MATHEMATICS FORM 1 – RATIO, PROFIT AND LOSS msomimaktaba, November 11, 2018August 17, 2024 RATIO, PROFIT AND LOSSRatio:- the ratio of number p and q is p:q or p ÷ q or p/qExamples:1. Joha and Siwenza shared 4,000 shillings between them. It Joha received 15,000 shillings and Siwenza got 25,000 shillings,Find the ratio of the amounts they received:Solution:Joha : Siwenza15,000: 25,000or15,000 = 15 = 325,000 = 25 = 53:5The ratio of the amount they received is 3:52. Express 6:4 in lowest termsSolution6 : 4 it can be written as6/4 = 3/2 6:4 in its lowest term is 3:2 3. Complete the following ratio(a) 9 : 24 = 3(b) 15 : 39 = 15Solution (a) 9 : 24 = 3Question:1. complete the following(a) 12 : __ = 3 : 7(b) 4 : 9 = ___ :63 solution (a) 12 : __ = 3 : 7(b) 4 : 9 = ___ :63 solution4 : 9 = ___ : 63Proportions 1. Divide 120 shillings the ratio 3 : 7Solution 3 + 7 = 10 3 x 120 = 36 shs 107 x 120 = 84 shs10 Therefore 3 of 120 is 36 shs 107 of 120 is 84 shs 102. divide 156 in the ratio of 3 : 4 : 5solution3 : 4 : 5 3 + 4 + 5 = 12+ 7 = 10 3 x 156 = 39 123 of 156 is 39124 x 156 = 52 124 of 156 is 52 125 x 156 = 65125 of 156 is 65 12Question1. Juma, Ali, Mary and Kalo have 300, 100, 500 and 600 shaves in a cooperative shop respectively. Divide 150,000 shs among them in the ratio of their shavesSolution300, 100, 500 and 600 = 300 + : 100 + : 500 + : 600= 3: 1: 5: 6= 3: 1 : 5: 6= 3 + 1 + 5 + 6 = 15= 3 x 150,000 ÷ 30,00015= Juma hot 30,000 shs1/15 x 150,000 = 10,000Ally got 10,000 shs 5/15 x 150,000 = 30,000Mary gor 50,000 shs 6/15 x 150,000 = 60,000Kalo got 60,000 shsProfit and lossProfit = Selling price – buying priceLoss = Buying price – selling price Percentage profit = profit made x 100%buying price Percentage loss = loss made x 100%buying priceExample:- 1. A ration is bought for 24,000 shs and sold for 36,000 shs. Find(a) The profit made(b) The percent profitSolution (a) Buying price = 24,000/=Selling price = 36,000/=The profit made= Selling price – buying price= 36,000/= – 24,000/== 12,000/= (b) The percentage profit= profit made x 100%Buying price= 12,000 x 100%24,000= ½ x 100%= 50%2. A dealer paid 800,000/=for a machine and solid it the following year for 600,000/=Find: (a) the loss made(b) the percentage loss(a) Solution The loss made= Buying price – selling priceBuying price = 800,000/=Selling price = 600,000/=800,000 – 600,000= 200,000/=(b) SolutionLoss made = 200,000/=Buying price =800,000/=The percentage loss= loss made x 100%Buying price=200,000 x 100%800,000= 1 x 100% 4 = 25%Question1.A house was sold a profit of 900,000. If its late of the profit is 37 1/2. Find the lost (buying price) of the houseSolution Profit made = 900,000 The percentage profit = 37 1/2 Buying price = ?profit made x 100% Buying price37 1/2 = 900,000 x 100%1 buying priceB = 900,000 x 100% 37 1/2= 900,00 x 100 75/2= 900,000 x 2 1 75= 18000000075= 2,400,000/=2. A dealer paid sh. 20,000 for 10 books. He sold 2/5 of them at 3000 each and the remaining at 2500 each. What was his percentage profit?Solution2/5 x 10 = 44 x 3,000 = 12,000 shs.1/3 – 2/5 = 3/53/5 x 10 = 66 x 2500 = 15,000 shs12,000shs + 15,000 shs= 27,000 shs Profit = Selling – Buying piece= 27,000 shs – 20,000 shs= 7,000 shs. Percentage profit = Profit madeBuying price7,000 x 100%20,000= 7000200= 35 % Percentage is 35% Question 1.Find the profit or loss when an article bought for(a) 4000 sh and sold at 48000/=(b) Sh. 1250 and sold at 1500/=Solutions (a) The profit made = Selling price – buying price buying price = sh 4000 selling price = sh 4800 profit = selling – buying= sh 4800 – 4000 = sh 800(b) buying price = 1250sh selling price = sh 1500 profit = selling – buying profit = sh 1500 – sh 1250 = 250 sh(c) buying price = 2000sh selling price =1400sh loss = buying – selling = 2000sh – 1400sh = 600sh2. Find the buying price of an article which is sold at:-(a) 98000sh at a profit of 40%(b) 30,000 sh at profit of 16%(c) 108,500 sh at profit of 7% (a) SolutionBuying price = ?Selling price = 98000 shPercentage profit = 40%But percentage profit= Profit made x 100%Buying price40% = Profit made x 100%Buying price40% = profit made100 Buying priceBut profit made = S. P – B .P2/5 = Selling – BuyingBusiness price2/5 = 98000 – BB2 X B = 5 ( 98000 – B)2B = 490000 – 5B2B + 5B = 490,0007B = 490,0007 7BB = 70,000 shTherefore Buying price = 70,000sh(b) SolutionBuying price (BP)Selling price (SP) = 30,000/=Percentage profit (PP) = 16%PP = PM X 100%BPPM = SP – BP 16% = (SP – BP )100%BPBP X 16% = (30,000 – BP X 100%)BP BP16%BP = (30,000 – BP) 100% 100%16BP = 30000 – BP100 14BP = 25(30,000 – BP)4BP = 750,000 – 25 BP4BP + 25BP = 750,00029BP = 750,00029 29BP = 750,000 29 BP = 25862.7Therefore the buying price of an article is 25862.07shs (c) 108,500 sh at profit of 7% Data given : selling price = 108,500sh profit percentage = 7% buying price = ? But the percentage profit = profit made x 100%buying price7% = profit made 100% buying price But the profit made = selling price – buying price7 = selling price – buying price 100 buying price7 = 108,000 – B 100 B7 x B = 100(108,500 – B)7B = 108,500 – 100B100B + 7B = 108,500107B = 108,500 107 107B = 10140.19shTherefore The buying price of an article is 10140.19sh3. find the article which is bought for 25,000 sh and sold at profit of 24%solutionData given ; Buying price = 25,000sh profit percentage = 24% selling price = ? but profit percentage profit made x 100% buying price24% = profit made x 100% buying price24 = profit made 100% buying priceBut profit made= selling price – buying6 = selling price – buying price 25 buying price6 = S.P – 25,000 25 25,0006 X 25,000 = 25 (S.P – 25,000) 150,000 = 25S.P – 625,000 625,000 + 150,000 = 25S.P 775,000 = 25S.P 25 2531,000 = S.PTherefore Selling price of the article is 31,000 sh4. A machine losting 18000/= is sold at profit of 40%. What is the selling price?solutionData given :buying price = 180,000 profit percentage = 40% selling price = ?40 x 180,000 100= 40 x 1,800 = 72,000 sh Then 180,000 sh + 72,000 sh = 252,000 shTherefore Selling price of machine is 252,000 sh5. At a clearance sale boots which are cost 30,000 sh each were sold at a loss of 25%. Calculate the loss and clearance priceSolutionData given :buying price = 30,000loss = 25% loss made = 25% x 30,000sh = 25 x 30,000sh 100= 7,500 shThe loss made of boots is 7,500 shClearance price (selling price) loss made = buying – selling price 7,500sh = 30,000sh – S 7,500 – 30,00 = – S -22,500 = -S -1 -1Therefore Clearance price is 22,500shSIMPLE INTEREST Simple interest (1): Is the amount of money paid as a profit.Principle (P): The money borrowed or deposited or lent.Rate (R) : Is the principle change per year.Time (T): Period I yearAmount (A): The sum of principle and interestInterest = Principle x made x Time100I = PRT100Examples:Find the simple interest on 18,000 sh for 2 year at rate of 6%Solution Data given:Principle = 18,000 shRate = 6%Time = 2 yearsInterest = ?I = PRT100I = 18,000 x 6 x 2100I = 180 X 12= 2,160 Therefore the simple interest is 2,160 shSolution:Data given:Simple interest = 90,000 shRate = 4 %Time = 5 yearsPrinciple =?I = PRT10090,000 = P x 4 x 510090,000 = 20P1 10020P x 1 = 90,000 x 10020 20P = 450,000 SHThe principle is 450,000 sh.EXERCISE 1. Find the percentage rate at which interest on :(a) 20,000 /= for 4 years is 2000/=SolutionData given:Interest = 2000/=Principle = 20,000/=Time = 4 yearsRate = ?I = PRT10020,000 = 200 x R x 410020,000 = 800 R800 80025 % = R:. Rate is 25% (b) 16, 0000 /= for 3 years is 9600/=Solution Data givenInterest = 96, 00/=Principle = 160,000/=Time = 3 yearsRate =?I = PRT1009600 = 160,000 x R x 31009600 = 4800 R4800 48002% = R:. The rate is 2% 2. Find the principle that would earn 54,000 /= in 10 years at rate of 5%SolutionData givenInterest = 54,000/=Rate = 5%Time = 10yearsPrinciple = ?I = PRT10054,000 = P x R 10210054000 = P x 22108000 = P:. The rate principle is 108 ,000/=1. Find the simple interest on 54,000/= inverted for 18 months at the rate of 12% per annumSolutionData givenInterest =?Principle = 54,000/=Rate = 12Time = 18 months= 1 years and 6 months= 1 years and halfI = PRT10054,000 x 12 x 1 ½100540 x 6 x= 3,240 =:. The simple interest is 3.240 /=EXERCISE 2 1. Find the simple interest on(a) 80,000 sh for 1 year at rate of 20% per annumSolutionData given:Principle = 80,000/=Rate = 20%Time = 1 yearInterest = ?But, I = PRT100I = 80,000 x 20 x 1100= 800 X 20= 16,000The simple interest is 1,600/=(a) 14,000/= for 1 year at rate of 3% per annulSolution:-Data given:Principle = 14,000/=Rate = 3%Interest = 1 yearBut, I = PRT100I = 14,000 x 3 x 1100= 140 x 3= 420/=The simple interest is 420/=2. Find the number of year (time in year ) in which the interest on:(a) 20,000/= at the rate of 4% is 4,800/=Solution:Data given:Principle = 20,000/=Rate = 4%Interest = 4,800/=But, I = PRT1004,800 = 20,000 x 4 x I1004,800 = 800T800 8006 = TThe time is 6 years(b) 50,000/= at the rate of 1% in 30,000/=Solution:Data given:Principle = 60,000/=Rate = 1%Interest = 30,000But, I = PRT10030,000 = 60,000 X 1 X T 10030,000 = 600T600 60050 = TTherefore The time is 50 years ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Basic Mathematics Study Notes Form 1 Basic Mathematics Study Notes Msomi Maktaba All Notes FORM 1MATHEMATICSPost navigationPrevious postNext postRelated Posts Form 6 Geography Study Notes Advanced Geography 2 (Geography Form 6) – 5.3 SUSTAINABLE USE OF FUEL AND POWER February 25, 2019August 8, 2019ALL NOTES FOR ALL SUBJECTS QUICK LINKS: AGRICULTURE O LEVEL PURE MATHEMATICS A LEVEL BAM NOTES A LEVEL BASIC MATH O LEVEL BIOLOGY O/A LEVEL BOOK KEEPING O LEVEL CHEMISTRY O/A LEVEL CIVICS O LEVEL COMPUTER(ICT) O/A LEVEL ECONOMICS A LEVEL ENGLISH O/A LEVEL COMMERCE O/A LEVEL ACCOUNTING A LEVEL… Read More FORM TWO MOCK – EXAMINATION: AUGUST 2020 | BASIC MATHEMATICS October 8, 2020PRESIDENT’S OFFICE REGIONAL ADMINISTRATION AND LOCAL GOVERNMENT ITILIMA DISTRICT COUNCIL FORM TWO MOCK- EXAMINATION: AUGUST 2020 041 BASIC MATHEMATICS TIME 2:30 HRS Tuesday 4TH August 2020 am Instructions Answer all questions 1. 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