Principal (P) = Rs 20000 Show
Rate (R) = 8% p.a. Period (T) = 2 years Hence, Simple interest (S.I.) = PRT / 100 = Rs (20000 × 8 × 2) / 100 We get, = Rs 3200 Now, Amount on compound interest A = P {1 + (R / 100)}n = RS 20000 {1 + (8 / 100)}2 On further calculation, We get, = Rs 20000 × (27 / 25) × (27 / 25) = Rs 32 × 729 = Rs 23328 Therefore, Compound interest = Final amount – (original) Principal = Rs 23328 – Rs 20000 We get, = Rs 3328 Hence, Difference in compound interest – simple interest = Rs 3328 – Rs 3200 = Rs 128
Answer (Detailed Solution Below) India's Super Teachers for all govt. exams Under One Roof Given: Laxmi borrowed = 3200 Rate of interest = 5% Time = 1 year Formula: A = P(1+ R/100)n Calculation: A = P(1+ R/100)n ⇒ 3200 [1 + (5/100)] ⇒ 3200 × (21/20) ⇒ 3360 CI = A – P ⇒ 3360 – 3200 ⇒ 160 Rs. ∴ CI for given sum is Rs. 160India’s #1 Learning Platform Start Complete Exam Preparation Trusted by 3.5 Crore+ Students A fixed deposit is a lumpsum investment having fixed tenure to maturity. You can open a fixed deposit with any bank or financial institution. It is a safe investment having a predetermined interest rate, which is a little higher than saving accounts. Since the time period of investment is fixed and no premature withdrawals are permissible it is also known as term deposits. What is the Taxability of Interest Income?The interest from fixed deposits is fully taxable. It comes under the head “Income from Other Sources” while filing an income tax return. In case of fixed deposits, the bank or financial institutions deduct tax at source at the end of each year when the interest is paid by them. The rate of tax deduction at source is 10% if the income from interest for each year exceeds Rs 10,000. [This limit has been increased to Rs 40,000 in Budget 2019]. However, if you don’t submit your PAN card, TDS @ 20% is deducted on your interest income. Any amount deducted as TDS can be verified with Form 26AS. The TDS can be deducted at the time of calculating taxable income or while paying the self-assessment tax to the Government. For example- If Mr X has a fixed deposit for 5 years, and has earned an interest of INR 50,0000 every year.So the bank is required to deduct TDS on the interest amount @10% as the interest amount exceeds INR 40,000. Don’t want your TDS to be deducted? Click here to read. What if TDS is not deducted?Even if TDS is not deducted every year, Add the interest income in your total income. Because if you don’t, then in the last year(i.e.5th Year), consequences will be –
Let me explain you with an example-Suppose Aakriti has an income of Rs. 9.5 Lakhs. hence she belongs to the 20 % tax bracket. she has an FD of Rs 2 lakh with a bank that gives her an 8 % interest per annum. So, the interest she earns on the FD for the current financial year is Rs 16,000. (Remember, banks tax FDs at 10 percent only.) Now, Aakriti is liable to pay tax on the interest she earns at the same tax rate as she pays for her gross income. vedanta EXCEL in MATHEMATICS 10Book Authors vedanta Vedanta Publication (P) Ltd. Vanasthali, Kathmandu, Nepal vedanta EXCEL in MATHEMATICS 10Book Authors All rights reserved. No part of this publication may Published by: Vanasthali, Kathmandu, Nepal Preface This “Teachers’ Manual of Vedanta EXCEL in MATHEMATICS BOOKS-9 and 10” EXCEL in MATHEMATICS has incorporated the applied constructivism which This manual helps the teachers to have the chapter-wise learning competencies, Grateful thanks are due to all Mathematics Teachers throughout the country who Last but not least, any constructive comments, suggestions and criticisms from the Authors Contents (Class 10) Page No. Unit Set 1 Competency Allocated teaching periods 8 - To find the relation between the sets and solve the related problems by demonstrating Learning Outcomes - To solve the verbal problems related to the cardinality of sets by using Venn diagram Level-wise learning objectives S.N. LEVELS OBJECTIVES - To find the cardinality of a set. terminologies relations in set notations - To solve the verbal problems on operations (union, - To solve the verbal problems on operations of sets by - To relate the problem related to set with other areas - To link various real life/ contemporary problems with Required Teaching Materials/ Resources 1 Vedanta Excel in Mathematics Teachers' Manual - 10 A. Problems including two sets Pre-knowledge: Check the pre-knowledge on cardinality of sets, relation of sets, operations of sets and Venn-diagram Teaching Activities 1. Present models of Venn-diagrams and recall the relations of sets as warm up. 2. Write any two disjoint sets on the board and tell the students to list the elements of A, B and A « B then n (A), n (B),n (A « B) and n (A) + n (B). Then draw the conclusion from the examples as n (A « B) = n (A) + n (B). 3. Select some students and ask whether they like coffee or tea. Denote the set of Bishnu Laxmi Dinesh the drinks by C « T. List the members of Rajendra Elina Imran set and discuss about the relation of the sets (overlapping or disjoint). Then find the cardinality of the sets. Use Venn diagram to show the relation of sets C and T. For example: If U = {Bishnu, Laxmi, Elina, Chandani, Shashwat, Rajendra, Imran, Hari, Pemba, Hritesh, Amar, Dinesh}, C = {Laxmi, Bishnu, Amar, Shashwat, Hari, Pemba}, T = {Amar, Dinesh, Laxmi, Hritesh, Chandani} and C « T= {Rajendra, Elina, Imran} Tell the students to find C«T, C ª T, n (C), n (T), n (C «T), n (CªT) and n(C « T) from Venn-diagram then tell them to try to establish the relations of n (C), n (T), n (CªT) and n (C« T). Finally, conclude that C ªT = {Laxmi, Amar}, C « T = {Laxmi, Bishnu, Amar, Shashwat, Hari, Pemba, Dinesh, Hritesh, Chandani} ?n (C ªT) = 2, n (C«T) = 9. Ask to students why n (C«T) is not equal to n (C) + n (T) in this case? Clarify that n (C«T) = n (C) + n (T) – n (CªT) by similar real life examples of overlapping sets. Also, establish the relation of n (U), n (C« T) and n (CªT). 1. List the following formula under discussion through examples in Venn-diagrams. Case-I: When A is a subset of B (i) n (AªB) = A BA U (iii) no (B) = n (B) – n (A) (i) n (AªB) = 0 A BU (ii) n (A« B) = n (A) + n (B) Case- III: When A and B are overlapping sets (i) n (A« B) = n (A) + n (B) – n (AªB) A BU (iii) n (only A) = no (A) = n (A) – n (AªB) Vedanta Excel in Mathematics Teachers' Manual - 10 2 (iv) n (only B) = no (B) = n (A) – n (AªB) Note: Note: effective rather than using formula. However, the given problems can be solved by Solution of selected questions from Vedanta Excel in Mathematics 1. In a group of students, the ratio of number of students who liked music and sports is 9: 7. Out of which 25 liked both the activities, 20 liked music only and 15 liked none of the activities. (i) Represent the above information in a Venn-diagram. (ii) Find the total number of students in the group. Solution: Let M and S denote the sets of students who liked music and sports respectively. Then n (M) = 9x, n (S) = 7x (say),n (Mª S) = 25, no (M) = 20, n(M « S)= 15 Hence, the total number of students in the group was 70. 2. In an examination 45% students passed in Science only, 25% passed in English only and 5% students failed in both subjects. If 200 students passed in English, find the total number of students by using a Venn-diagram. Solution: 3 Vedanta Excel in Mathematics Teachers' Manual - 10 Let S and E denote the sets of students who passed in Science and English respectively. Then n (U) = 100%, no (S) = 45%,no (E) = 25%, n(S « E)= 5% and n (E) = 200 Again, 45% x 25% Also, n (E) =no (E) + n (S ªE) = 25% + 25% = 50% Then, by question, n (E) = 200 or, 50% of y = 200 ?y = 400 Hence, the total number of students was 400. 3. There are 1200 students in a school. They are allowed to cast vote either only for X or for Y as their school prefect. 50 of them cast vote for both X and Y and 24 didn’t cast the vote. The candidate Y won the election with majority of 56 more votes than X. (i) How many students cast the vote? (ii) How many valid votes are received by X candidate? (iii) Show the result in Venn-diagram. Solution: Let A and B denote the sets of students who cast vote the candidates X and Y respectively. Then, n (U) = 1200, n (A ª B) = 50,n(A « B )= 24, no (B) = no (A) + 56 (i) n (A «B) = n (U) - n(A « B )= 1200 – 24 = 1176. Hence, 1176 students cast the vote. (ii) n (U) = no (A) + no (B) + n (AªB) + n(A « B ) A U 4. Due to the heavy rain fall during a few monsoon days, 140 households were victimized Solution: Vedanta Excel in Mathematics Teachers' Manual - 10 4 Let F and S denote the sets of households who got food support and shelter respectively. The number of households who got food support only is 30. = Rs 25,50,000 ?n(F « S)= 40. 5. In a class the ratio of number of students who passed maths but not science and those (ii) Draw a Venn-diagram to show all result. Solution: Let M and S denote the sets of students who passed maths and science respectively. Then, no (M) + no (S) = 80, n (M«S) = 100 Let no (M) = 3x and no (S) = 5x. Then no (M) + no (S) = 80 or, 3x + 5x = 80 or, 8x = 80 ?x = 10 ?no (M) = 3x = 310 = 30 and no (S)= 5x = 510 = 50 Also, let n (M ª S) = 2y and n(M « S) = y Now, n (M«S) = no (M) + no (S) + n (MªS) or, 100 = 30 + 50 + 2y ?y = 10 (i) n (U) = no (M) + no (S) + n (MªS) + n(M « S) M U Hence, the total number of students is 110. (ii) Drawing the Venn-diagram to show above result. 30 20 50 5 Vedanta Excel in Mathematics Teachers' Manual - 10 Extra Questions 1. In class X Jamboree of a school, students performed their various talent shows. Out of 80 parents participating in the Jamboree; 34 preferred dance but not comedy shows, 23 preferred comedy but not dance shows and 13 parents preferred the shows other than these. (i) How many parents preferred both the shows? (ii) How many parents preferred at most one show? (iii) Show the above information in a Venn-diagram. [Ans: (i) 10 (ii) 70] 2. In a group of 125 students, the ratio of student who like tea to the number of students who like coffee is 4:5.If 20 of them like both the drinks and 10 of them like none of them then find: (i) How many of them like only one drink? (ii) Find the ratio of number of students who like and don’t like both the drinks. (iii) Draw a Venn-diagram to represent the above information. [Ans: (i) 95 (ii) 2:1] 3. In a school 32 teachers like either milk or curd or both. The ratio of number of teacher who like milk to the number of teacher who like curd is 3:2 and 8 teachers like both milk and curd. Find: (i) How many teachers like milk only? (ii) How many teachers like curd only? (iii) Show the above information in a Venn-diagram. [Ans: (i) 16 (ii) 8] 4. In a group of 50 students 20 like only Math and 15 like only science. If the member of students who do not like any of the two subjects is double of the number of students who like both subjects, find the number of students who like at most one subject by using a Venn-diagram. [Ans: 45] 5. Out of 200 students in a class, 150 like football and 120 like cricket. If the number of students who like only cricket is one-third of the number of students who like football only,, then using a Venn-diagram, find the number of students who like: (i) Both games (ii) Football only (iii) None of these games [Ans: (i) 105 (ii) 45 (iii) 35] B. Problems Including Three Sets Pre-knowledge: Cardinality relation of union of two sets, verbal problems involving two sets, set operations on three sets, Venn-diagrams of three sets Teaching Activities 1. Create a sound learning environment through a real life problem based on three sets by the use of Venn-diagram. 2. By visualization of relations of sets in Venn-diagrams, clarify the following cardinality relations of union of three sets. A B CU (i) n (AªBª C) = 0 (ii) n (A« B« C) = n (A) + n (B) + n (C) Case- II: When A, B and C are overlapping subsets of a universal set U. (i) n (A B« C) = n (A) + n (B) + n (C) – n (AªB) – n (BªC) – n (CªA) + n (Aª BªC) Vedanta Excel in Mathematics Teachers' Manual - 10 6 (ii) n (A« B« C) = no (A) + no (B) + no(C) + no(AªB) + no (BªC) + no (CªA) + n (Aª BªC) (iii) n (A« B« C) = no (A) + n (A«B) = no (B) + n (B«C) = no (C) + n (C«A) (iv) no (A) = n (A) – n (AªB) –n (CªA) + n (Aª BªC) A BU (vi) no (C) = n (C) – n (BªC) –n (CªA) + n (Aª BªC) (vii)no (A) = n (A) – n o (AªB) –n o (CªA) – n (Aª BªC) (viii) no (B) = n (B) – n o (AªB) –n o (BªC) – n (Aª BªC) C (x) n(A « B « C ) = n(U) – n(A« B« C) 3. Explain the following useful terminologies in solving verbal problems. (i) Number of people who like all three activities = n (Aª BªC) (ii) Number of people who like at least one activity/either A or B or C = n (A« B« C) (iii) Number of people who like only/exactly one activity = no (A) + no (B) + no (C) = no(AªB) + no (BªC) + no (CªA) (vii) Number of people who don’t like any activity / neither A nor B nor C = n(A « B « C ) Solution of selected questions from Vedanta Excel in Mathematics 1. In a survey of 700 tourists who arrived in Nepal during ‘Visit Nepal 2020’, 350 preferred to go trekking, 400 preferred rafting and 250 preferred forest safari. Likewise 200 preferred trekking and rafting, 110 preferred rafting and forest safari and 100 preferred forest safari and trekking. If 50 tourists preferred all these activities, by drawing a Venn-diagram find how many tourists preferred none of these activities? Solution Let T, R and F denote the sets of tourists who preferred trekking, rafting and forest safari respectively. Then, n (U) = 700, n (T) = 350, n (R) = 400, n (F) = 250, n (T ª R) = 200, n (R ª F) = 110, n (F ª T) = 100 and n (T ª Rª F) = 50, n (T ∪ R ∪ F) =? Let the number of tourists who preferred none of these activities T RU Now, illustrating the above information in a Venn-diagram 150 Thus, 60 tourists who preferred none of these activities x 90 2. Of the total students in an examination, 40% students passed in Mathematics, 45% in Science and 55% in Nepali. 10% students passed in both Mathematics and 7 Vedanta Excel in Mathematics Teachers' Manual - 10 Science, 20% passed in Science and Nepali and 15% in Nepali and Mathematics. (i) Find the percentage of students who passed in all the three subjects? (ii) Show the above information in a Venn-diagram. Solution Let M, S and N denote the sets of students who passed in Mathematics, Science and Nepali respectively. Then, n (U) = 100 % = n (MS«N), n (M) = 40%, n (S) = 45%, n (N) = 55%, n (MªS) = 10%, n (SªN) = 20%, n (NªM) = 15% and n (MªSªN) =? (i) Now, n (M S« N) = n (M) + n (S) + n (N) – n (MªS) – n (SªN) – n (NªM) + n (Mª SªN) or, 100 % = 40% +45% + 55% - 10% - 20%-15%+ n (Mª SªN) SU ? n (Mª SªN) = 5% M Hence, the 5% students passed in all the three subjects. 25% N 3. In a survey, people were asked what types of movies they like. It was found that 75 liked Nepali movie, 60 liked English, 40 liked Hindi, 35 liked Nepali and English, 30 liked Nepali and Hindi, 20 liked Hindi and English, 10 liked all three and 25 people were found not interested in any types of movies. (i) How many people did not like only Hindi films? (ii) How many people did not like only Nepali or Hindi films? Solution: Let N, H and E denote the sets of people who liked Nepali, Hindi and English movies respectively. Then, n (N) = 75, n (E) = 60, n (H) = 40, n (NªE) = 35, n (NªH) = 30, n (HªE) = 20, n (Nª HªE) = 10 and n(N ª H ª E) Now, N EU 20 10 From Venn-diagram, 25 H (i) No. of people who did not like only Hindi films = n o (N) + no (E) + n (NªE) (ii) No. of people did not like only Nepali or Hindi films = no (E) = 15 4. In a group of students, 20 study Economics, 18 study History, 21 study Science, 7 Vedanta Excel in Mathematics Teachers' Manual - 10 8 Solution Let E, H and S denote the sets of students who study Economics, History and Science respectively. Then, n (E) = 20, n (H) = 18, n (S) = 21, no (E) = 7, no(S) = 10, no(EªS) = 6 and no (SªH) = 3 Let the number of students who study all the subjects n (EªHªS) be x. E U E U 10 S 10 S (ii) From Venn-diagram, n(S) = 10 + 6 + 3 + x or, x = 2 (ii) Also, the number of students who study only Economics and History Extra Questions 1. In a survey of 160 farmers of a village, it was found that 60 farmers has buffalo farming, 70 have cow farming and 80 have goat farming. Similarly, 25 have buffalo as well as cow farming, 30 have cow as well as goat farming and 15 have goat as well as buffalo farming. If 10 farmers have all these farming then by drawing a Venn-diagram find: (i) How many farmers have only one of these farming? (ii) How many farmers have exactly two of these farming? (iii) How many farmers have other than these farming [Ans: (i) 100, (ii) 40 (iii) 10] 2. In a survey of 2000 Indian tourists who arrived in Nepal, 65% wished to visit Pashupati, 50% wished to visited Chandragiri and 45% wished to visit Manakamana. Similarly, 35% wished to visit Pashupati and Chandragiri, 25% to Chandragiri and Manakamana and 20% to Manakamana and Pashupati. If 5% wished to visit none of these places, find the number of tourists who wished to visit all these three places. Also, show the above data in a Venn-diagram. [Ans: 300] 3. Among examinees in an examination, 40% obtained A+ grade in Science, 45% in Math and 55% in Social Studies. Similarly, 10% obtained A+ grade in Math and Science, 20% in Science and Social Studies and 15% in Social Studies and Math. If every student obtained A+ grade in at least one subject. (i) If 300 students were surveyed, how many students did obtain A+ grade in only one subject? (ii) Find the percentage of students who obtained A+ in all the three subjects? (iii) Show the above information in a Venn-diagram. [Ans: 5%, (ii) 195] 9 Vedanta Excel in Mathematics Teachers' Manual - 10 Unit Tax and Money Exchange 2 Allocated teaching periods 7 mathematical instruction and logical thought. Learning Outcome activities. Level-wise learning objectives S.N. LEVELS OBJECTIVES 1. Knowledge (K) - To define value added tax selling price and VAT rate are given. and selling price including VAT. - To describe the VAT exempted goods or services. - To solve the problems on VAT. revaluation of currency is given - To mathematize the contextual probems on VAT and solve Required Teaching Materials/ Resources Vedanta Excel in Mathematics Teachers' Manual - 10 10 A. Value Added Tax (VAT) Pre-knowledge: cost price (C.P.), marked price (M.P.), selling price (S.P.) , income tax etc. Teaching strategies 1. Recall cost price (C.P.), marked price (M.P.), selling price (S.P.) and income tax as warm up. 2. Paste/show the different types of taxes in the colourful chart paper and explain with appropriate examples. 3. List the following formulae under discussion (i) Discount amount = M.P. – S. P. (ii) Discount amount = Discount % of M.P. (iii) Rate of discount = Discount amount × 100% (v) S.P. = M.P. – Discount% of M.P. = M.P. (1 – Discount %) (vi) S.P. = M.P. (1 – D1%)(1 – D2%) when two successive discount rates D1% (vii) VAT amount = S.P. including VAT – S. P. excluding VAT (viii) VAT amount = VAT% of S.P. (ix) Rate of VAT = VAT amount × 100% (xi) S.P. with VAT = S.P. + VAT% of S.P. = S.P. (1 + VAT %) 4. Discuss on the local tax, transportation cost and profit /loss Cost price (C.P.) Marked price (M.P.) + Profit – Discount Selling price (S.P.) + VAT S.P. including VAT Project Work Divide the students into the groups of 3 members each and give the work: price, discount rate/ discount amount, VAT rate/ VAT amount and selling price including 11 Vedanta Excel in Mathematics Teachers' Manual - 10 Solution of selected questions from Vedanta Excel in Mathematics 1. A trader bought some electric blenders for Rs 4,000 per piece. He sold each blender Solution: S.P. with VAT = Rs 4,332 S.P. of the blender = Rs 3800. Since C.P.> S.P Thus, his loss = C.P.-S.P = Rs 4000-Rs 3800= Rs 200 2. A supplier purchased a photocopy machine for Rs 2,00,000 and spent Rs 1,500 for Solution Here, For supplier: Cost price (C.P.) of photocopy machine = Rs 2,00,000 = Rs 2,00,000 + Rs 1,500 + Rs 500 = Rs 2,02,000 Selling price (S.P.) of the photocopy machine at 10% For customer: C.P. of the photocopy machine with 13% VAT = Rs222200 + 13% of Rs 222200 = Rs 2,51,086 3. Mr. Jha purchased a bicycle costing Rs 5,600 from a dealer at 5% discount and sold Solution: noodles allowing 5% discount, find his/her profit percent. Now, profit percent = S.P. – C.P. × 100% = 1.1875x – x × 100% = 18.75% Vedanta Excel in Mathematics Teachers' Manual - 10 12 5. A trader fixed the price of cosmetic items 30% above the cost price. When he/she sold and item at 25% discount, there was a loss of Rs 15. Find the cost price and marked price of the item. Solution: Let the cost price of the cosmetic item (C.P.) be Rs x. Then, M.P. = Rs x + 30% of Rs x = Rs 1.3 x S.P. = MP – D% of MP = Rs 1.3x – 25% of Rs 1.3x = Rs 0.975x According to question Loss amount = C.P. – S.P. or, Rs 15 = x – 0.975x or, 15 = 0.025x ?x = 600 Hence, the required cost price of the item is Rs 600 and its marked price is Rs 1.3×600 = Rs 780 6. A shopkeeper sold an article at 20% discount and made a loss of Rs 90. If he had sold it at 5% discount, he would have gained Rs 90. Find the cost price and the marked price of the article. Solution: Let the marked price of the article (M.P.) be Rs x. Now, S.P. in 20% discount = M.P. – D% of M.P. = x – 20% of x = Rs 0.8x ?C.P. = S.P. + loss = Rs (0.8x + 90) … (i) Again, S.P. in 5% discount = M.P. – D% of M.P. = x – 5% of x = Rs 0.95x ?C.P. = S.P. – profit = Rs (0.95x - 90) … (ii) Equating equations (i) and (ii), we get 0.8x + 90 = 0.95x – 90 or, 0.15x = 180 ?x = 1200 Again, putting the value of x in equation (i), we get C.P. = Rs (0.8×1200 + 90) = Rs 1050 Hence, the required cost price of the article is Rs 1050 and marked price is Rs 1200. 7. An article after allowing a discount of 20% on its marked price was sold at a gain of 20%. Had it been sold after allowing 25% discount, there would have been a gain of Rs 125. Find the marked price of the article. Solution: Let the marked price of the article (M.P.) be Rs x. ?C.P. = 100 S.P. = 100 × 0.8x = 80x = Rs 2x … (i) ?C.P. = S.P. – profit = Rs 3x – 125 = Rs 3x – 500 … (ii) Hence, the required marked price of the article is Rs 1500. 8. A shopkeeper purchased a bicycle fro Rs 5,000 and the marked its price a certain percent above the cost price. Then he sold it at 10% discount. If a customer paid Rs 6,356.25 with 13% VAT to buy it, how many percentage is the marked price above the cost price? 13 Vedanta Excel in Mathematics Teachers' Manual - 10 Solution: Let the selling price of the bicycle (S.P.) be Rs x. Now, S.P. with VAT = S.P. + VAT% of S.P. or, Rs 6,356.25 = x + 13% of x ?x = Rs 5625 Let the marked price of the bicycle (M.P.) be Rs y. Also, S.P. = M.P. – D% of M.P. or, Rs 5625 = y – 10% of y ?y = Rs 6250 Again, difference between M.P. and C.P. = Rs 6,250 – Rs 5,000 = Rs 1,250 9. After allowing 20% discount on the marked price of a mobile, 15% VAT was levied and sold it. If the difference between the selling price with VAT and selling price after discount is Rs 1,800, find the marked price of the mobile. Solution: Let the marked price of the mobile be Rs x Then, S.P. after 20% discount = M.P. – D% of M.P = x – 20% of x = Rs 0.8x Again, S.P. with 15% VAT = S.P. + VAT% of S.P. = Rs0.8x + 15% of Rs 0.8x = 0.92x According to the question, S.P. with VAT – S.P. after discount = Rs 1,800 or, 0.92x – 0.8x = Rs 1,800 ?x = Rs 15,000 Hence, the marked price of the mobile is Rs15,000. 10. When an article was sold at a discount of 10%, a customer paid Rs 9,153 with 13% Solution Discount rate = 10%, VAT rate = 13%, S.P. with VAT = Rs 9,153 Profit percent = 8% ?y = 7500 = Rs 1500 Again, M.P. of the article is more than C.P. by Difference ×100% = Rs 1500 ×100% = 20% amount 11. In the peak season of winter days, a retailer marked the price of an electric heater Solution: Vedanta Excel in Mathematics Teachers' Manual - 10 14 Also, C.P. = 100 × S. P. = 100 × Rs 3600 = Rs 3000 Case – II: Profit percent = 12% We have, new S.P. = 100 + profit% × C. P. = 100 + 12 × Rs 3000 = Rs 3360 Also, new discount = M.P. – S.P = Rs 4000 – Rs 3360 = Rs 640 New Discount rate = Discount × 100% = = Rs 640 × 100% = 16% Hence, the discount should be increased by 16% - 10% = 6% in summer season. 12. A retailer allowed 4% discount on her/his goods to make 20% profit and sold a refrigerator for Rs 10,848 with 13% VAT. By how much is the discount percent to be increased so that she/he can gain only 15% profit? Solution: Let the S.P. without VAT be Rs x. Now, S.P. with VAT = S.P. + VAT% of S.P. or, Rs 10,848 = x + 13% of x ?x = Rs 9,600 Let the marked price of the goods (M.P.) be Rs y. Also, S.P. = M.P. – D% of M.P. or, Rs 9,600 = y – 4% of y ?M.P. = y = Rs 10,000 Again, S.P. without VAT = C.P. + P% of C.P. or, 9,600 = C.P. + 20% of C.P. ?C.P. = Rs 8,000 Now, new profit% = 15% ?New S.P. = C.P. + 5% of C.P. = Rs 8,000 + 15% of Rs 8,000 = Rs 9,200 And, new discount = M.P. – new S.P. = Rs 10,000 – Rs 9,200 = Rs 800 Hence, the discount should be increased by 8% - 4% = 4%. 13. A supplier sold a scanner machine for Rs 41, 400 with 15% VAT after allowing 10% discount on it’s marked price and gained 20%. By how much the discount percent to be reduced to increase the profit by 4%? Solution: Let the S.P. without VAT of scanner machine be Rs x. Now, S.P. with VAT = S.P. + VAT% of S.P. or, Rs 41,400 = x + 15% of x ?x = Rs 36,000 Let the marked price of the scanner machine (M.P.) be Rs y. Also, S.P. = M.P. – D% of M.P. or, Rs 36,000 = y – 10% of y ?M.P. = y = Rs 40,000 Again, S.P. without VAT = C.P. + P% of C.P. or, 36,000 = C.P. + 20% of C.P. ?C.P. = Rs 30,000 Now, new profit% = 20% + 4% = 24% ?New S.P. = C.P. + 24% of C.P. = Rs 30,000 + 25% of Rs 30,000 = Rs 37,200 And, new discount = M.P. – new S.P. = Rs 40,000 – Rs 37,200 = Rs 2,800 Hence, the discount should be decreased by 10% - 7% = 3%. 14. Mrs. Gurung allowed 10% discount on her fancy items to make 25% profit and sold a lady bag for Rs 5,085 with 13% VAT. Due to excessive demands of her items, she decreased the discount percent by 2%. By how much was her profit percent increased? 15 Vedanta Excel in Mathematics Teachers' Manual - 10 Solution: Let the S.P. without VAT of lady bag be Rs x. Now, S.P. with VAT = S.P. + VAT% of S.P. or, Rs 5,085 = x + 13% of x ?x = Rs 4,500 Let the marked price of lady bag (M.P.) be Rs y. Also, S.P. = M.P. – D% of M.P. or, Rs 4,500 = y – 10% of y ?M.P. = y = Rs 5,000 Again, S.P. without VAT = C.P. + P% of C.P. or, 4,500 = C.P. + 25% of C.P. ?C.P. = Rs 3,600 Now, new discount% = 10% - 2% = 8% ?New S.P. = M.P. – D% of M.P. = Rs 5,000 – 8% of Rs 5,000 = Rs 4,600 And, new profit = new S.P. –C.P. = Rs 4,600 – Rs 3,600 = Rs 1,000 Hence, the profit should be increased by 27.78% - 25% = 2.78%. 15. Mrs. Dhital makes a profit of 50% of the cost of her investment in the transaction of her cosmetic items. She further increases her cost of investment by 25% but the selling price remains same. How much is the decrease in her profit percent? Solution: Let the cost of her investment be Rs x. Now, S.P. with 50% profit = C.P. + P% of C.P. ?S.P.1 = x + 50% of x = Rs 1.5x Hence, the profit should be decreased by 50% - 20% = 30%. 16. A retailer hired a room in a shopping mall at Rs 45,000 rent per month and started a business of garments. He spent Rs 20,00,000 to purchase different items in the first phase and marked the price of each item 30% above the cost price. Then he allowed 10% discount on each item and sold to customers. His monthly miscellaneous expenditure was Rs 15,000 and the item of worth 10% of the investment remained as stock after two months. Find his net profit or loss percent. Solution: Here, the amount of investment = Rs 20,00,000 Stocks after two months = 10% of Rs 20,00,000 = Rs 2,00,000 ?The investment excluding stocks = Rs 20,00,000 – Rs 2,00,000 = Rs 18,00,000 Now, M.P. of items = Rs 18,00,000 + 30% of Rs 18,00,000 = Rs 23,40,000, discount percent = 10% S.P. of items = M.P. – D% of M.P. = Rs 23,40,000 – 10% of Rs 23,40,000 = Rs 21,06,000 Gross profit = Rs 21,06,000 – Rs 18,00,000 = Rs 3,06,000 Again, rent of room in 2 months = 2×Rs 45,000 = Rs 90,000 Miscellaneous expenditure in 2 months = 2×Rs 15,000 = Rs 30,000 ?Total expenditure = Rs 90,000 + Rs 30,000 = Rs 1,20,000 Now, net profit = Gross profit – total expenditure = Rs 3,06,000 – Rs 1,20,000 = Rs 1,86,000 Hence, the required net profit is 10.33% Vedanta Excel in Mathematics Teachers' Manual - 10 16 17. The marked price of a calculator is Rs 1,000 and a shopkeeper allows discount which Solution: or, Rs 836.20 = Rs (1000 – 20x) + x (1000 – 20x) or, 836.20 = 1000 – 10x – x2 or, 4181 = 5000 – 50x – x2 or, x2+ 50x – 819 = 0 or, x2+ 63x – 13x – 819 = 0 or, x (x + 63) – 13( x + 63)= 0 or, (x + 63) (x – 13) = 0 Either, x + 63 = 0 ? x= -63 which is not possible because rate of VAT cannot be negative. Extra Questions 1. If the marked price of a camara is Rs. 3200, a shopkeeper announces a discount of 8%. [Ans: 3,238.40] 2. A shopkeeper sold his goods for Rs. 9944 after allowing 20% discount levying 13% value 3. A tourist bought a Nepali Thjanka at 15% discount on the marked price with 13% Value 4. A mobile price is tagged Rs. 5,000. If a customer gets 12% discount and certain percent 5. A retailer allowed 20% discount and sold a mobile set for Rs 4,520 with 13% VAT and B. Money Exchange Pre-knowledge rates of money exchange, revaluation or devaluation of currencies and commission. 17 Vedanta Excel in Mathematics Teachers' Manual - 10 Note: exchange centre buys the foreign currency and so uses the buying rate. with Nepali rupees then the bank or money exchange centre sells the foreign currency Solution of selected questions from Vedanta Excel in Mathematics 1. Manoj Regmi is going to visit Thailand for his family trip. He estimated to exchange Solution: Here, US $1 = NPR 114.50 ? US $4000 = 4000× NPR 114.50 = NPR 4,58,000 Now, commission = 1.5% of NPR 4,58,000 = NPR 6870 ?Required Nepali rupees with commission = NPR (4,58,000 + 6870) = NPR 4,64,870 2. Nirmal Shrestha bought some EURO (€) for NPR 3,20,000 at the exchange rate of € 1 = NPR 128 to visit a few European countries. Unfortunately because of his Visa problem, he cancelled his trip. Within a week Nepali rupees is devaluated by 2%. He again exchanged his EURO into Nepali rupees after a week. How much did he gain or loss? Solution: 3,20,000 When Nepali rupees is devaluated by 2% then €1 = NPR 128+2% of NPR 128 = NPR 130.56 Again, €2500 = 2500×NPR 130.56 = NPR 3,26,400 Hence, his profit amount = NPR. 3,26,400 – NPR 3,20,000 = NPR 6400 After a few days, Nepali currency was revaluated by 10% and he exchanged his Solution: 5,50,000 When Nepali rupees is revaluated by 10% then $1 = NPR 110 – 10% of NPR 110 = NPR 99 Again, $ 5000 = 5000×NPR 99 = NPR 4,95,000 Hence, his loss amount = NPR. 5,50,000 – NPR 4,95,000 = NPR 55,000 4. Mrs. Magar wants to buy a book online. She finds a publisher in London selling the book for £ 15. This publisher is offering free transportation on the product. She finds the same book from a publisher in New York for $ 17 with transportation fee of $ 2. Which publisher should she buy the book from? (Exchanged rate: £ 1 = NPR 140 and $ 1 = NPR 113.00) Vedanta Excel in Mathematics Teachers' Manual - 10 18 Solution: = NPR 2,100 The cost of the book in New York with transportation fee = $ 17 + $ 2 = 19 $ We have, 1 $ = NPR 113 ?19 $ = 19×NPR 113 = NPR 2,147 5. Mr. Gurung bought 10 Tola gold in Hong Kong for KHD$ 3800 per Tola and brought Solution ? cost of 10 Tola gold = 10 HKD$ 3800 = HKD$ 38000 = HKD$ 47500 Extra Questions 1. If Rs. 500 (IC) = Rs. 800 (NC), $ 45 = Rs. 4645.35 (NC), how many US dollars can be 2. Sulochana is going to Australia for her higher study and needs 4000 AUD and pays 2% 3. A business man exchanged Rs 7,00,000 into pound sterling at the rate of ₤1 = NPR 140. 4. A merchant purchased 600 pieces of Nepali Pasmina at Rs 2,400 per piece. He exported 19 Vedanta Excel in Mathematics Teachers' Manual - 10 Unit Compound Interest 3 5. Jasmin wishes to buy a book online. She finds a publisher in London selling the book - To solve and test the behavioural arithmetic problems on daily life activities using the mathematical instruction and logical thought. Learning Outcome - To take the information of system of calculation of compound interest from financial company and solve the related simple problems Level-wise learning objectives S.N. LEVELS OBJECTIVES 1. Knowledge (K) - To recall simple interest interest interest - To differentiate compound interest from simple interest known. 3. Application (A) - To calculate the difference between compound and - To find the sum when difference / sum between the - To solve real life problems based on yearly and half yearly - To prepare a record on interest systems of nearby financial Vedanta Excel in Mathematics Teachers' Manual - 10 20 4. High Ability (HA) - To derive the formula of annual compound amount/ - To compare among simple interest, yearly and half-yearly - To evaluate the difference percentage of yearly and half - To justify the annual instalments with calculation. Required Teaching Materials/ Resources Bank account forms, charts of interest rates, model for balance sheet, multi-media Pre-knowledge: principal, rate of interest, simple interest Teaching strategies 1. Recall simple interest and ask the formulae of simple interest. 2. Explain with daily life problems about the difference between simple interest and compound interest. R inductive method based on simple interest as shown in the following table. BOOK ko page 35 ma xa 4. Encourage the students to tell the formula for finding the compound interest compounded annually as R 5. Discuss with students on the formula C.A. = P 1 + 2 R 2Tof semi-annual 6. Ask the students to tell the formula for finding the compound interest compounded semi- by the discussion on the meaning of half year/semi-annual and rate of interest during it. 7. Make clear on the points which are mentioned below. (i) According to the yearly compound interest system, the interest is added to the principal at the end of every year and a new principal is obtained for the next year. (ii) According to the half-yearly compound interest system, the interest is added to the principal at the end of every 6 months and a new principal is obtained for the next 6 month period. (iii) The yearly compound interest is equal to simple interest only up to the first year on the same principal (sum) at the same rate of interest. Then after the compound interest is always more than simple interest on the same sum at the same rate for the same time duration. (iv) The half-yearly compound interest is equal to simple interest / annual compound interest only up to the first 6 months on the same principal (sum) at the same rate of interest. Then after the half yearly compound interest is always more than yearly compound interest / simple interest on the same sum at the same rate for the same time duration. Thus, simple interest < yearly compound interest < half yearly compound interest 21 Vedanta Excel in Mathematics Teachers' Manual - 10 (v) The difference between yearly compound interest and simple interest or the difference between half-yearly compound interest and yearly compound interest/simple interest determine the gain/profit or loss. Note: 1. When P is principal, R1%, R2% and R3% are rates of annual compound interest for the 2. When P is principal, R1%, R2% and R3% are rates semi-annual compound interest for the first, second and the third years respectively then (i) Semi-annual compound amount (C.A.) for 3 years (ii) Semi-annual compound interest (C.I.) for 3 years 3. When principal = P, rate of annual compound interest = R% , time duration = a fraction or T years and M months then Annual compound amount (C.A.) = P 1+ R T 1 + MR 4. When principal = P, rate of semi-annual compound interest = R% , time duration = a fraction in 2T or 2T half-years and M months then Annual compound amount (C.A.) = P 1+ R 2T 1 + MR Give some examples related to above formula and make a sound classroom atmosphere to solve the problems in groups or individually. Solution of selected questions from Vedanta Excel in Mathematics 1. Pratik borrowed Rs 1,50,000 from Bishu at the rate of 21% per year. At the end of nine months how much compound interest compounded half-yearly should he pay? Solution: + 3 months; rate (R) = 21% p.a. 2 Now, half-yearly C.I.= P 1 + R 2T 1 + MR –1 = 15000 1 + 21 2T 1+ 3×21 –1 Hence, he should pay Rs 24451.88 interest in nine months. 2. The difference between the annual and semi-annual compound interest on a sum of Vedanta Excel in Mathematics Teachers' Manual - 10 22 Now, annual C.I. = P 1 + R T 1 + MR –1 =x 1 + 18 1 1+ 6×10 –1 = 0.155x By question, semi-annual C.I. - Annual C.I. = Rs 63 or, 0.157625x – 0.155x = Rs 63 or, 0.002625x = 63 ?x = 24000 Hence, required sum is Rs 24,000. ainn1d21thyeeasrusm, i.nterest 3. If a sum of money becomes Rs 1,40,450 in 1 year and Rs 1,48,877 Solution: Let the required principal be Rs P and the rate of the interest be R% p.a. According to the first condition: Time duration (T) = 1 year, semi-annually C.A. = Rs 1,40,450 We have, CI = P 1 + R 2T or, Rs 1,40,450 = P 1 + R 2 … (i) We have, CI = P 1 + R 2T or, Rs 1,48,877= P 1+ R 2 × 3 or, Rs 1,48,877= P 1 + R 3 … (ii) Rs 1,40,450 = P 1 + 12 2 ? P = Rs 125000 4. The compound interest calculated yearly on a certain sum of money for the second year is Rs 1,320 and for the third year is Rs 1,452. Calculate the rate of interest and the original sum of money. Solution: Here, the difference between the C.I. of two successive years = Rs 1,452 – Rs 1,320 = Rs 132 ?Rs 132 is the interest on Rs 1,320 for 1 year. C.A. for the first year = P 1+ R T=P 1+ 10 1 = 1.1 P By question, 0.11P = Rs 1,320 ?P = 12000 Hence, the rate of interest is 10% p.a. and the original principal is Rs 12,000. 5. The compound compounded in 1 year and 2 years are Rs 450 an Rs 945 respectively. Find the rate of interest compounded yearly and the sum. Solution: Let the required principal be Rs P and the rate of the interest be R% p.a. According to the first condition: 23 Vedanta Excel in Mathematics Teachers' Manual - 10 Time duration (T) = 1 year, yearly C.I. = Rs 450 We have, CI = P 1 + R T– 1 or, Rs 450 = P 1 + R 1– 1 … (i) Time duration (T) = 2 years, yearly C.I. = Rs 945 We have, CI = P 1 + R T– 1 or, Rs 945 = P 1 + R 2– 1 … (ii) or, Rs 945 = P 1 + R 2– 1 oSru,b2s.t1it=uti2n+g th1Re00val?ueRof=R1i0n equation (i), we get Hence, the required sum is Rs 4,500 and rate of interest is 10% p.a. 6. Krishna lent altogether Rs 10,000 to Gopal and Radha for 2 years. Gopal agrees to pay simple interest at 12% p.a. and Radha agrees to pay compound interest at the rate of 9% p.a. If Radha paid Rs 596.70 more than Gopal as interest, find how much did Krishna lend to each. Solution: Let the money lent to Gopal (P1) = Rs x. Then, the money lent to Radha (P2) = Rs (10,000 – x) ?PrSin.Ic.=ipalP(1P×1)10=T0×RsRx,=raxte×o1f20s×0im1p2le=in0te.2r4esxt (R) = 12 % p.a. and time (T) = 2 years. Principal (P2) = Rs (10000 – x), rate of compound interest (R) = 9 % p.a. and time (T) = 2 years. According to question, C.I. – S.I. = Rs 596.70 or, (1881 – 0.1881x) – 0.24x = Rs 596.70 or, 1284.3 = 0.4281x ?x = 3000 So, money lent to Gopal = x = Rs 3,000. The money lent to Radha = 10000 – x = 10000 – 3000 = Rs 7,000 7. Suntali deposited Rs 9,000 altogether in her saving account and fixed deposit Solution: account (P2) = Rs (9,000 – x) Vedanta Excel in Mathematics Teachers' Manual - 10 24 ?C.I.1= P1 1+ R T– 1 =x 1 + 5 1 – 1 = 0.05x For fixed deposit account; Principal (P2) = Rs (9000 – x), rate of half-yearly compound interest (R) = 10 % p.a. and time According to question, C.I.2 – C.I.1 = Rs 160 or, 922.5 – 0.1025x – 0.05x = Rs 160 So, sum deposited in saving account = x = Rs 5,000. The sum deposited in fixed deposit account = 9000 – x = 9000 – 5000 = Rs 4,000 8. A bank has fixed the rate of interest 10% p.a. semi-annually compound interest in account X and 12% p.a. compound interest in account Y. If you are going to deposit Rs 50,000 for 2 years, in which account will you deposit and why? Give your reason with calculation. Solution: For account X Principal (P) = Rs 50,000, rate (R) = 10% p.a., time (T) = 2 years, semi-annual C.I. =? We have, half yearly C.I. = P 1+ R 2T – 1 = Rs 50,000 1+ 10 2×2 – 1 = Rs 10,775.31 Principal (P) = Rs 50,000, rate (R) = 12% p.a., time (T) = 2 years, annual C.I. =? We have, yearly C.I.= P 1 + R T –1 = Rs 50,000 1 + 12 2 –1 = Rs 12,720 Hence, I would deposit in account Y because it gives Rs 1,944.69 more interest. 9. A person deposited Rs 80,000 in a bank at the rate of 12% p.a. interest compounded semi-annually for 2 years. After one year, the bank revised its policy to pay the interest compounded annually at the same rate. What is the percentage difference between the interests of the second year due to the revised policy? Give reason with calculation. Solution: For the first year Principal (P) = Rs 80,000, rate (R) = 12% p.a., time (T) = 1 year, semi-annual C.I. =? We have, half yearly C.I. =P 1+ R 2T – 1 = Rs 80,000 1 + 12 2×1 – 1 = Rs 9,888 Principal (P) = Rs 80,000 + Rs 9,888 = Rs 89,888, rate(R) = 12% p.a., time (T) = 1 year, annual C.I. =? We have, yearly C.I.= P 1 + R T–1 = Rs 89,888 1+ 12 1 –1 = Rs 10786.56 25 Vedanta Excel in Mathematics Teachers' Manual - 10 ?Half yearly C.I. – yearly C.I. = Rs 11110.16– Rs 10786.56 = Rs 323.60 Hence, due to the policy revised, the interest in the second year is decreased by 2.91% because the interest is calculated annually. 10. A housewife deposited Rs 10,000 on saving account at 5% p.a. interest compounded yearly and another sum on fixed deposit account at 8% p.a. interest compounded half yearly. After one year, the interest on fixed deposit account was Rs 152.80 more than the interest on the saving account; find the total amount of money in her two accounts at the end of the year. Solution: For the saving account Principal (P) = Rs 10,000, rate (R) = 5% p.a., time (T) = 1 year, yearly C.I. =? We have, yearly C.I. = P 1 + R T –1 = Rs 10,000 1 + 5 1 –1 = Rs 500 Let principal (P) = Rs x, rate(R) = 8% p.a., time (T) = 1 year, half-yearly C.I. =? We have, half yearly C.I.=P 1 + R 2T – 1 =x 1 + 8 2–1 = Rs 0.0816x or, 0.0816x – 500 = 152.80 ?x = 8000 deposit = Rs 8,000 + 0.0816×Rs 8000 = Rs 8,652.80 Hence, the total amount of money in two account at the end of the year is Rs 10,500 + Rs 8,652.80 = Rs 19,152.80 Extra Questions 1. According to yearly compound interest, the sum becomes Rs 3240 in 3 years and Rs 3888 in 4 years. How much is the amount more than the principal in 2 years? [Ans: Rs 1875, 20%, Rs 825] 2. A sum of Rs 150000 amounts to Rs 2,62,500 at a certain rate of simple interest in 5 years. Find the sum of money that amounts to Rs 1,98,375 at the same rate of compound interest in 2 years. [Ans: Rs 1,50,000] 3. A bank has fixed the rate of interest 10% p.a. semi-annually compound interest in account M and 12% per annum annually compound interest in account N. If you are going to deposit Rs 80,000 for 2 years in the same bank, in which account will you deposit and why? Give your reason with calculation. [Ans: account N] Vedanta Excel in Mathematics Teachers' Manual - 10 26 Unit Population Growth and Compound Depreciation 4 Competency Allocated teaching periods 7 - To solve and test the behavioural arithmetic problems on daily life activities using the mathematical instruction and logical thought. Learning Outcome - To solve the simple problems related to population growth and compound depreciation Level-wise learning objectives S.N. LEVELS OBJECTIVES - To define population growth - To recall the formula of annual compound amount 1. Knowledge (K) - To write the formula of finding the population after T years - To define compound depreciation - To tell the formula for finding the depreciated value after T years - To calculate the population after T years when population growth rate is given - To find the rate of population growth rate. - To calculate the time duration when initial population, population after T years and population Understanding (U) growth rate are given 2. - To find out the value of item after T years when annual depreciation rate is given - To find out the original value before T years when annual depreciation rate and the depreciated value are given - To find the depreciation rate and time interval - To solve the problems based on initial population, population after T years, no. of deaths, no. of in- 3. Application (A) migrants and no. of out-migrants. - To solve the real life problems based on depreciation - To relate the behavioural problems with compound 4. High Ability (HA) appreciation and depreciation Required Teaching Materials/ Resources A. Population growth Pre-knowledge: annual compound amount, annual compound depreciation 27 Vedanta Excel in Mathematics Teachers' Manual - 10 Teaching strategies 1. Ask about population of the students’ village or city or municipality whether it is increasing or decreasing, increment of number of students of a school, numbers of vehicles etc. 2. Make clear that the population of a certain places increases in a certain annual rate and the population after T years can be calculated as the annual compound amount. If Po = initial/previous population, T = time duration, R% = annual population growth formula works effectively or not by finding the total population of each year up to 2/3 years and answer what is obtained by using formula. 4. Discuss in group about value of an article/ land, height of plant, number of bacteria instead of population. 5. Encourage the students to tell the formula for finding the population after T years in the following cases and conclude the result. Case-1: When P is initial population, R1%, R2% and R3% are rates of annual population growth for Increased population in 3 years = P 1 + R1 1 + R2 1+ R3 2 –1 If Po = initial/previous/original population, T = time duration, R% = annual population Note: in out 1. Population growth rate per year = annual birth rate – annual deathrate 2. Population growth rate per year = annual birth rate + annual immigration rate 3. Time duration (for example) from beginning of 2076 B.S. to beginning of 2077 B.S. = from end of 2075 B.S. to the beginning of 2077 B.S. = 1 year but from the beginning of 2076 B.S. to the end of 2077 B.S. = 2 years Project Work Give the following project works in groups or individual years and at present then find the growth rate of students. in the world and compare their growth rate of population of those countries. Vedanta Excel in Mathematics Teachers' Manual - 10 28 Solution of selected questions from Vedanta Excel in Mathematics 1. The population of a town increases every year by 10%. At the end of two years, if 5,800 people were added by migration and the total population of the town became 30,000, what was the population of the town in the beginning? Solution Population at the end of 2 years (PT) = 30,0000, time (T) = 2 years, population growth rate or, 30,000 = Po 1 + R 2 + 5800 2. In the beginning of 2074 B.S., the population of a town was 1,00,000 and the rate of growth of population is 2% every year. If 8,000 people migrated there from different places in the beginning of 2075 B.S., what will be the population of the town in beginning of 2077 B.S.? Solution Here, Population in the beginning of 2074 B.S. (Po) = 1,00,000 Number of in-migrants in the beginning of 075 B.S. (Min) = 8,000, time (T) = 1 year Again, population in the beginning of 2075 B.S. (Po) = 1,10,000 Population in the beginning of 2077 B.S. (PT) =?, time (T) = 2 years Hence, the population of the town in beginning of 2077 B.S. will be 114444. Extra Questions 1. The population of a municipality increases every year by 5%. At the end of two years, if 198 people were added by migration and the total population of the municipality became 9,900, what was the population of the municipality in the beginning? [Ans: 8800] 2. The population of a village increases every year by 5%. At the end of two years, if 1,025 3. In the beginning of 2017 A.D., the population of a metropolitan city was 2,00,000 and the rate of growth of population is 2% every year. If 16,000 people migrated there from different places in the beginning of 2018 B.S., what will be the population of the town at the end of 2018 A.D.? [Ans: 2,28,888] 29 Vedanta Excel in Mathematics Teachers' Manual - 10 B. Depreciation Pre-knowledge: annual compound amount, population growth Teaching strategies machinery items, building etc. decreases for some time. time. The process of lowering the value of an item at a certain rate every year is known If Po = initial/previous/original value, T = time period for depreciation, R% = rate of Po is initial value, R1%, R2% and R3% are rates of annual depreciation for the first, second PT= Po 1 – R1 1 – R2 1 – R3 Depreciated amount = P 1– 1 + R1 1 + R2 1 + R3 Solution of selected questions from Vedanta Excel in Mathematics 1. Mr. Ghising bought a tripper for Rs 28,08,000. He used it for transporting construction Solution: Here, the original cost of tripper (Po) = Rs 28,08,000, time (T) = 2 years Now, depreciated cost (PT) = Po 1 – R T = Rs 28,08,000 1 – 15 2 = Rs 20,28,780 Again, profit in 2 years = Rs 7,80,000 The value of tripper with profit = Rs 20,28,780 + Rs 7,80,000 = Rs 28,08,780 Since, the original cost of tripper = Rs 28,08,000 ? Profit = Rs 28,08,780 – Rs 28,08,000 = Rs 780 Hence, his profit is Rs 780. Extra Questions 2. Mr Dhurmus bought a tractor and used it for 3 years and earned Rs 2,25,000. If he sold 3. A man buys a piece of land for Rs 800000 in a rural municipality and immediately invests Vedanta Excel in Mathematics Teachers' Manual - 10 30 Unit Mensuration (I): Area of Plane Surface 5 Competency Allocated teaching periods 7 - To solve the problems related to area of triangle (right angled, equilateral, isosceles, scalene), quadrilateral (rectangle, square, parallelogram, rhombus, trapezium) Learning Outcomes equilateral, isosceles, scalene), quadrilateral (rectangle, square, parallelogram, rhombus, Level-wise learning objectives S.N. LEVELS OBJECTIVES - To define plane surface angled, equilateral, isosceles, scalene) - To find the perimeter of triangle - To find the area of triangle when the ratio of sides and 3. Application (A) perimeter are given - To estimate the cost of specific purpose based on area of triangle - To relate the problem related area of triangle and the 4. High Ability (HA) cost/ rent for some more extent - To link various real life/ contemporary problems with area of triangles and solve Required Teaching Materials/ Resources Teaching learning strategies Teaching Activities the formulae to find the perimeter of triangle, rectangle, square etc and units of perimeter. clarify about area of plane figures. 31 Vedanta Excel in Mathematics Teachers' Manual - 10 3. Discuss upon the area of place surface as amount of space inside its boundary. Recall the formula to find the area of triangle, rectangle, square, parallelogram etc. and units of area. 3 three sides as s(s – a) (s – b) (s – c) where a, b and c are the lengths of sides of the Solution of selected questions from Vedanta Excel in Mathematics 1. The area of isosceles triangle is 240 cm2. If the base of the triangle is 20 cm, find the length of its equal sides. Solution: Here, the area of the isosceles triangle (A) = 240 cm2 length of base (b) = 20 cm, equal sides (a) = ? Now, Area of an isosceles triangle (A) = b 4a2 – b2 2304 = 4a2 or, a2 = 676 ? a = 26 Hence, the length of equal sides are 26 cm and 26 cm 2. Derive that the area of an equilateral triangle is 43(side)2. A Let ABC be an equilateral triangle in which DC = a2 – a2 = 1 3 a unit Hence, the area of the equilateral triangle is 43(side)2. 3. Derive that the area of an isosceles triangle is 1 q 4p2 – q2 Solution: Let ABC is an isosceles triangle in which Ba Vedanta Excel in Mathematics Teachers' Manual - 10 32 Now, BD = CD = 1 BC = 1 q In 'ABD; AD = AB2 – BD2 [By Pythagoras theorem] = p2 – q 2 = 1 4p2 – q2 = 1 q 4p2 – q2 sq.unit 4. If x, y, z are the three sides and s is the semi - perimeter of a triangle derive that its area is s(s x) (s y) (s z) X Let XYZ be a triangle in which YZ = x, ZX = y and XY = z. y ?s X=WxA Y2yZ bze drawn such that XW = h = height of 'XYZ, Y a W x−a Z YW = a. ? WZ = x a. Now, in right angled 'XYW, XW2 = XY2 YW2 or, h2 = z2 a2 ... (i) Also, in right angled 'XWZ; XW2 = XZ2 = WZ2 or, h2 = y2 (x a)2 ... (ii) From (i) and (ii), we get, z2 a2 = y2 (x a)2 or, z2 a2 = y2 x2 2ax a2 h2 = z2 – x2 y2 z2 2 = z + x2 y2 z2 z– x2 y2 z2 = 2xz + x2 y2 z2 2xz – x2 y2 z2 = (x + z)2 – y2 y2 – (x2 – 2xz z2) = (x + z)2 – y2 y2 (x z)2 = (x + z + y) (x + z – y) (y + x z) (y x z) = 2s (2s 2y) (2s 2z) (2s 2x) [? x y z = 2s] ?h = 16s (s – y) (s – z) (s – x) = 2 s(s – x) (s – y) (s – z) Area of ' ABC = 1 YZ × XW = 1 u x u 2 s(s – x) (s – y) (s – z) = s(s – x) (s – y) (s – z) sq.unit D 21 cm C 5. The adjoining figure is a trapezium ABCD. Find 20 cm Solution: Here in trap. ABCD; CD // BA, AB = 16 cm, DC = 21 cm and 33 Vedanta Excel in Mathematics Teachers' Manual - 10 AC = 20 cm, ABC = 90°. D 21 cm C Now, in right angled 'ABC, BC = AC2 – AB2 20 cm B i) Area of trap. ABCD = 1 u BC (DC AB) = 6 cm u 37 cm = 222 cm2 ii) Construction: AE A DC is drawn where E is on DC. In right angled 'AED; AD = AE2 – DF2 = (12cm)2 – (5cm)2 = 13 cm ? Required length of BC is 12 cm and that of AD is 5 cm. 6. Two adjacent sides of a parallelogram are 8 cm and 6 cm and its area is 448 cm2. Find the length of its diagonal . DC Here, area of ABCD = 48cm2, AD = 8 cm and BC = 6 cm. ? 6 cm Also, In 'ABC, AB(c) = 8 cm, BC(a) = 6 cm and AC(b) = ? ? semi - perimeter (S) = a b c = b 14 cm Again, Area of 'ABC = s(s – a) (s – b) (s – c) or, 24 = b 14 b 14 6 b 14 b b 14 8 or, 24 = b 14 b 2 14 b b 2 = (196 b2) (b2 4) on squaring Both sides, we get or, 9216 = 196b2 784 b4 4b2 = 200b2 784 b4 or, b4 200b2 10000 = 0 or, (b2 100)2 = 0 or, b2 = 100 ? b = 10 Hence, the length of diagonal AC (b) = 10 cm. 7. A garden is in the shape of a rhombus whose each side is 15 m and its one of two diagonals is 24 cm. Find the area of the garden. DC Here, in rhombus side AB = 15m, diagonal AC = 24 m O Inrightangled'AOB,OB= AB2 – OA2 = (15 m)2 – (12 m)2 Vedanta Excel in Mathematics Teachers' Manual - 10 34 ?BD = 2 u OB = 2 u 9 m = 18 m Hence, the area of the rhombus ABCD is 216 m2. 8. The perimeter of a triangular garden is 18 m. If its area is 135 m2 and one of the three sides is 8 m, find the remaining two sides. A Let the unknown sides of the triangle be a and b. b Also, a b c = 18 m or, a b 8 m = 18 m ? b = (10 a)m ... (i) B or, 135 = 9 u (9 – a) u (a – 1) u 1 135 = 9(9a 9 a2 a) or, 15 = 10a 9 a2 or, a2 10a 24 = 0 or, a2 6a 4a 24 = 0 or, (a 6) (a 4) = 0 ? 6 or 4 When a = 6, from (i); b = (10 6) = 4 When a= 4, from (i); b = (10 4) = 6 Hence the required remaining sides of triangle are 6 m and 4 m or 4 m and 6 m 9. The perimeter of a right angled triangle is 24 cm and its area is 24 cm2. Find the sides of the triangle. Solution: Let p, b and h be the perpendicular base and hypotenuse of the right angled triangle. Then perimeter of the triangle = 24 cm or, p b h = 24 cm or, p b = 24 h Squaring on both side, we get p h (p b)2 = (24 h)2 or, p2 2pb b2 = 576 48h h2 or, h2 2pb = 576 48h h2 [? p2 + b2 = h2] or, 2pb + 48h = 576 ... (i) b Putting the value of pb in equation (i) we get 2 u 48 48h = 576 ? h = 10 Again p b h = 24 or, p b 10 = 24 or, p b = 14 ? b = 14 p ..... (iii) Putting the value of b in equn (ii), we get p(14 p) = 48 or, 14p p2 = 48 or, p2 14p 48 = 0 or, p2 8p 6p 48 = 0 or, (p 8) (p 6) = 0 ? p = 8 or 6. When p = 8, from (iii), b = 14 8 = 6 When p = b, from (iii), b = 14 6 = 8 Hence, the required sides of the triangle are 8 cm, 6 cm, 10 cm or 6 cm, 8 cm, 10 cm. 10. An umbrella is made up of 6 isosceles triangular pieces of cloths. The measurement 35 Vedanta Excel in Mathematics Teachers' Manual - 10 Solution: Here, base of each triangular piece (b) = 28 cm Two equal sides of each triangular piece (a) = 50 cm Now, the area of each isosceles triangular piece of cloth = 1 b 4a2 b2 = 1 u 28 4(50)2 282 = 672 cm2 Again, the rate of cost of cloth (R) = Rs 0.50 per sq.cm ? Total cost of the cloth (T) = Area of cloth (A) u rate (R) = 4032 u Rs 0.50 = Rs 2016 11. A real state agent divided the adjoining triangular 50 m 20 m 105 m Solution: Now, semi-perimeter (s) = a b c = 50 m 85 m 105 m = 120 m ? Area of triangular land (A1) = s(s – a) (s – b) (s – c) = 120(120 – 50) (120 – 85) (120 105) = 120 u 70 u 35 u 15= 2100 sq m 16=8021 u 32 (20 85) = 1680 sq.m2 We have, 1 aana = 31.79 m2 Again, the rate of cost of land (R) = Rs 2,10,000 per aana. ? Total cost of shaded 'kitta' = Area of land (A) u rate (R) = 13.211708 u Rs 2,10,000 = Rs 2774457.38 Hence, the cost of shaded 'kitta' is Rs 2774457.38 D 100 m C parallel sides are 100 m and 40 m and the non - parallel 56 m 56 m sides are 56 m and 52 m. Find the cost of ploughing the field at Rs 7.50 per sq.m Solution: A 40 m B CD = 100 m, BC = 52 m and AD = 56 m Construction : BE // AD to meet CD at E and BF A CD at F are 52 m Vedanta Excel in Mathematics Teachers' Manual - 10 36 In 'BCE, semi- perimeter (s) = a b c = 52 m 60 m 56 m = 84 m ? Area of 'BCE = 84(84 – 52) (84 – 60) (84 56) = 1344 m2 Rate of ploughing the field (R) = Rs 7.50 Per m2 ? Total cost of ploughing the field (T) = area of field (A) u Rate (R) = 3136 u Rs 7.50 = Rs 23,520 Hence, the cost of ploughing the field is Rs 23,520 Extra Questions 1. The length equal sides of an isosceles triangular plotare 17 feet each. If the area of the plot is 120square feet, find the length of its base. [Ans: 16 feet or 30 feet] 2. The perimeter of aright angled triangle is 24 cm and its area is 24 cm2, find the sides of the triangle. [Ans: 6 cm, 8 cm and 10 cm] 3. An umbrella is made up by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 15 cm, 41 cm and 28 cm. much cloth is required for the umbrella? If the rate of cost of the cloth is 44 paisa per sq. cm, find the total cost of the cloth to make the umbrella. [Ans: 1260 cm2, Rs 554.40] 4. The sides of a triangular field are 20 ft., 507 ft. and 493 ft. find the perimeter and area of the field. Estimate the cost of fencing it with 4 rounds by the wire at Rs 6.25 per ft. and panting the vegetable at Rs 20 per square feet. [Ans: 1020 ft, 3570 sq.ft.; Rs 25,500, Rs 71,400] 37 Vedanta Excel in Mathematics Teachers' Manual - 10 Unit Mensuration (II): Cylinder and Sphere 6 Competency Allocated teaching periods 10 - Solving the problems related to surface area and volume of regular solid cylinder, sphere and hemisphere Learning Outcomes - To solve the problems related to surface area and volume of regular solid cylindrical, spherical and hemispherical objects. Level-wise learning objectives S.N. LEVELS OBJECTIVES - To define cylinder - To tell the formula of finding the volume, curved surface 1. Knowledge (K) area and total surface area of cylinder - To recall the definition of great circle of the sphere - To list the formula of finding the surface area and volume of sphere and hemisphere - To find the volume of cylinder (U) - To find the volume and surface area of sphere - To find the volume and total surface area of combined - To determine the unknown dimension of cylinder when 3. Application (A) - To find the unknown dimension of sphere and hemisphere - To estimate the cost - To relate the volumes when a metallic sphere is melted - To link various real life/ contemporary problems with area 4. High Ability (HA) and volume of cylinder and sphere - To relate the curved surface areas of cylinder and sphere. the given measurements. Required Teaching Materials/ Resources Graph-board, pencil, scissors,volleyball, marbles, models of cylinder, models of sphere and hemisphere, definitions and formulae in colourful chart-paper, ICT tools, etc. Vedanta Excel in Mathematics Teachers' Manual - 10 38 A. CYLINDER Teaching Activities 1. Ask the students about the real life examples cylindrical, spherical and hemispherical objects. 2. Discuss upon the definition, curved surface area and total surface area of cylinder by presenting the models of cylinder or folding the rectangular sheet of r paper in the cylindrical form. 3. Explain the formulae of perimeter and area of circular base, volume and h - Volume (V) = Area of circular base (A) length (h) = Sr2h - Curved surface area (C.S.A) = Perimeter of circular base height (h) = 2Srh - Total surface area (T.S.A.) = C.S.A. + 2A = 2Srh + 2Sr2 = 2r(r + h) 4. Explain the formulae of volume and curved/total surface area of the hollow cylinder - Volume (V) = Area of annular base (A) height (h) = S (R2 - r2)h = Sh (R+ r) (R – r) - Curved surface area (C.S.A) = internal C.S.A. + external C.S.A. R r - Total surface area (T.S.A.) = C.S.A. + area of 2 annular base = 2Sh(R + r) + 2S (R+ r) (R – r) h B. SPHERE AND HEMISPHERE Teaching Activities 1. With manipulative models, discuss on curved surface, total surface of sphere and hemisphere. r 2. Use measuring cylinder and sphere of equal radii and height to find the volume formula of sphere 3. Use thread and two hemispheres of equal radii to explain about the surface area of the sphere. 4. To motivate the students to find the following formulae under discussion (i) Sphere centre - Circumference of the great circle = 2Sr r - Area of great circle = Sr2 Surface area of sphere = 4Sr2 Volume of sphere = 4 Sr3 - Curved surface area of hemisphere = 2Sr2 - Total surface area of hemisphere = 2Sr2 + r2 = 3Sr2 - Give the project work to the students to write the formulae of surface are and volume of the cylinder, sphere and hemisphere with clear figures in the colourful chart- paper and present in the classroom. 39 Vedanta Excel in Mathematics Teachers' Manual - 10 Solution of selected questions from Vedanta Excel in Mathematics 1. A cylindrical water tank contains 4,62,000 litres of water and its radius is 3.5 m, find the height of the tank. Solution: Here, = 462000 m3 [? 1000 l = 1m3] = 462 m3 radius (r) = 3.5 m height (h) = ? Now Volume (V) = Sr2h Hence the height of the tank is 12 m. 2. If the perimeter of the great circle is S cm, find the volume of the hemisphere. Solution: Here, perimeter of the great circle (C) = S1 cm Volume of hemisphere (V) = 2 Sr3 = 2 S u 13 3. The total surface area of a hemisphere is 243Scm2, find its volume. Solution: TSA of hemisphere = 243 S cm2 or, 3Sr2 = 243 S cm2 r2 = 81 ? r = 9 cm Now, 2 2 4. A solid metallic sphere of radius 7 cm is cut into two halves. Find the total surface area of the two hemispheres so formed. Solution: 7 cm 7 cm 7 cm T.S.A of a hemisphere = 33Sur 2272 ? T.S.A. of two hemisphere = 2 u 462 cm2 = 924 cm2 5. The area soufrcfaucreveadresaurifsa9c2e4acrmea2,offinadstohliedvcoylluimndeeorfisth32e of the total surface area. If Solution: Let r and h be the radius of base and height of a cylinder respectively. Now 2 ? h = 2r ..... (i) Vedanta Excel in Mathematics Teachers' Manual - 10 40 Also, total sruface area = 924 cm2 or, 2Sr(r h) = 924 or, r2 = 49 ? r = 7 and h = 2r = 2 u 7 cm = 14 cm 22 6. By how many times the volume of a cylinder increase when its diameter is doubled. Solution: L?WehtVeV1n1=tbheSert2hrhaed=viuoSslxui2msydecouou.fubtnlheiedt c,..yt.hl(iien)ndreardwiuhsenofintserwadciyulsinids exr units and height y units. y units. HL?eetVnV2ce2=btheSert2hhveo=lvuoSml(u2emxb)ee2 couofymn=eew44cStiyxml2iyenscduteh.ruatnoitfso=rig4inuaSlxc2yylicnud.uernwit h=en4Vth1e [From ... (i)] 7. A hollow cylindrical metallic pipe is 21 cm long. If the external and internal diam- eters of the pipe are 12 cm and 8 cm respectively, find the volume of metal used in making the pipe. Solution: Here, the external diameter of the pipe (D) = 12 cm the internal diameter of the pipe (d) = 8 cm ? internal radius (r) = 8 cm = 4 cm Now, Volume of the material (V) = Sh(R r) (R trh) e=p2i7p2e is2113(260 c4m) 3(.6 4)= 1320 8. The internal radius of a cylindrical bucket of height 50 cm is 21 cm. It is filled with water completely. If the water is poured into a rectangular vessel with interval length 63 cm and breadth 44 cm and it is completely filled with water, find the height of the vessel. Solution: Here, the internal radius of the bucket (r) = 21 cm The height of the bucket (h) = 50 cm 22 ? Volume of water = interval volume of bucket = 69,300 cm3 Again, Volume of the rectangular vessel = Volume of water or, l u b u h = 69,300 or, 63 u 44 u h = 69,300 9. An iron pipe of internal diameter 2.8 cm and uniform thickness 1 mm is melted and 41 Vedanta Excel in Mathematics Teachers' Manual - 10 Solution: Here, the internal diameter of an iron pipe (d) = 2.8 cm ? the external radius of the pipe (R) = 1.4 cm 0.1 cm = 1.5 cm Length of pipe = h cm (say) Now, Volume of the material (V) = Sh(R r) (R r) = Sh(1.5 1.4) (1.5 1.4) = Sh u 2.9 u 0.1 = 0.29Sh Again, Volume of cylindrical rod = Volume of material used to make the pipe or, Sr2h = 0.29Sh [Length of pipe and rod are equal] or, r2 = 0.29 ? r = 0.54 cm and d = 2r = 2 u 0.5 cm = 1.08 cm Hence; the diameter of the rod is 1.08 cm. 10. A solid iron sphere of diameter 42 cm is dropped into a cylindrical drum partly filled with water. If the radius of the drum is 1.4 m. by how much will the surface of the water be raised ? Solution: iron sphere (d) = 42 cm ? radius (r) = 42 cm = 21 cm Let height of water surface raised in the drum be h cm. Then, volume of water displaced by the sphere = Volume of sphere or, Sr2h = 12348 S or, 140 u 140 h = 12348 = 0.63 cm = 0.63 u 10 mm = 6.3 mm Hence. the surface of water is raised by 6.3 mm. 11. A cylindrical jar of radius 6 cm contains water. How many iron solid spheres each of radius 1.5 cm are required to immerse into the jar to raise the level of water by 2 cm. Solution: Here, radius of a cylindrical jar (r) = 6 cm height of water level raised by iron spheres (h) = 2 cm Now, Volume of water displaced by the iron sphere (V1) = Sr2h = S u 62 u 2 = 72S cm3 ? Required number of iron spheres (N) = Volume of displaced water (V1) = 72 S cm3 = 16 12. Given figure in a solid composed of a cylinder with hemisphere at one end. If the total surface area and height of the solid are 770 sq.cm and 14 cm respectively, find the height of the cylinder. 14 cm Let r and h be the common radius of base and height of the cylinder respectively. Then, height of hemisphere = radius of hemisphere = r Now, total height of combined solid = r h = 14 cm Vedanta Excel in Mathematics Teachers' Manual - 10 42 Also T.S.A. of solid = C.S.A of hemisphere C.S.A of cylinder Area of base or, 770 = 2Sr2 2Srh Sr2 or, 770 = Sr (2r 2h r) or, 770 = 22 r [3r 2(14 r)] [From (i)] or, r2 28r 245 = 0 or, r2 35r 7r 245 = 0 or, r(r 35) 7(r 35) = 0 or, (r 35) (r 7) = 0 Eithere, r 35 = 0 ? r = 35 which is impossible as the radius can not be negative or, r 7 = 0 ? r = 7 and h = (14 7) cm = 7 cm Hence, the height of the cylinder is 7 cm. 13. A combined solid made up of a cylinder of radius 3 cm and length h cm and a hemisphere with the same radius as the cylinder has volume 792 cm3. Find the value of h. Solution: Here, radius of cylinder = radius of hemisphere (r) = 3 cm height of cylinder = h cm Volume of combined solid = 792 cm3 Now, volume of combined solid = Volume of cylindr Volume of hemisphere or, 792 = or, 28 = h 2 or, h = 26 Hence, the height of the cylindrical part is 26 cm. 14. A roller of diameter 112 cm and length 150 cm takes 550 complete revolution to level a compound. Calculate the cost of levelling the compound at Rs 9 per square meter. Solution: Here, diameter of the roller = 112 cm 51160500cmcm==150160.5 Now, area of covered by the roller in 1 resolution = C.S.A. of the roller ? Area covered by the roller in 550 revolution = 550 u 5.28 m2 = 2904 m2 ? Area of the compound = 290 m2 Again, rate of levelling the compound (R) = Rs 9 per m2 ? Cost of levelling the compound (T) = A u R = 2904 u 9 = Rs 26,136 Hence, the required cost of levelling the compound is Rs 26,136. 15. The external and internal radii of hollow cylindrical metallic vessel 56 cm long are 10.5 cm and 10.1 cm respectively. Find the cost of metal contained by the vessel at Rs 2 per cubic cm. Also, find the cost of polishing its outer surface at 20 paisa per square cm. 43 Vedanta Excel in Mathematics Teachers' Manual - 10 Solution: Here, the internal radius of teh cylindrical Vessel (r) = 10.1 cm the external radius of the vesel (R) = 10.5 cm the length of the vessel (h) = 56 cm Now, Volume of metal contained by the vessel (V) = Sh(R r) (R r) = 22 u 8 u 20.6 u 0.4 = 1450.24 cm3 ? Cost of metal contained by the vessel (T) = V u R = 1450.24 u Rs 2 = Rs 2900.48 ? Cost of polishing the outer surface (T) = A u R = 3696 u 20 paisa = 73920 paisa = Rs 739.20 16. Agatehastwocylindricalpillarswithahemispherical end on a top of each pillar. The height of each pillar is 9.96 cm and the height of each cylinder is 9.75 m. 9.96 m at Rs 500 per sq.m. 9.96 m Solution: Here, For a pillar height of cylindrical part (h) = 9.75 m height of hemispherical part = radius of base (r) = 9.96 m 9.75 m = 0.21 m Now, C.S.A of a pillar = C.S.A of hemisphere C.S.A of cylinder ? Total surface area of two pillars = 2 u 13.1472 m2 = 26.2944 m2 ? Cost of colouring the surface of the both pillar (T) = A u R = 26.2944 u Rs 500 = Rs 13,147.20 Extra Questions 1. The total surface and curved surface area of a cylindrical can are 748 cm2 and 440 cm2 Vedanta Excel in Mathematics Teachers' Manual - 10 44 Unit Mensuration (II): Prism and Pyramid 7 Competency Allocated teaching periods 11 - Solving the problems related to surface area and volume of regular solid objects (prism, pyramid, cylinder, sphere, hemisphere and cone) Learning Outcomes - To solve the problems related to surface area and volume of regular solid objects (prism, pyramid, cylinder, sphere, hemisphere and cone) Level-wise learning objectives S.N. LEVELS OBJECTIVES - To define prism of cone - To find the volume of triangular prism - To apply the related formula to solve the problems of pyramid and cuboid etc. 4. High Ability - To estimate the cost and quantity related the solid objects Required Teaching Materials/ Resources Teaching learning strategies 45 Vedanta Excel in Mathematics Teachers' Manual - 10 A. PRISM Teaching Activities 1. With manipulative models of prisms, discuss upon the prism and its base and lateral surface. 2. Discuss with models on volume of prism as the product of cross-sectional area and the height i.e., V = A h Rectangle 1 area (L.S.A) as the product perimeter of base and height Rectangle 2 B. PYRAMID Teaching Activities O discuss upon the pyramid and its base and lateral surface. hl 2. List the following formulae: a Area of base (A) = a2 1 1 Area triangular faces or lateral surface area (L.S.A) == 421 al = 2al C. CONE Teaching Activities 1. Using manipulative models of cone describe about the cone and its base and curved surface. 2. Explain the relation among r, h and l as l2 = r 2 + h 2 3. List the following formulae: - Area of base (A) = Sr2 1 1 - Curved surface area (C.S.A) = = Srl - Total surface area (T.S.A.) = Area of base (A) + C.S.A. = Sr2+ Srl = Sr (r + l) 4. Using manipulative models of combined solid ask and explain the volume and surface area of the solid. Solution of selected questions from Vedanta Excel in Mathematics 1. The area of cross section of a triangular prism is 126 cm2 and its volume is 5040 Vedanta Excel in Mathematics Teachers' Manual - 10 46 What is the compound interest on Rs 31250 at 8% per annum for 2 years?=38637−31250=Rs. 7387.
What is the compound interest on 3000 rupees at 6% per annum for 2 years?Detailed Solution
∴ The compound interest for 2 years is Rs. 1320.
What is compound interest on an amount of Rs 1000 at 8% per annum for 2 years?=1124.864−1000=Rs. 124.864.
What will be the compound interest on 5000 at 8% per annum for 2 years?5832∴ C.I. = A - P = Rs. 5832 - 5000= Rs. 832 (c)
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