The Essence of Mathematics Skills Textbook- Math Primary_2

51 5 is called quotient. Hence, dividend ÷ divisor = quotient Relationship of Multiplication and Division Piles of limes 3 piles Each pile has 4 limes Total limes 12 limes The symbolic expression is 3× 4 = 12. Total limes 12 limes Total limes 12 limes Divided into 3 piles Each pile has 4 limes Each pile has 4 limes Piles of limes 3 piles The symbolic expression is 12 The symbolic expression is 12 ÷ 3 = 4. ÷ 4 = 3. Multiplicand multiplier product dividend divisor quotient 3× 4= 12 12 ÷ 3= 4 12 ÷ 4= 3 dividend divisor quotient From the example above, it can be seen that there is a relationship between multiplication and division, i.e. 1. Multiplication can be changed into division as follows: 1.1 Multiplication changes to division when - multiplicand becomes divisor or quotient - multiplier becomes quotient or divisor

52 - product becomes dividend 1.2 Multiplication expression can change to division expression 2. Division can change back to multiplication as follows: dividend ÷ divisor = quotient divisor × quotient = dividend Relationship between multiplication and division 1. Check if the quotient is correct by using: divisor × quotient = dividend Example 10 ÷ 2 = Method: 10 ÷ 2 = 5 Answer: 5 Checking the answer, 5 × 2 = 10; thus, the answer is correct. 2. Find the quotient more conveniently and quickly by using the times table or reciting multiplication. Example 15 ÷ 5 = Method: 15 ÷ 5 = 3 Answer 3 Checking the answer, 5 × 3 = 15; thus, the answer is correct. Analysis Divisor × Quotient = Dividend From the times table, 5 × 3 = 15. Hence, 15 divided by 5 results in the quotient of 3. How to use the multiplication times table to find the quotient In the top row of the multiplication times table which is the multiplicand, change it to the divisor. At the left-most of the multiplication times table which is the multiplier, change it to the divisor. In the multiplication times table, the number in the table which is the product, change it to the dividend.

53 multiplicand quotient × 1 2 3 4 5 6 7 8 9 10 11 12 1 2 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 divisor 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 multiplier 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144 Example: 15 ÷ 3 =  Approach: Step 1: Look for number 3 at the leftmost column. Step 2: From 3, follow the same row horizontally from left to right until number 15. Step 3: From 15, follow up the column until the top row to find Example: 4n2um÷ b7er=5. Approach: SHteenpc1e:, Lthoeokqufootriennutmobfe1r57÷in3 tihse5leftmost column. Step 2: From 7, follow the same row horizontally from left to right until number 42. Step 3: From 42, follow up the column until the top row to find Likewise, to afibnoHndvuetemnh.cebeeq,rut6ho.etieqnutostioefnot tohfer4p2a÷ir7siosf6.numbers, use the same approach as the example Forms of division Long division Short division 3. 1. 4 ) 96 96 ÷ 4 = 24 24 2. 4. 96 24 4 = 24 4 ) 96 80 4 × 20 16 16 4 × 4 00

54 Division with a dividend that is up to three-digit in number and the quotient has no remainder. When the divisor is a single digit number: Example: What is the result of 184 ÷ 8 ? Method 1: Use the long division (the symbol for long division is “ “) The division expression is 184 ÷ 8 =  Method: Analysis: 23 8 × 20 1. 20 is the largest number that 8)184 multiplies by 8 and the product is not over 184. 160 2. Subtract the product 8 × 20 which is 24 160 from 184. The remainder is 24. 2 4 8× 3 3. 3 is the number that when multiplied by 8one gets the exact number 24. 00 Subtract 24 from 24 which is the dividend, 0 is obtained. Answer: 23 Hence, the quotient altogether is 23. Checking the answer, 8× 23 = 184 Thus, the answer is correct. Method 2 Use the long division (shortcut) The division expression is 184 ÷ 8 =  Analysis: This method uses the divisor to divide the dividend digit by digit. Method: 023 1. Divide 1 which is the greatest value Answer: 8 ) 1 8 64– digit by 8 first. Since 8 is greater than 1 1, combine it with the next digit, 24 resulting in 18. 2 4– 2. Divide 18 by 8. From the times table, 00 8 × 2 = 16 which is the closest number to 18 and not exceeding 18. Place the 23 quotient of 2 over 8 which is the ten‟s Checking the answer 8× 23 = 184. Thus, place of the dividend. Subtract 16 the answer is correct. from 18, the remainder is 2. Take 4 down vertically to be 24. 3. Divide24by 8. From the times table, 8 × 3 = 24. Thus, the quotient is 3. Place 3 over 4 which is the dividend, then subtract 24 from 24, resulting in 0. Hence, the quotient altogether is 23.

55 Method 3 Use the short division (the symbol for short division is “ ) “) The division expression is 184 ÷ 8 =  Analysis The analysis process is the same as for Method: 8 ) 1 8 4 the long division method and either 023 method 1 or 2 can be adopted as convenient. If learners can recite the Answer: 23 times table accurately and understand the Checking the answer 8× 23 = 184; thus, the answer is correct. long division well, mental math can be used and the quotient can be found quickly. When the divisor is a two-digit number, the suitable method is as follows: Example Find the result of 7,936 ÷ 31. Method 1: Long division Method 2: Long division (shortcut) The division expression is 7,936 ÷ 31 = The division expression is 7,936 ÷ 31 =   Method: Method: Answer: 256 256 200 31 ) 7 9 3 6 31 × 31) 7 9 3 6 Answer: 6 2 0 –0 62 – 31 × 50 1 7 3 –06 31 × 6 173 1 5 5 –66 1 5 5– 1 8 1 8 1 8 66– 000 1 8 256 000 23 Checking the answer 31 × 256 = 7,936 Checking the answer 31 × 256 = 7,936 Thus, the answer is correct. Thus, the answer is correct. Method 3 Short division (break the divisor into factors) This method is used when the divisor can be factored into single digit factors, then each factor is used to divide the dividend. This method simplifies the division and reduces solving time. Example 1,218 ÷ 21 =  Analysis: Method: 21 = 3 × 7 1. 1,218 divided by 3 equals 406. 3)1218 7 )4 0 6 2. Dividing 406 by 7 equals the exact 58 amount of 58. Hence, the quotient Answer: 58 Checking the answer 58 × 7 × 3 = 1,218 is 58. Thus, the answer is correct.

56 When the divisor is a three-digit number: Example Divide 34,932 by 246 Method 1: Long division Method 2: Long division (shortcut) The division expression is 34,932 ÷ 246 =  The division expression is 34,932 ÷ 246 =  Method: Method: 142 142 246) 3 4 9 3 02– 246 × 100 246) 3 4 9 3 2 2 4 6 0 246 × 40 246 – 1 0 3 3 –20 246 × 2 9 8 4 1033 4 9 –22 9 8 4– 4 9 000 4 9 22– Answer: 142 4 9 000 Answer: 142 Checking the answer 246 × 142 = 34,932 Checking the answer 246 × 142 = 34,932 Thus, the answer is correct. Thus, the answer is correct. The examples shown are all vertical divisions. The horizontal division is used sometimes in solving mathematical problems and in such a case the symbolic expression form is generally used. Exercise 20 a. Find the answers to the following questions. (1) How many times can 6 be subtracted from 20? (2) How many times can 6 be subtracted from 24? (3) How many times can 7 be subtracted from 35? (4) There are 24 balls. Putting 8 balls into each basket, how many baskets will be used? (5) If a 54-meter rope is cut into pieces of 6 meters each, how many pieces will there be? b. Fill in the missing numbers in the following expressions. (1) 6 ÷ 2 =  (2) 15 ÷ 5 =  (3) 48 ÷ 8 =  (4) 7 ÷  = 1 (5) 25 ÷ = 5 (6) 54 ÷  = 6 (7)  ÷ 2 = 4

57 (8)  ÷ 7 = 7 (9)  ÷ 8 = 10 c. Find the quotient of the following expressions by long division. (1) 84 ÷ 4 (2) 784 ÷ 7 (3) 2,600 ÷ 13 (4) 27,600 ÷ 24 (5) 985,472 ÷ 32 d. Find the quotient of the following expressions by short division. (1) 96 ÷ 6 (2) 99 ÷ 9 (3) 726 ÷ 6 (4) 968 ÷ 8 (5) 200 ÷ 25 When the divisor is a single digit number: Example Find the result of 137 ÷ 5. Method 1: Long division Method2: Short division Method: Method: 27 5 )1 3 7 remainder 2 27 5) 1 3 07– Answer 27 remainder 2 1 0 5 × 20 3 75– 3 2 5× 7 The quotient is 27 with a remainder of 2. Answer: 27 remainder 2 Checking the answer, (5 × 27) + 2 = 135 + 2 Checking the answer, (5 × 27) + 2 = 135 + 2 = 137 = 137 Hence, the answer is correct. Hence, the answer is correct. When the divisor is a two-digit number: Method 1: Long division (shortcut) Method 2: Short division Method: Method: 21 32 = 4 × 8 32) 6 9 2 – 4 )6 9 2 8 )1 7 3 64 21 remainder 5 52 The actual remainder is 5 × 4 = 20. 3 2– The quotient is 21 with a remainder 20 of 20. Answer: 21 with a remainder of 20 The quotient is 21 with a remainder of 20. Answer: 21 with a remainder of 20 Checking the answer, (32 × 21) + 20 Checking the answer, (21 × 8 + 5) ×4 = 672 + 20 = 692 = 173 × 4 = 692 Hence, the answer is correct. Hence, the answer is correct.

58 To find the actual remainder of the short division use the following method. 5 is not the actual remainder since before using 8 to divide, 4 was used as the divisor. The value of the remaining number is thus reduced by 4 times. It must, therefore, be 4 times 5 resulting in the product of 20 which is the actual remainder. If both divisions have remainders, the first remainder shall also be added. Example: 1,526 ÷ 28 =  Method to find the actual remainder Method: 28 = 4 × 7 1. Find the remainder of the last 4)1526 4 × 3 = 12. division, i.e. 2. Add 2 which is the first remainder. 12 + 2 7) 3 8 1 remainder 2 5 4 remainder 3 Actual remainder is (3 × 4) + 2 = 14. equals 14 which is the actual remainder. Answer: 54 with a remainder of 14 SCtheepcdW1kiirvhneiesgunitolhttnshe.eiandn(is5vw4ise×or7,r)i+s 3a=th3r8ee1-digit number, the 2sirmespulletsstimn e(3t8h1o×d 4is) +th2e=lo1n,5g26. Step Hence, the answer is correct. Example 52,148 ÷ 462 =  Method 112 462 ) 5 2 1 4 8 – 462 – 594 462 – 1328 924 404 The quotient is 112 with a remainder of 404 Answer 112 remainder 404 ExCerhceicskein21g the answer, (462 × 112) + 404 = 51,744 + 404 = 52,148. Hae.ncFei,ntdhethaenasnwsweerrissfcroomrrethcet.following questions. (1) 9 ÷ 2 =  remainder (1) 25 ÷ 5 =  remainder (2) 75 ÷ 7 =  remainder

59 (3) 100 ÷ 9 =  remainder (4) 50 kilograms of rambutan are split equally, 8 kilograms, into each round bamboo basket. The remaining rambutan is for the children. How many kilograms can the children get? (5) 495 ducks are fed and divided equally for selling 7 times. How many ducks are sold each time and how many are left? b. Find the quotient and check the answer. (1) 20 ÷ 3 (2) 35 ÷ 4 (3) 82 ÷ 2 (4) 150 ÷ 12 (5) 1,031 ÷ 51 (6) 28,023 ÷ 145 Division problem exercises The following examples are related to daily life and may include addition, subtraction, multiplication and division together in one problem or only some of these operations. Example: 7 workers are hired to dig a pond. Together they receive 12,460 baht as wage. If the wage is divided equally, how much does a worker receive? The division expression is 12,460 ÷ 7 =  Solution: The workers received wages of 12,460 baht The wages are divided equally for 7 workers Hence, each worker receives 7 ) 12,460 baht 1,780 baht Answer: 1,780 baht Example: 8,460 pens are filled into boxes with 250 pens per box. How many boxes will be filled? The division expression is 8,460÷250 =  Solution: There are 8,460 pens Each box contains 250 pens So, the pens fill 33 boxes 250) 8460 750 960 750 210 Hence, 33 boxes are filled with a remainder of 210 pens. Answer: 33 boxes with 210 pens left. – Example: Goods are sold for 1,789 baht in the first week, for 1,826 baht in the second week, and for 2,310 baht in the third week. On average, for how much are goods sold per week? The expression is (1,789 + 1,826 + 2,310) ÷ 3 =  Method: The first week‟s sale of goods 1,789 baht The second week‟s sale of goods 1,826 baht + The third week‟s sale of goods 2,310 baht Total of three-week sale 5,925 baht Hence, on average, weekly revenue from the sale of goods is 5,925÷ 3 = 1,975 baht. Answer: 1,975baht

60 Problem exercises of addition, subtraction, multiplication and division Problems related to daily life may need to be solved with a combination of addition, subtraction, multiplication, and division, as in the following example: Example: Ming sold 2 cows with a weight of 186 kilograms and 174 kilograms respectively for 38 baht a kilogram. Then, he buys 100 mango sprouts at the cost of 25 baht each. How much money does he have left? The math expression is (186 + 174) ÷ 38 – (100 × 25) =  Solution: The first cow weighs 186 + kilograms The second cow weighs 174 kilograms Total weight of 2 cows 3 6 0 kilograms The cows are sold, per kilogram, for 3 8 baht 2880 × baht 1080 baht Total earnings from selling cows 1 3 6 8 0 + baht baht The price of each mango sprout 25 Number of mango sprouts bought 1 0 0 units Total money spent for mango sprouts 2500 × baht The money gained from selling cows 13680 baht Pay for mango sprouts 2500 – baht Hence, the remaining money is baht 11180 Answer: 11, 180 baht Exercise 22 Write the symbolic expression and solve the following problems by showing the method used. 1. On average Mr. Sinchai earns 3,670 baht per day from repairing motorcycles. At the end of a month, after deducting equipment costs and other expenses, how much money does he have left? 2. There are 3,144 refugees staying at a refugee center. The cost of supporting all of the refugees is 141,480 baht per day. How much is spent per head? 3. People in a province have suffered from a drought with critical-severity for 428 villages and medium-severity for 82 villages. The government supports each critically effected village

61 with food and plant seeds amounting to 50,000 baht, and 37,000 baht for each medium effected village. How much is the government spending for this support? 4. Somjai catches fish for wholesale at the market and gets 79,600 baht for the fish from the first pond and 83,400 baht from the second pond. Then, he buys a water pump for 37,500 baht and 5,000 fry at 7 baht each. How much money does he have left? 5. An Agricultural Cooperative has 3,796 shares. At year-end, the net profit is 318,864 baht. How much dividend will shareholders get per share? Topic 8: Factors of Counting Numbers and Factoring Definition of factors Consider the following statements: 30 can be exactly divided by 5. We say that 5 is a factor of 30. 24 can be exactly divided by 8. We say that 8 is a factor of 24. 19 cannot be exactly divided by 6. We say that 6 is not a factor of 19. A factor of a counting number is the counting number that can exactly divide such number. Exercise 23 Answer the following questions. (1) Is 4 a factor of 20? Why? (2) Is 3 a factor of 18? Why? (3) Is 7 a factor of 37? Why? (4) Is 9 a factor of 45? Why? (5) Which counting numbers have 2 as a factor – 2, 5, 8, 9, 12, 14? (6) Which counting numbers have 3 as a factor – 2, 3, 6, 15, 20, 24? (7) Give examples of the counting numbers from 21 to 39 that have 5 as a factor. (8) Give examples of the counting numbers from 15 to 40 that have 6 as a factor.

factors of 8 62 exactly divide 8 Consider the counting numbers that are factors of 8: 1 exactly divides 8. Thus, 1 is a factor of 8. 2 exactly divides 8. Thus, 2 is a factor of 8. 4 exactly divides 8. Thus, 4 is a factor of 8. 8 exactly divides 8. Thus, 8 is a factor of 8. No other counting numbers can exactly divide 8. Hence, 8 has 4 factors: 1, 2, 4, and 8. Consider the counting numbers that are factors of 5: 1 exactly divides 5. Thus, 1 is a factor of 5. 5 exactly divides 5. Thus, 5 is a factor of 5. No other counting numbers can exactly divide 5. Hence, 5 has 2 factors: 1 and 5. Exercise 24 a. Answer the following questions. (1) What counting numbers can exactly divide 12? (2) What counting numbers are the factors of 12? (3) What counting numbers can exactly divide 18? (4) What are the factors of 18?

63 Factor flower Factors of 18 Fill in the factors Fill in the factors of 22 in the petals. of 24 in the petals. Factors of 22 Factors of 24 Topic 9: Prime Number and Prime Factor 9.1 Prime Number Consider the factors of the following numbers: 2 has 2 factors i.e. 1 and 2. 3 has 2 factors i.e. 1 and 3. 5 has 2 factors i.e. 1 and 5. 11 has 2 factors i.e.1 and 11. Each of the counting numbers above has only 2 different factors i.e. 1 and itself. A counting number that has only 2 factors, 1 and itself, is called a prime number.

64 2 is a prime number since 2 has only 2 factors, i.e. 1 and 2. 7 is a prime number since 7 has only 2 factors, i.e. 1 and 7. 8 is not a prime number since 8 has more than 2 factors, i.e. 1, 2, 4 and 8. Exercise 25 Answer the following questions. (1) Is 13 a prime number? Why? (2) Is 15 a prime number? Why? (3) For the counting numbers from 20 to 30, which numbers are prime numbers? (4) For the counting numbers from 50 to 60, which numbers are prime numbers? (5) For the counting numbers from 90 to 100, which numbers are prime numbers? 9.2 Prime factor The number 12 has 6 factors, i.e. 1, 2, 3, 4, 6 and 12, but only 2 and 3 are prime numbers. Thus, 2 and 3 are prime factors of 12. When a factor is a prime number, it is called prime factor. The factors of 8 include 1, 2, 4 and 8. The prime factor of 8 is 2. The factors of 30 include 1, 2, 3, 5, 6, 10, 15 and 30. The prime factors of 30 are 2, 3 and 5. Exercise 26 Answer the following questions: What are the factors of the following numbers and which ones are the prime factors? (1) The factors of 9 include___________________________________________________ The prime factors of 9 are _____________________________________________ (2) The factors of 22 include___________________________________________________ The prime factors of 22 are _____________________________________________

65 (3) The factors of 36 include___________________________________________________ The prime factors of 36 are _____________________________________________ (4) The factors of 50 include ___________________________________________________ The prime factors of 50 are______________________________________________ (5) What are prime factors of 37? __________________________________ 4. Factorization Factorization is writing a number in the form of the product of the factors. The factors of 12 are 1, 2, 3, 4, 5, 6, and 12. We can write the number in the form of the product of each factor. For instance, 12 = 1 × 12, or 12 = 2 × 6, or 12 = 3 × 4 Exercise 27 a. Write the following numbers as the product of two factors which neither of them is 1. (1) 21 (2) 24 (3) 28 (4) 36 (5) 49 (6) 51 (7) 63 (8) 81 (9) 72 (10) 90 b. Are we able to write the following numbers 11, 13, 17, 23, 29 as the product of two factors when neither of them is 1? Why? Topic 10: Prime factorization Consider writing 12 as the product of two factors which neither of them is 1: 12 = 2 × 6 or 12 = 3 × 4 Since 6 and 4 are not prime factors, we can further write 6 and 4 as the product of the factors as follows. 12 = 2 × 6 or 12 = 3 × 4 = 2×2×3 = 3 ×2×2 When we write 12 = 2 × 2 × 3, or 12 = 3 × 2 × 2, it can be seen that 12 is written as the product of prime factors.

66 Writing a number in the form of the product of prime factors is called prime factorization. Example: Find the prime factorization of20 Method: 20 = 4 × 5 = 2×2×5 The prime factorization of 20 is 2 × 2 × 5. Answer: 20 = 2 × 2 × 5 Example: Find the prime factorization of48 Method: 48 = 3 × 16 = 3×2×8 = 3×2×2×2×2 The prime factorization of 48 is 3 × 2 × 2 × 2 × 2, or 3 × 24. Answer: 48 = 3 × 2 × 2 × 2 × 2 or 3 × 24 The multiplication of a number by itself several times e.g. 2 × 2 × 2 × 2 can be written in an exponentiation form 24 which is read as two to the power of four. Exercise 28 (10) 100 Find the prime factorization of the following numbers: (1) 6 (2) 14 (3) 28 (4) 35 (5) 36 (6) 52 (7) 45 (8) 60 (9) 72 Prime factorization by way of division To find the prime factorization of 20, we take the lowest prime number that can exactly divide 20 and than the next higher prime number that can exactly divide the resulting quotient to divide that quotient. Doing this repetitively until the last quotient is the prime number. We can write 20 as the product of every divisor and the last quotient which are all prime numbers. Example: Find the prime factorization of 20 Solution 2 ) 20 2 ) 10 5 The prime factorization of 20 is 2 × 2 × 5.

67 Exercise 29: Find the prime factorization of the following numbers. (1) 27 (2) 39 (3) 42 (4) 56 (6) 68 (6) 96 (7) 250 (8) 216 Finding the product by using factors Finding the product of two numbers may be done by writing one number as the product of its factors and applying the associative property of multiplication. Example: Find the product of 97× 35. Checking the answer by vertical Method: 97 × 35 = 97 × (5 × 7) 97 × multiplication. 35 = (97 × 5) × 7 485 = 485 × 7 = 3,395 2,910 + 3,395 Answer 3,395 Answer 3,395 Exercise 30 Find the product by using factors and checking the answers with different methods. (1) 46 × 36 (2) 92 × 48 (3) 126 × 45 (4) 218 × 28 (5) 118 × 25 (6) 256 × 32 Topic11: G.C.D. and L.C.M. 11.1 Finding G.C.D. Common Divisor We know that the factors of any number can exactly divide that number. For instance, the factors of 12 are 1, 2, 3, 4, 6, and 12. Thus, we may call each of the factors of 12 a divisor of 12. Consider the divisors of 8 and 12: The divisors of 8 are 1, 2, 4, 8. The divisors of 12 are 1, 2, 3, 4, 6, 12.

68 8 and 12 have 1, 2, and 4 as the same divisors. We call 1, 2, and 4 the common divisor or common factor of 8 and 12. A counting number that exactly divides at least two numbers is called the common divisor of those numbers. Example: Find the common divisors of 9, 15, and 21. Method: The divisors of 9 are 1, 3, 9. The divisors of 15 are 1, 3, 5, 15. The divisors of 21 are 1, 3, 7, 21. The common divisors of 9, 15, and 21 are 1, 3. Answer: 1 and 3 Exercise 31 Find the common divisor of the following numbers. (1) 12, 18 (2) 16, 24 (3) 27, 36 (4) 10, 21 (5) 8, 14, 18 (6) 10, 20, 30 Greatest Common Divisor (G.C.D.) The divisors of 16 are 1, 2, 4, 8, 16. The divisors of 20 are 1, 2, 4, 5, 10, 20. The common divisors of 16 and 20 are 1, 2, 4. The common divisor of 16 and 20 that has the greatest value is 4. The common divisor with the greatest value is called the Greatest Common Divisor, or G.C.D. Hence, the common divisor that has the greatest value or G.C.D. of 16 and 20 is 4. Example: Find the G.C.D. of 18 and 27. Method: The divisors of 18 are 1, 2, 3, 6, 9, 18. The divisors of 27 are 1, 3, 9, 27. The greatest common divisor or G.C.D. of 18 and 27 is 9. Answer: 9

69 Exercise 32 Find the G.C.D. of the following numbers. (1) 10, 14 (2) 9, 12 (3) 14, 28 (6) 18, 24 (4) 8, 27 (5) 16, 28 (9) 18, 27, 63 (7) 6, 4, 22 (8) 10, 20, 30 Finding the G.C.D. by prime factorization To find the G.C.D. of numbers, we may apply prime factorization and use the product of the common prime factors. For example, to find the G.C.D. of 18 and 27, factorize 18 and 27 as follows. 18 = 2 × 3 × 3 27 = 3 × 3 × 3 The largest number which exactly divides 18 and 27 is the number in the form of 3 × 3, i.e. G.C.D. of 18 and 27 is 3 × 3 = 9. Try another example, find the G.C.D. of 40 and 30. 40 = 2 × 2 × 2 × 5 30 = 2 × 3 × 5 The greatest number that can exactly divide 40 and 30 is the number in the form of 2 × 5, i.e. the G.C.D. of 40 and 30 is 2 × 5 = 10. Example: Find the G.C.D. of 16, 24, and 28. Method: 16 = 2 × 2 × 2 × 2 24 = 2 × 2 × 2 × 3 28 = 7 × 2 × 2 G.C.D. of 16, 24, and 28 is 2 × 2 = 4 Answer: 4 Exercise 33 Find the G.C.D. of the following numbers: (1) 16, 36 (2) 15, 25 (3) 26,34 (6) 42, 64 (4) 12, 27 (5) 35, 21 (9) 18, 27, 54 (7) 49, 56, 63 (8) 15, 30, 45

70 Finding the G.C.D. by way of division To find the G.C.D. of several numbers, we can use division in the same way as to find prime factors. For instance, to find the G.C.D. of 12, 18, and 24, we can do the following: (1) Find the prime number that is a common 2 ) 12, 18, 24 divisor of 12, 18, and 24, e.g. take 2 to divide 12, 18, 6, 9, 12 and 24, resulting in the quotients of 6, 9, and 12, respectively. (2) Find the prime number that is a common 2 ) 12, 18, 24 divisor of 6, 9, and 12, which is 3 and use it to divide 3) 6, 9, 12 6, 9, and 12, resulting in the quotients of 2, 3, 4. 2, 3, 4 (3) Find the prime number that is a common 2 ) 12, 18, 24 divisor of 2, 3, and 4, but no such prime number exists. 3) 6, 9, 12 Hence, the greatest common divisor or 2, 3, 4 the G.C.D. of 12, 18, and 24 is the product of all of the The G.C.D. of 12, 18, and 24 is 2 × 3 = 6. common factors which is 2 × 3 = 6. Answer: 6 Example Find the G.C.D. of 15, 25, and 35. Method: 5 ) 15, 25, 35 Example: 3, 5, 7 The G.C.D. of 15, 25, and 35 is 5. Answer: 5 Find the G.C.D. of 24, 60, 36. Method: 2 ) 24, 60, 36 Answer: 12 2 ) 12, 30, 18 แบบฝกึ หดั ที่ 35 3 ) 6, 15, 9 2, 5, 3 The G.C.D. of 24, 60, and 36 is 2 × 2 × 3 = 12.

71 Practice with the following exercises: a. Find the G.C.D. of the given numbers: (1) 21, 35, 42 (2) 27, 63, 81 (3) 10, 25, 30 (6) 20, 15, 45, and 40 (4) 24, 32, 64 (5) 16, 20, 36 (7) 24, 12, 60, and 48 (8) 28, 14, 70, and 84 b. Factor flower Finding the L.C.M. Common multiple The numbers with 4 as a factor are 4, 8, 12, 16, 20, 24, 28, 32, 36, ...... The numbers with 6 as a factor are 6, 12, 18, 24, 30, 36, 42, 48, 54, ....... The numbers that have both 4 and 6 as factors are 12, 24, 36,... We call the numbers 4 and 6 the common factor of 12, 24 and 36. The common multiple of any two numbers or more is the counting number that has those numbers as the factors. Example: Find the common multiple of 3 and 4. Method: The numbers with 3 as factor are 3, 6, 9, 12, 15, 18, 21, 24, 27, 36, ...... The numbers with 4 as factor are 4, 8, 12, 16, 20, 24, 28, 32, 36, ....... The common multiples of 3 and 4 are 12, 24, 36, ........ Answer: 12, 24, 36

72 Exercise 34 Find the common multiple of the following numbers: (1) 2, 3 (2) 4, 8 (3) 6, 9 (4) 10, 15 (5) 4, 6, 8 (6) 10, 15, 20 Least Common Multiple (L.C.M.) The numbers with 4 as a factor are 6, 12, 18, 24, 30, 36, 42, 48, 54, .... The numbers with 8 as a factor are 8, 16, 24, 32, 40, 48, 56, 64, 72, … The common multiples of 6 and 8 are 24, 48, ..... The common multiple of 6 and 8 that has the smallest value is 24. We call the common multiple that has the smallest value the least common multiple or L.C.M. Hence, the least common multiple or the L.C.M. of 6 and 8 is 24. Example: Find the L.C.M. of 4 and 6. Method: The numbers with 4 as a factor are 4, 8, 12, 16, 20, 24, 28, 32, 36, ...... The numbers with 6 as a factor are 6, 12, 18, 24, 30, 36, 42, 48, 54, ....... The common multiples of 4 and 6 are 12, 24, 36, ........ The least common multiple or the L.C.M. of 4 and 6 is 12. Answer: 12 Exercise 35 Find the L.C.M. of the following numbers: (1) 5, 6 (2) 2, 4 (3) 6, 9 (4) 10, 15 (5) 4, 6, 8 (6) 8, 10, 20 Finding L.C.M. by factorization To find the L.C.M. of numbers, we may use factorization. For example, to find the L.C.M. of 4 and 6, the factorization of 4 and 6 is 4 = 2 ×2 6 = 2 ×3 Hence, the smallest number that has 4 and 6 as its factors is 12. 12 = 2 × 2 × 3

73 2 × 2 × 3 is obtained by the following method: 4 =2×2 6 =2×3 2 × 2 × 3 = 12 Hence, the L.C.M. of 4 and 6 is 12. Example: Find the L.C.M. of 15 and 21. Method: 15 = × 5 3 21 = 3 ×7 The L.C.M. of 15 and 21 is 3 × 5 × 7 = 105. Answer: 105 Example: Find the L.C.M. of 18, 24. Method: 18 = × 3 × 3 24 =2 × 2 × 2 × 2 × 3 The L.C.M. of 18 and 24 is 2 × 3 × 3 × 2 × 2 = 72. Answ2er: 72 Example: Find the L.C.M. of 8, 10, 12. Method: 8 = 2×2×2 10 = 2 × 5 12 = 2 × 2 × 3 The L.C.M. of 8, 10, 12 is 2 × 2 × 2 × 5 × 3 = 120. Answer: 120 Exercise 36 Find the L.C.M. of the following numbers: (1) 6, 10 (2) 30, 50 (3) 6, 9, 15 (6) 15, 45 (4) 15, 20, 30 (5) 12, 20 (7) 8, 14, 16 (8) 12, 48, 60

74 Finding the L.C.M. by long division To find the L.C.M. of numbers, we may use the long division, e.g. we can find the L.C.M. of 8, 10, and 12 as follows: (1) Find the prime number that is the common 2 ) 8, 10, 12 divisor of 8, 10, and 12 or of at least 2 numbers e.g. divide 8, 4, 5, 6 10, and 12 by 2, and the quotients are 4, 5, and 6 respectively. (2) Find the prime number that is the common 2 ) 8, 10, 12 divisor of 4, 5, and 6, or at least 2 numbers, e.g. 2, since 3) 4, 5, 6 it exactly divides 4 and 6 but cannot exactly divide 5. Write 5 2, 5, 3 as the above line. (3) Find the prime number that is the common 2 ) 8, 10, 12 divisor of 2, 5, and 3, or of at least 2 numbers but no 2) 4, 5, 6 prime number exists. 2, 5, 3 Hence, the smallest numbers are 8, 10, and 12, or The L.C.M. of 8, 10, and 12 is 2 × 2 × 2 × 5 × 3 = 120. the factorization is 2 × 2 × 2 × 5 × 3 = 120. Answer: 120 Example: Find the L.C.M of 12, 16, 18. Method: 2 ) 12, 16, 18 2 ) 6, 8, 9 3 ) 3, 4, 9 1, 4, 3 The L.C.M. of 12, 16, and 18 is 2 × 2 × 3 × 1 × 4 × 3 = 144. Answer: 144 Example: Find the L.C.M of 18, 24. Method: 2 ) 18, 24 3 ) 9, 12 3, 4 The L.C.M. of 18, 24 is 2 × 3 × 3 × 4 = 72. Answer: 72

Exercise 37 75 (4) 4, 12, 24, Find the L.C.M of the following numbers. (1) 16, 24 (2) 15, 45 (3) 9, 36, 24 (7) 14, 2 32 (5) 20, 28 (6) 16, 30, 48

76 Lesson 2 Fraction Main content Fraction reading and writing, fraction comparison, fraction addition, fraction subtraction, fraction multiplication, fraction division, and fraction problem solving. Expected learning outcome 1. Be able to define fractions and their types as well as read fractions 2. Be able to write fractions in reduced terms, mixed fractions and improper fractions 3. Be able to compare and order fractions 4. Be able to add and subtract fractions, and apply fraction knowledge to problem solving 5. Be able to multiply fractions and apply fraction multiplication knowledge to problem solving 6. Be able to divide fractions and apply fraction division knowledge to problem solving 7. Be able to add, subtract, multiply and divide fractions, and apply the knowledge to problem solving Content scope Topic 1: Meaning and characteristic of fractions and how to read fractions Topic 2: Writing fractions in reduced terms, mixed fractions and improper fractions Topic 3: Comparing fractions Topic 4: Adding and subtracting fractions and problem exercises Topic 5: Multiplying fractions and problem exercises Topic6: Dividing fractions and problem exercises Topic 7: Adding, subtracting, multiplying and dividing mixed fractions and problem exercises

77 Topic 1: Meaning and characteristic of fractions and how to read fractions (1) Fractions are a way of representing division of a whole into equal parts. For example: This circle represents one whole. It is divided into 4 equal parts. parts. The shaded area is 1 part of the 4 quarter” It can be written as which reads “one- This is a rectangle. It is divided into 5 equal parts. The shaded area is 2 parts of the 5 parts. fifths” It can be written as which reads “two- There are 3 types of fractions 1. Proper fractions have a numerator which is less than the denominator such as 1 , 3 , 2 , 11 4 5 7 15 2. Improper fractions have a numerator which is larger than the denominator such as 7 , 12 35 3. Mixed fractions are a combination of an integer and a proper fraction such as 3 1 , 5 4 , 11 5 2 7 12

78 Exercise 1 a. Write a fraction representing the shaded section of each rectangle below. (1) (2) (3) (4) (5) b. Write each fraction out in words. Example: reads two-thirds (1) 5 (2) 4 (3) 7 (4) 1 (5) 6 68 9 7 7 c. Write each fraction in numbers. (1) Five-eights = __________________________________ (2) Two-thirds = __________________________________ (3) Seven-ninths = __________________________________ (4) Six-sevenths = __________________________________ (5) Three-fifths = __________________________________ (2) Fractions represent parts of a whole which is equally divided. For example: There are 6 eggs and they are equally divided into 3 parts. So, each part is of all the eggs. Each part of the eggs is one- third of all the eggs. It can be written as of all the eggs or of 6 eggs or one part equals 2 eggs. 1 There are 6 oranges and they are equally divided into 2 piles. Each pile represents of 2 all the oranges. 1 So, there are a total of 2 piles of oranges. It can be written as of 6 oranges 2 or each pile is equal to 3 oranges.

79 Reading fractions Mixed fraction Improper fraction Picture Numerical Words Numerical Words 1. One and Five- one-quarter quarters 2. Two and Seventeen- one-eighth eighths 3. One and Eleven- five-sixths sixths Topic2: Writing fractions in reduced terms, mixed fractions and improper fractions 2.1Below are illustrations of fractions in reduced terms. 2= 1 42 4= 2 84 2= 1 63

80 2.2 Mixed fractions – Study the following fractions: 3 1 , 5 3 , 4 10 , 7 21 , 2 7 13 35 10101 135 2.3 Improper fractions – Study the mixed fractions in 2.2 above and convert them into improper fractions. 3 1 = …………………… , 5 3 =…………………, 4 10 = ……………………. 2 7 13 7 21 = …………………… , 10 101 = …………………… 35 135 Topic 3: Comparing fractions Fraction comparison is done by comparing a fraction with another fraction. If the values of two fractions are equal, use the equal sign (=). If the values of two fractions are not equal, use the less than sign (<)or the greater than sign (>). 3.1 Comparing fractions which have like denominators but are unequal by using<or > < 13 44 3.2 Fractions having an equal value. = 12 24

81 Exercise 2 Compare the fractions below and correctly fill in the  with< (less than), >(greater than) or= (equal) signs. (1)  (2 )  (3)  (4)  (5)  (6)  (7)  (8)  (9)  (10)  (11)  (12)  (13)  (14)  (15)  (16)  (17)  (18)  (19)  (20)  Topic 4: Adding and subtracting fractions and problem exercises Adding and subtracting fractions Adding and subtracting fractions with like denominators Steps for adding and subtracting fractions with like denominators: (1) Sum the numerators; (2) Use the same denominator. Example: Subtraction Addition 3 2 1 141 1 4444 1 41 1 1 4444 Method: 1  1 = 1  1 = 2 Method: 4  3 = 43 = 1 44 4 4 44 44 Exercise 3 Fill in the blanks with the correct answers. (1) 2 + 3 =  (2) 4 + 5 =  77 55 (3) 4 3 = 72 = + (4) - 88 77 (5) 6 - 5 =  (6) 8 2 = - 10 10 97

82 (7) 2  3  2 =  (8) 1   2  2  =  999 4 4 (9)  2  4  - 3 =  (10)  3  2   1 =  7 7 7 5 5 (11)  2  2  + 3 =  (12) 4   2  2  =  7 7 7 5 5 5 Creating equivalent fractions 1) To raise fractions to higher terms, multiply both the numerator and denominator by the same number. Example: == == 2) To decrease fractions to lower terms, divide both the numerator and denominator by the same number. Example: == == Adding and subtracting fractions with unlike denominators Adding and subtracting fractions with unlike denominators by raising fractions to higher terms. To add or subtract fractions with unlike denominators, start with raising fractions to higher terms in order to create a like denominator. Then, add or subtract the numerators of the fractions. Raising fractions to higher terms means increasing both the numerator and denominator of a fraction without changing its value. For example: Example: Find the sum of 1 + 2 How to make the denominators of 1 and 2 53 53 Method: 1 + 2 = 13 + 2  5 equal? Analysis 5 3 53 35 The denominators are 5 and 3and 5 cannot be = 3 + 10 evenly divided by 3. So, multiply 3 by both 15 15 the numerator and denominator of 1 and = 3 10 5 15 multiply 5 by both the numerator and denominator of 2 . The products are 3 and = 13 3 15 15 10 . The fractions now have a common Answer 13 15 15 denominator.

83 Example: Find the difference of 5 - How to find a common denominator of 5 and 1 ? 7 1 73 3 Analysis Method: 5 - 1 = 53 - 1 7 The denominators are 7 and 3and 7 cannot be 7 3 73 37 evenly divided by 3. So, multiply 3 by both the numerator and denominator of 5 and = 15 - 7 7 21 21 multiply 7 by both the numerator and = 15  7 denominator of 1 . The products are 15 and 21 3 21 =8 7 . The fractions now have a common 21 21 Answer 8 denominator. 21 Exercise 7 (2) 2 + 2 Find the answers. 64 (1) 1 + 2 (4) 1 + 3 25 37 (3) 1 + 3 (6) 11 - 8 58 12 13 (5) 1 + 2 45 Commutative property of fraction addition Example: Prove if 2 + 1 equals 1 + 2 55 55 Method: 2 + 1 = 2 1 1 + 2 = 12 55 5 55 5 =3 = 3 55 Therefore 2 + 1 = 1 + 2 5 55 5

84 Example: Prove if 1 + 3 equals 3 + 1 27 72 Method: 1 + 3 = 1 7 + 3 2 3 + 1 = 3 2 + 1 7 2 7 27 72 7 2 72 27 = 7+6 = 6+7 14 14 14 14 = 76 = 67 14 14 = 13 = 13 14 14 Therefore 1 + 3 = 3 + 1 2 772 Analysis When 2 fractions are added, the sum is the same regardless of the orders of the addends. Therefore, it can be concluded that fraction addition has a commutative property. Exercise 5 (2) 12 + 19 =  + 12 25 27 25 Correctly fill the fractions in the blanks  (4) 13 +  = 25 + 13 (1) 5 + 7 = 7 +  25 29 25 9 13 13 (6) 21 +  = 19 + 21 (3) 2 + 8 = 8 +  91 87 91 5 27 27 (5)  + 5 = 5 + 11 12 12 23 Associative property of fraction addition Example: Prove if  1  2  + 3 equals 1 +  2  3  5 7 7 5 7 7 Method:  1  2  + 3 =  1 7  2  5  + 3 1 +  2  3  = 1 + 5 5 7 7 57 75 7 5 7 7 5 7 =  7  10  + 3 = 1 7 + 55  35 35  7 57 75 = 17 + 3 = 7 + 25 35 7 35 35 = 17 + 35 = 32 35 75 35 = 17 + 15 = 32 35 35 35  1  2  + 3 equals 1 +  2  3  5 7 7 5 7 7

85 Analysis Which method is more convenient? Solving 1 +  2  3  is easier. As 5 7 7 numerators of like fractions can be added up immediately, we should calculate 2  3  5 before adding the sum with 1 which has a different denominator. 77 7 5 The associative property of addition can be applied to fraction addition to make the operation simpler. Conclusion: When adding up 3 fractions, changing the fraction grouping does not change the answers. This confirms that fraction addition has an associative property. Exercise 6 Correctly fill fractions in the blanks  (1)  2  7  + 1 =  +  7  1  5 9 9 9 9 (2)  1  2  + 3 = 1 + 2   3 7 7 3 7   (3)  3  2  + 3 = 3 +   3  4 9 9 4  9  (4)   1  + 3 = 3 +  1  3   8 8 4 8 8 Fraction addition and subtraction problem exercises Adding fractions Example: Malai grows vegetables on 5 of the land. Malee grows vegetables on 2 of the 88 land. How much is the land on which they grow vegetables? Analysis: According to the problem, Malai grows vegetables on 5 of the land and Malee 8 grows vegetables on 2 of the land. The denominators of the two fractions are the 8 same. To sum the fractions, add the numerators of both fractions and use the same denominators. Method: Symbolic expression 5 + 2 =  88 Malai grows vegetables on 5 of the land 8 Malai grows vegetables on 2 of the land 8 Both of them grow vegetables on 5 + 2 = 5  2 of the land 88 8 = 7 of the land 8 ตอบ 7 ของแปลง 8

86 Exercise 7 Write a symbolic expression and the method to apply for each problem. (1) Uthai had 4 of a bag of fertilizer. Winai had 3 of a bag of fertilizer. Supon 99 had 1 of a bag of fertilizer. Put together, how much fertilizer did they have? 9 (2) Preecha had 2 litre of green beans. Anand had 3 litre of green been. How 77 many green beans did they have combined? (3) Flour was added to 3 of a cup and sugar was added to 2 of the cup. How much 66 of the ingredients were added? (4) Kanlaya planted morning glory on 2 of the plot, cabbage on 1 of the plot and 55 lettuce on 1 of the plot. How much of the plot did she plant on? 5 (5) Mana planted roses in a pot. He filled 3 of the pot with soil and 1 of the pot 55 with fertilizer. How much of the pot was filled Subtracting fractions Example: Mana‟s rope was 4 metre and he used 3 metre for box wrapping. What 99 is the length of his remaining rope? Analysis:When subtracting fractions with the same denominator, subtract numerators of the fractions and use the same denominator. Method: Symbolic expression 4  3 =  99 Mana‟s rope is 4 metre 9 He cuts the rope off at 3 metre 9 The length of the remaining rope is 4  3 = 4  3 metre = 99 9 Exercise 8 1 metre 9 Answer: 1 metre 9 Write a symbolic expression and method for solving each problem. (1) Orathai had 9 of a bag of fertilizer. She then used 1 of the bag on vegetables 13 13 and 2 on mango trees. How much was left? 13

87 (2) Saijai grew vegetables on 6 of the land. Suda grew vegetables on 9 of the 11 11 land. Who grew more vegetables and by how much? (3) Winai had seeds which covered 11 of the box. After sowing, he had 5 of the 12 12 box left with seeds. How much seed did he use? (4) A road of 8 kilometer must be repaired. If 7 kilometer is repaired, how 15 15 many kilometer is left to be repaired? (5) Wirach walked 4 kilometres to the school. Weena walked 2 kilometres to the 77 school. How much farther did Wirach walk compared to Weena? Adding and subtracting fractions of different types. You have learned how to add and subtract proper fractions. This section deals with how to add and subtract mixed fractions and improper fractions. Examples are as follows: Example: Find the sum of 2 7 + 3 1 Analysis 88 1. Sum the integers. In this example, Method: 2 7 + 3 1 = 2 + 3 + 7 + 1 the integers are 2 and 3. Then, sum 88 88 =5+ 8 the fractions. 2. Reduce 8 to the simplest form. 8 8 =5+1 =6 8 ÷ 8 =1 Answer: 6 88 Example: Find the sum of 12 + 11 Analysis 5 10 1. 12 equals 5  5  2 Method: 12 + 11 =  5  5  2  + 10 + 1 5 555 5 10  5 5 5  10 10 11 equals 10 + 1 =1+1+ 2+1+ 1 10 10 10 5 10 2. Reduce 5 to the simplest form =3+ 22 + 1 10 5 2 10 5 5= 1 =3+ 4 + 1 10 5 2 10 10 =3+ 5 10 =3+ 1 2 = 31 2 Answer: 3 1 2

88 Example: Find the difference between 8 3 Analysis 1. Subtract the integers. Then, 7 add or subtract the remaining fractions as applicable. and 5 2 2. Simplify 7 . 21 21 Method: 8 3  5 2 = 8 – 5 + 3  2 7 21 7 21 7 7 =1 =3+ 3 2 21 7 3 7 21 = 3 +  3  3   2  7 3  21 =3+ 9  2 21 21 =3+ 7 21 =3+ 1 3 = 31 3 Answer: 3 1 3 Exercise 9 (2) 9 3  6 5 =  Find the answers. 88 (1) 3 2  7 3 =  (4) 12 3  5 1 =  35 7 14 (3) 37 + 45 =  (6) 12 8  4 1 =  6 12 9 18 (5) 48 3  30 1 =  5 10 Topic 5: Multiplying fractions and problem exercises Multiplying fractions When multiplying fractions, you must multiply numerators by numerators and multiply denominators by denominators. Then, simplify the fractions. Example: 1 of 1 =  25 Method: 1 of 1 = 1  1 2 5 25 = 11 25 =1 10 Answer: 1 10

89 Example: 4  5 =  Analysis The product of multiplying numerators 56 together and denominators together is 20 . Method: 4  5 = 45 30 5 6 56 Then, reduce the product to the simplest form by dividing both the numerator and = 20 the denominator by 10. The answer is 2 . 30 3 = 20 10 30 10 =2 3 Answer: 2 3 Exercise 10 Multiply the fractions and reduce the products to the simplest form. (1) 6  8 =  (2) 12  26 =  79 13 20 (3) 14  9 =  (4) 11  7 =  27 15 14 15 (5) 15  18 =  18 27 Multiplying fractions by integers Multiplying a fraction by an integer means adding the same fraction for the number of time indicated by its integer. For example: Exercise 11 Find the answers. (1) 7  6 =  (2) 9  5 =  (3) 6 3 =  8 11 7 (4) 5  7 =  (5) 12  4 =  (6) 11  3 =  6 13 15 (7) 13  4 =  (8) 10 8 =  12 17 Fractions of integers Fractions of integers can be interpreted and calculated the same way as for the multiplication of fractions by integers. For example: Example: How much is 3 of 50 baht? 5 Method: 3 of 50 baht = 3  50 baht 55 = 3 50 baht 5 = 150 baht 5 = 30 baht Answer: 30 baht

90 Exercise 12 Find the answers of the following problems: (1) The land is 400 square wah. Pineapples are planted on 5 of it. How many square 8 wah of the land are the pineapples planted on? (2) A high school has 1,200 students. Male students are 3 of the students. How many 4 male students are there? (3) A rope was 500 metre long and 2 of the rope was sold. How many metres of the 25 rope was sold? (4) There were 200 chickens. 1 of the chickens died of a contagious disease. How 20 many chickens died? (5) There are 75 mango trees in the orchard. 1 of them are Kaew mangoes. How many 3 Kaew mango trees are there in the orchard? Topic 6: Dividing fractions and problem exercises Dividing integers by fractions 11 11 2  1 =  2  2  ÷  1  2  22 22 2  1 2 1 There are 2 rai of land The land is equally divided into plots =  2  2  ÷ 1 of 1 rai.  1 2 = 2 2 So, the land is divided into 4 parts. 1 Therefore 2  1 = 2  2 21 =4

91 Dividing fractions by integers 1 1  2 =  1  1  ÷  2  1  3 3 3 2  2 11 66 =  1  1  ÷ 1 There is a piece of land of 1 rai. 3 2 3 = 11 It is equally divided into 2 parts 32 So, each part equals 1 rai. Therefore 1  2 = 1  1 6 3 32 =1 6 Dividing fractions by fractions 4 5 4  2 =  4  5  ÷  2  5  5 5 5 2 5 2 =  4  5  ÷ 1 5 2 22 = 45 55 52 There is a land of 4 rai. Therefore 4  2 = 4  5 5 55 52 =2 It is equally divided into parts of 2 rais 5 each. So, the plot of land is divided into 2 parts. “Dividing fractions means splitting fractions into equal, smaller amounts”. There are 3 types of fraction division which are dividing integers by fractions, dividing fractions by integers and dividing fractions by fractions. The division principles are as follows: Dividing integers by fractions To divide an integer by a fraction, you must multiply the integer by the reciprocal of the fraction. Example: 6  2 =  3 Method: 6  2 = 6  3 312 = 63 2 = 18 2 =9 Answer: 9

92 Explanation (1) The reciprocal of 2 is 3 . 32 (2) Multiply 3 by 6. Start with multiplying the nominators, that is, 2 multiplying 3 by 6. The product is 18. As 6 is an integer, its denominator is 1. The product of denominator multiplication is 2 because 2  1 equals 2. (3) 18 is an improper fraction and can be simplified by dividing 18 by 2 2. The result is 9. Dividing fractions by integers To divide a fraction by an integer, you must multiply the fraction by the reciprocal of the integer. Example: 8  4 =  9 Method: 8  4 = 8  4 9 91 = 81 94 = 81 94 =8 36 = 84 36  4 =2 9 Answer: 2 9 Explanation (1) Convert 4 which is an integer into a fraction by putting the integer as the numerator and 1 as the denominator. (2) The reciprocal of 4 is 1 . Multiply the fraction by 8 . The product is 8 . 14 9 36 (3) Simplify 8 by dividing the fraction by 4 which is the highest 36 common divisor of the numerator and the denominator. The result is 2 . 9

93 Dividing fractions by fractions To divide a fraction by a fraction, you must multiply the first fraction by the reciprocal of the second fraction. Example: 2  3 =  5 10 Method: 2  3 = 2  10 5 10 5 3 = 2 10 53 = 20 15 = 20  5 15  5 =4 3 = 11 3 Answer: 11 3 Explanation (1) The reciprocal of 3 is 10 . Multiply the fraction by 2 . The product is 10 3 5 20 . 15 (2) Reduce 20 to the simplest form by dividing the fraction by 5 which is the 15 highest common divisor of the numerator and the denominator. The result is 4 . 3 (3) Change 4 into a mixed fraction by dividing 4 by 3. The result is 11 . 33 Example: 3 4  3 3 =  54 Method: 3 4  3 3 = 19  15 5 45 4 = 19  4 5 15 = 19  4 5 15 = 76 75 = 11 75 Answer: 1 1 75

94 Explanation(1) Converted into improper fractions, 3 4 equals 5 19 and 3 3 equals 15 . 54 4 (2) The reciprocal of 15 is 4 . Multiply the reciprocal by 19 . The 4 15 5 product is 76 . 75 (3) 76 can be converted into a mixed fraction as 1 1 . 75 75 Remark: To divide mixed fractions by fractions or by mixed fractions, you can apply the same method used in dividing fractions by fractions. Start with the conversion of mixed fractions into improper fractions. Then, divide them in the same way you generally divide fractions. Fraction division problem exercises Fraction division problem exercises have the same characteristics as fraction subtraction problem exercises because division is a shorter and faster way for repeated subtraction. Example: A father has an amount of land of 22 1 rai. If he equally divides and gives the 2 land to his 3 children, how much of the land will each child receive? Symbolic expression: 22 1  3=  2 Method: A father has a land of 22 1 rai. 2 The land is equally divided and given to his 3 children. Each child will get 22 1  3 = 45  3 rai 2 21 = 451 rai 23 = 45  3 rai 6 3 = 15 rai 2 = 7 1 rai 2 Answer: 7 1 rai 2 Exercise 13 Part 1: Show the method to solve the problem exercises and find the answers. 1. 4  1 =  7. 2  5 = . 4 59 2. 5  5 =  8. 4  1 =  7 5 10 3. 8  8 =  9. 2 1  5 =  9 36

95 4. 14  7 =  10. 3  2 1 =  15 10 15 5. 1  1 =  11. 5 5  7 1 =  44 84 6. 1  1 =  12. 21 3 15 3 =  22 57 Part 2: Show the method to solve the problem exercises. 1. Multiplication of two numbers yields 54 . If one number is 9 , what is 55 15 the other number? 2. There are 36 3 bags of the rice left. The remaining rice can feed 4 immigrants in a camp for 6 days. How much rice is used each day? 3. 30 volunteers can dig 5 of a well in 5 days. How deep can they dig in 1 6 day? 4. There is oil which fills 63 of a tank. If we divide the oil and put it in 112 cans, how many cans does the oil fill if each can is 3 of the tank? 16 Topic 7: Adding, subtracting, multiplying and dividing mixed fractions and problem exercises In some cases, a mathematical problem may require mixed operations which are addition, subtraction, multiplication and division, or has a parenthesis or “of”. The order of operations is as follows: (1) Start with calculating numbers in the parenthesis. (2) Change “of” to“”sign and perform the multiplication first. (3) Multiplication and division have the same order. (4) Addition and subtraction have the same order. Example 1:  3  5  ÷ 7 1 =  4 6 2 Method:  3  5  ÷ 7 1 =  3 3  5  2  ÷ 15 4 6 2 43 62 2 =  9  10  ÷ 15 12 12  2 = 19  15 12 2 = 19  2 12 15 = 38 180 = 38  2 180  2 = 19 90 Answer: 19 90

96 (1) Begin with the addition of fractions in the parenthesis:  3  5  yields 19 . 12 4 6 (2) After obtaining a common fraction by calculating all the fractions in the parenthesis, divide it by 7 1 after having changed it into an improper fraction. 2 Example2:  25  4  + 5 1  2 3  =   28 5   2 4  Method:  25  4  + 11  11 = 100  11 4   28  5   2 4  140  2 11 = 5  11 4 7 2 11 = 5  44 7 22 = 52 7 = 25 7 Answer: 2 5 7 Explanation (1) In the first parenthesis, the multiplication of the numerators or 25  4 yields 100 and the multiplication of the denominators or 28  5 yields 140. Then, reduce the fraction by dividing 20 which is the highest common divisor of 100 and 140 into both the numerator and denominator. The result is 5 . 7 (2) As for the second parenthesis, change the fractions into improper fractions. The results are 11 and 11 . Then, inverse the divisor from 24 11 to 4 and multiply it by 11 . The result is 44 . 4 11 2 22 44 ivided by 22 is 2. (3) Add 5 and 2. The answer is 2 5 . 77

97 Example3: A gardener harvests 122 1 kilograms of mangoes from the first mango tree and 134 1 24 kilograms from the second mango tree. If the harvested mangoes are equally divided into 3 piles, how much does each pile weigh in kilograms? Symbolic expression: (122 1 134 1 ) ÷ 3 =  24 Method: A gardener harvests from the first mango tree 122 1 kilograms of mangoes 2 He harvests from the second mango tree 134 1 kilograms of mangoes 2 He harvests from both mango trees = 122 1 134 1 kilograms of mangoes 24 = 245  537 kilograms of mangoes 24 = 245 2  537 kilograms of mangoes 22 4 = 490  537 kilograms of mangoes 44 = 1027 kilograms of mangoes 4 The mangoes are divided into 3 equal piles. Each pile weighs = 1027  3 kilograms 41 = 1027  3 kilograms 41 = 1027 kilograms 12 = 85 7 kilograms 12 Answer: 85 7 kilograms 12 Exercise 14 Part 1: Show the method to solve the following problem exercises: 1. 1 5  2   1 =   8 3 4 2.  3  2   1 =  4 5 5 3. 7   4  2  =   7 14  4. 2 3  10 2  6 =  5 7 

98 5. 5 1  2 3   7 1 =   2 4 3 6.  1  7    2 4  1  =   8 8   5 14  7.  35  4    2  10  =   36 5   3 12  8. 15 5 12 1  25 of 9 =  6 3 54 100 Part 2: Write the following problems in the form of a symbolic expression and find the answers: 1. I bought durians, mangosteens and rambutans. Their total weight is 10 1 kilograms. If the rambutans weigh 3 1 kilograms, the mangosteens weigh 42 3 2 kilograms, how much did the durians weigh? 3 2. The first rope is 12 9 metres long and the second rope is 25 1 metres long. 93 If they are connected, how long will the ropes be? 3. A road is 60 1 kilometers long. If I bike at the speed of 15 1 kilometers per 28 hour, how much time will it take for me to pass on the road? 4. There is a plot of land of 50 rai. If the land is divided into plots of 1 1 rai each, 4 how many plots will there be? 5. On the first day, the construction workers constructed 1 of the entire road. 3 On the second day, they constructed 1 of the entire road and there were 5 kilometres 2 left. How long is the road in kilometres? 6. Boonyod has an income of 5,400 baht per month. He pays 1 of his income for 9 rent and 1 of his income for food. How much money does he have left? 3

99 Lesson 3 Decimals Main content Reading and writing decimals, writing decimals in an expanded form, decimal comparison, decimal ordering, decimal estimation, relationship between decimals and fractions, decimal addition, subtraction, multiplication and division, and problem solving. Expected learning outcome 1. Be able to define, write and read decimals 2. Be able to identify place value and values of digits in each decimal place 3. Be able to write decimals in an expanded form 4. Be able to compare and order decimals 5. Be able to convert decimals into fractions and convert fractions and integers into decimals 6. Be able to round decimals to whole numbers, the tenths place and the hundredths place 7. Be able to add and subtract decimals and apply the knowledge to problem solving 8. Be able to multiply and divide decimals and apply the knowledge to problem solving Content scope Topic 1: Meaning of decimals and how to read and write decimals Topic 2: Place value and values of digits in each decimal place Topic 3: Writing decimals in an expanded form Topic 4: Comparing and ordering decimals Topic 5: Relationship between decimals and fractions Topic 6: Estimating decimal values Topic 7: Adding and subtracting decimals and problem solving exercises Topic 8: Multiplying and dividing decimals and problem solving exercises

100 Topic 1: Meaning of decimal and how to read and write decimals 1.1 One decimal digit Decimals represent fractions with denominators of 10, 100, 1,000, 10,000, etc. The place value is indicated by the decimal point (.). For example: The above rectangle is equally divided into 10 parts. The shaded section comprises 7 parts and can be expressed in fraction as 7 and in decimal form as 0.7 10 1.2 When reading decimals, begin with the integer to the left of the decimal point. Then, read the decimal point as “and” followed by the digits to the right of the decimal point. For example: 0.2 reads zero point two 0.53 reads zero point five three 3.48 reads three point four eight 72.316 reads seventy-two point three one six 1.3 When writing decimals, the number to the left of the decimal point is the integer. The first number to the right of the decimal point is called “the first decimal place” and it shows the number of tenths. For example: The shaded section accounts for 4 parts out of 10 equal parts or 4 . It can be expressed 10 in decimal form as 0.4 which reads “zero point four”. Likewise, if a rectangle is divided into 100 equal parts and 79 parts out of 100 parts are shaded, the shaded parts can be expressed in fraction as 79 and in decimal as 0.79 which reads “zero point seven nine”. 100