = –2n – 2 or, –2n = – 6 ∴ n = 3 Checking Second mean (g2 ) = ar2 = 1 × (-2)2 = 4 Last mean (g5 ) = ar5 = 1 × (-2)5 = -32 which are given in the question.

Vedanta Excel in Mathematics - Book 10 152 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series (ii) Also, r = 1 b n+1 a or, 2 = 1 b 5+1 1 or, 2 = b 1 6 or, 26 = b ∴ b = 64 Hence, the last term is 64. Example 8: The arithmetic and geometric means of two positive numbers are 50 and 40 respectively. Find the numbers. Solution: Let a and b be two unequal positive numbers. Then, A.M. = 50 or, a + b 2 = 50 or, a + b = 100 ∴ b = 100 – a … (i) Also, G.M. = 40 or, ab = 40 ∴ ab = 1600 …(ii) Putting the value of b in equation (ii), we get a(100 – a) = 1600 or, 100a – a2 = 1600 or, a2 – 100a + 1600 = 0 or, a2 – 80a – 20a + 1600 = 0 or, a(a – 80) – 20(a – 80) = 0 or, (a – 80) (a– 20) = 0 ∴a = 80 or 20 When a = 80, from (i); b = 100 – 80 = 20 When a = 20, from (i); b = 100 – 20 = 80 Hence, required numbers are 80 and 20 or 20 and 80. Example 9: A commercial bank pays the interest at a certain rate per year so that the annual balances form a geometric progression. Mr. Limbu deposits Rs 20,000 in the bank. If his balance in the beginning of the fifth year becomes Rs 29,282; answer the following questions: (i) What is the common factor that is useful on calculating the balances? (ii) What will be the balance in the beginning of 2nd, 3rd and 4th years? (iii) Again, remaining means are; g1 = ar = 1 × (–2) = –2 g3 = ar3 = 1 × (–2)3 = –8 g4 = ar4 = 1 × (–2)4 = 16 Hence, the remaining means are –2, –8 and 16. Alternatively, a = A.M. + (A.M.)2 – (G.M.)2 = 50 + (50)2 – (40)2 = 50 + 900 = 50 + 30 = 80 b = A.M. – (A.M.)2 – (G.M.)2 = 50 – (50)2 – (40)2 = 50 – 900 = 50 – 30 = 20

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 153 Vedanta Excel in Mathematics - Book 10 Sequence and Series Solution: Here, principal (a) =Rs 20,000, balance in the beginning of 5th year (b) = Rs 29,282 and number of geometric means (n) = 3 (i) Now, common factor (r) = 1 b n+1 a = 1 29282 n+1 20000 = (1.4641) 1 4 = 1.1 (ii) Again, the balance in the beginning of 2nd year =g1 = ar = Rs 20,000 × 1.1 = Rs 22,000 The balance in the beginning of 3rd year = g2 = ar2 = Rs 20,000 × (1.1)2 = Rs 24,200 The balance in the beginning of 4th year = g3 = ar3 = Rs 20,000 × (1.1)3 = Rs 26,620 Thus, the balances in the beginning of 2nd, 3rd and 4th years are Rs 22,000, Rs 24,200 and Rs 26,620 respectively. EXERCISE 7.3 General section 1. a) Define geometric mean. b) What is the geometric mean between any two numbers a and b? c) What is the geometric mean between 3 and 27? 2. a) If there are ‘n’ geometric means between the numbers a and b, what will be the common ratio (r)? b) If there are 4 geometric means between the numbers 2 and 64, what is the common ratio (r)? 3. a) In a G.S. if there are 5 means between a and b, how many terms are there? b) Identify the second mean in the geometric sequence 1, 3, 9, 27, 81? c) What is the last mean in the geometric sequence 8, 4, 2, 1, 1 2 , 1 4 , 1 8 ? d) A G.P. has first term a, common ratio r and n number of means, what is the last mean? 4. Find the geometric mean between the following two numbers. a) 2 and 8 b) 12 and 3 c) – 5 and – 20 d) – 54 and – 6 e) 8 and 50 f) 2 3 and 8 27 g) 2 and 8 h) 10 +1 and 10 – 1 5. Find the 2nd term from the following geometric sequences. a) 1st term = 4 and 3rd term = 25 b) 1st term = 27 and 3rd term = 12 Alternative process for finding r: No. of years (n) =3 + 2 = 5 Now, balance in the beginning of the fifth years (t5 ) = 29282 or, ar4 = 29282 or, 20000 r4 = 29282 or, r4 = 1.4641 ∴ r = 1.1

Vedanta Excel in Mathematics - Book 10 154 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series 6. Find the term as indicated in the following geometric sequences. a) 5th term = 8, 7th term =18, find the 6th term b) 4th term =–3, 6th term = –75, find the 5th term 7. Find the value of x if: a) 3, x, 12 are in a G.P. b) 11, x, 44 are in a G.P. 8. a) If the geometric mean between the numbers x and 8 is 4, find the value of x. b) If 15 is the geometric mean between the numbers 5 and y, find the value of y. c) If the geometric mean between x and 6 is 12, find the value of x. Also, find the arithmetic mean between them. d) If the geometric mean between 10 and x is 50, find the value of x. Also, find the arithmetic mean between them. Creative section–A 9. Insert the geometric means between the given terms. a) 3 geometric means between 2 and 162. b) 4 geometric means between 3 and 96. c) 5 geometric means between 8 and 1 8 d) 3 geometric means between 9 4 and 4 9 . 10. a) If 8, x, y, 27 are in G.P., find: (i) the common ratio (ii) the values of x and y. b) If 125, m, n, –1 are in G.P., find: (i) the common ratio (ii) the values of m and n. c) If 6, p, q, r, 96 are in G.S., find: (i) the common ratio (ii) the values of p, q and r. d) If 1 8 , x, y, z, 2 are in G.S., find: (i) the common ratio (ii) the values of x, y and z. 11. a) There are n geometric means between 4 and 64. If the third mean is 32, find: (i) the common ratio (ii) the value of n (iii) the remaining means b) There are n geometric means between 48 and 3. If the second mean is 12, find: (i) the common ratio (ii) the value of n (iii) the remaining means 12. a) There are n geometric means between 1 and 64. If the ratio of the first mean to last mean is 1: 16, find: (i) the common ratio (ii) the value of n (iii) the means b) n geometric means are inserted between 15 and 480. If the ratio of the first mean to last mean is 1: 8, find: (i) the common ratio (ii) the value of n (iii) the means 13. a) There are three geometric means between a and b. If the first and third means are 25 and 625 respectively, find: (i) the first term a and the last term b (ii) the remaining means b) There are four geometric means between a and b. If the first and last means are 12 and 324 respectively, find: (i) the first term a and the last term b (ii) the remaining means Creative section–B 14. a) The arithmetic mean and geometric mean of two numbers are 5 and 4 respectively. Find the numbers. b) Find the two numbers whose arithmetic mean is 13 and geometric mean is 12.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 155 Vedanta Excel in Mathematics - Book 10 Sequence and Series 15. a) A commercial bank pays the interest at a certain rate per year so that the annual balances form a geometric progression. Mrs. Sharma deposits Rs 10,000 in the bank. If her balance in the beginning of the fifth year becomes Rs 14,641; answer the following questions: (i) What is the common factor that helps on calculating the balances? (ii) What will be the balances in the beginning of 2nd, 3rd and 4th years? b) Mr. Hamal lands a job as an officer in a company with starting salary Rs 50,000. An agreement on increment of his salary, forming a geometric progression, is made with the management so that he will have Rs 1,03,680 salary in fifth year. (i) What is the common factor that is used on calculating his salaries? (ii) What will be his salaries in the 2nd, 3rd and 4th years of his job? 7.10 Sum of the first n terms of an Geometric Series Let 1, 2, 4, 8, …be a geometric sequence. Then, 1 + 2 + 4 + 8+ … is the geometric series. A series whose terms are in Geometric progression is called Geometric series. Let a = first term, r = common ratio, l = last term and n = number of terms in a G.P. Then, the first n terms of are; a, ar, ar2 , …, arn – 1 Let Sn denote the sum of n terms of the G.P. Now, Sn = a + ar + ar2 +… + arn –1 … (i) Multiplying both sides by r, we get r Sn = ar + ar2 +ar3 + … + arn … (ii) Subtracting equation (i) from equation (ii), we get r Sn = ar + ar2 +ar3 + …+ arn –1 + arn Sn = a + ar + ar2 +… …+ arn –1 (–) (–) (–) (–) (–) (–) rSn – Sn = arn – a or, (r – 1)Sn = a (rn – 1) ∴ Sn = a(rn – 1) r – 1 We know, the last term (l) = arn – 1 ∴ Sn = a(rn – 1) r – 1 = arn – a r – 1 = arn – 1 . r – a r – 1 = lr – a r – 1 Facts to remember 1. When r = 1, the formula for finding the sum of first n terms of a G.P., Sn = a + a + … + a = n.a 2. When r ≠ 1, then there arise the following cases. (i) Case - I: When a, r and n are known, Sn = a(rn – 1) r – 1 (ii) Case – I: When a, r and l are known, Sn = lr – a r – 1 www.geogebra.org/classroom/de9qvgj7 Classroom code: DE9Q VGJ7 Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 10 156 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Worked-out Examples Example 1: Find the sum of the series 1 + 3 + 9 + …up to 6 terms Solution: Here, the first term (a) = 1, the common ratio (r) = 3 1 = 3 and the no. of terms (n) = 6 Sum of first 6 terms (S6 ) =? Now, Sn = a(rn – 1) r – 1 = 1(36 – 1) 3 – 1 = 728 2 = 364 Hence, the required sum is 364. Example 2: Find the sum of the first 8 terms of the series: –128 + 192 – 288 + …….. Solution: Here, the first term (a) = –128, then, the common ratio (r) = 192 –128 = – 3 2 and the no. of terms (n) = 8 Sum of the first 8 terms (S8 ) =? Now, Sn = a(rn – 1) r – 1 = –128[(– 3 2 ) 8 – 1] – 3 2 – 1 = –128( 6561 256 – 1) – 5 2 = 256 × 6305 256 5 = 1261 Hence, the required sum is 1261. Example 3: Find the sum of the geometric series: 3 + 6 + 12 + …..+ 384. Solution: Here, the first term (a) = 3, Alternatively: Last term (tn) = 384 or, arn – 1 = 384 or, 3 × 2n – 1 = 384 or, 2n – 1 = 128 or, 2n – 1 = 27 or, n – 1 = 7 ∴ n = 8 Now, Sn = a(rn – 1) r – 1 = 3(28 – 1) 2 – 1 = 3 × 255 = 765 Common ratio (r) = 6 3 = 2 Last term (l) = 384 Sum (Sn) =? Now, Sn = lr – a r – 1 = 384 × 2 – 3 2 – 1 = 765 Hence, the sum of the given series is 765. Example 4: In a geometric progression; the common ratio is 2, the last term is 448 and the sum is 889. Find: (i) the first term (ii) the number of terms. Alternatively, Last term (l) = t6 = ar5 = 1 × 35 = 243 ∴ Sn = lr – a r – 1 = 243 × 3 – 1 3 – 1 = 728 2 =364

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 157 Vedanta Excel in Mathematics - Book 10 Sequence and Series Solution: Here, common ratio (r) = 2, the last term (l) = 448 and sum (Sn) = 889, (i) Now, Sn = lr – a r – 1 or, 889 = 448 × 2 – a 2 – 1 or, 889 = 896 – a or, a = 7 Hence, the first term is 7. Example 5: Given geometric series is 3 – 6 + 12 – 24 + … (i) The sum of the terms of the series is – 63, how many terms are there? (ii) What is the last term of the series? Solution: Here, first term (a) = 3, common ratio (r) = – 6 3 = –2 Sum of terms (S6 ) = – 63 (i) Now, Sn = a(rn – 1) r – 1 or, – 63 = 3[(–2)n – 1] –2 – 1 or, – 63 = –(–2)n + 1 or, (–2)n= 64 or, (–2)n= (–2)6 ∴ n = 6 Hence, the required number of terms is 6. Example 6: The third term of a geometric progression is 12 and its sixth term is 96. Find: (i) the first term and common ratio (ii) seventh term (iii) the sum of the first 8 terms. Solution: Let ‘a’ and ‘r’ be the first term and the common ratio of a G.P. respectively (i) 3rd term (t3 ) = 12 ∴ar2 = 12 … (i) 6th term (t6 ) = 96 ∴ ar5 = 96 … (ii) Now, dividing equation (ii) by equation (i), we get ar5 ar2 = 96 12 or, r3 = 8 ∴r = 2 Also, putting the value of ‘r’ in equation (i), we get a × 22 = 12 ∴a = 3 Hence, the first term of the GP is 3 and the common ratio is 2. (ii) The 7th term (t7 ) = ar6 = 3 × 26 = 192 (iii) Sum of the first 8 terms (S8 ) = a(rn – 1) r – 1 = 3(28 – 1) 2 – 1 = 765 Hence, the sum of first 8 terms of the GP is 765. (ii) Again, last term (tn) = 448 or, arn – 1 = 448 or, 7 × 2n – 1 = 448 or, 2n – 1 = 64 or, 2n – 1 = 26 or, n – 1 = 6 ∴ n = 7 Hence, the series has 7 terms. (ii) Again, last term (tn) = t6 = ar5 = 3× (–2)5 = 3× (–32) =–96 Hence, the last term is –96. Checking: When a = 3 and r = 2: t 3 = ar2 = 3 × 22 = 12 t 6 = ar5 = 3 × 25 = 96, which are given in the question.

Vedanta Excel in Mathematics - Book 10 158 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Example 7: A man borrows Rs 24,200 without interest and repays the loan in 5 installments, each installment being treble the preceding one. Find: (i) the first installment (ii) last installment Solution: Let, the first installment be Rs a. Then the sequence of installments is a, 3a, 9a, … Common ratio (r) = 3a a = 3 and the sum of 5 installments (S5 ) =Rs 24,200 (i) Now, Sn = a(rn – 1) r – 1 or, 24,200 = a(35 – 1) 3 – 1 or, 24,200 = 121a ∴ a = 200 Hence, the first installment is Rs 200. Example 8: A student surveys the number of pens sold in two stationery shops during a week. In the first shop, 60 pens are sold in the first day and 6 more pens are sold in each day as comparison of previous day. Similarly, in the second shop, only 5 pens are sold in the first day and the number of pens sold in each day is double than that of the number of pens sold in the previous day. Based on this context, answer the following questions. (i) How many pens are sold in the first stationery? (ii) How many pens are sold in the second stationery? (iii) Which stationery shop has sold more pens and by how many? Solution: (i) For the first stationery: number of pens sold in 1st day (a) = 60, common difference (d) = 6 Number of days (n) = 7, total number of pens sold (Sn) =? Now, Sn = n 2 [2a + (n – 1) d] ∴ S7 = 7 2 [2× 60+ (7 – 1) × 6] = 7 2 × 156 = 546 Hence, the first stationery sold 546 pens during the week. (ii) Again, for the second stationery; number of pens sold in 1st day (a) = 5 common ratio (r) = 2 Number of days (n) = 7 total number of pens sold (Sn) =? We know, Sn = a(rn – 1) r – 1 = 5(27 – 1) 2 – 1 = 635 Hence, the second stationery sold 635 pens during the week. (iii) Difference = 635 – 546 = 89 Hence, the second stationery sold 89 more pens than first stationery. (ii) Again, last installment (t5 ) = ar4 = 200 × 34 = 16,200 Hence, the last installment is Rs 16,200.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 159 Vedanta Excel in Mathematics - Book 10 Sequence and Series EXERCISE 7.4 General section 1. a) What is the sum of first n terms of a geometric series having first term a and common ratio r? b) What is the sum of the terms of a GP when its first term a, common ratio r and last term l are known? 2. Find the sum of following geometric series. a) 1 + 2 + 4+ … to 6 terms b) 3 + 6 + 12 + … to 8 terms c) 2 – 6 +18 – … to 7 terms d) – 5 + 10 – 20 + … to 10 terms e) 16 + 24 + 36 + … to 5 terms f) 5000 + 4000 + 3200 + … to 5 terms 3. a) The first term of a geometric series is 4 and the common ratio is 2, find the sum of its first 8 terms. b) The first term of a GP is 11 and the common ratio is 3, find the sum of its first 6 terms. 4. Find the sum of the first 6 terms of each of the following geometric sequence: a) 1st term = 4 and 2nd term = 8 b) 1st term = 5 and 2nd term = 15 c) 1st term = 64 and 2nd term = 32 d) 1st term = 243 and 2nd term = 81 5. Calculate the sum of all the terms of the following geometric progressions. a) 1 + 3 + 9 + … + 243 b) 3 + 6 +12 + 24 + … + 384 c) 2 + 6 + 12 + … + 486 d) 0.005 + 0.05 + 0.5 + … + 500 Creative section–A 6. a) In a geometric progression; the common ratio is 2, the last term is 320 and the sum is 635. Find: (i) the first term (ii) number of terms. b) A geometric series has the common ratio 3, the last term is 486 and the sum is 728. Find: (i) the first term (ii) number of terms. c) A geometric series has the first term 9, common ratio is –2 and the sum is 99. Find: (i) the last term (ii) number of terms. 7. a) Given geometric series is 25 + 50 + 100+ … Answer the following questions. (i) If the sum of the terms of the series is 775, how many terms are there? (ii) What is the last term of the series? b) Study the given series and answer the questions: 64 + 96 + 144 + …. (i) If the sum of the terms of the series is 844, how many terms are there? (ii) What is the last term of the series? 8. a) The third term of a geometric progression is 8 and its sixth term is 64. Find: (i) the first term and common ratio (ii) the eighth term (iii) the sum of first 10 terms. b) In a G.P., the second term is 48 and its fifth term is 6. Find: (i) the first term and common ratio (ii) the fourth term (iii) the sum of first 6 terms.

Vedanta Excel in Mathematics - Book 10 160 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Sequence and Series Creative section–B 9. a) Sayara uploaded a music video in YouTube at 6 a.m. and got 50 views in the first hour, 100 views in second hour, 200 views in the third hour and so on. How many views did it get (i) till 4 p.m. of the same day? (ii) till 6 p.m. of the same day? b) A frog hops 16 cm, then 8 cm, then 4 cm, and so on. (i) How far will the frog travel in four hops? (ii) How far will the frog travel in five hops? 10. a) There are 10 varieties of birds in a zoo, the number of each variety is double the number of another variety. If the total number of birds is 5,115, find: (i) the number of birds in the first variety (ii) the number birds in the last variety b) A man borrows Rs 18,900 without interest and repays the loan in 6 installments, being each installment double the preceding one. Find: (i) the first installment (ii) the last installment 11. a) A man surveys the number of mobile sets sold in two shops and gets the following information. Shop I: sells 30 mobile sets in the first day, 35 mobile sets in the second day, 40 mobile sets in the third day and increasing by 5 mobiles each day for a week. Shop II: sells 3 mobile sets in the first day, 6 mobile sets in the second day, 12 mobile sets in the third day and continuing to double for a week. (i) How many mobile sets does the shop-I sell during a week? (ii) How many mobile sets does the shop-II sell during a week? (iii) Which shop sells more mobile sets in the week and by how many? b) A National Lottery is offering prizes in a new competition. The winner may choose one of the following options: Option I: Rs. 250 in the first week, Rs 450 in the second week, Rs 650 in the third week, increasing by Rs 200 each week for a total of 10 weeks. Option II: Rs 10 in the first week, Rs 20 in the second week, Rs 40 in the third week, continuing to double for a total of 10 weeks. (i) What total amount do you receive if you select option-I? (ii) What total amount do you receive if you select option-II? (iii) Which option do you prefer to choose? Justify your answer. Project Work and Activity Section 12. a) Draw an equilateral triangle having each side 16 cm. Mark the mid points of each side and join them to form a new equilateral triangle. Also, join the mid points of the sides of each of the new equilateral triangles to from another equilateral triangle. Continue this process upto six steps. Find t he sum of the perimeter of equilateral triangles formed in this process.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 161 Vedanta Excel in Mathematics - Book 10 Sequence and Series b) Draw a square with each side 20 cm. Join the mid points of its each side to form a new square. Also, join the mid points of the sides of each of the new squares to from another square. Continue this process to form 5 squares. Find the sum of the area of all the squares. OBJECTIVE QUESTIONS Tick the correct alternative. 1. The arithmetic mean (A.M.) between two numbers a and b is given by (A) a + b 2 (B) a – b 2 (C) ab (D) a b 2. The common difference (d) of an A.P. having n means between any two numbers a and b is given by (A) d = b – a n – 1 (B) d = b – a n + 1 (C) d = a + b n – 1 (D) d = a – b n + 1 3. The last mean or nth mean (mn) of an A.P. is given by (A) a + nd (B) a + (n – 1)d (C) b – d (D) Both (A) and (C) 4. When the first term a and common difference d are known, the sum (Sn) of first n terms is given by (A) n 2 [a + n – 1 d] (B) n 2 [2a + n – 1 d] (C) n 2 [2a + d – 1 n] (D) n 2[2a + n + 1 d] 5. The sum of first n terms of an A.P. having the first term a and last term l is ……….. (A) Sn = n a + l2 (B) Sn = n a - l2 (C) Sn = 2 a + ln (D) Sn = 2 a - ln 6. The geometric mean (G.M.) between two numbers a and b is given by (A) a + b 2 (B) a – b 2 (C) ab (D) a b 7. The common ratio (r) of a G.P. having n means between any two numbers a and b is given by (A) ab n (B) r = ab n + 1 (C) r = a b n + 1 (D) r = b a n + 1 8. The last mean or nth mean (gn) of a G.P. is given by (A) arn (B) br (C) arn – 1 (D) Both (A) and (B) 9. What is the sum of first n terms of a G.P. whose first term a and common ratio r are known? (A) Sn = a(rn – 1) r – 1 (B) Sn = a(rn + 1) r + 1 (C) Sn = a(rn – 1) r + 1 (D) Sn = a(rn + 1) r – 1 10. The sum of first n terms of a G.P. having the first term a and last term l is ……….. (A) Sn = lr – a r – 1 (B) Sn = lr – a r + 1 (C) Sn = ar – l r – 1 (D) Sn = r – 1 lr – a

Vedanta Excel in Mathematics - Book 10 162 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Unit 8 Quadratic Equations 8.1 Quadratic equations – review Let’s consider two equations x + 3 = 0 and x + y = 7. In x + 3 = 0, the highest power of the variable x is 1. So, it is called a first degree equation of one variable. A first degree equation is also called a liner equation. In x + y = 7, the highest power of each variable is 1. Therefore, it is also a first degree equation with two variables x and y. On the other hand, let’s consider another equation x2 + 2x + 3 = 0. In this equation, the highest power of the variable x is 2. So, it is called a second-degree equation. A second degree equation of one variable is also called a quadratic equation. Every quadratic equation can be expressed in the form ax2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. Therefore, ax2 + bx + c = 0 is known as the standard form of a quadratic equation. Facts to remember 1. The standard form of quadratic equation ax2 + bx + c = 0 which has the variables both in degree 2 and 1 is known as adfected quadratic equation. For example, x2 + 5x + 6 = 0, 2x2 + 3x – 2 = 0, 3x2 – 4x = 0, etc. are adfected quadratic equations. 2. The quadratic equation of the form ax2 + c = 0, where the middle term with variable of degree 1 is missing is known as pure quadratic equation. For example: x2 – 4 = 0, 2x2 = 9, etc. are pure quadratic equations. 3. The values of variable that satisfy the given quadratic equation are called roots or zeroes or solutions of the given quadratic equation. 8.2 Solution of quadratic equations A quadratic equation is a second degree equation. Therefore, we obtain two solutions (or roots) of the variable from a quadratic equation. In other words, a quadratic equation has two roots. There are various methods of solving quadratic equations. Here, we discuss three methods. (i) Factorization method (ii) Completing square method (iii) By using formula Arab mathematician Abraham bar Hiyya Ha-Nasi, often known by the Latin name Savasorda, is famed for his book ‘Liber Embadorum’ published in 1145 AD (CE) which is the first book published in Europe to give the complete solution of a quadratic equation.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 163 Vedanta Excel in Mathematics - Book 10 Quadratic Equations (i) Solving a quadratic equation by factorization method In this method, we follow the following steps to solve the quadratic equations. • Clear all fractions or brackets (if necessary). • Express the equation in the form ax2 + bx + c = 0 or ax2 – b = 0. • Factorize the expression on the left by splitting the middle term and express as the product of two linear factors. • Solve each of linear factors by applying zero factor property. Facts to remember 1. Zero factor property: If a.b = 0, then either a = 0 or b = 0 or both. In other words, if the product of two expressions is zero; then one or both expressions is zero. 2. Square root property: If k ≥ 0, the solutions of x2 = k are x = ± k Worked-out Examples Example 1: Solve: x2 – 9 = 0 Solution: Here, x2 – 9 = 0 or, x2 – 32 = 0 or, (x + 3) (x – 3) = 0 Either, x + 3 = 0 i.e., x = – 3 Or, x – 3 = 0 i.e., x = 3 ∴x = ± 3 Alternative process x2 – 9 = 0 or, x2 = 9 or, x2 = (± 3)2 ∴ x = ± 3 Example 2: Solve: x2 – 3x = 10 Solution: Here, x2 – 3x = 10 or, x2 – 3x – 10 = 0 or, x2 – (5 – 2)x – 10 = 0 or, x2 – 5x + 2x – 10 = 0 or, x (x – 5) + 2 (x – 5) = 0 or, (x – 5) (x + 2) = 0 Either, x – 5 = 0 i.e., x = 5 Or, x + 2 = 0 i.e., x = – 2 ∴ x = 5 or, x = – 2 Checking the solutions When x = 3 then x2 – 9 = 0 or, 32 – 9 = 0 or, 9 – 9 = 0 or, 0 = 0 (True) When x =– 3 then x2 – 9 = 0 or, (–3)2 – 9 = 0 or, 9 – 9 = 0 or, 0 = 0 (True) So, x = 3 or x =– 3 are the correct solutions. Remember, (+3)2 = 9 and (– 3)2 = 9. Therefore, 9 = (± 3)2 Checking the solutions When x = 5; x2 – 3x = 10 or, 52 – 3 × 5 = 10 or, 25 – 15 = 10 or, 10 = 10 (True) When x = – 2; x2 – 3x = 10 or, (– 2)2 – 3(– 2) = 10 or, 4 + 6 = 10 or, 10 = 10 (True) So, x = 5 or x =– 2 are the correct solutions.

Vedanta Excel in Mathematics - Book 10 164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations Example 3: Solve: 9x2 + 4= 15x Solution: Here, 9x2 + 4 = 15x or, 9x2 – 15x + 4 = 0 or, 9x2 – 12x – 3x + 4 = 0 or, 3x (3x – 4) – 1(3x – 4) = 0 or, (3x – 4) (3x – 1) = 0 Either, 3x – 4 = 0 i.e., x = 4 3 Or, 3x – 1= 0 i.e., x = 1 3 ∴ x = 4 3 or x = 1 3 Example 4: Solve: 2 3 x2 + 13x – 5 3 = 0 Solution: Here, 2 3 x2 + 13x – 5 3 = 0 or, 2 3 x2 + (15 – 2) x – 5 3 = 0 or, 2 3 x2 + 15x – 2x – 5 3 = 0 or, 3 x (2x + 5 3 ) – 1(2x + 5 3 ) = 0 or, (2x + 5 3 ) ( 3x – 1) = 0 Either, (2x + 5 3 ) = 0 i.e., x = –5 3 2 Or, ( 3x – 1) = 0 i.e., x = 1 3 Hence, x = –5 3 2 or x = 1 3 (Note: Please replace the variable in the given equation by –5 3 2 and 1 3 , the equation is satisfied. So, these are the required solutions.) Example 5: Find the quadratic equation whose roots are –2 and 3. Solution: Since the roots of the required quadratic equation are –2 and 3, x = –2 and x = 3 x + 2 = 0 and x – 3 = 0 ∴ (x + 2) (x – 3) = 0 or, x2 – 3x + 2x – 6 = 0 or, x2 – x – 6 = 0 Hence, x2 – x – 6 = 0 is the required quadratic equation. Example 6: Solve: 11 – x x + 3 = 5 x + 1 Solution: Here, 11 – x x + 3 = 5 x + 1 or, (11 – x) (x + 1) = 5(x + 3) or, 11x + 11 – x2 – x = 5x + 15 Checking the solutions When x = 4 3 ; 9x2 + 4= 15x or, 9 (4 3) 2 + 4 = 15 × 4 3 or, 16 + 4 = 20 or, 20 = 20 (True) When x =1 3 ; 9x2 + 4 = 15x or, 9 (1 3) 2 + 4 = 15 × 1 3 or, 1 + 4 = 5 or, 5 = 5 (True) So, x = 4 3 or x = 1 3 are the correct solutions. Investigating two numbers: Product = 2 3 × 5 3 = 10 × 3 = 30 Difference = 13 Required numbers are 15 and 2.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 165 Vedanta Excel in Mathematics - Book 10 Quadratic Equations or, –x2 + 10x – 5x + 11 – 15 = 0 or, –x2 + 5x – 4 = 0 or, x2 – 5x – 4 = 0 or, x2 – 4x – x + 4 = 0 or, x(x – 4) – 1(x – 4) = 0 or, (x – 4) (x – 1) = 0 Either, x – 4 = 0, i.e. x = 4 or, x – 1 = 0, i.e. x = 1 Hence, 4 and 1 are the required solutions of the given equation. Example 7: Solve: 3x – 2 2x – 3 = 3x – 8 x + 4 Solution: Here, 3x – 2 2x – 3 = 3x – 8 x + 4 or, (3x – 2) (x + 4) = (2x – 3) (3x – 8) or, 3x2 + 12x – 2x – 8 = 6x2 – 16x – 9x + 24 or, 3x2 + 10x – 8 = 6x2 – 25x + 24 or, 3x2 – 35x + 32 = 0 or, 3x2 – 32x – 3x + 32 = 0 or, x(3x – 32) –1(3x – 32) = 0 or, (3x – 32) (x – 1) = 0 Either, x – 32 = 0, i.e. x = 32 3 = 102 3 or, x – 1 = 0, i.e. x = 1 Hence, 102 3 and 1 are the required solutions of the given equation. Example 8: By using factorization method, solve : x – 1 x – 2 + x – 3 x – 4 = 31 3 ; x ≠ 2, 4 Solution: Here, x – 1 x – 2 + x – 3 x – 4 = 31 3 or, (x – 1) (x – 4) + (x – 3) (x – 2) (x – 2) (x – 4) = 10 3 or, x(x – 4) – 1(x – 4) + x(x – 2) –3(x – 2) x(x – 4) –2(x – 4) = 10 3 or, x2 – 4x – x + 4 + x2 – 2x – 3x + 6 x2 – 4x – 2x + 8 = 10 3 or, 2x2 – 10x + 10 x2 – 6x + 8 = 10 3 or, 10x2 – 60x + 80 = 6x2 – 30x + 30 Taking LCM of denominators Cross-multiplying

Vedanta Excel in Mathematics - Book 10 166 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations or, 10x2 – 60x + 80 – 6x2 + 30x – 30 = 0 or, 4x2 – 30x + 50 = 0 or, 2 (2x2 – 15x + 25) = 0 or, 2x2 – 15x + 25 = 0 or, 2x2 – (10 + 5)x + 25 = 0 or, 2x2 – 10x – 5x + 25 = 0 or, 2x (x – 5) – 5 (x – 5) = 0 or, (x – 5) (2x – 5) = 0 Either, x – 5 = 0 i.e., x = 5 Or, 2x – 5 = 0 i.e., x = 5 2 ∴ x = 5 or x= 5 2 are the required solutions. Example 9: The number of volleyball games that must be scheduled in a league with n teams is given by G (n) = n2 – n 2 , where each team plays with every other team exactly once. A league schedules 15 games. How many teams are in the league? Solution: Here, total number of scheduled games in a league is given by G (n) = n2 – n 2 Now, total number of games = 15 or, n2 – n 2 = 15 or, n2 – n = 30 or, n2 – n – 30 = 0 or, n2 – (6 – 5) – 30 = 0 or, n2 – 6n + 5n – 30 = 0 or, n (n – 6) + 5 (n – 6) = 0 or, (n – 6) (n + 5) = 0 Either, n – 6 = 0 i.e., n = 6 Or, n + 5 = 0 i.e., n = – 5 But the number of teams cannot be negative. Hence, there are 6 teams in the league. Exercise 8.1 EXERCISE 8.1 General section 1. Answer the following questions. a) Define quadratic equation with an example. b) What is pure quadratic equation? Give an example. c) What is adfected quadratic equation? Give an example. d) What do you mean by roots of a quadratic equation? 2. Use the zero factor property to solve each equation. a) (x + 1) (x – 1) = 0 b) (x + 2) (x – 3) = 0 c) (x – 3) (2x + 1) = 0 d) (2x – 5) (5x – 2) = 0 e) x (2x – 1) = 0 f) 3x (4 – 3x) = 0 Expressing as ax2 + bx + c = 0 Splitting middle term

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 167 Vedanta Excel in Mathematics - Book 10 Quadratic Equations 3. Solve each equation: a) x2 = 9 b) 2x2 = 8 c) 9y2 = 16 d) 4y2 = 25 e) x2 – 36 = 0 f) x2 – 81 = 0 g) 25x2 – 64 = 0 h) 49y2 – 100 = 0 i) 2x 3 = 6 x j) 3x 4 = 4 3x k) 5 x2 – 36 5 = 0 l) 64x 7 – 7 x = 0 4. Find the quadratic equations which have the following roots. a) 3 and 5 b) –2 and 3 c) –4 and –2 d) 4 and –3 Creative section – A 5. Solve each equation by factorization method: a) x2 – 4x = 0 b) 2x2 – 10x = 0 c) 3x2 – 12x = 0 d) x2 + 3x + 2 = 0 e) x2 + 7x + 12 = 0 f) x2 – 3x + 2 = 0 g) x2 – 8x + 15 = 0 h) x2 – x – 12 = 0 i) 2x2 – 3x + 1 = 0 j) 3x2 – 10x – 8 = 0 k) 5x2 + 8x – 21 = 0 l) 2x2 + 21 = 17x 6. Solve these equations by factorization method: a) 3 2x2 + 5x + 2 = 0 b) 2 3x2 – 8x + 2 3 = 0 c) 6x2 – 7x – 5 6 = 0 Creative section – B 7. Solve these equations by factorization method: a) x – 1 x – 2 = 6 x b) x 2x – 3 = 2 x – 1 c) x 2x + 1 = x 4x – 1 d) 2 x + 1 = 3x 5x – 1 e) x + 2 x + 3 = 2x – 3 3x – 7 f) x – 1 x – 2 = 2x – 5 3x – 10 g) x + 4 3x – 4 = 2x – 1 2x + 1 h) 3x – 8 x – 2 = 5x – 2 x + 5 8. Solve these equations by factorization method: a) x + 2 x – 2 + x – 2 x + 2 = 31 3 b) x + 3 x – 3 + x – 3 x + 3 = 41 4 c) x x – 1 + x – 1 x = 21 2 d) 3 x – 1 + 2 x – 3 = 2 x e) x – 1 x – 2 + x – 3 x – 4 = 31 3 f) 1 (x – 1) (x –2) + 1 (x – 2) (x –3) = 2 3 9. a) While solving the equation (x + 2) (x – 3) = 6, Mr. Pravesh, a student of class X, solved it as either, x + 2 = 6 i.e., x = 4 or, x – 3 = 6 i.e., x = 9 (i) Is there error in his reasoning? (ii) What is the correct way of solving this equation? b) Joshika is studying in class X in a Secondary School. She solved the quadratic equation (x – 1) (x – 2) = 12 as either, x – 1 = 12 i.e., x = 13 or, x – 2 = 12 i.e., x = 14 (i) What error do you get in her reasoning? (ii) What is the correct way of solving this equation? 10. a) The number of hand shacking is given by H (x) = x(x – 1) 2 where each person shakes the hand with other people exactly once. If a group of people make 45 hand shakings in a meeting, how many people are there in the party? b) The sum of first n even number is given by S (n) = n2 + n. How many even numbers should be added to make their sum 110?

Vedanta Excel in Mathematics - Book 10 168 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations (ii) Solving a quadratic equation by completing square method In this method, we transpose the constant term (c) to R.H.S. and L.H.S. is expressed as a perfect square expression. Study the following example. Let’s take the standard form of quadratic equation. ax2 + bx + c= 0 or, ax2 + bx = – c (Transposing c to R.H.S.) or, ax2 a + b a x = – c a (To make the coefficient of x2 unity, dividing both sides by a) or, x2 + b a x = – c a or, x2 + b a x + b 2a 2 = b 2a 2 – c a or. x2 + 2. b 2a x + b 2a 2 = b2 4a2 – c a or, x + b 2a 2 = b2 – 4ac 4a2 or, x + b 2a = ± b2 – 4ac 2a or, x = – b 2a ± b2 – 4ac 2a = – b ± b2 – 4ac 2a Thus, the required roots of x are – b + b2 – 4ac 2a and – b – b2 – 4ac 2a . Worked-out Examples Example 1: Solve: x2 – 11x + 24 = 0 Solution: x2 – 11x + 24 = 0 or, x2 – 11x = – 24 or, x2 – 11x + 11 2 2 = 11 2 2 – 24 or,x2 – 2.x. 11 2 x + 11 2 2 = 121 4 – 24 or, x – 11 2 2 = 25 4 or, x – 11 2 2 = ± 5 2 2 or, x – 11 2 =± 5 2 Taking +ve sign, x – 11 2 = 5 2 or, x = 5 2 + 11 2 = 16 2 = 8 Taking –ve sign, x – 11 2 = –5 2 or, x = –5 2 + 11 2 = 6 2 = 3 Hence, 8 and 3 are the required solutions. (Adding the square of half of the coefficient of x to both sides) www.geogebra.org/classroom/s5aabztw Classroom code: S5AA BZTW Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 169 Vedanta Excel in Mathematics - Book 10 Quadratic Equations Example 2: Solve: 4x2 – x – 3 = 0 by completing the square. Solution: 4x2 + x – 3 = 0 or, 4x2 + x = 3 or, 4x2 4 + x 4 = 3 4 or, x2 + x 4 + 1 8 2 = 1 8 2 + 3 4 or, x2 + 2.x. 1 8 + 1 8 2 = 1 64 + 3 4 or, 2 x + 1 8 = 49 64 or, x + 1 8 = ±7 8 Hence, –1 and 3 4 are the required solutions. Taking +ve sign, x + 1 8 = 7 8 or, x = 7 8 – 1 8 = 6 8 = 3 4 Taking –ve sign, x + 1 8 = – 7 8 or, x = – 7 8 – 1 8 = – 8 8 = – 1 Example 3: Solve 2 x – 1 = 3 x + 1 + 4 x + 2 by completing the square. Solution: 2 x – 1 = 3 x + 1 + 4 x + 2 or, 2 x – 1 = 3 (x + 2) + 4 (x + 1) (x + 1) (x + 2) or, 2 x – 1 = 7x + 10 x2 + 3x + 2 or, (x – 1) (7x + 10) = 2(x2 + 3x + 2) or, 5x2 – 3x – 14 = 0 or, 5x2 5 – 3x 5 = 14 5 [Dividing each term by 5] or, x2 – 3 5 x = 14 5 or, x2 – 3 5 x+ 3 10 2 = 3 10 2 + 14 5 or, x2 – 2.x. 3 10 + 3 10 2 = 9 100 + 14 5 or, 2 x – 3 10 = 2 ±17 10 or, x – 3 10 = ±17 10 Hence, 2 and – 7 5 are the required solution. Taking +ve sign, x – 3 10 = 17 10 or, x = 17 10 + 3 10 = 2 Taking –ve sign, x – 3 10 = – 17 10 or, x = – 17 10 + 3 10 = –7 5 EXERCISE 8.2 General section 1. Express both sides of these equations in square forms. Then, solve: a) x2 + 2.x. 1 + 12 = 4 b) x2 + 2.x.3 + 32 = 16 c) x2 – 2. x. 4 + 42 = 25 d) x2 – 2.x.5 + 52 = 81

Vedanta Excel in Mathematics - Book 10 170 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations e) x2 + 2.x. 1 2 + 1 2 2 = 4 9 f) x2 – 2.x. 3 5 + 3 5 2 = 4 25 2. Express both sides of these equations in square forms. Then, solve. a) x2 + x + 1 2 2 = 1 2 2 + 6 b) x2 – 7x + 7 2 2 = 7 2 2 – 12 c) x2 – 3x + 3 2 2 = 3 2 2 + 10 d) x2 – 10x + 52 = 52 – 21 e) x2 – x + 1 2 2 = 1 2 2 + 20 f) x2 + 5x + 5 2 2 = 5 2 2 + 24 Creative section - A 3. Solve each equation by completing the square: a) x2 – 3x = –2 b) x2 + 2x = 8 c) x2 + 3x = 10 d) x2 – 2x – 3 = 0 e) x2 – 7x + 12 = 0 f) x2 + 3x – 28 = 0 g) 2x2 – 7x + 3 = 0 h) 3x2 + 5x – 2 = 0 i) 6x2 – 5x – 6 = 0 j) 3x2 = 10x – 3 k) 5x2 = 8x + 21 l) 15x2 – 2ax = a2 m) x2 4 + 2 = 3x 2 n) 9x + 1 x = 6 o) x + 3 x + 2 = 3x – 7 2x – 3 p) 3x – 8 x – 2 = 5x – 2 x + 5 q) 2x x – 4 + 2x – 5 x – 3 = 81 3 r) 2x – 1 2x + 1 – 2x + 1 2x – 1 = –22 3 (iii) Solving a quadratic equation by using formula We obtained the following two roots of x while solving the quadratic equation of the form ax2 + bx + c = 0 by completing the square. x = – b + b2 – 4ac 2a and x = – b – b2 – 4ac 2a These roots can be written as x = – b ± b2 – 4ac 2a where, a is the coefficient of x2 , b is the coefficient of x and c is the constant term. We use this formula to find the required solutions of the given quadratic equation. For the quadratic equation ax2 + bx + c = 0, a ≠ 0, the expression b2 – 4ac is called discriminant. It is generally denoted by 'D'. ∴ D = b2 - 4ac. Facts to remember The roots of the quadratic equation ax2 + bx + c = 0; a ≠ 0 are x = – b ± b2 – 4ac 2a . www.geogebra.org/classroom/js5v5b6k Classroom code: JS5V 5B6K Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 171 Vedanta Excel in Mathematics - Book 10 Quadratic Equations Worked-out Examples Example 1: Solve 16x2 – 9 = 0 by using the formula. Solution: 16x2 – 9 = 0 or, 16x2 + 0.x + (–9) = 0 Here, a = 16, b = 0 and c = 9 \ x = – b ± b2 – 4ac 2a = – 0 ± 02 – 4 × 16 × (–9) 2 × 16 = ±24 32 =± 3 4 Example 2: Solve 2x2 + 5x – 12 = 0 by using formula. Solution: 2x2 + 5x – 12 = 0 or, 2x2 + 5x + (–12) = 0 Here, a = 2, b = 5 and c = –12 \ x = – b ± b2 – 4ac 2a = – 5 ± 52 – 4 × 2 × (–12) 2 × 2 = – 5 ± 11 4 Taking +ve sign, Taking –ve sign, x = – 5 + 11 4 = 6 4 = 3 2 x = – 5 – 11 4 = – 16 4 = –4 \ x = 3 2 and – 4 Example 3: Solve x – 2 x + 2 + x + 2 x – 2 = 2x + 6 x – 3 by using the formula. Solution: x – 2 x + 2 + x + 2 x – 2 = 2x + 6 x – 3 or, (x – 2)2 + (x + 2)2 (x + 2) (x – 2) = 2x + 6 x – 3 or, x2 – 4x + 4 + x2 + 4x + 4 x2 – 4 = 2x + 6 x – 3 or, 2x2 + 8 x2 – 4 = 2x + 6 x – 3 or, (x2 – 4) (2x + 6) = (x – 3) (2x2 + 8) or, 2x3 + 6x2 – 8x – 24 = 2x3 + 8x – 6x2 – 24 or, 12x2 – 16x = 0 or, 4 (3x2 – 4x) = 0 or, 3x2 + (–4) x + 0 = 0 Here, a = 3, b = – 4 and c = 0 \ x = – b ± b2 – 4ac 2a = – (–4) ± (–4)2 – 4 × 3 × 0 2 × 3 = 4 ± 4 6 Taking +ve sign, Taking –ve sign, x = 4 + 4 6 = 4 3 x = 4 – 4 6 = 0 6 = 0 \ x = 4 3 and 0.

Vedanta Excel in Mathematics - Book 10 172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations Example 4: The pilot of a plane may run out of runway if the landing speed is too fast, and the plane may stall if the speed is too slow. The appropriate landing speed of a plane is determined by D = 0.1x2 – 3x + 22, where x is average landing speed in feet per second and D is the distance in feet. (i) What is the approximate landing speed if the run way is 800 feet long? (ii) How long does it take to land on the airport? Solution: (i) Here, 0.1x2 – 3x +22 = D and distance (D) = 800 \ 0.1x2 – 3x +22 = 800 i.e., 0.1x2 – 3x – 778 = 0 Comparing it with ax2 + bx + c = 0, we get a = 0.1, b = – 3 and c = –778 Now, using quadratic formulas, we get, x = – b ± b2 – 4ac 2a = – (–3) ± (–3)2 – 4(0.1)(-778) 2(0.1) = 3 ± 9 + 311.2 0.2 = 3 ± 17.9 0.2 Taking (+) ve sign, we get x = 3 + 17.9 0.2 = 104.5 Taking (–) ve sign, we get x = 3 – 17.9 0.2 = –74.5 But the speed cannot be negative. Hence, the landing speed of the plane is approximately 104.5 feet per second. (ii) Time taken for landing the airplane = Distance travelled Average speed = 800 104.5 = 7.66 So, the plane takes approximately 7.66 seconds to land on the airport. Exercise 8.3 EXERCISE 8.3 General section 1. Answer the following questions. a) What are the roots of the quadratic equation ax2 + bx + c = 0; a ≠ 0? b) What are the roots of the quadratic equation px2 + qx + r = 0; a ≠ 0? c) What are the roots of the quadratic equation x2 – 25? 2. Replace a, b, and c by their given values and evaluate x = – b ± b2 – 4ac 2a . a) a = 1, b = 3, c = 2 b) a = 1, b = 5, c = 6 c) a = 1, b = 1, c = –6 d) a = 2, b = –5, c = 2 e) a = 3, b = – 10, c = 8 f) a = 2, b = – 13, c = 15

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 173 Vedanta Excel in Mathematics - Book 10 Quadratic Equations Creative section – A 3. Solve each equation by using the formula: a) x2 – 4 = 0 b) x2 = 9 c) 2x2 – 32 = 0 d) 50x2 = 8 e) x2 – x = 0 f) 2x2 – 6x = 0 g) 6x2 = –3x h) 15x2 = – 10x 4. Solve these equations by using the formula: a) x2 – 3x + 2 = 0 b) x2 – 5x + 6 = 0 c) x2 + 2x – 3 = 0 d) x2 + 2x – 8 = 0 e) 3x2 – 5x + 6 = 2(x2 + 3) f) 4x(x + 7) + 10 = 2(x2 + 5) g) 2x2 + 3x + 1 = 0 h) 3x2 – 2x – 21 = 0 i) 3x2 + 10 = –11x j) (x + 1) (2x + 3) = 4x2 – 22 k) x 3 + 2 x = 7 3 l) 4x2 = 4 15 x + 3 Creative section – B 5. Solve these equations by factorization method: a) x + 3 2x – 7 = 2x – 1 x – 3 b) 1 x – 2 = 2 (x – 3) x (x – 1) c) x x + 1 + x + 1 x = 13 6 d) x + 2 x – 2 – x – 2 x + 2 = 5 6 e) x + 1 x – 1 – x – 1 x + 1 = 22 3 f) 1 x – 2 + 2 x – 1 = 6 x g) x + 3 x + 2 + x – 3 x – 2 = 2x – 3 x – 1 h) x – 1 x + 1 + x + 2 x – 2 = x – 3 x + 3 + x + 4 x – 4 6. a) A cottage industry of a locality produces a certain number of pottery articles in a day. The total cost of production of articles is given as C = 2x2 – 5x, where x is the number of articles produced in a particular day and C is the production cost of articles. (i) How many pottery articles should be produced on the day to make Rs. 12,400 as the cost of production? (ii) What is the rate of cost of each pottery article? b) Sunayana is driving a car from Kathmandu to Pokhara with her family. The distance covered by the car is given by D = 27x2 + x, where x is the time in hours and D is the distance travelled by car in km. (i) How long does it take to reach at a place of distance 110 km? (ii) What is the average speed of the car? Project Work and Activity Section 7. a) Make the quadratic equations in x and solve to get the following values of x. (i) x = ±1 (ii) x = ±2 (iii) x = ±7 (iv) x = ±9 (v) x = 2, 3 (vi) x = 3, –2 (viii) x = –3, 2 (viii) x = –3, –2 b) Write one quadratic equation of each of the following forms and solve them. (i) x2 + ax + b = 0 (ii) x2 + ax – b = 0 (iii) x2 – ax – b = 0 (iv) x2 – ax + b = 0

Vedanta Excel in Mathematics - Book 10 174 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations 8.3 Application of Quadratic equations A certain type of real life problems in verbal form can be expressed and solved by the method given by quadratic equation. Follow these working steps to solve the verbal problems on quadratic equations: (i) Consider the unknown quantity as a variable such as x, y, z, a, b, c, etc. (ii) Express the verbal problem in a quadratic equation form. (iii) Solve the quadratic equation by any of the methods i.e., factorization, completing the square or by quadratic formula. In some cases, only one solution of the quadratic equation may be meaningful and it will be the required value. Any value which does not satisfy the condition of the problem should be rejected. Here, we shall discuss on solutions of various types of verbal problems related to quadratic equation. Worked-out Examples Type I: Verbal Problems Based on Numbers Facts to remember 1. The consecutive natural numbers: x, x + 1, x + 2, ……. 2. The consecutive odd/even numbers: x, x + 2, x + 4, ……. 3. The sum of squares of x and y: x2 + y2 4. The reciprocal of a natural number x: 1 x 5. In two-digit number formed by x and y i.e., 10x + y; sum of digits = x + y, product of digits = xy Example 1: If 4 is subtracted from two-third of square of a natural number, the result is 20. Find the number. Solution: Let the required natural number be x. According to the given condition, 2 3 x2 – 4 = 20 or, 2 3 x2 = 24 or, 2x2 = 72 or, x2 = 36 or, x2 = (± 6)2 or, x = ± 6 Hence, the required natural number is 6. Answer checking: When x = 6; 2 3 x2 – 4 = 20 or, 2 3 × 62 – 4 = 20 or, 24 – 4 = 20 or, 20 = 20 (True) Hence, 6 is the required natural number.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 175 Vedanta Excel in Mathematics - Book 10 Quadratic Equations Example 2: The product of two consecutive even numbers is 168. Find the numbers. Solution: Let the required consecutive even numbers be x and x + 2 respectively. According to the given condition, x (x + 2) = 168 or, x2 + 2x – 168 = 0 or, x2 + 14x – 12x – 168 = 0 or, x (x + 14) – 12 (x + 14) = 0 or, (x + 14) (x – 12) = 0 Either x + 14 = 0, i.e., x = – 14 (rejected) or, x – 12 = 0, i.e., x = 12 So, the first even number = 12 The next consecutive even number = x + 2 = 12 + 2 = 14. Example 3: If the sum of two numbers is 9 and their product is 18, find the numbers. Solution: Let one of the numbers be x. Then, the other number is (9 – x). According to the given condition, x (9 – x) = 18 or, 9x – x2 = 18 or, x2 – 9x + 18 = 0 or, x2 – 6x – 3x + 18 = 0 or, x (x – 6) – 3 (x – 6) = 0 or, (x – 6) (x –3) = 0 Either, x – 6 = 0 i.e. x = 6 or, x – 3 = 0 i.e. x = 3 When x = 6, then 9 – x = 9 – 6 = 3 When x = 3, then 9 – x = 9 – 3 = 6 Hence, the required numbers are 6 and 3. Example 4: The product of the digits in a two-digit number is 54. When 27 is added to the number the digits interchange their places. Find the number. Solution: Let the digit at tens place be x and ones place be y. Then, the number so formed = 10x + y When the digits interchange their places, the number so formed = 10y + x. Now, according to the first condition, xy = 54 … (i) Also, according to the second condition, (10x + y) + 27 = 10y + x or, 10x + y – 10y – x = – 27 or, 9x – 9y = – 27 or, 9 (x – y) = – 27 or, x – y = –3 or, x = y – 3 ... (ii) Substituting the value of x from equation (ii) in equation (i), we get, (y – 3) y = 54 or, y2 – 3y – 54 = 0 Answer checking: Product of 12 and 14 = 12 × 14 = 168; which is given in the question. Thus, the answer is correct. Answer checking: Sum of 6 and 3 = 9 Product of 6 and 3 = 18; which are given in the question. Thus, the answer is correct.

Vedanta Excel in Mathematics - Book 10 176 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations or, y2 – (9 – 6)y – 54 = 0 or, y2 – 9y + 6y – 54 = 0 or, y (y – 9) + 6 (y – 9) = 0 or, (y – 9) (y + 6) = 0 Either, y – 9 = 0, i.e. y = 9 or, y + 6 = 0, i.e. y = –6 (rejected) When y = 9, then x × 9 = 54, i.e. x = 6 Hence, the required number is 10x + y = 10 × 6 + 9 = 69. Example 5: A number consisting two digits is equal to four times the sum and three times the product of its digits. (i) What is the number? (ii) By what percent should the number be increased to make the number formed by reversing its digits? Solution: (i) Let the digit at tens place be x and ones place be y. Then, the number so formed = 10x + y Now, according to the first condition, 10x + y = 4 (x + y) or, 10x + y = 4x + 4y or, 6x = 3y or, y = 2x … (i) According to the second condition, 10x + y = 3xy ... (ii) Substituting the value of y from equation (i) in equation (ii), we get, 10x + 2x = 3x (2x) or, 12x = 6x2 or, 6x2 – 12x = 0 or, 6x (x – 2) = 0 Either, 6x = 0, i.e. x = 0 or, x – 2 = 0, i.e. x = 2 When x = 0, then y = 2 × 0 = 0 (rejected) When x = 2, then y = 2 × 2 = 4 Hence, the required number is 10x + y = 10 × 2 + 4 = 24. (ii) Again, the number formed by reversing the digits of 24 is 42. Difference = 42 – 24 = 18 ∴ Increased percent = Difference 24 × 100% = 18 24 × 100% = 75% Hence, the number should be increased by 75% to make its reversing number. Type II: Verbal Problems Based on Ages Facts to remember If the present age of a person is x years, then (i) Before T years, he/she was (x – T) years old. (ii) After T years, he/she will be (x + T) years old. Answer checking: In 69, product of digits = 6 × 9 = 54 When 27 is added to 69; 69 + 27 = 96, the digits are reversed; which are given in the question. Thus, the answer is correct. Checking answer: 4 times the sum of digits = 4 (2 + 4) = 24 3 times the product of digits = 3 × 2 × 4 = 24; which are given in the question. Hence, the answer is correct.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 177 Vedanta Excel in Mathematics - Book 10 Quadratic Equations or, y2 – (9 – 6)y – 54 = 0 or, y2 – 9y + 6y – 54 = 0 or, y (y – 9) + 6 (y – 9) = 0 or, (y – 9) (y + 6) = 0 Either, y – 9 = 0, i.e. y = 9 or, y + 6 = 0, i.e. y = –6 (rejected) When y = 9, then x × 9 = 54, i.e. x = 6 Hence, the required number is 10x + y = 10 × 6 + 9 = 69. Example 5: A number consisting two digits is equal to four times the sum and three times the product of its digits. (i) What is the number? (ii) By what percent should the number be increased to make the number formed by reversing its digits? Solution: (i) Let the digit at tens place be x and ones place be y. Then, the number so formed = 10x + y Now, according to the first condition, 10x + y = 4 (x + y) or, 10x + y = 4x + 4y or, 6x = 3y or, y = 2x … (i) According to the second condition, 10x + y = 3xy ... (ii) Substituting the value of y from equation (i) in equation (ii), we get, 10x + 2x = 3x (2x) or, 12x = 6x2 or, 6x2 – 12x = 0 or, 6x (x – 2) = 0 Either, 6x = 0, i.e. x = 0 or, x – 2 = 0, i.e. x = 2 When x = 0, then y = 2 × 0 = 0 (rejected) When x = 2, then y = 2 × 2 = 4 Hence, the required number is 10x + y = 10 × 2 + 4 = 24. (ii) Again, the number formed by reversing the digits of 24 is 42. Difference = 42 – 24 = 18 ∴ Increased percent = Difference 24 × 100% = 18 24 × 100% = 75% Hence, the number should be increased by 75% to make its reversing number. Type II: Verbal Problems Based on Ages Facts to remember If the present age of a person is x years, then (i) Before T years, he/she was (x – T) years old. (ii) After T years, he/she will be (x + T) years old. Answer checking: In 69, product of digits = 6 × 9 = 54 When 27 is added to 69; 69 + 27 = 96, the digits are reversed; which are given in the question. Thus, the answer is correct. Checking answer: 4 times the sum of digits = 4 (2 + 4) = 24 3 times the product of digits = 3 × 2 × 4 = 24; which are given in the question. Hence, the answer is correct. Example 6: The present ages of a father and his son are 35 years and 12 years respectively. Find how many years ago the product of their ages was 210. Solution: Let the required number of years is x years. ∴The age of the father before x years = (35 – x) years The age of the son before x years = (12 – x) years. According to the problem, (35 – x) (12 – x) = 210 or, 420 – 35x – 12x + x2 = 210 or, x2 – 47x + 210 = 0 or, x2 – 42x – 5x + 210 = 0 or, x (x – 42) – 5 (x – 42) = 0 or, (x – 42) (x – 5) = 0 Either, x – 42 = 0 i.e. x = 42 But, the value does not satisfy the condition. So, it is rejected. or, x – 5 = 0 i.e. x = 5 So, 5 years ago, the product of their ages was 210. Example 7: The difference of the ages of two brothers is 4 years and the product of the number of their ages is 221. Determine the ages of two brothers. Solution: Let the age of the elder brother be x years and the age of younger brother be y years. Now, according to the first condition, x – y = 4 or, x = y + 4 … (i) Also, according to the second condition, xy = 221 … (ii) Substituting the valye of x from equation (i) in equation (ii), we get, (y + 4) y = 221 or, y2 + 4y – 221 = 0 or, y2 + 17y – 13y – 221 = 0 or, y (y + 17) – 13 (y + 17) = 0 or, (y + 17) (y – 13) = 0 Either, y + 17 = 0 i.e. y = – 17 But age cannot be negative. So, it is rejected. or, x – 13 = 0 i.e. x = 13 Again, substituting the value of x in equation (i), we get x = 13 + 4 = 17 Hence, the required ages of the brothers are 13 years and 17 years. Example 8: 6 years ago a man’s age was 6 times the age of his son. The product of the number of the present ages of the father and his son is 396. What are their present ages? Answer checking: 5 years ago, The age of father = 35 – 5 = 30 The age of son = 12 – 5 = 7 The product of their ages = 30 × 7 = 210; which is given in the question. Thus, the answer is correct. Answer checking: Difference = 17 – 13 = 4 Product = 17 × 13 = 221; which are given in the question. Thus, the answer is correct.

Vedanta Excel in Mathematics - Book 10 178 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations Solution: Let the present age of the father be x years and the present age of the son be y years. 6 years ago, the age of the father = (x – 6) years 6 years ago, the age of the son = (y – 6) years According to the first condition, (x – 6) = 6(y – 6) or, x = 6y – 30 … (i) According to the second condition, x × y = 396 … (ii) Substituting the value of x from equation (i) in equation (ii), we get, (6y – 30) × y = 396 or, 6y2 – 30y – 396 = 0 or, 6(y2 – 5y – 66) = 0 or, y2 – 11y + 6y – 66 = 0 or, y(y – 11) + 6(y – 11) = 0 or, (y – 11) (y + 6) = 0 Either, y – 11 = 0, i.e. y = 11 Or, y + 6 = 0, i.e. y = – 6, But, age cannot be negative. So, y = – 6 is rejected. Now, substituting the value of y in equation (i), we get, x = 6 × 11 – 30 = 36 So, the present ages of the father and his son are 36 years and 11 years respectively. Type III: Verbal Problems Based on Mensuration Facts to remember 1. The perimeter of rectangle (P) = 2 (l + b) and its area (A) = l × b 2. The area of path with uniform width of d units running outside the rectangular field having length l units and breadth (b) is given by A = 2d (l + b + 2d) 3. The area of path with uniform width of d units running inside the rectangular field having length l units and breadth (b) is given by A = 2d (l + b – 2d) Example 9: A room is 2 m longer than its breadth. If the area of the floor of the room is 120 m2 , find its length and breadth. Solution: Let the length of room be l m and its breadth = b m. Now, according to the first condition, l = b + 2 …(i) Also, according to the second condition, l × b = 120 … (ii) Substituting the value of l from equation (i) in equation (ii), we get, (b + 2)b = 120 or, b2 + 2b – 120 = 0 or, b2 + 12b – 10b – 120 = 0 or, b (b + 12) – 10 (b + 12) = 0 or, (b + 12) (b – 10) = 0 Either, b + 12 = 0 i.e. b = – 12 Answer checking: 6 years ago, father’s age = 36 – 6 = 30 6 years ago, son’s age = 11 – 6 = 5 ∴Father’s age was 6 times son’s age. Now, product = 36 × 11 = 396; which are given in the question. Thus, the answer is correct. (b + 2) m 120 sq. m b m

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 179 Vedanta Excel in Mathematics - Book 10 Quadratic Equations But the breadth of the room cannot be negative. So, it is rejected. or, b – 10 = 0 i.e. b = 10 Again, substituting the value of b in equation (i), we get l = 10 + 2 = 10 + 2 = 12 So, the length and the breadth of the room are 12 m and 10 m respectively. Example 10: The area of a rectangular field is 2200 sq. meters and perimeter is 188 m. (i) Find the length and breadth of the field. (ii) Out of length or breadth, which one is to be decreased by what percent to make it a square? Solution: (i) Let the length of room be x m and its breadth be y m. Now, according to the first condition, perimeter of the field (P) = 188 m or, 2 (l + b) = 188 or, x + y = 94 ∴ y = 94 – x … (i) Also, according to the second condition, l × b = 2200 ∴ x × y = 2200 … (ii) Substituting the value of y from equation (i) in equation (ii), we get, x (94 – x) = 2200 or, x 2 – 94 x + 2200 = 0 Comparing it with ax 2 + bx+ c = 0, we get a = 1, b = – 94 and c = 2200 Using quadratic formula; we get x = – b ± b2 – 4ac 2a = – (–94) ± (–94)2 – 4(1)(2200) 2(1) =94 ± 8836 – 8800 2 = 94 ± 6 2 Taking (+) sign, we get, x = 94 + 6 2 = 50 Taking (–) sign, we get, x = 94 – 6 2 = 44 When x = 50, y = 94 – 50 = 44 When x = 44, y = 94 – 44 = 50 Assuming that the length is longer the breadth, Thus, length of the field (l) is 50 m and the breadth (b) = 44 m (ii) Again, difference between length and breadth = 50 m – 44 m = 6 m Percentage of difference = Difference length × 100% = 6 m 50 m × 100% = 12% Hence, the length of the field should be decreased by 12% to make it a square field. x m y m Alternative method: x 2 – 94 x + 2200 = 0 or, x 2 – (50 + 40) x + 2200 = 0 or, x 2 – 50x – 40x + 2200 = 0 or, x (x – 50) – 40 (x – 50) = 0 or, (x – 50) (x – 40) = 0 Either, x – 50 = 0 i.e., x = 50 Or, x – 44 = 0 i.e., x = 44

Vedanta Excel in Mathematics - Book 10 180 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations Example 11: The area of a rectangular room is 45 sq.m. If the length had been 3 m less and the breadth 1 m more, it would have been a square. Find the length and the breadth of the room. Solution: Let the length and the breadth of the rectangular room be l m and b m respectively. From the first condition, l × b = 45 ... (i) From the second condition, l – 3 = b + 1 (since the room becomes a square) or, l = b + 4 ... (ii) Substituting the value of l from equation (ii) in equation (i) we get, (b + 4) × b = 45 or, b2 + 4b – 45 = 0 or, b2 + 9b – 5b – 45 = 0 or, b (b + 9) – 5 (b + 9) = 0 or, (b + 9) (b – 5) = 0 Either, b + 9 = 0, i.e. b = – 9 which is impossible and rejected. or, b – 5 = 0, i.e. b = 5 When b = 5, then l × b = 45, i.e. l = 9 So, the required length and breadth of the room are 9 m and 5 m respectively. Example 12: A rectangular park is 130 m long and 96 m wide. A footpath of uniform width of x m runs all around inside the park. (i) Find the area of the path in terms of x. (ii) If the area of the path is 888 sq. meters, what is its width? Solution: Here, the length if the park (l) = 130 m, the breadth of the park (b) = 96 m, the width of the inner path (d) = x m (i) Now, area of the inner path (A) = 2d (l + b – 2d) = 2x (130 + 96 – 2x) m2 = (452x – 4x2 ) m2 (ii) Also, area of the path (A) = 888 m2 or, 452x – 4x2 = 888 [Using (i)] or, 4 (113x – x2 ) = 888 or, 113x – x2 = 222 or, x2 – 113x + 222 = 0 or, x2 – (111 + 2)x + 222 = 0 or, x2 – 111x – 2x + 222 = 0 or, x (x – 111) – 2 (x – 111) = 0 or, (x – 111) (x – 2) = 0 Either, x – 111= 0 i.e., x = 111 which is impossible as it is wider than the breadth. or, x – 2 = 0 i.e., x = 2 Hence, the path is 2 m wide. l m b m (l – 3) m (b + 1) m 130 m x 96 m Answer checking: Area of path = 2d (l + b – 2d) = 2 × 2 (130 + 96 – 2 × 2) m2 = 888 m2 which is given in the question. Thus, the answer is correct.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 181 Vedanta Excel in Mathematics - Book 10 Quadratic Equations Type IV: Verbal Problems Based on Geometry Facts to remember Pythagoras theorem, p2 + b2 = h2 Example 13: The sides of a right-angled triangle containing the right-angle are less than its hypotenuse by 2 cm and 4 cm respectively. Find the lengths of the sides of the triangle. Solution: Let the hypotenuse, perpendicular and base of the right angled triangle be h, p, and b respectively. From the given condition, p = h – 2 and b = h – 4 By using Pythagoras theorem in the right angled triangle, h2 = p2 + b2 or, h2 = (h – 2)2 + (h – 4)2 or, h2 = h2 – 4h + 4 + h2 – 8h + 16 or, h2 – 12h + 20 = 0 or, h2 – 10h – 2h + 20 = 0 or, h (h – 10) – 2 (h – 10) = 0 or, (h – 10) (h – 2) = 0 Either, h – 10 = 0, i.e. h = 10 cm or, h – 2 = 0, i.e. h = 2 cm which is impossible. Now, when h = 10 cm, p = (10 – 2) cm = 8 cm Also, when h = 10 cm, b = (10 – 4) cm = 6 cm So, the required lengths of the sides of the triangle are 10 cm, 8 cm, and 6 cm respectively. Type V: Verbal Problems Based on Distance, Speed and Time Facts to remember 1. Distance travelled (D) = Speed (S) × Time (T) 2. Time (T) = Distance (D) Speed Distance Speed Time Example 14: The distance between two sub-metropolitan cities P and Q is 360 km. A bus travels from city P to city Q at a uniform speed. The bus takes 1 hour less for the journey if its speed is increased by 4 km per hour from its usual speed. (i) What is the usual speed of the bus? (ii) If the speed of the bus is 6 km/hr less than the usual speed, how long more time would it take for the journey? Solution: (i) Let the usual speed of the bus be x km/hr. A B C p b h

Vedanta Excel in Mathematics - Book 10 182 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations Then, time taken by the bus = Distance (D) Speed = 360 x hrs. [Given, distance = 360 km] When the speed is increased by 4 km/hr, then new speed = (x + 4) km/hr Time taken by the bus = 360 x + 4 hrs. According to the condition given in the question, 360 x = 360 x + 4 + 1 or, 360 x – 360 x + 4 = 1 or, 360(x + 4 – x) x(x + 4) = 1 or, x2 + 4x = 1440 or, x2 + 4x – 1440 = 0 or, x2 + 40x – 36x – 1440 = 0 or, x (x + 40) – 36(x + 40) = 0 or, (x + 40) (x – 36) = 0 Either, x + 40 = 0 i.e., x = – 40 [Rejected as the speed cannot be negative] or, x – 36 = 0 i.e., x = 36 Hence, the usual speed of the bus is 36 km/hr. (ii) Again, time taken by the bus in usual speed (T) = 360 x = 360 36 = 10 hrs. If the speed is 6 km/hr less than its usual speed, new speed = (36 – 6) km/hr = 30 km/hr And, time taken by the bus = 360 30 = 12 hrs Difference of time taken = 12 hrs – 10 hrs = 2 hrs Hence, the bus takes 2 hrs more time to for the same journey if its speed is decreased by 6 km/hr than the usual speed. Type VI: Miscellaneous Verbal Problems Example 15: Last week Mrs. Yadav bought some vegetables for Rs 150. This week the rate of cost of vegetables increases by Rs 5 per kg and she can buy 1 kg less vegetables for the same amount of money. (i) What was the rate of cost of vegetable last week? (ii) By what percent is the rate of cost of vegetables increased? Solution: (i) Let the rate of cost of vegetables in last week be Rs x per kg. Then, the quantity of vegetables purchased in last week = 150 x kg In this week, the increased rate of cost of vegetables = Rs (x + 5) per kg Alternative method: Let the usual speed = x km/hr and time taken = y hrs. Then, xy = 360 ∴y = 360 x … (i) Again, (x + 4) (y – 1) = 360 or, (x + 4) ( 360 x – 1) = 360 or, 360x – x2 + 1440 – 4x = 360x or, x2 + 4x – 1440 = 0 or, (x + 40) (x – 36) = 0 Either, x + 40 = 0 i.e., x = – 40 [Rejected as the speed cannot be negative] or, x – 36 = 0 i.e., x = 36

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 183 Vedanta Excel in Mathematics - Book 10 Quadratic Equations Then, the quantity of vegetables purchased in this week 150 x + 5 kg According to the question, 150 x – 150 x + 5 = 1 or, x2 + 5x – 750 = 0 or, (x + 30) (x – 25) = 0 or, x = –30 or 25 But, rate of cost cannot be negative. So, the rate of cost of vegetables in last week was Rs 25 per kg. (ii) Again, percentage of increased amount of rate of cost = Rs 5 Rs 25 × 100% = 20% Hence, the rate of cost of vegetables is increased by 20% in this week. Exercise 8.4 EXERCISE 8.4 General section 1. The area of the following figures are given, find the value of x. a) b) c) 2. a) If 6 is added to the square of a positive number, the sum is 70. Find the number. b) If 5 is added to twice the square of a natural number, the sum is 23. Find the number. c) If 8 is subtracted from thrice the square of a natural number, the difference is 40. Find the number. d) If 4 is subtracted from the half of the square of a natural number, the result is 14. Find the number. e) If 10 is added to two-third of the square of a natural number, the sum is 34, find the number. f) If 17 is subtracted from the twice of the square of a positive number, the result is 111. Find the number. g) If the double of a natural number is equal to the square of the number, find the number. 3. a) The sum of a positive number and its square is 30, find the number. b) If a natural number is subtracted from its square, the difference will be 12, find the number. c) If the product of two consecutive natural numbers is 56, find the numbers. d) The product of two consecutive even numbers is 80. Find the numbers. e) The product of two consecutive odd numbers is 63. Find the numbers. 4. a) When a natural number is decreased by 6, the result is 40 times the reciprocal of the number. Find the number. b) If the sum of a number and 21 times its reciprocal is 10, find the number. 60 cm2 40 cm2 18 cm2 (x + 3) cm (x + 2) cm x cm x cm 2x cm (x – 2) cm

Vedanta Excel in Mathematics - Book 10 184 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations 5. a) The product of the number of the ages of a man 5 years before and 5 years hence is 875. Find the present age of the man. b) Gita’s age is twice the age of Rita. If the product of the number of their ages is 200, find their ages. Creative section – A 6. a) If the sum of two numbers is 10 and their product is 24, find the numbers. b) The product of two positive numbers is 56. If their difference is 10, find the numbers. c) A number exceeds another number by 3 and their product is 40. Find the numbers. d) Divide 11 into two parts so that their product will be 24. 7. a) The sum of two numbers is 10 and the sum of their squares is 68. Find the numbers. b) The difference of two numbers is 4 and the sum of their squares is 58. Find the numbers. c) If the sum of the squares of two consecutive natural even numbers is 100, find the numbers. d) The sum of the squares of two consecutive natural odd numbers is 74, find the numbers. 8. a) In a two-digit number, the product of the digits is 18 and the sum is 9. Find the number. b) The sum of digits of two digit number is 7 and their product is 12. Find the numbers. c) In a two-digit number, the digit at tens place is 1 more than the digit at unit place. If the product of the digits is 20, find the number. d) A two-digit number is such that the product of the digits is 24. When 45 is added to the number, the digits interchange their places. Find the number. e) A number consists of two-digits whose product is 18. If 63 is subtracted from the number, the digits are interchanged. Find the number. f) A two-digit number is four times the sum and three times the product of its digits. Find the number. g) A number of two-digits is equal to four times the sum of the digits. If the product of the digits is 8, find the number. h) The product of digits in a two-digit number is 18. The number formed by interchanging the digits of the number will be 27 more than the original number. Find the original number. i) In a number of two digits, the square of the sum of its digits is 100. If 2 is subtracted from three times of the number, the digits are reversed. Find the number. j) The square of the sum of the digits of a two-digit number is 64. If 3 is added to four times the number, the digits are reversed. Find the number. 9. a) The present ages of two brothers are 15 years and 22 years respectively. After how many years will the product of the number of their ages be 408? b) The present ages of a mother and her daughter are 30 years and 14 years respectively. Find how many years ago the product of the number of their ages was 192.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 185 Vedanta Excel in Mathematics - Book 10 Quadratic Equations 10. a) The difference of the present ages of two sisters is 5 years. If the product of their ages is 204, find the ages of the two sisters. b) The sum of the present ages of a brother and his younger sister is 16 years. If the product of their ages is 63, find their ages. c) Sunayana is 7 years younger than Bishwant. If the product of their ages is 144, find their present ages. d) The difference of the ages of two brothers is 4 years. 5 years ago, if the product of their ages was 96, find their present ages. e) The difference of the ages of two sisters is 5 years. After 4 years, if the product of their ages becomes 104, find their present ages. f) The difference of the ages of a sister and her younger brother is 6 years. The numerical value of the product of their ages is equal 4 times the sum of their ages. Find the present ages of the sister and her brother. g) The daughter’s age is 4 years more than the son’s age. 2 years ago, the product of their ages was 32. Find the present ages. h) The product of the present ages of two sisters is 150. 5 years ago, the elder sister was twice as old as her younger sister. Find their present ages. i) One year hence, a father’s age will be 5 times the age of his son. The product of the present ages of the father and his son is 145. Find their present ages. j) The present age of mother is equal to the square of the age of her daughter after one year. If the age of daughter in 10 years hence will be 1 year less than the age of her mother in 10 years ago, find their present ages. 11. a) The length of a room is 2 m longer than its breadth. If the area of the room is 63 sq.m, find the length and the breadth of the room. b) The breadth of a rectangular garden is 4 m shorter than its length and the area of the garden is 192 sq.m. Find the perimeter of the garden. c) If the area of a rectangular courtyard is 63 sq.m and its perimeter is 32 m, find the length and the breadth of the courtyard. d) A rectangular meadow has an area of 247 sq.m and its perimeter is 64 m. Find the length and the breadth of the meadow. e) A rectangular piece of land is 40 m long and 25 m wide. A path of uniform width and 426 m2 area surrounds the land from outside. Find the width of the path. f) The height of a tank is 2 m and its length is 3 m more than its breadth. If 80 cu.m of water can be stored in the tank, find the length and the breadth of the tank. 12. a) If the sides of a right-angled triangle are (x – 2) cm, x cm and (x + 2) cm, find the length of each of its side. b) The perimeter of a right-angled triangle is 12 cm and its hypotenuse is 5 cm. Find the other two sides of the triangle. c) The area of a right-angled triangle is 24 sq. cm. If the perpendicular of the triangle is 2 cm shorter than its base, find two sides of the triangle.

Vedanta Excel in Mathematics - Book 10 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Quadratic Equations d) The sides of a right-angled triangle containing the right-angle are less than its hypotenuse by 5 cm and 10 cm respectively. Find the length of the sides of the triangle. Creative section – B 13. a) Rajani is studying in class VI in a school. Her age, consisting of two digits, is equal to four times the sum and six times the product of the digits of the age. (i) How old is she now? (ii) If the age of her brother Rajan is formed by reversing the digits of her age, by what percent is the age of Rajan more than that of Rajani? b) Shashwat is an engineer in a hydro company. His age, consists of two digits, is equal to four times the sum and three times the product of the digits of the age. (i) How old is he now? (ii) If the age of his mother is formed by just interchanging the digits of his age, by what percent is the age of mother more than his age? 14. a) A school has a rectangular play-ground of dimensions 40 m × 30 m. A rectangular basketball court is constructed in it so that the area of grass strip of uniform width surrounding the court is 1000 square meters. (i) What is the width of the grass strip? (ii) By what percent is the length of the court more than its breadth? b) Mr. Mahato has a rectangular field. The area of the field is 720 sq. m. and perimeter is 108 m. (i) Find the dimensions of his field. (ii) By what percent is the longer side of the field to be decreased to make it a square? 15. a) Mrs. Magar bought a certain kilograms of vegetables for Rs 120 last week. This week, the rate of cost of vegetables decreases by Rs 20 per kg and she can buy 1 kg more vegetables for the same amount of money. (i) What was the rate of cost of vegetable last week? (ii) By what percentage is the rate of cost of vegetables decreased? b) A piece of cloth costs Rs 800. If the piece was 5 m longer and the rate of cost of cloth per metre was Rs 8 less, the cost of the piece would have remained unchanged. (i) How long is the piece and the original rate of cost per meter? (ii) If the piece of cloth is 4 m shorter for the same cost, by what percent is the rate of cost per meter increased? 16. a) Some students of school planned a picnic. The budget for the food was Rs 7,200. As four of them unable to join the picnic, the cost of the food for each students increased by Rs 60. (i) How many students went for the picnic? (ii) If 1 more student joins the picnic than the planning, by what percent would the cost for each student be decreased?

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 187 Vedanta Excel in Mathematics - Book 10 Quadratic Equations b) A new-year party bill for a number of people is Rs 4,800. If there were 2 people more, the bill each person had to pay would have reduced by Rs 80. (i) Find the number of people attending the party. (ii) If there were 2 people less, by what percent would the bill of each person be increased? c) A bus travels 360 km at a uniform speed. If the usual speed has been 5 km/hr more, it would have taken 1 hour less for the same journey. (i) What is the usual speed of the bus? (ii) If the speed of the bus is 4 km/hr less than the usual speed, how long would it take for the journey OBJECTIVE QUESTIONS Tick the correct alternative. 1. The second degree algebraic equations are ……………. equations. (A) linear (B) quadratic (C) cubic (D) bi-quadratic 2. Which of the following expressions is not the quadratic equation? (A) x2 – 9 = 0 (B) x (x + 5) + 6 = 0 (C) x2 – x = 0) (D) x (x + 1) = x2 + 2 3. Identify the pure quadratic equation. (A) x2 = 25 (B) x2 – 7x + 6 = 0 (C) x (x + 8) + 15 (D) x2 – 3(x – 2)=0 4. The equation having roots 3 and – 2 is (A) x2 + x – 6 = 0 (B) x2 + 5x + 6 = 0 (C) x2 – x – 6 = 0 (D) x2 –5x – 6 = 0 5. What are the roots of the quadratic equation ax2 + bx + c = 0; a ≠ 0? (A) – b ± b2 – 4ac 2a (B) – a ± b2 – 4ac 2b (C) – c ± c2 – 4ab 2c (D) – b ± 4ac – b2 2a 6. What are the roots of the quadratic equation x2 – 81 = 0? (A) 3 (B) 9 (C) – 9 (D) ± 9 7. The product of two consecutive even numbers is 24, what is the smaller number? (A)2 (B) 4 (C) 6 (D) 8 8. What is the side length of a square if the numerical values of its area and perimeter area equal? (A)1 unit (B) 2 units (C) 4 units (D) 8 units 9. In a two-digit number, the sum of digits is 5 and the product of the digits is 6, the number is … (A)56 (B) 32 (C) 23 (D) Both (B) and (C) 10. The product of ages of a boy before 2 years and after 2 years is 45, what is his present age? (A)5 years (B) 7 years (C) 9 years (D) 10 years

Vedanta Excel in Mathematics - Book 10 188 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Unit 9 Rational Expressions Classwork - Exercise 9.1 Rational expressions–Looking back 1. Let’s choose and tick only the polynomials from the given expressions as quickly as possible. a) 2x3 – 5x2 + 3x + 7 b) x3 +8x2 + x – 3 c) 2 x4 – x d) y2/3 + y + 1 2. Let’s write the value of ‘x’ for each of the expressions given below which makes it undefined. a) 2 x b) 4 x – 7 c) x – 6 x – 9 d) x + 1 3x – 6 e) 3 x2 – 25 3. Let’s write the given rational expressions in the lowest form. a) 4x7 6x4 b) 35a2 b4 7a3 b4 c) 3x – 12 4x – 16 d) y – x x – y e) p3 + q3 p2 – q2 The numbers 1 2 , 2 3 , 4 7 , 11 15, –2 5 , 5 –6 , –2, 4, 0 …… and so on are rational numbers. The rational numbers can be expressed in the form p q , where p and q are integers and q ≠ 0. Similarly, x 2 , 4a2 b 5 , a + b a – b etc, are a few examples of the algebraic rational expressions. However, a + b a – b is not a rational expression. Facts to remember 1. An expression written in the form p(x) q(x) , where p(x) and q(x) are polynomials and q(x) ≠ 0 is called rational expression. It is the ratio of two polynomials. 2. The rational expression p(x) q(x) is said to be in the lowest form if the H.C.F. of p(x) and q (x) is 1. 3. The expression p(x) q(x) becomes undefined when q (x) = 0. 9.2 Simplification of rational expressions In case of addition and subtraction of rational expressions with a common denominator, we should simply add or subtract the numerators. Then, the sum is reduced to its lowest terms.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 189 Vedanta Excel in Mathematics - Book 10 Rational Expressions For example, x x2 – y2 – y x2 – y2 = x – y x2 – y2 = x – y (x + y)(x – y) = 1 x + y In case of addition and subtraction of rational expressions with unlike denominators, at first, we should find the L.C.M. of the denominators. Then, the L.C.M. is divided by each denominator and the quotient is multiplied by the corresponding numerator as like the simplification of unlike fractions in arithmetic. Let’s learn the process of simplification from the following illustrations. Worked-out Examples Example 1: Simplify: a) 1 x – y – x + y x2 – y2 b) a2 + ab + b2 a + b + a2 – ab + b2 a – b Solution: a) 1 x – y – x + y x2 – y2 = 1 x – y – x + y (x + y) (x – y) = 1 x – y – 1 x – y = 1 – 1 x – y = 0 Example 2: Simplify: a) x xy – y2 + y xy – x2 b) 1 (x – y)(x – z) – 1 (z – x) (y – z) Solution: a) x xy – y2 + y xy – x2 = x y(x – y) + y x(y – x) = x y(x – y) + y –x(x – y) = x y(x – y) – y x(x – y) = x2 – y2 xy(x – y) = (x + y) (x – y) xy(x – y) = x + y xy Example 3: Simplify 2 (a – 2) (a – 3) + 2 (a – 1) (3 – a) + 1 (1 – a) (2 – a) Solution: 2 (a – 2) (a – 3) + 2 (a – 1) (3 – a) + 1 (1 – a) (2 – a) b) a2 + ab + b2 a + b + a2 – ab + b2 a – b = (a – b) (a2 + ab + b2 ) + (a + b) (a2 – ab + b2 ) (a + b) (a – b) = a3 – b3 + a3 + b3 a2 – b2 = 2a3 a2 – b2 b) 1 (x – y)(x – z) – 1 (z – x)(y – z) = 1 (x – y)(x – z) – 1 – (x – z)(y – z) = 1 (x – y)(x – z) + 1 (x – z)(y – z) = y – z + x – y (x – y)(x – z) (y – z) = x – z (z – y)(x – z) (y – z) = 1 (x – y) (y – z)

Vedanta Excel in Mathematics - Book 10 190 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Rational Expressions = 2 (a – 2) (a – 3) – 2 (a – 1) (a – 3) + 1 (a – 1) (a – 2) = 2(a – 1) – 2(a – 2) + a – 3 (a – 2) (a – 3) (a – 1) = 2a – 2 – 2a + 4 + a – 3 (a – 2) (a – 3) (a – 1) = a – 1 (a – 2) (a – 3) (a – 1) = 1 (a – 2) (a – 3) Example 4: Simplify 1 x2 – 5x + 6 + 1 x2 – 3x + 2 + 2 x2 – 8x + 15 Solution: 1 x2 – 5x + 6 + 1 x2 – 3x + 2 + 2 x2 – 8x + 15 = 1 x2 – 3x – 2x + 6 + 1 x2 – 2x – x + 2 + 2 x2 – 5x – 3x + 15 = 1 x(x – 3) – 2(x – 3) + 1 x(x – 2) – 1(x – 2) + 2 x(x – 5) – 3(x – 5) = 1 (x – 3) (x – 2) + 1 (x – 2) (x – 1) + 2 (x – 5) (x – 3) = x – 1+ x – 3 (x – 3) (x– 2) (x – 1) + 2 (x–5) (x – 3) = 2(x – 2) (x – 3) (x– 2) (x – 1) + 2 (x–5) (x – 3) = 2(x – 5) + 2(x – 1) (x – 3) (x– 1) (x – 5) = 2x – 10 + 2x – 2 (x – 3) (x– 1) (x – 5) = 4(x – 3) (x – 3) (x– 1) (x – 5) = 4 (x– 1) (x – 5) Example 5: Simplify x x – y + y x + y + 2xy y2 – x2 Solution: x x – y + y x + y + 2xy y2 – x2 = x x – y + y x + y – 2xy x2 – y2 = x(x + y) + y(x – y) (x – y) (x + y) – 2xy x2 – y2 = x2 +xy + xy – y2 x2 – y2 – 2xy x2 – y2 = x2 +2xy – y2 – 2xy x2 – y2 = x2 – y2 x2 – y2 = 1

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 191 Vedanta Excel in Mathematics - Book 10 Rational Expressions Example 6 : Simplify 1 1 – p + p2 – 1 1 + p + p2 – 2p 1 – p2 + p4 Solution: 1 1 – p + p2 – 1 1 + p + p2 – 2p 1 – p2 + p4 = 1 + p2 + p – (1 + p2 – p) (1 + p2 – p) (1 + p2 + p) – 2p 1 + p4 – p2 = 1 + p2 + p – 1 – p2 + p (1 + p2 ) 2 – p2 – 2p 1 + p4 – p2 = 2p 1 + 2p2 + p4 – p2 – 2p 1 + p4 – p2 = 2p 1 + p4 + p2 – 2p 1 + p4 – p2 = 2p(1 + p4 – p2 ) – 2p(1 + p4 + p2 ) (1 + p4 + p2 ) (1 + p4 – p2 ) = 2p + 2p5 – 2p3 – 2p – 2p5 – 2p3 (1 + p4 ) 2 – (p2 ) 2 = –4p3 1 + 2p4 + p8 – p4 = –4p3 1 + p4 + p8 Example 7 : Simplify y – 2 y2 – 2y + 4 + y + 2 y2 + 2y + 4 – 16 y4 + 4y2 + 16 Solution: y – 2 y2 – 2y + 4 + y + 2 y2 + 2y + 4 – 16 y4 + 4y2 + 16 = (y – 2) (y2 + 2y + 4) + (y + 2) (y2 – 2y + 4) (y2 – 2y + 4) (y2 + 2y + 4) – 16 y4 + 4y2 + 16 = y3 – 23 + y3 + 23 [(y2 + 4) – 2y] [(y2 + 4) + 2y] – 16 y4 + 4y2 + 16 = 2y3 (y2 + 4)2 – (2y) 2 – 16 y4 + 4y2 + 16 = 2y3 y4 + 8y2 + 16 – 4y2 – 16 y4 + 4y2 + 16 = 2y3 y4 + 4y2 + 16 – 16 y4 + 4y2 + 16 = 2y3 – 16 y4 + 4y2 +16 = 2(y3 – 8) (y2 – 2y +4) (y2 + 2y + 4) = 2(y – 2) (y2 + 2y + 4) (y2 – 2y +4) (y2 + 2y + 4) = 2(y – 2) y2 – 2y +4

Vedanta Excel in Mathematics - Book 10 192 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Rational Expressions Example 8 : Simplify x2 – (y – z) 2 (x + z) 2 – y2 + y2 – (x – z) 2 (x + y) 2 – z2 + z2 – (x – y) 2 (y + z) 2 – x2 Solution: x2 – (y – z) 2 (x + z) 2 – y2 + y2 – (x – z) 2 (x + y) 2 – z2 + z2 – (x – y) 2 (y + z) 2 – x2 = (x + y – z) (x – y + z) (x + y + z) (x – y + z) + (y + x – z) (y – x + z) (x + y + z) (x + y – z) + (z + x – y) (z – x + y) (x + y + z) (y + z – x) = x + y – z x + y + z + y – x + z x + y + z + z + x – y x + y + z = x + y – z + y – x + z + z + x – y x + y + z = x + y + z x + y + z = 1 Example 9: Prove that: 1 m + 1 + m m2 – 1 + m2 1 – m4 = 1 1 – m4 Solution: L.H.S. = 1 m + 1 + m m2 – 1 + m2 1 – m4 = 1 m + 1 + m (m + 1) (m – 1) + m2 1 – m4 = m –1 – m (m + 1) (m – 1) – m2 1 – m4 = –1 m2 – 1 – m2 1 – m4 = –1 m2 – 1 – m2 (1 + m2 ) (1 – m2 ) = 1 + m2 – m2 (1 + m2 ) (1 – m2 ) = 1 1 – m4 = R.H.S. Proved Example 10: Simplify: 1 x + 1 + 2 x2 + 1 + 4 x4 – 1 Solution: 1 x + 1 + 2 x2 + 1 + 4 x4 – 1 = 1 x + 1 + 2 x2 + 1 + 4 (x2 + 1) (x2 – 1) = 1 x + 1 + 2(x2 –1) + 4 (x2 + 1) (x2 – 1) = 1 x + 1 + 2x2 – 2 + 4 (x2 + 1) (x2 – 1) = 1 x + 1 + 2x2 + 2 (x2 + 1) (x2 – 1)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193 Vedanta Excel in Mathematics - Book 10 Rational Expressions = 1 x + 1 + 2(x2 + 1) (x2 + 1) (x2 – 1) = 1 x + 1 + 2 (x + 1) (x – 1) = x – 1 + 2 (x + 1) (x – 1) = x + 1 (x + 1) (x – 1) = 1 x – 1 Example 11: Bina and Suresh were asked to simplify 3(x + 2) x2 – 4 – 1 x – 2 . (i) Bina said that to convert to like fractions, the LCM of denominators was x – 2. Do you agree with her? Explain and simplify it. (ii) Suresh said that the LCM can also be taken as x2 – 4. Show that Suresh’s method is also correct. Solution: (i) Here, 3(x + 2) x2 – 4 = 3(x + 2) (x + 2) (x –2) = 3 x – 2 We agree with Bina because 3 x – 2 and 1 x – 2 are like fractions. Now, 3(x + 2) x2 – 4 – 1 x – 2 = 3 x – 2 – 1 x – 2 = 3 – 1 x – 2 = 2 x – 2 (ii) 3(x + 2) x2 – 4 – 1 x – 2 = 3(x + 2) (x + 2) (x – 2) – 1 x – 2 = 3(x + 2) –1(x + 2) (x + 2) (x – 2) = 3x + 6 – x – 2 (x + 2) (x – 2) = 2x + 4 (x + 2) (x – 2) = 2(x + 2) (x + 2) (x – 2) = 2 x – 2 We get the same answer by Suresh method too. Hence, Suresh method is also correct. EXERCISE 9.1 General section 1. Simplify: a) 1 m – n – m + n m2 – n2 b) x + 1 x + 2 + x – 2 x2 – 4 c) 4x x2 – 1 – x + 1 x – 1 d) a2 + 1 a + 1 + a2 – a + 1 a2 – 1 e) x x + y – xy x2 + 2xy + y2 f) 1 a – x + x a2 – 2ax + x2

Vedanta Excel in Mathematics - Book 10 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Rational Expressions 2. Simplify: a) a2 + ab + b2 a + b + a2 – ab + b2 a – b b) p2 + pq + q2 p + q – p2 – pq + q2 p – q c) x + a x + b – x – b x – a d) x – 4 x + 6 – x – 6 x + 4 e) m + 2 m – 2 – m – 2 m + 2 f) y + 1 y + 5 – y – 1 y + 5 3. Simplify: a) 1 ab – b2 + 1 ab – a2 b) x xy – y2 + y xy – x2 c) p 2p – 4 + 2 2p – p2 d) c2 cd – d2 + d2 cd – c2 e) n2 7n – 49 + 49 7n – n2 f) 1 (x – y) (x – z) + 1 (z – x) (z – y) 4. Prove that: a) x – 1 x – 3 – x2 – x + 2 x2 –2x – 3 = 1 x + 1 b) 1 1 – a2 + a 1 + a3 = 1 (1 – a) (1 + a3 ) c) a2 – 3ab + 2b2 a2 – 4b2 – 4ab2 4a2 b + 8ab2 = a – b a + b d) 2 x + 11 + 44 121 – x2 = 2 x – 11 Creative section-A 5. Simplify: a) 1 (x – 2) (x – 3) + 1 (x – 2) (x – 1) – 2 (x – 3) (x – 1) b) 1 (x – 3) (x + 2) + 3 (x + 2) (4 – x) + 2 (x – 3) (x – 4) c) 2 (a – 3) (a – 4) (a – 5) + a – 1 (3 – a) (a – 4) + a – 2 (5 – a) (a – 3) d) x – 1 (2x – 1) (x + 2) + 3 (x + 2) (x – 1) – 1 (1 – x) (1 – 2x) 6. Simplify: a) 1 x2 – 5x + 6 – 2 x2 – 4x + 3 – 1 x2 –3x + 2 b) 1 a2 – 5a + 6 – 2 a2 – 4a + 3 – 1 a2 – 3a + 2 c) x – 1 x2 – 3x + 2 + x – 2 x2 – 5x + 6 + x – 5 x2 – 8x + 15 d) 2a – 6 a2 – 9a + 20 – a – 1 a2 – 7a + 12 – a – 2 a2 – 8a + 15 7. Simplify: a) x x + 2 + x x – 2 – 4x x2 – 4 b) 2 a + b – 2 a – b + 4a a2 – b2

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195 Vedanta Excel in Mathematics - Book 10 Rational Expressions c) x + y x – y – x – y x + y + 4xy x2 + y2 d) a + b a – b – a – b a + b – 2ab a2 – b2 e) x + 1 x – 1 + x – 1 x + 1 – 2x2 x2 – 1 f) x + y x – y + x – y x + y – 2 (x2 – y2 ) x2 + y2 8. Simplify: a) 1 1 + x + x2 – 1 1 – x + x2 – 2x 1 + x2 + x4 b) x + 3 x2 + 3x + 9 + x – 3 x2 – 3x + 9 – 54 x4 + 9x2 + 81 c) x – y x2 – xy + y2 + x + y x2 + xy + y2 – 2y3 x4 + x2 y2 + y4 d) a + 2 1 + a + a2 – a – 2 1 – a + a2 – 2a2 1+ a2 + a4 e) 2a + b 4a2 + 2ab + b2 + 2a – b 4a2 – 2ab + b2 – 2b3 16a4 + 4a2 b2 + b4 f) 3x – 1 9x2 – 3x + 1 – 3x + 1 9x2 + 3x + 1 + 54x3 81x4 + 9x2 + 1 9. Simplify: a) a2 – (b – c) 2 (a + c) 2 – b2 + b2 – (a – c) 2 (a + b) 2 – c2 + c2 – (a – b) 2 (b + c) 2 – a2 b) (a – b) 2 – c2 a2 – (b + c) 2 + (b – c) 2 – a2 b2 – (c + a) 2 + (c – a) 2 – b2 c2 – (a + b) 2 10. Simplify: a) 1 (1 – x ) – 1 (1 + x ) + x (1 – x) b) 2 x + y + 3 x – y – 5 x – y x – y c) 1 8 (1 – a) – 1 8 (1 + a) + 2 a 8 (1 – a) d) 1 8 ( x – 1) + 1 8 ( x + 1) + 2 x 8 (x – 1) 11. Prove that: a) 1 a + 1 – a a2 – 1 + a2 1 – a4 = 1 1 – a4 b) 1 x – 1 – x x2 – 1 – x2 x4 – 1 = 1 x4 – 1 c) 1 p + 1 + 2 p2 + 1 + 4 p4 – 1 = 1 p – 1 d) 1 a + x + 2a a2 + x2 + 4a3 x4 – a2 = 1 x – a Creative section-B 12. a) Manish and Ayusha were asked to simplify w w + 1 – w – 1 w2 – 1 (i) Manish said that to convert to like fractions, the LCM of denominators was (w + 1). Do you agree with him? Explain and simplify it. (ii) Ayusha said that the LCM can also be taken as w2 – 1. Show that Ayusha’s method is also correct.

Vedanta Excel in Mathematics - Book 10 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Rational Expressions b) Rekha and Bindu were asked an expression 1 a + 2 + a2 – 5a + 6 a2 – 4 to simplify. (i) Rekha said that to convert into like fractions, the LCM of denominators was a + 2. Do you agree with her reasoning? Explain and simplify it. (ii) Bindu said that the LCM can also be taken as a2 – 4. Show that Bindu’s method is also correct. OBJECTIVE QUESTIONS Tick the correct alternative. 1. For what value of x, the expression 4 – x 3x – 3 becomes undefined? (A) 4 (B) 3 (C) 1 (D) 9 2. What value of x makes the expression x2 – 4 x2 – 8x – 12 undefined? (A) 2 (B) 6 (C) 4, 6 (D) 2, 6 3. The rational expression – 3 – 2x 4x + 5 is equivalent to (A) –3 + 2x –4x – 5 (B) –3 – 2x 4x + 5 (C) 2x – 3 4x + 5 (D) 3x + 3 –4x + 5 4. The simplified expression of x2 – x x3 – x is (A) x – 1 x2 – 1 (B) 1 x – 1 (C) 1 x + 1 (D) x – 1 x + 1 5. The simplified form of x2 – 1 x + 1 × x + 2 x2 – x is (A) x + 2 x (B) x + 1 x (C) x x + 2 (D) x – 1 x + 1 6. The area of a rectangular park is (x + 4) (x + 5) 2x + 8 sq. meter. If it is x – 5 2 meter long, how wide is it? (A) x + 5 x – 5 m (B) x – 5 x + 5 m (C) x – 4 x + 5 m (D) x – 5 x + 4 m 7. The sum of x x2 – 16 + 4 16 – x2 m is equal to (A) 1 x – 4 (B) 1 x + 4 (C) x + 4 (D) 1 4 – x 8. The rational expression a – 1 a – 2 – a + 1 a + 2 is equal to (A) 4a a2 – 4 (B) 2a a2 + 4 (C) 2a a2 – 4 (D) a2 – 1 a2 – 4 9. The expression 1 m – n – 2n m2 – n2 is simplified to (A) 1 m – n (B) 1 m + n (C) m – 3n m2 – n2 (D) m m2 – n2 10. The length of sides of a triangle are 2 x , 3 2x and 1 x + 1 . The perimeter of the triangle is (A) 7x + 9 2x(x + 1) (B) 7x + 10 2x(x + 1) (C) 9x + 7 2x(x + 1) (D) 9x + 5 2x(x + 1)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197 Vedanta Excel in Mathematics - Book 10 Unit 10 Exponential Equations 10.1 Indices –Looking back 1. Let’s observe the growth of infected people with a certain virus in the following figure. The number of people infected by a sick person with virus in a few steps are : 1, 2, 4, 8, 16, … i.e., 2o , 21 , 23 , 24 , … and so on. It seems that the virus spreads in an exponential form. 2. Let’s answer the following questions. a) The value of (3a)o , a ≠ 0 is …..…… b) The index of 6 in 36 is …..…… c) In 0.001, the power of 10 is ………… d) The value of 8 27 1 3 is ….........…… 10.2 Indices- review Let's multiply an algebraic term x four times by itself. It is written as x × x × x × x and the product is shown in the abbreviated form which is x4 . In this case, x is called the base and 4 is called the index or exponent. Thus, the index refers to the power to which a number is raised. In the example x4 , the base x is raised to the power 4. Indices is the plural form of index. 10.3 Laws of indices There are certain verified rules which are used in the operations of indices of algebraic terms. Product rule, quotient rule, power rule, etc. are the examples of such rules. The verifications of these rules have been already discussed in our earlier grades. The table given below shows the summary of these rules.

Vedanta Excel in Mathematics - Book 10 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Exponential Equations Laws of indices at a glance (i) Product law am × an = am + n, where 'm' and 'n' are positive integers (ii) Quotient law am ÷ an = am – n when m > n am ÷ an = 1 an – m when m < n (iii) Power law of indices (am) n = am × n, (ab) m = ambm, a m b = am bm (iv) Law of negative index a– m = 1 am or am = 1 a–m (v) Law of zero index a° = 1, b° = 1, x° = 1 and so on (vi) Root law of indices n am = m n a 10.4 Exponential equation Let’s take an equation 2x = 8. In this equation, the unknown variable ‘x’ is the base and 2 is its coefficient. Now, take another equation 2x = 8. In this case, the unknown variable ‘x’ is the exponent of the base 2. Such an equation in which the unknown variable appears as an exponent of a base is known as exponential equation. For example: 2x = 8, 9y + 1 = 3y + 5, 5x + 5x+1 = 25 etc. are exponential equations. Let’s follow the activities given below. Activity 1 1. Let’s fill up the table and find the value of x which satisfies the equation 2x = 8. x 0 1 2 3 4 2x 2o = 1 21 = 2 22 = 4 23 = 8 24 = 16 From the above table, 2x = 8 is true when x = 3. Thus, required value of x is 3. 2. Let’s fill up the table then find the value of x which makes the equation 3x + 1 = 1 true. x -2 -1 0 1 2 3x+1 From the above table, 3x+1 = 1 is true when x = ……….. Thus, required value of x is ………..

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 199 Vedanta Excel in Mathematics - Book 10 Exponential Equations 3. Let’s fill up the table and identify value of x which satisfies the equation 4x + 1 = 8x . x 0 1 2 3 4 4x+1 8x From the above table, 4x +1 = 8x is true when x = ……….. Thus, required value of x is ……….. Activity 2 Suppose Mr. Gurung deposits a sum of Rs 5,000 in a commercial bank at 10% p.a. interest compounded annually. In how many years, will he withdraw an amount of Rs 6,655? Here, principal (P) = Rs. 5,000, rate (R) = 10% p.a. and compound amount (C.A.) = Rs. 6,655 We know, C.A. = R 100 T P 1 + or, Rs. 6,655 = Rs. 5000 10 100 T 1 + or, Rs.6655 Rs. 5000 = 10 100 T 1 + or, 1.331 = (1.1)T or, (1.1)3 = (1.1)T or, 3 = T Hence, required time is 3 years. Facts to remember 1. An equation in which the unknown variable appears as an exponent of a base is known as exponential equation. For example: 3x = 81, 5y + 3 = 25y etc. 2. The following axioms are useful while solving the exponential equations: (i) If ax = ab , then x = b (ii) If ax = 1, then ax = a° and x = 0 3. While solving an exponential equation, we should simplify the equation till the equation will be obtained in the forms ax = ab or ax = 1. Expressing 1.331 as the power of 1.1

Vedanta Excel in Mathematics - Book 10 200 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Exponential Equations Worked-out Examples Example 1: Solve: a) 2x = 32 b) 52x = 1 25 c) 6x – 2 – 1 = 0 Solution: a) Here, 2x = 32 b) Here, 52x = 1 25 c) Here, 6x – 2 – 1 = 0 or, 2x = 25 or, 52x = 1 52 or, 6x – 2 = 1 ∴ x = 5 or, 52x = 5-2 or, 6x – 2 = 6o or, 2x = - 2 or, x – 2 = 0 ∴ x = -1 ∴ x = 2 Example 2: Solve: a) 3x + 3x +1 = 36 b) 5y+2 – 3× 5y + 1 = 10 c) ( 2) x– 3 = ( 2 4 ) x+1 Solution: Here, 3x + 3x +1 = 36 b) Here, 5y+2 – 3× 5y + 1 = 10 c) Here, ( 2) x– 3 = ( 2 4 ) x+1 or, 3x + 3x × 31 = 36 or, 5y × 52 – 3× 5y × 5 = 10 or, 2x – 3 2 = 2x + 1 4 or, 3x (1 + 3) = 36 or, 5y (25 – 3 × 5) = 10 or, x – 3 2 = x + 1 4 or, 3x × 4 = 36 or, 5y × 10 = 10 or, 4x – 12 = 2x + 2 or, 3x = 9 or, 5y = 1 or, 2x = 14 or, 3x = 32 or, 5y = 50 ∴ x = 7 \ x = 2 ∴ y = 0 Example 3: Solve: a) 2x + 1 × 3x – 2 = 48 b) 23x – 5 ax – 2 = 2x – 2 a1 – x Solution: a) Here, 2x + 1 × 3x – 2 = 48 b) Here, 23x – 5 ax – 2 = 2x – 2 a1 – x or, 2x × 21 × 3x × 3–2 = 48 Dividing both sides by 2x – 2 a1 – x, we get or, 2x × 3x × 2 × 1 32 = 48 or, 23x – 5 ax – 2 2x – 2 a1 – x = 2x – 2 a1 – x 2x – 2 a1 – x or, (2 × 3)x = 216 or, 23x – 5 – x + 2 ax – 2 – 1 + x = 1 or, 6x = 63 or, 22x – 3 a2x – 3 = 1 ∴ x = 3 or, (2a)2x – 3 = (2a)o or, 2x – 3 = 0 or, 2x = 3 ∴ x = 3 2 10.5 Solving exponential quadratic equations An exponential quadratic equation is an equation where the variable appears in the exponent. Therefore, while solving an exponential equation, we should simplify the equation till the equation will be obtained in the forms ax = ab or ax = 1. Example 4: Solve 5x + 1 5x = 25 1 25