CSS Primary Standard “Mathematics” 151 Property tax = ? Property tax = 0.8% of 9000000 = 0.8 9000000 100 = Rs. 72000 Exercise 7a Classroom Activity: If you buy a bat for Rs. 4000 and sold it for Rs. 4300 how much profit % you would earn? Solution: Cost price = Rs. 4000 Sale price = Rs. 4300 Profit = 4300 – 4000 = Rs. 300 Profit % = Profit ×100 Cost Price = 300 100 4000 = 7.5 % What would be to the principal value if the markup value is Rs. 670 for 6 months at 8% per annum. Solution: Principal = ? Markup rate (R) = 8 % per annum Markup = Rs. 670 Time period = 6 12 = 1 2 year By using formul: Markup = RPT 100 670 = 1 8 P 2 100 P = 670 100 2 8 = Rs. 16750 Q1. Convert the following fraction into percentage.
CSS Primary Standard “Mathematics” 152 (i) 17 20 Step-1 17 20 = 17 5 20 5 = 85 100 Step-2 85% (ii) 440 100 = 440% (iii) 15 8 900 Step-1 15 8 900 = 7215 900 Step-2 7215 9 900 9 = 801067 100 Step-3 = 801067% (iv) 13 8 Step-1 13 25 8 25 = 325 200 Step-2 325 2 200 2 = 162.5 100 Step-3 = 160.5% (v) 14 4 49 Step-1 14 210 210 100 4 49 49 49 100 = 21000 4900 Step-2 21000 49 4900 49
CSS Primary Standard “Mathematics” 153 = 428.57 100 Step-3 = 428.57% (vi) 155 75 Step-1 175 155 4 75 75 4 = 620 3000 Step-2 620 3 300 3 = 206.67 100 Step-3 = 206.67% (vii) 54 200 Step-1 54 200 Step-2 54 2 200 2 = 27 100 Step-3 = 27% (viii) 18 4 77 Step-1 18 326 4 77 77 326 100 77 100 = 32600 7700 Step-2 32600 77 7700 77 = 423.38 100 Step-3 423.38% (ix) 4 50 250
CSS Primary Standard “Mathematics” 154 Step-1 4 12504 50 250 250 12504 12504 2 250 250 2 = 25008 500 Step-2 25008 5 500 5 = 5001.6 100 Step-3 5001.6% (x) 7 19 Step-1 7 100 700 19 100 1900 Step-2 700 19 36.84 1900 19 100 Step-3 36.84% (xi) 2 11 25 Step-1 2 277 11 25 25 = 277 4 25 4 = 1108 100 Step-2 1108% (xii) 3 50 Step-1 3 3 2 50 50 2 = 6 100 Step-2 6% Q2. Convert the following percentage to fractions. (i) 7% = 7 100 (ii) 83%
CSS Primary Standard “Mathematics” 155 = 83 100 (iii) 450% = 450 100 = 1 2 5 1 4 4 10 2 (iv) 47% = 47 100 (v) 33% = 33 100 (vi) 750% = 750 100 = 5 1 7 7 10 2 (vii) 154% = 154 100 = 77 50 = 27 1 50 (viii) 210% = 210 100 = 21 10 = 1 2 10 (viii) 574% = 574 100 = 287 50 4 10 45 40 5 7 10 75 70 5 1 50 77 50 27 2 10 20 20 1 5 50 287 250 37
CSS Primary Standard “Mathematics” 156 = 37 5 50 Week 6 Exercise 7b Classroom Activity: Ask your mother how much tola gold is she have and calculate the zakat she would pay after one year as per recent rates of gold. Solution: Let mothers has gold = 20 tola Rate of gold per tola = Rs. 54000 Value of gold = Rs. 54000 × 20 = Rs. 1080000 Rate of zakat = 2.5% Amount of zakat = 2.5 % of 1080000 = 2.5 1080000 100 = Rs. 27000 She will pay Rs. 27000 zakat after one year. If your father has 5 kanal tube well irrigated land. What is the yield and what usher will he pay? Solution: Let Total land = 5 kanal Total yield = 85000 kilograms. Sale price = Rs. 100 Total sale price = Rs. 8500000 Usher = 5% of 8500000 = 5 8500000 100 = Rs. 425000 Q1. Convert the following from percentage to decimal. (i) 65% = 75 100 = 0.75 (ii) 165%
CSS Primary Standard “Mathematics” 157 = 165 100 = 1.65 (iii) 390% = 390 100 = 3.9 (iv) 86% = 86 100 = 0.86 (v) 145% = 145 100 = 1.45 (vi) 870% = 870 100 = 8.7 (vii) 114% = 114 100 = 1.14 (viii) 750% = 750 100 = 7.5 (ix) 5455% = 5455 100 = 54.55 (x) 6745% = 6745 100 = 67.45 Q2. Convert the following decimal fractions into percentage.
CSS Primary Standard “Mathematics” 158 (i) 0.06% = 0.06 × 100 = 6% (ii) 8.76% = 8.76 × 100 = 876% (iii) 0.76% = 0.76 × 100 = 76% (iv) 876.40% = 876.40 × 100 = 87640% (v) 0.04961 = 0.0491 × 100 = 4.961% (vi) 0.74896 = 0.74896 × 100 = 74.896% (vii) 0.7476 = 0.7476 × 100 = 74.76% (viii) 0.005 = 0.005 × 100 = 0.5% Exercise 7c Q1. Express the following into fractions. (i) 15% of 240 12 3 15 10 8 24 0 36 0 (ii) 25% of 500 = 574 100 500 = 125
CSS Primary Standard “Mathematics” 159 (iii) 4% of 150 3 2 4 10 2 15 0 6 0 (iv) 350% of 1000 = 350 100 10 00 = 3500 (v) 17% of 105 21 17 100 20 5 357 20 10 (vi) 10% of 1000 = 10 100 10 00 = 100 Q2. Convert the following into percentage. (i) 24 10 Step-1 = 24 24 10 10 10 100 = 240 100 Step-2 = 240% (ii) 2 11 25 Step-1 = 2 277 11 25 75 = 277 4 25 4 = 1108 100 Step-2 = 1108%
CSS Primary Standard “Mathematics” 160 (iii) 3 50 Step-1 = 3 2 50 2 = 6 100 Step-2 = 6% (iv) 2 8 10 Step-1 = 2 82 8 10 10 = 82 10 10 10 = 820 100 Step-2 = 820% (v) 6 7 8 Step-1 = 6 62 7 8 8 = 62 25 1550 8 25 200 Step-2 = 1550 2 775 200 2 100 Step-3 = 775% (vi) 7 50 Step-1 = 7 7 2 50 50 2 = 14 100 Step-2 = 14% Q3. Find the rate percent in all the below questions.. (i) 8 out of 32.
CSS Primary Standard “Mathematics” 161 25 4 100 8 32 25% (ii) 44kg of sugar out of 66kg of sugar. = 44 100 66 = 66.67% (iii) Rs. 70 out of Rs. 150. 20 70 15 3 100 0 140 3 46.67% (iv) 40 ons out of 60 ons. 2 40 60 3 100 200 3 6 .67% 6 (v) 90 students out of 150. 3 20 90 15 5 100 0 60% Q4. What percent of 1 day is 45 minutes? Solution 1 day = 1440 minutes Let the required percent is x% Then x% percent of 1 day = 46 minutes x% = 46 23 1440 720 = 0.032 x = 0.032 100 = 3.3%
CSS Primary Standard “Mathematics” 162 Q5. What percent of 1 year is 1 day? Solution 1 year = 364 days Let the required percent is x% Then according to given condition x% percent of 1 year = 46 minutes x% × 364 days = 1 day x% = 1 364 = 0.6.00274 x = 0.00274 100 = .274% Q6. 5 member were absent out of 22 members in a meeting. Find percentage of absent members? Solution Total member = 22 No. of absent members = 5 According to given condition 5 members absent out of 22 = 5 22 Percentage of absent member = 50 5 22 100 11 = 250 11 = 22.73% Q7. Ibrahim saves 27% of his pocket money, if he gets Rs.3100 per month. How much he save every month? Solution Ibrahim gots pocket money = Rs. 3100 27% of his pocket money = 27% of 3100 = 27 100 3100 = 837
CSS Primary Standard “Mathematics” 163 Ibrahim save every month = Rs. 837 Q8. Due to having crises in banking sectors, bank retrenched 97 out of its 700 employs. What percentage of employs was retrenched? Solution Bank retrenched 97 out of 700 employs = 97 100 Percentage of retrenched employs = 97 100 100 = 13.86% Q9. A survey was conducted in which 120 persons were asked which cellular company provides good internet speed. Internet speed (i) 24 said Jazz (ii) 36 said Ufone (iii) 54 said Telenor Rest said they are satisfied with services of Zong. Write these results in percentages. Solution Persons who said Zong provide good internet speed = 120 – (24 + 36 + 54) = 6 Percentage of Jazz costumers = 2 5 120 100 = 20% Percentage of Ufone costumers = 3 36 12 0 100 = 30% Percentage of Telenor costumers = 9 5 54 120 10 6 0 = 45% Percentage of Zong costumers = 5 6 120 10 2 0 = 5%
CSS Primary Standard “Mathematics” 164 Q10. A company finds that % 1 4 4 of their cars made in the year 2012 to 2016 has some problem in ABS. The company made 28000 cars in this time duration. How many cars were defective? Solution Cars having problem = 1 4 4 % = 17 4 % Company made the cars = 28000 17 4 % of 28000 = 17 4 % × 2800 70 17 4 100 28000 = 1190 No. of defected cars = 1190 cars. Exercise 7d Q1. A book seller purchased 20 dozens of pens at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was is percentage profit? Solution Cost price of 1 dozen pens = Rs. 375 Selling price of 1 pen = Rs. 33 Selling price of 1dozen pens = 12 × 33 = Rs. 396 Profit = 396 – 375 = Rs. 21 Profit % = Profit 100 Cost price % = 21 100 375 = 5.6% Q2. 100 bananas are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. What is percentage profit or lose? Solution Cost price of 100 bananas = Rs. 350 Cost price of 1 banana = Rs. 350 100 = Rs. 3.5 Selling price of 1 dozen (12) bananas = Rs. 48
CSS Primary Standard “Mathematics” 165 Selling price of 1 banana = Rs. 48 12 = Rs. 4 Profit = selling price – cost price = 4 – 3.5 = Rs 0.5 Profit % = Profit 100 Cost price % = 0.5 100 3.5 % = 14.29% Q3. Anoral lost Rs. 28 by selling a hockey for Rs. 322. What was the loss percentage? Solution Sale price = Rs. 322 Lost = Rs. 28 Cost price = 322 + 28 = Rs. 350 Loss % = Loss 100 Cost price % = 28 100 350 % = 8% Q4. A used car was bought for Rs. 1,50,000 and sold for Rs. 1,20,000. Find loss percentage? Solution Loss = cost price – sale price = 1,50,000 – 1,20,000 = 30,000 Loss % = Loss 100 Lost price % = 20 30,000 1,50,000 100 5 % 20% Q5. At Ramdan bazar the price of dates was reduce 12% Rs. 440 per kg. What was the original price of dates? Solution
CSS Primary Standard “Mathematics” 166 After 12% reduction the price of 1 kg dates 440 Let original price is 10% After reduction price is (100 – 12)% = Rs. 88% 88% of price of dates = Rs. 440 % of price of dates = 440 88 100% of the price of dates = 440 100 80 % Original price of dates = Rs. 500 Q6. During summer mega sale on electricity brand reduce the prices of all its goods by 15%. Calculate the original selling price of the LCD, which was sold for Rs. 23800 during sale? Solution Let original selling price = 100% During sale 85% of selling price = Rs. 23800 1% of selling price = 23800 85 100% of selling price = 23800 100 85 % = Rs. 28000 Original selling price = Rs. 28000 Q7. Fid the marked price of jeans and t-shirt, given the amount of discount Rs. 49? Solution (i) Discount = 17% of total = 49 = 17 100 × total = 49 Total = 49 17 × 100 = 288.33 (ii) Discount = 20 33 3 % = 101 3 % 101 3 % of total = Rs. 270 101 300 × total = Rs. 270
CSS Primary Standard “Mathematics” 167 Total = 270 × 300 = 801.98 Week 7 Review Exercise 7 Q 1: Choose the correct answer and fill the circle: i. Which of the following equal to 4 25 ? 106 8 to 100 16% none ii. 21.8% is equal to: 218 to 100 0.218 0.218 1000 none iii. In a class, Ms. Shazia gave an A grade to 15 out of every 100 students and Ms. Nazia gave an A grade to 3 out of 20 students. What percent of each teacher’s student received an A? 15% 20% 70% none iv. What is percentage of 3 to 20? 35% 15% 30% 40% v. What is the percent of 7 8 ? 78.5 87.5 81.5 86.5 vi. Which of the following is equal to 36%: 7 25 5 8 6 25 none vii. Maria sold a bicycle for Rs. 24900 at a profit Rs. 600. Find the price at which she bouth. 24300 34000 24250 none viii. Danyal sold a book worth Rs. 850 at a loss of Rs. 180. Find the selling price of book? Rs. 700 Rs. 650 Rs. 670 none ix. If a chair was sold for Rs. 990 at the loss of Rs. 110. Find cost price of chair. Rs. 1200 Rs. 1100 Rs. 1250 Rs. 1150 Q2 If a 20 pound birthday cake of 3 flavor divide into 125 equal pieces. 30 are of chocolate, 20 are of strawberry, the remaining pieces are of coffee in the cake. What is the percentage of coffee flavor in cake? Solution
CSS Primary Standard “Mathematics” 168 The pieces of coffee flavor = 125 – (30 + 20) = 75 Percentage of coffee flavor = 15 4 75 125 100 25 % 60% Q3 In census of a village, there are 9500 people are resident, 7% of them were failed to provide the complete data. Calculate the number of people who give complete data. Solution People are resident = 9500 Number of people failed to provide data = 7% of total = 7 100 9500 = 665 Number of people give complete data = 9500 – 665 = 8835 Q4 The cost price of 20 Articles is same as the selling price of x article. If the profit is 25% then what is value of x? Solution Let cost of 1 article = Rs. 1 Cost of x articles = x Sale price of x articles = 20 Profit = Sale price – Cost price = 20 – x Profit% = Profit 100 Cost price % 25 = 20 – x 100 n 25x = 2000 – 100x 125x = 2000 x = 2000 125 x 16 Q5 In a medical store the profit is 320% of cost, if the lost increases by 25% but the selling price remain constant, approximately what percentage of selling price is profit? Solution Let Cost price be 100% Profit = 320%
CSS Primary Standard “Mathematics” 169 Profit = Sale price – Cost price Sale price = Profit + Cost price = 320 + 100 Sale price 420 If cost price increased by 25% The cost price = 100 + 25 = 25 Now profit = Sale price – Cost price = 420 – 125 = 295 Profit % of selling price = 295 100 420 70.23% Q6 What decimal of an hour in second? Solution 1 hours = 60 minutes 1 minute 60 seconds So 1 hour = 3600 seconds Decimal of an hour in second = 1 3600 = 0.000278 Q7 Fill in the blanks? (i) Cost price = Rs. 2400 Profit = Rs. 400 Sale price = ? Loss = ? As Profit = Sale price – Cost prie Sale price = Cost price + Profit 2400 + 400 Sale price = 2800 Loss = 0 (ii) Cost price = Rs. 1900 Loss = Rs. 300
CSS Primary Standard “Mathematics” 170 Sale price = ? Loss = Cost price – Sale price Sale price = Cost price – Loss = 1900 – 300 Sale price = 1600 Profit = 0 (iii) Sale price = Rs. 2900 Profit = Rs. 100 Cost price = ? Profit = Sale price – Cost price Cost price = Sale price – Profit = 2900 – 1300 Cost price = 2800 Loss = 0 (iv) Sale price = Rs. 1590 Loss = Rs. 60 Loss = Cost price – Sale price 60 = Cost price – 1590 Cost price = 60 + 1590 Cost price = 1650 Profit = 0 (v) Cost price = Rs. 4100 Profit = Rs. 300 Sale price = ? Profit = Sale price – Cost price Cost price = Profit – Cost price = 300 – 4100 = Rs. 4400 Loss = 0 (vi) Cost price = Rs. 1200 Loss = Rs. 180 Sale price = ?
CSS Primary Standard “Mathematics” 171 Loss = Cost price – Sale price Sale price = Cost price – Loss = 1200 – 180 = Rs. 1020 Profit = 0 Unit: 8 Interdiction of Algebra Teacher Objectives: ☻ To introduce a constant as a symbol having a fixed numerical value. ☻ To variable as a quantity which can take various numerical values. ☻ To literal as an unknown number represented by an alphabet. ☻ To algebraic expression as a combination of constants and variable connected by the sign of fundamental operations. ☻ To introduce polynomial as an algebraic expression in which the powers of variables are all whole numbers. ☻ To recognize a monomial, a binomial and a trinomial as a polynomial having one term, two terms and three terms respectively. ☻ To explain the addition of two or more polynomials. ☻ To explain the subtraction of a polynomial from other polynomial. ☻ To explain the product of monomial with monomial, monomial with binomial/trinomial and binomial with binomial/trinomial. ☻ To explain the simplification of algebraic expression involving addition, subractions and multiplication, subtraction and multiplication. ☻ To introduce and explain the verification of algebraic identities such as (x + a) (x + b) = x2 + (a + b) x + ab, (a + b)2 = (a + b) (a + b) = a2 + 2ab + b2 , (a – b)2 = (a – b) (a –b) = a2 – 2ab + b2 , a 2 – b 2 = (a – b) (a + b). ☻ To explain the factorization of an algebraic expression by algebraic. ☻ Identities. ☻ To explain the factorization of an algebraic expression by making groups. Learning Outcomes: ☻ Explain the term algebra as an extension of arithmetic in which letters replace the numbers. ☻ Know that: a sentence is a set of words making a complete grammatical structure and
CSS Primary Standard “Mathematics” 172 conveying full meaning. sentences that are true or false are known as statements. a statement must be either true or false but not both. a sentence that does not include enough information required to decide whether it is true or false is known as open statement (e.g., D + 2 = 9) a number that makes an open statement true is said to satisfy statement (e.g. D = 7 makes the statement D + 2 = 9 true). use English alphabet x in the open statement D + 2 = 9 to modify it to x + 2 = 9. ☻ Define variables as letters used to denote numbers in algebra. ☻ Know that any numeral, variable or combination of numerals and variables connected by one or more of the symbols “+” and “ –” is know as an algebraic expression (e.g. x+2y). ☻ Know that x, 2y and 5 are called the terms of the expression x+ 2y +5. ☻ Know that the symbol or number appearing as multiple of a variable used in algebraic term is called its coefficient (e.g. in 2y, 2 is the coefficient of y). ☻ Know that the number, appearing in algebraic expression, independent of a variable is called a constant term (e.g. in x + 2y + 5, number 5 is a constant terms). ☻ Differentiate between like and unlike terms. ☻ Know that: like terms can be combined to give a single term. addition or subtraction cannot be performed with unlike terms. ☻ Add and subtract given algebraic expressions. ☻ Simplify algebraic expressions grouped with brackets. ☻ Evaluate and simplify an algebraic expression when the values of variables involved are given. Teacher materials. CSS Primary Standard Mathematics Book 6. Writing Board. Marker. Eraser. Exercise 8a Classroom Activity: = 8x – 10 – 5x + 7 = (8 – 5) x + (–10 + 7) = 3x – 3
CSS Primary Standard “Mathematics” 173 Subtract 7a – 3 from 10a – 7 by vertical method. 10a – 7 7a – 3 – + 3a – 4 Q1. Tell whether these are sentence or words simply written together. (i) Red rose love. Simply word together (ii) Thanks God! Sentence (iii) Six is an odd number Sentence (iv) Baby sleep Word together (v) Urdu is our national language Sentence (vi) Water flows from lower level to higher level Sentence Q2. Find out whether the following statements are “true” “false” or “an open statement”. (i) 16 × 2 = 11 False (ii) 17 15 False (iii) 17x – 10 = 23 Open (iv) 6 × 4 = 24 True (v) 5 × 5 = 25 True (vi) 8 + 2 = y Open (vii) 4 2 False (viii) x + 14 = 17 Open (ix) (12 13)<6 True Q3. Put “>” “<”or “=” in the blanks to make sentence true. (i) 7x – 2x _>_4x (ii) 13xy + 10xy _>_22xy (iii) – 30x2 _<_30x2 (iv) 13 × 10 _=_130 (v) 18xyz + 23xyz _<_173xyz (vi) 180x2 y – 140x2 y_=_40x2 y (vii) 112xy + 114xy _>_203xy
CSS Primary Standard “Mathematics” 174 Exercise 8b Q1. Write down all the variables, constants and coefficient from the given algebraic expression. (i) x + 2y – 3y. Variables x,y,z Coefficients 1,2,–3 Constant (ii) 4x + 5y – 4. Variables x,y Coefficients 1,2 Constant –4 (iii) a + 4b – 3b2 . Variables a,b Coefficients 1,4, –3 Constant (iv) 3x + 4x2 – 16. Variables x Coefficients 3,4 Constant –16 (v) 17xy + 14x2 – 17x. Variables x,y Coefficients 17,–14,17 Constant (vi) 14x2 + 15z – 4x2 . Variables x,y Coefficients 14,15,–4 Constant (vii) 27 + 163 – 4x. Variables x Coefficients 16,–14 Constant 27 (viii) 15xyz + 3x3 + 1. Variables x,y,z Coefficients 15,3 Constant 1 Q2. Write down separately the terms of following algebraic expression.
CSS Primary Standard “Mathematics” 175 (i) 3a + b – 2c 3a, b, –2c (ii) –4xyz + 3xy + z –4xyz, –3xy, z (iii) 16x3 + 4xy –fgk 16x3 , 4xy, –fgk (iv) 4x – 1 x + 3x2 4x, –1 x , 3x2 (v) abc + 13fgh – af2 – bg2 – ch2 abc, 13fgh, –af2 , –bg2 , –ch2 (vi) 18 1 x – 14x2 y 2 –kgk2 18 1 x , 14x2 y 2 , –kgk2 (vi) –5abc + 7bcd + 3abd –5abc, 7bcd, 3abd Q3. Find the like terms in the following algebraic expression. (i) x + 3ny – 14x2 – 6xy 3xy, –6xy (ii) a 2 b + ab + b2 a + 2ab42 ab No like term (iii) a 2 – 2a2 b + 4a2 b – 5a2 a 2 , 5a2 –2a2 b, 4a2 b (iv) xyz + x2 y + 2xyz + 8x2 y xyz, 2xyz and x 2 y, 8x2 y (v) x + y + 4x – 3y x, 4x and y, 3y (vi) x 2 y + 2xy2 + 3x2 y + 5xy2 x 2 y, 3x2 y and 2xy2 , 5xy2 Q4. Translate the each of following word expression into an algebraic expression. (i) Five times the number x increased by six.
CSS Primary Standard “Mathematics” 176 5x + 6 (ii) Subtract 29 from half of 10. 1 2 b – 2a (iii) Sum of number 1 and square of x. 7 +x2 (iv) Five times a number which is 7 more than p. 5x = 7 + p (v) Total number of arranges in x crates where each crate contains y orange. xy Week 8 Exercise 8c Q1. Simplify the following by addition. (i) 3x + 4x + 10x + 6y. = 17x + 6y (ii) 18x2 – 10x2 + 13x – 4x. = 8x2 + ax (iii) 3a2 y + 4a2 y + 10a2 b. = 7a2 y + 10a2 b (iv) 13x2 y – 18x2 y + 14x2 y – 26x2 y. = –5x2 y – 12x2 y = – 17x2 y (v) 10x3 y 3 + 10y2 – 100y. = 10x3 y 3 + 10y2 – 100y (vi) ax + bx + cx + dx. = (a + b + c + d)x (vii) 88xyz + 10xyz – 10xy + 20xy = 98xyz + 10xy Q2. Find the sum of the following expressions. (i) 3x2 + 4x + 2, 4x2 – 5x + 4, 7x2 + 8x – 1. 3x2 + 4x + 2 4x2 – 5x + 4 7x2 + 8x – 1 14x2 + 7x + 5 (ii) –3abc, +12abc, –7abc. –3abc
CSS Primary Standard “Mathematics” 177 +12abc –7abc 2abc (iii) 2m 2 + mn + n2 , 3m2 – 3mn + 4n2 , –m 2 + mn – 2n2 . 2m 2 + mn + n2 3m2 – 3mn + 4n2 –m 2 + mn – 2n2 4m2 – mn + 3n2 (iv) p 2 + 10pq + 3q2 , 2p2 – 3pq + q2 , 3p2 + pq – 2p2 . p 2 + 10pq + 3q2 2p2 – 3pq + q2 3p2 + pq – 2p2 6p2 – 8pq + 2q2 (v) 17a2 – ab + b2 , 8a2 – 3ab +2b2 , 4a2 + 8ab + 4b2 . 17a2 – ab + b2 8a2 – 3ab +2b2 4a2 + 8ab + 4b2 29a2 – 4ab + 27b2 (vi) 3x4 – 4x2 + 5x + 1, x4 – 2x2 +3x + 4, 4x4 + 3x2 + 4x – 5. 3x4 – 4x2 + 5x + 1 x 4 – 2x2 +3x + 4 4x4 + 3x2 + 4x – 5 8x4 – 5x2 + 6x + 0 Q3. Find the subtract of the following expressions. (i) –4x from x + 1. x + 1 –4x 5x + 1 (ii) 5x2 – 4x3 + 4x2 – 3x from 1 5x2 – 4x2 + 4x3 – 3x from 1 9x2 + 4x3 – 3x from 1 _9x2 4x3 3x –9x2 – 4x3 + 3x + 1 4x3 – 9x2 + 3x + 1 (iii) a 2 b + 2b2 c + ac2 from 4a2 b – 2b2 c + 6ac2
CSS Primary Standard “Mathematics” 178 4a2 b – 2b2 c +6ac2 _a 2 b 2b2 c ac 2 3a2 b – 4b2 c + 5ac2 (iv) x 2 + 4xy from 3x2 + 4xy 3x2 + 4xy _x 2 4xy 2x2 (v) 2x + 3 from –x –x _2x 3 –3x –3 (vi) 3x4 + 15xy – 13xyz from 7x4 – 10xy + 10xyz 7x4 – 10xy + 10xyz _3x4 15xy 13xyz 4x4 – 25xy + 23xyz (vii) 18x2 yz – 13xy – 10xy from 10x2 yz + 10xy 18x2 yz – 3xy from 10x2 yz + 10xy 10x2 yz + 10xy _18x2 yz 3xy –8x2 yz + 13xy Q4. Simplify the following expressions. (i) (a + b) + (2a – b) = a + 2a + b – b = 3a + 0 = 3a (ii) (x2 y + 4) – (13x2 y + 14y2 – 6) = x 2 y + 4 – 13x2 y – 14xy2 + 6 = x 2 y – 13x2 y – 14y2 + 4 + 6 = –12x2 y – 14y2 + 10 (iii) (14a2 – 3a) – {(–3a + 6a2 ) – (29 – 8a2 )} = (14a2 – 3a) – {–3a + 6a2 – 29 + 8a2 } = 14a2 – 3a + 3a – ba2 + 29 – 8a2 = 14a2 – 6a2 – 8a2 – 3a + 3a + 29 2 2 = 14a – 14a – 3a + 3a + 29 = 29 (iv) {(16m2 – 3n) – (2m2 + 5n)}
CSS Primary Standard “Mathematics” 179 16m2 – 3n – 2m2 – 5n = 16m2 – 2m2 – 3n – 5n = 14m2 – 8n (v) (10x2 yz + 2yz) – [3yz + {2x2 y + z + 5y2 – (2yz + 6)}] = (10x2 yz + 2yz) – [3yz + {2x2 y + z + 5y2 – 2yz + 6}] = 10x2 yz + 2yz – 3yz – 2x2 y + z + 5y2 + 2yz + 6 = 10x2 yz + 2yz – 3yz + 2yz – 2x2 y + 5y2 – 5y2 – z + 16 = 10x2 yz + yz – 2x2 y – 5y2 – z + 16 Exercise 8d Q1. If x = 2, evaluate the following. (i) x + 6 Put x = 2 = 2 + 6 = 8 (ii) 16x2 Put x = 2 = 16(2)2 = 16(4) = 64 (iii) 2x + 1 4x = 2(2) + 1 4(2) = 24 + 1 8 = 33 8 (iv) 14x2 – 13x = 14(2)2 – 13(2) = 14(4) – 26 = 30 (v) 16 x + 18 x = 16 2 + 18 2 = 8 + 9 = 17 Q2. Evaluate the following for x = 1 y = 2. (i) 16x2 y + 4y2 . Put x = 1, y = 2 = 16(1)2 (2) + 4(2)2
CSS Primary Standard “Mathematics” 180 = 16(1) + 4(4) = 16 + 16 = 32 (ii) x + y + 14xy. Put x = 1, y = 2 = 1 + 2 +14(1)(2) = 1 + 2 + 28 = 31 (iii) 3x – 4xy + 15x2 . = 3(1) – 4(1)(2) + 15(1)2 = 3 – 8 + 18 = 18 – 8 = 10 (iv) 2x – 1 5 y. = 2(1) – 1 5 (2) = 2 – 2 5 = 10 2 5 = 8 5 = 1.6 (v) 10x 2 – xy + y2 . = 10(1)2 – (1)(2) + 22 = 10 – 2 + 4 = 12 Q3. Evaluate the following for x = 2, y = 3, z = –1. (i) 2x3 + 4xy – 3z. = 2(2)3 + 4(2)(3) – 3(–1) = 2(8) + 4(6) + 3 = 16 + 24 + 3 = 43 (ii) 4x3 – 2y2 – 2z. = 4(2)3 – 21(3)2 – 2(–1) = 4(8) – 2(9) + 2 = 32 – 18 + 2 = 16 (iii) 15x2 + 10z – 4xz –1 2y. = 15(2)2 + 10(–1) – 4(2)(–1) + 2 (z) = 15(4) + 10 + 8 + 6 = 60 – 10 + 8 + 6
CSS Primary Standard “Mathematics” 181 = 64 (iv) 41x + 10xy – 4xz – 17xyz. = 41(2) + 10(2)(3) – 17(2)(3)(–1) = 82 + 10(6) + 17(6) = 82 + 60 + 102 = 244 (v) 14xyz – 4z + 10xyz. = 14(2)(3)(–1) – 4(–1) + 10(2)(3)(–1) = –14(6) + 4 – 10(6) = –84 + 4 – 60 = –140 (vi) 105x + 14xy – 100y. = 105(2) + 14(2)(3) – 100(–1) = 210 + 14(6) + 100 = 210 + 4 + 100 = 394 (vii) 10x2 y + 10z – 18z2 . = 10(2)2 (3)2 + 10(–1) – 18(–1)2 = 10(4)(9) + 10 – 18 = 10(36) + 10 – 18 = 360 + 10 – 18 = 332 Q4. If a = 1, b = 2, c = –1. (i) 4abc. = 4(1)(2)(–1) = –8 (ii) 2a + 2b2 – 3c. = 2(1) + 2(2)2 – 3(–1) = 2 + 2(4) + 3 = 2 + 8 + 3 = 13 (iii) a 2 – 2b2 + 2c. = (1)2 –2(2)2 + 2(–1) = 1 – 2(4) –2 = 1 – 8 – 2 = –9 (iv) (a + b)2 – (b – c)2 . = (1 + 2)2 – (2 + 1)2
CSS Primary Standard “Mathematics” 182 = 9 – 9 = 0 (v) 3a 2 – 14c2 . = 3(1)2 – 14(–1)2 = 3(1) – 14(1) = 3 – 14 = –11 (vi) 19b2 + 16ab – 3c2 . = 19(2)2 + 16(1)(2) – 3(–1)2 = 19(4) + 16(2) – 3 = 76 + 32 – 3 = 105 Review Exercise 8 Q1. Choose the correct answer and fill the circle. i. What is added to 2a3 – 2a2 + 3a – 1 to get a3 + 3a2 – 3a + 1? –a 3 + ba2 – 6a + 2 –2a3 + 3a2 – 3a + 2 –a 3 – ba2 + 6a + 2 4a3 + 3a2 – 3a + 2 ii. 4x + 3y is an algebraic: expression sentence equation term iii. x 2 yz and xy2 z 2 are: constants like terms unlike terms sentence iv. Z is called: constant variable coefficient expression v. The coefficient in 4x2 is: 14x2 n 2 4 2 vi. 4n in this expression how many terms are used: 13 3 6 1 vii. If x = 6, y = 5, the value of x + 2y is: 12 14 16 18 viii. The sum of 3a2 + 5a2 + 7a2 is: 10a2 12a2 15a2 20a2 ix. 4x is which type of algebraic: polynomial monomial binomial trinomial Q2. Write T or F if statement is true or false respectively. (i). 13<6 is true statement. T (ii). 3x2 and 3x are like terms. F
CSS Primary Standard “Mathematics” 183 (iii). The exponent in 3x2 + 1 is 1. F. (iv). 16x + 14y = 20 is false statement. F (v). In open statement, the sentence must be true. T Q3. Evaluate each of following algebraic expression. (i). 13x – 4y + 2z when x = 2, y = –3, z = –5 = 13(2) – 4(–3) + 2(–5) = 26 + 12 – 10 = 38 – 10 = 28 (ii). 18(x – 2y) when x = 5, y = z = 18(5 – 2(3)) = 18(5 – 6) = 18(–1) = –18 (iii). 4 5 a – 5 8 b when a = 10, b = 16 2 2 4 5 (10 5 ) – 8 (16 ) = 8 – 10 = –2 (iv). 3 9 b + 18 3 c 2 when b = 27, c = 6 3 18 2 27 6 9 3 ( ) ( ) 3 12 3 9 ( 27 18 ) 3 ( 36 ) = 9 + 216 = 225 (v). x + 14xy – z 2 where x = 2, y = 1, z = 2 = 2 + 14(2)(1) – (2)2 = 2 + 28 – 4 = 26 Q4. Translate each of the following algebraic expression into words. (i). 3a 3 times of a. (ii). 14a – 16
CSS Primary Standard “Mathematics” 184 Sum of 14 times of a and b or 14 times a is increased by b. (iii). 3x + 4y Sum of 3 times a and 4 times y. (iv). (4x + 6) 5 Four times x increased by 6, divided by 6. (v). 1(a + 4) A is increased by 4 (vi). 13x + 14y + 13 13 times x is increased by 14 times y and thirteen. Q5. Find the sum of following. (i). 4x2 + 4x – 3x, 12x + 4x, 3x3 – 10x 4x2 + x, 16x, 3x2 + 10x 4x2 + x 16x 3x2 – 10x 7x2 + 7x (ii). 18z2 – 14z, 10z3 + 15y – 2z2 , 4z3 – 10z 18z2 – 14z 10z3 + 15y – 2z2 4z3 – 10z 14z3 + 16z2 – 9z (iii). 100p2 – 14p +13, 50p2 + 28p, 14p3 – 10p2 100p2 – 14p +13 50p2 + 28p 14p3 – 10p2 14p3 + 140p2 + 14p + 13 (iv). 18a2 – 14a2 , 13a4 – 11a3 + 100a2 , a 3 + 10a2 18a2 – 14a2 13a4 – 11a3 + 100a2 a 3 + 10a2 13a4 + 8a3 + 96a2 (v). 117q3 + 14q2 – 10q, 100q4 – 3q2 , 10q2 – 14q 117q3 – 14q2 – 10q 100q4 – 3q2 10q2 – 14q
CSS Primary Standard “Mathematics” 185 100q4 + 117q3 + 21q2 – 24q (vi). 71s2 – 14s – 14, 18s2 – 10s + 11s, 118s2 – 14s + 14 71s2 – 14s – 14, 18s2 + s, 118s2 – 14s + 14 71s2 – 14s – 14 18s2 + s 118s2 – 14s + 14 207s2 – 27s (vii). 118x3 – 174x2 , 148x4 – 114x3 + 100x2 , 114x + 10x2 118x3 – 174x2 148x4 – 114x3 + 100x2 –114x + 10x2 148x4 + 4x3 – 184x2 + 114x Q6. Simplify the following. (i). (3 + x) + (2x – x) = 3 + x + 2z – x = 3 + x – x + 2z = 3 + 2z (ii). (p2 q + 4) – (13p2 q + 14p2 – 6) = p 2 q + 4 – 13p2 q – 14p2 + 6 = p 2 q – 13p2 q – 14p2 + 4 + 6 = p 2 q – 13p2 q – 14p2 + 10 = –12p2 q – 14p2 + 10 (iii). –2[3c –{d – 2(c + d)}] = –2[3c – {d – 2c – 2d}] = –2[3c – d + 2c + 2d] = –2[3c + 2c – d + 2d] = –2[5c + d] = 10c – 2d (iv). (14x2 – 3x) – (–3x + 6x2 ) = 14x2 – 3x + 3x – 6x2 = 14x2 – 6x2 – 3x + 3x = 8x2 (v). (16a2 – 3b) – (–3a2 + 5b) = 16x2 – 3b + 3a2 – 5b = 16a2 + 3a2 – 3b – 5b = 19a2 – 8b (vi). (10p2 qr – 2pq) – (2p2 qr + 5q2 – 2qr + 6)
CSS Primary Standard “Mathematics” 186 = 10p2 qr – 2pq – 2p2 qr – 5q2 + 2qr – 6 = 10p2 qr –2p2 qr – 2pr – 5q2 + 2qr – 6 = 8p2 qr – 2pq –5q2 + 2qr – 6 (vii). 10{q2 + 3r(p2 + r) + 3(r2 + rp2 )} = 10{q2 + 3rp2 + 3r2 + 3r2 + 3rp2 } = 10{q2 + 3rp2 + 3rp2 + 3r2 + 3r2 } = 10{q2 + 6rp2 + 6r2 } = 10q2 + 60rp2 + 60r2 (viii). (17x2 – 2x2 – 3xy2 ) – (2x2 + 12x2 – xy 2 ) = (5x2 – 3x2 y 2 ) – (14x2 – xy 2 ) = 5x2 – 3xy2 – 14x2 – xy 2 = 5x2 – 14x2 – 3xy2 + xy2 = –9x2 – 2xy2 Unit: 9 Linear Equation Teacher Objective: ☻ To introduce an algebraic equation. ☻ To explain the difference between algebraic expression and algebraic equation. ☻ To explain the construction of linear expression and linear equation with one variable. ☻ To explain the solution of simple linear equation involving fractional and decimal coefficients. ☻ To explain the solution of real life problems involving linear equation. Learning Outcomes: At the end of this Unit, the students will be able to: ☻ Define an algebraic equation. ☻ Differentiate between algebraic equation and algebraic equation. ☻ Define linear equation in one variable. ☻ Construct linear expression and linear equation in one variable. ☻ Solve simple linear equations involving fractional and decimal coefficient like 1 2 x + 5 = x – 1 3 ☻ Solve real life problems involving linear equations.
CSS Primary Standard “Mathematics” 187 Teaching Material: CSS primary standard Mathematics Book 7. Writing Board. Marker. Eraser. Exercise 9a Q1. Write each sentences an algebraic equation. (i) Twice a number, decreased by twenty nine, is seven. 2x – 2a = 7 (ii) Thirty two is twice a number increased by 8. 32 = 2x + 8 Or 2x + 8 = 32 (iii) Twelve is sixteen less than four times a number. 12 = 4x – 16 Or 4x – 16 = 12 Q2. Write each an algebraic equation as sentence. (i) x + 13 = 24. A number increased by thirteen is twenty four. (ii) 25b = 200 Twenty five times a number is two hundred. (iii) y + 43 = 82 A number increased by forty thee is eighty two. (iv) z – 18 = 43 A number decreased by eighteen is forty three. (v) 350 – d = 280 Three hundred fifty decreased by a number is two hundred eighty. (vi) 15 p = 3 Fifteen divided by a number is three. (vii) p + 10 = 40 A number increased by ten is forty. (viii) 8–1 p = 4 Eight decreased by one divided by a number is four. Q3. Which of the following equations are true or false for given value of variables?
CSS Primary Standard “Mathematics” 188 (i) 3y –7 = 0 where y = 7 3 7 3 3 –7 = 0 7 – 7 = 0 0 = 0 True (ii) 4 + m = 1 where m = 9 4 + 9 = 1 13 1 False (iii) 4 5 h – 1= –1 5 where h = 1 4 5 (1) – 1 = –1 5 4 5 – 1 = –1 5 4 –5 5 = –1 5 True (iv) 3 2 z – 5 = 2z + 4 where z = 8 4 3 (8) 2 – 5 = 2(8) + 4 12 – 5 = 16 + 4 7 + 10 False (v) f + 8 = 5f – 9 where f = 6 6 + 8 = 5(6) – 9 14 21 False Exercise 9b Q1. Solve the following equations to satisfy the equation. (i) 13x – 10 = 29 Step-1 13x – 10 = 29 13x – 10 + 10 = 29 + 10 (Add 10 on both sides) 13x = 39 Step-2 13x 13 = 29 13 (Divide by 13 on both sides) x = 3 (ii) 29x – 50 = 95
CSS Primary Standard “Mathematics” 189 Step-1 29x – 50 = 95 29x – 50 + 50 = 95 + 50 (Add 50 on both sides) 29x = 145 Step-2 29x 29 = 5 145 29 (Divide by 13 on both sides) x = 5 (iii) 4 + y = 3y Step-1 4 + y = 3y – y (Subtracting by on both sides) 4 = 2y Step-2 2y 2 = 4 2 (Divide by 2 on both sides) y = 2 (iv) 17 – y = 10 Step-1 17 – y = 10 17 – y – 17 = 10 – 17 (Subtracting by 17 on both sides) – y = –7 Step-2 –y 7 = –7 7 (Divide by –7 on both sides) y = 7 (v) 14 + z = 12 Step-1 14 + z – 14= 10 – 14 z = –2 (vi) 100 + z = 2z Step-1 100 + z – z = 2z – z (Subtracting by 3 on both sides) 100 = z Z = 100 Q2. Which of the following equation are true or false? For the given value of the variables. (i) 4x + 1 = 9 for x = 2 4x + 1 = 9 4(2) + 1 = 9 8 + 1 = 9 9 = 9 True (ii) 13x – 1 = 14 for x = 10 13(10) – 1 = 14 130 – 1 = 14 129 14 False
CSS Primary Standard “Mathematics” 190 (iii) 13x + 4 = 13 for x = 3 13(3) + 4 = 13 39 + 4 = 13 43 13 (iv) 18x + 10 = 46 for x = 11 18(11) + 10 = 46 198 + 10 = 46 208 46 False (v) 20x – 10 = 30 for x = 10 20(10) – 10 = 30 200 – 10 = 30 190 30 False (vi) 180x + 10 = 70 for x + 10 180(10) + 10 = 70 1800 + 10 = 70 1810 70 False Exercise 9c Q1. Write any equivalent equation of the following equations. (i) x + 4 = 0 2x + 8 = 0 (Multiplying both sides by 2) (ii) 2x – 10 = 0 4x – 20 = 0 (Multiplying both sides by 2) (iii) x – 42 – 9 3x – 12 = –27 (Multiplying both sides by 3) (iv) 4x – 6 = 6 + 4 8x – 12 = 2x + 8 (Multiplying both sides by 2) (v) 8x + 2 = 4x 16x – 4 = 8x (Multiplying both sides by 2) Q2. Solve the following equation and find the value of variables. (i) 0.8x + 0.6 = 1.6x – 1.4 0.8x + 0.6 + 1.4 = 1.6x – 1.4 + 1.4 (Add by 1.4 on both sides) 0.8x + 2 = 1.6x 0.8x + 2 – 1.6x = 1.6x – 1.6x (Subtract by 1.6 on both sides) (0.8 – 1.6)x + 2 = 0 – 0.8x + 2 = 0
CSS Primary Standard “Mathematics” 191 0.8x 2 0.8 0.8 (Divide by 0.8 on both sides) X = 2.5 (ii) 2x – 10 = 4 2x – 10 + 10 = 4 + 10 (Add by 10 on both sides) 2x = 14 2x 14 2 2 (Divide by 2 on both sides) x = 7 (iii) 3x + 27 = 33 3x + 27 – 27 = 33 – 27 (Subtract by 21 on both sides) 3x = 6 3x 6 3 3 (Divide by 2 on both sides) x = 2 (iv) 4x – 16 = –36 4x – 16 + 16 = 36 + 16 (Add by 16 on both sides) 4x = 20 4x 20 4 4 (Divide by 4 on both sides) x = –5 (v) 4x – 6 = 2x + 2 4x – 6 + 6 = 2x + 2 + 6 (Add by 2 on both sides) 4x = 2x + 8 4x – 2x = 2x + 8 – 2x (Subtract by 2x on both sides) 2x = 8 2x 8 2 2 (Divide by 2 on both sides) x = 4 (vi) 1.4x + 6.6 = 3.6x 1.4x + 2.4 + 4.2 = 3.6x – 4.2 + 4.2 (Add by 4.2 on both sides) 1.4x + 6.6 = 3.6x 1.4x – 6.6 – 1.4x = 3.6x – 1.4x (Subtract by 1.4 on both sides) 6.6 = 2.2x 6.6 2.2x 2.2 2.2 (Divide by 2.2 on both sides) 3 = x x = 3
CSS Primary Standard “Mathematics” 192 (vii) 4x 2 – 5 3 = 2x + 4 12x–10 15 = 2x + 4 12x – 10 = 15(2x + 4) (Multiply by 15 on both sides) 12x – 10 = 30x + 60 12x – 10 – 60 = 30x + 60 – 60 (Subtract by 60 on both sides) 12x – 70 = 30x 12x – 70 – 12x = 30x – 12x (Subtract by 12x on both sides) –70 = 18x –70 18x 18 18 (Divide by 18 on both sides) –35 9 = x x = –35 9 (viii) 4 2 x– 5 3 = 2x + 4 12x–20 15 = 2x + 4 12x – 20 = 15(2x + 4) (Multiply by 15 on both sides) 12x – 20 = 30x + 60 12x – 20 – 60 = 30x + 60 – 60 (Subtract by 60 on both sides) 12x – 80 = 30x 12x – 80 – 12x = 30x – 12x (Subtract by 12x on both sides) –80 = 18x –80 18x 18 18 (Divide by 18 on both sides) –40 9 = x x = –40 9 (ix) 4 3 1 1 x – x 5 5 5 5 4x – 3 4 1 5 5 4x – 3 = x + 1 (Multiplying both sides by 5)
CSS Primary Standard “Mathematics” 193 4x – 3 + 3 = x + 1 + 3 (Add 3 on both sides) 4x = x + 4 4x – x = x + 4 – x (Subtract x on both sides) 3x = 4 3x 4 3 3 (Divide by 3 on both sides) x = 4 3 (x) 6 3 1 1 1 x 10 6 x 5 5 5 5 11 3 51 31 x x 5 5 5 5 11x 3 51 31x 5 5 11x + 3 = 51 + 31x (Multiplying by 5 on both sides) 11x + 3 – 51 = 51 + 31x – 51 (subtract by 51 on both sides) 11x – 48 = 31x 11x – 48 – 11x = 31x – 11x (Subtract by 11x on both sides) 48 = 20x –48 20x 20 20 (Divide by 20 on both sides) –12 5 = x x = –12 5 (xi) 1.6 – 2x = 1.4 1.6 – 2x – 1.6 = 1.4 – 1.6 (Subtract by 1.6 on both sides) –2x = –0.2 –2x 0.2 –2 –2 (Subtract by –2 on both sides) x = +0.1 Q3. A women is now three times as old as his son. In 10 years’ time, the sum of their ages will be 76. How old was the women when her son was born? Solution Let son’s age = x Then women’s age = 3x
CSS Primary Standard “Mathematics” 194 After 10 years Son’s age = x + 10 Women’s age = 3x + 10 According to given condition (x + 10) + (3x + 10) = 76 x + 3x + 10 + 10 = 76 4x = 20 = 76 4x = 76 – 20 4x = 56 x = 56 4 x = 14 Then women’s age = 3x = 3(14) = 42 Q4. Bilal buy a novel and note book and pays Rs. 00 for both. The moved costs Rs. 80 more than twice the cost of note book. Find the cost of novel? Solution Let cost of note book = x Then cost of novel = 2x + 80 According to given condition x + 2x + 80 = 500 3x + 80 = 500 3x = 500 – 80 3x = 420 x = 420 3 x = 140 Now the cost of novel = 2x + 80 = 2(140) + 80 = 280 + 80 = Rs. 360 Q5. The length of a rectangular pool is 6cm more than its width and the perimeter of rectangle is 36cm. What is its length and width? Solution Let width = x Then length = x + 6
CSS Primary Standard “Mathematics” 195 Perimeter = 36 2(2x + 6) = 36 2x + 6 = 18 2x = 18 – 6 2x = 12 x = 6 Hence width = x = 6cm Length = x + 6 = 6 + 6 = 12cm Q6. Two friends have Rs. 8000. If one of them had four times as much as the other, how much money did each of them have? Solution Let share of one = x Then share of other = 4x According to given condition x + 4x = 8000 5x = 8000 x = 8000 5 x = 1600 Hence share of one = x = 1600 share of other = 4x = 4(1600) = 6400 Q7. Age of a father is 3 times the age of his daughter. If father is 18 years old then his daughter born. Then find the age of father and daughter? Solution Let daughter’s age = x Then father’s age = 3x At time of birth, the difference between their ages are is i.e 3x – x = 18 2x = 18 x = 18 2 x = 9 Hence daughter’s age = x = 9 years father’s age = 3x = 3(9) = 27 years Review Exercise 9 Q1. Choose the correct answer and fill the circle.
CSS Primary Standard “Mathematics” 196 i. A mathematical sentence with an equality sign is called: algebra expression equation equality ii. A/an ______ is a phase: sentence variable expression symbol iii. When left side is equal to right side,______is completed: equation sentence linear equation none iv. Seven times a number increased b 2: 2x + 7 7x + 2 6 – 2 7 none v. Three forth of a number subtracted from 6: 6 – 3 4 x 4 – 6 3 3 4 3 4 – 6x all vi. A number divided by 6 decreased by 5: x 5 – 6 6 x – 5 x 5 + 6 all vii. Five less than three times a number is 46: 5 – 3x = 46 3x – 5 = 46 5x = 46 3 none of these viii. Twice number is 28: x + 2 = 28 28 = x 2x = 28 2 2x = 28 ix. Equation 2m – 9 = 3 for m = ? 1 3 5 6 Q2. What should be the value, which balance the following equations. (i). 28 2 a – a = 1 9 5 a – a = 1 9a – 5a 5 = 1 4a 5 = 1 4a = 5 × 1 (Multiplying by 5 on both sides) 4a = 5 a = 5 4 (Divide by 4 on both sides) (ii). 6a – 3 = 15a 6a – 3 – 6a = 15a – 6a (Subtract by 6a on both sides)
CSS Primary Standard “Mathematics” 197 –3 = 9a –3 = 9a 9 –3 3 = a a = –3 3 (iii). 0.4a + 9.5 = 9.99 0.4a + 9.5 – 9.5 = 9.99 – 9.5 (Subtract by 9.5 on both sides) 0.4a = 0.49 a = 0.49 0.4 = 1.225 (iv). 1.65a + 385 = –a 1.65a + 3.85 + a = –a + 9 (Add a on both sides) (1.65 + 1)a + 3.85 = 0 2.65a + 3.85 = 0 2.65a + 3.85 – 3.85 = 0 – 3.85 2.65a = –3.85 a = – 3.85 2.65 = –1.436 Q3. Solve the each of the following equations. (i). 8x + 13 = 0 8x + 13 – 13 = 0 – 13 8x = –13 x = –13 8 (ii). 3 4 5 x – 5 = 0 23 5 x – 5 = 0 23 5 x – 5 + 5 = 0 + 5 23 5 x = 5 23 5 x × 5 = 5 × 5 (Multiply by 5 on both sides) 25x = 25
CSS Primary Standard “Mathematics” 198 x = 25 23 (iii). 5(x – 4) = 3x + 2 5x – 20 = 3x + 2 5x – 20 + 20 = 3x + 2 + 20 5x = 3x + 22 5x – 3x = 3x + 22 – 3x 2x = 22 2x 22 2 2 x = 11 (iv). 2x + 0.05 = 0.8x + 1.25 2x + 0.05 – 0.05 = 0.5x + 1.25 – 0.05 2x = 8x + 1.2 2x – 8x = 8x + 1.2 – 8x (2 – 0.8)x = 1.2 1.2x = 1.2 1.2x 1.2 1.2 1.2 x = 1 (v). 5 2 n = 4 3 + 2n 5 2 n – 2n = 4 3 + 2n – 2n 5 –2 2 n = 4 3 5–4 2 n = 4 3 1 2 n = 4 3 n = 4 3 × 2 n = 8 3 (vi). 12n + 32 = – 4 12n + 32 – 32 = –4 –32 12n = –36
CSS Primary Standard “Mathematics” 199 12n 12 = –36 12 n = –3 Q4. Area of right angled triangle is 1 2 b × c square meter. If the area is equal to 15 square meter b = 5 then c = ?. Solution As Area = 1 2 b × c 15 = 1 2 5 × c 15 × 2 = 1 2 5 × c × 2 30 = 5c 30 5 = 5c 5 6 = c c = 6 Q5. Find the number where five times when added to 12 give the result 17. Solution Let the required number x According to given condition 5x + 12 = 17 5x + 12 – 12 = 17 – 12 5x = 5 5x 5 = 5 5 x = 1 Q6. Find three consecutive in which x is middle whose sum is 99. Find the number. Solution If the middle number = x Then first number = x – 1 3 rd = number = x + 1 Now according to given condition x – 1 + x + x + 1 = 99
CSS Primary Standard “Mathematics” 200 3x = 99 3x 3 = 99 3 x = 33 Now first number = x – 1 = 33 – 1 = 32 Middle number = x = 33 Tird number = x + 1 = 33 + 1 = 34 Q7. Find the three consecutive even number where sum is 54. Solution Let first even number = 2x 2 nd even number 2x + 2 3 rd even number = 2x + 4 According to given condition Sum of consecutive even number = 54 2x + 2x + 2 + 2x = 4 = 54 6x + 6 = 54 6x = 54 – 6 6x = 48 x = 48 6 = 8 x = 8 Now first even number = 2x = 2(8) = 16 2 nd even number = 2x + 2 = 2(8) + 2 = 16 + 2 = 18 3 rd even number = 2x + 4 = 2(8) + 4 = 16 + 4 = 20 Unit: 10 Geometer Teaching Objectives