www.jimmymaths.com [email protected] 1 www.jimmymaths.com [email protected] About the Author Jimmy Ling - Director and Math tutor of Grade Solution Learning Centre - Online Math Coach at jimmymaths.com - Graduated from NUS with degree in Mathematics - More than 10 years of teaching experience 2 www.jimmymaths.com [email protected] Foreword Dear Parents, Does your child always struggle with problem sums? Even though he can score well in Paper 1, he is not able to do most of the questions when it comes to Paper 2. He appears to understand each topic well but he can’t solve the questions accurately during exams. He stares into each question for a long time, not knowing how to start. He also spends too much time on one question, and hence, have insufficient time for the rest. And when he approaches you for help, you - Are at a loss on how to solve the question yourself Tried your best to explain to him but he cannot understand Managed to explain to him the solutions, but when he met a similar question next time, he is stuck again. At a loss of what to do, you buy lots of assessment books and past year papers to drill your child on problem sums. He couldn’t do most of the questions and ended up copping the solutions blindly. And even after spending a lot of time drilling different types of Math questions, he still does not get the results he deserves. Your child begins to hate Math even more. It is frustrating, isn’t it? 3 www.jimmymaths.com [email protected] Well, I got good news for you. I created this guide for parents to learn how to teach your child to solve problem sums effectively. Over my past 10 years of teaching, I realised that most high scoring students have received immense support from their parents ever since they are very young. Their parents have an effective way to explain to them and nurture them into problem-solvers. You can do the same too! If you can give your child proper guidance and continuous support, you can expect to see improvements in a short period of time. In the first half of the book, you will learn empowering strategies which you can use to guide your child effectively in problem sums. In the second half of the book, I will share the Must-Know problem solving heuristics or concepts so that you can relearn to teach your child. By applying the strategies taught in this book, your child will turned into a more competent, confident, and independent problem-solver. So read on, and may you empower your child to succeed in overcoming Math problem sums one day! 4 www.jimmymaths.com [email protected] How Can Your Child Improve in Math Many parents have approached me to coach their child. If your child needs help too, feel free to enrol in my online courses or any of our tuition classes. (Sorry, I don’t do one-to-one tuition) Here are a few things to note: - We don’t let your child practice any questions first before understanding the concept. We simplify concepts in ways which your child can understand first. - We don’t do drilling of problem sums. Instead, we classify questions carefully to their heuristic or concepts, and let your child master each of them. We will also do revision constantly to revise the concepts again. - We don’t teach students to stick to one method of solving problem sums. We will teach a variety of methods so that your child has more tools to use during exams. Here are 2 ways which you can seek help from us… 5 www.jimmymaths.com [email protected] Online Courses and Tuition Classes Online Courses Find out more by clicking below: https://jimmymaths.com/our-courses/ - Learn from recorded videos - Get access to lots of common exam questions to ensure sufficient practice - Get unlimited support and homework help If you have any questions, you can email us at [email protected] Group Tuition Classes Together with a team of tutors, we conduct Math tuition at our tuition centre, Grade Solution Learning Centre, located near Kovan MRT. We also have a new branch near Bugis MRT. You can find out more details below. Contact Us Call: 6280 6841 Website: https://gradesolution.com.sg 6 www.jimmymaths.com [email protected] Contents Page Chapter Page Number Why Does Your Child Struggle in Problem Sums? The 3 Steps Approach to Solve Any Problem Sum Questions are the Answers Helping Your Child To Practice Effectively Remainder Concept (Branching) Remainder Concept (Model) Equal Fractions Concept Model Drawing Constant Part Constant Total Constant Difference Everything Changed Excess and Shortage Concept Difference Concept Grouping Concept Number × Value Concept Guess and Check/Assumption Concept Simultaneous Equations/ Units and Parts Concept Working Backwards Concept Double If Concept Conclusion 8 10 13 15 17 18 19 22 24 26 28 30 32 35 37 40 43 45 48 50 53 7 www.jimmymaths.com [email protected] Why Does Your Child Struggle in Problem Sums? Some parents thought that when their child is weak in problem sums, it means he doesn’t understand the chapter well. However, from my experience, most students have a good understanding of the chapter. What most students lack, are the critical skills to comprehend the question, link the question to the right concept, and how to apply the concept correctly. That is why most of the students can do well in paper 1 but crashed in paper 2. This problem-solving skill is not innate. Your child is not born with it. It can be nurtured, trained and mastered. If you throw someone into a swimming pool without teaching him how to swim first, he will drown. But if he has received the right and sufficient training, he can be the next Joseph Schooling. Unfortunately, most schools do not have a structure to teach problem solving skills. 8 www.jimmymaths.com [email protected] Most students are taught the core concepts of each chapter. But when it comes to problem sums, most students are given questions to do without proper explanations of the concepts behind it and they are expected to know how to do. Without understanding of concepts, your child will likely copy solutions and memorise workings. Remember that being good in Math is not about how much your child remember. It is testing how competent his skills are in solving questions quickly and accurately. And these problem solving skills are so important, that once your child acquire them, he will be able to solve questions on his own. He will find Math more fun instead of finding it a pain. As a parent, you stand in the first line of support for your child. When your child faces difficulty, chances are he will approach you first instead of his teachers or friends. So it is also critical for you to acquire these skills so that you can impart to your child effectively. 9 www.jimmymaths.com [email protected] The 3 Steps Approach to Solve Any Problem Sum Let me share with you the 3 steps process which I have used to help hundreds of students to improve in their problem sums in just a few months. Extract Link Apply 1) Extract You need to train your child to extract the keywords in every question. Examples of common keywords: “more than” “less than” “twice the” “equals to” “at first” “in the end” “of the remainder” “has X times more” “has X times less” Keywords also include all the numbers. All the numbers given in the question are important in solving the question. After your child has identified the keywords, you need to make sure your child is able to comprehend their meaning and know exactly 10 www.jimmymaths.com [email protected] what the question mean. The best way to do this is to ask guiding questions which I will share more on later. 2) Link Using the information given in the question, you need to train your child to link to the right heuristic or concept. “Of the remainder” can be linked to branching. “More than/Less than” and “Before-After” scenarios can be linked to model drawing. Fraction questions which involve “equal amounts” can be linked to equal fractions concept. “In the end” and “Find the number at first” can be linked to working backwards. “Age” questions can be linked to constant difference. “Internal transfer” questions can be linked to constant total. In order for your child to learn how to link properly, your child needs to understand how it works too. For example, for internal transfer questions, you can give your child some erasers and yourself some erasers. Ask your child to count the total erasers at the start. Then give some erasers to your child and ask your child to count the total erasers in the end. The number of erasers should remain the same, unless your child break the eraser into different parts. 11 www.jimmymaths.com [email protected] 3) Apply After your child has extracted the key information and link to the right concept, you will need to train your child on how to apply the concept. For example, Under branching, your child needs to know when to multiply and divide the fractions. Under equal fractions concept, your child needs to know how to make the numerators equal. Under units and parts, your child needs to know how to use Algebra to solve equations as well as the cross multiplication method. After your child has applied the concept to solve the question, make sure your child check his answer by checking against the question again. Often, I saw students who worked rigorously and find one unit, but didn’t give what the question is asking for. What a waste! Remember this E-L-A approach and use it to train your child. With this approach, you will be able to identify your child’s weaknesses and guide him accordingly. 12 www.jimmymaths.com [email protected] Questions are the Answers When your child ask you a question, you do not have to answer with the solutions right away. Instead, a more effective way is to ask him back with more questions, and guide him to solve the question himself. However, your questions need to give clues on what he can do. If not, he will just stare at you with a blank face, with no idea what to answer you. I call this type of questions, “Guiding Questions.” Your questions can help to connect his ideas together, so that he won’t get confused. When he answer your questions, he put his thoughts into words and have a better picture of the question. On top of that, it enables you to check your child’s understanding. You can tell from his answers whether he really knows or he is pretending to know. Most importantly, your questions should empower your solve the question himself. Here are some guiding questions: When Your Child is Reading the Question - What do you understand from this sentence? 13 www.jimmymaths.com [email protected] - What are the keywords in this question? - What is the question asking for? When Your Child is Solving the Question: - How should you represent this information? By model? Or by ratio? - What method should you use? - Why do you want to do it this way? After Your Child has Solved the Question: - How do you check your answer? - Can you use another method to solve this question? - Can you use this method for other types of problem sums? What types? This process may take a longer time to help your child to solve the question. But it enables him to solve the question himself with lesser help. And when your child is able to solve himself, his confidence will spike and he will be able to solve similar questions next time without your help. 14 www.jimmymaths.com [email protected] Helping Your Child To Practice Effectively In a short moment, you are going to learn the important concepts for solving different types of problem sums. Reading and understanding the examples is only half of problem sums mastery. Your child needs to put in practice too. When it comes to practice, many parents make their child practice past year papers. And when their child got stuck, he will refer to the solutions. Unfortunately, most of the solutions lack details and skip many crucial workings. Without a full understanding, your child will simply copy down the solutions, with a false hope that he have understood the problem. When it comes to practice, I believe “Less is More.” Your child do not have to practice many questions. Instead, your child needs to practice the right kind of question and ensure that he fully understands each one. And here’s how you can guide your child. Whenever, he tries a question, ask him to write down the heuristic or concept beside the question. 15 www.jimmymaths.com [email protected] Concept is different from chapter. Do not simply write down “Fraction” beside the question. Instead, write down “Equal Fractions Concept.” You can ask guiding questions as explained before to guide your child to identify the correct concept used. In the end, check his solutions and highlight those concepts which your child is weak in. Then, refer to other practice papers, and select those questions which test the same concepts. Alternatively, just change the numbers in the question yourself and make your child do again. Make your child do those questions until he is confident of solving them himself independently. And remember, do not rush through the workings. Make sure your child understands the rationale behind each step. And the best way to do this is to write down the reason why he is doing this step. Though this takes up more time, it ensures your child understands the flow fully and prevents careless mistakes too. It is better to spend 30 mins on one question and ensuring a full understanding, than to spend 30 mins on 10 questions by just skimming through the solutions. 16 www.jimmymaths.com [email protected] Remainder Concept (Branching) Example 1 3 1 of his money on books and of the remainder on a wallet. 5 3 What fraction of his total money did he have left? John spent 3 (books) 5 Remainder 1 1 (wallet) 3 3 2 5 5 1 2 Leftover 1 3 3 Fraction of Money Left = 2 2 4 (Answer) 3 5 15 Example 2 (Continued from Example 1) If John had $16 left, how much did he have at first? 4 units = $16 15 units = $16 ÷ 4 × 15 = $60 (Answer) Note: If your child is weak in branching or has not learnt branching yet, you can use model to explain to your child instead. However I strongly encourage students to use branching to solve “Remainder” questions as it can save more time. 17 www.jimmymaths.com [email protected] Remainder Concept (Model) Step 1: Draw a model and cut it into 3 units. Step 2: Label the books and remainder. Books Remainder Step 3: Cut each unit into 3 units Books Remainder Step 4: Find the number of units for the wallet and leftover 1 3 Wallet 6 2 units Leftover 6 – 2 = 4 Books Wallet Leftover Step 5: Find the fraction that represent the leftover 4 15 18 www.jimmymaths.com [email protected] Equal Fractions Concept This is one of the most commonly tested concept which most students stumble. In this concept, you need to make the numerators equal (Not the denominator). 5 2 of males is equal to of females, what is the ratio of 7 3 males to females? When we say Step 1: Make the numerator the same 5 10 7 14 2 10 3 15 Step 2: Compare the denominator Males : Females = 14 : 15 Example 1 5 2 of the male dancers is equal to 7 3 of the female dancers. How many more female dancers are there? 87 people are dancing in a hall. Step 1: Make the numerator the same 19 www.jimmymaths.com [email protected] 5 10 7 14 2 10 3 15 Step 2: Compare the denominator Males : Females = 14 : 15 Step 3: Find the total units 29u = 87 Step 4: Find 1 unit 1u = 3 (Answer) Example 2 Lena spent $16 on a pair of jeans. She spent a pizza. 1 of her remaining money on 6 1 of her money was left. How much money had she at first? 2 Step 1: Use branching method to find the fraction for leftover $16 (jeans) Remainder 1 (pizza) 6 Leftover 1 1 5 6 6 20 www.jimmymaths.com [email protected] Step 2: Compare the remainder with the total 1 5 of Remainder = of Total 6 2 Step 3: Make the numerator the same 1 5 2 10 Step 4: Compare the denominator Remainder : Total = 6 : 10 =3:5 Step 4: Find the difference between remainder and total and equate it to the amount for jeans 5u – 3u = 2u 2u = $16 Step 5: Find 1 unit 1u = $8 Step 6: Find the amount of money at first. 5u = $40 (Answer) 21 www.jimmymaths.com [email protected] Model Drawing Concept Model drawing is used extensively from Primary 1 to Primary 5.For students in P6.,I will recommend students to learn Algebra instead. Model Drawing is useful when you see the word “more than” or “less than” in Fractions questions. Example 1 Jenny had 150 more cupcakes than Denise. After Jenny gave away cupcakes and Denise ate 1 of her 5 3 of hers, Jenny had 208 more cupcakes than 4 Denise. How many cupcakes did Denise have at first? Solutions: Step 1: Draw models for Jenny and Denise J 150 D Step 2: Change both fractions to the same denominator 1 4 5 20 3 15 4 20 Step 3: Cut both models into 20 units. J 20u D 20u 150 22 www.jimmymaths.com [email protected] Step 4: Find the leftover portion for the ‘150’. 4 150 120 5 Step 5: Draw models for the leftovers. J D 16u 120 5u 208 Step 6: Find 1 unit. 11u = 208 – 120 1u = 8 20u = 160 cupcakes (Answer) 23 www.jimmymaths.com [email protected] Constant Part Concept Constant part means one of the parts remained the same while the other part changed. In this case, you will need to make the part which remained the same to be equal to each other. Example 1 Ali and Billy have money in the ratio of 5 : 6. After Billy spent $16, the ratio became 3 : 2. How much money does Billy have in the end? Important Thought: Step 1: Make the ratio for Ali the same Before: A:B =5:6 = 15 : 18 24 www.jimmymaths.com [email protected] After: A:B = 3: 2 = 15 : 10 Step 2: Find the difference between Billy’s starting amount and ending amount 18u – 10u = 8u Step 3: Find 1 unit 8u = $16 1u = $2 Step 4: Find the amount for Billy in the end 10u = $20 (Ans) 25 www.jimmymaths.com [email protected] Constant Total Concept Constant Total means the total remained the same. Usually, this concept applies for question related to “Internal Transfer”. If A transfer an amount to B, A will decrease while B will increase by the same amount. So the total will stay the same. Example 2 Ali and Billy have money in the ratio of 5 : 4. After Ali gave Billy $20, they have an equal amount of money. How much money does Billy have in the end? Important Thought: Step 1: Make the total for Ali and Billy to be the same Before: A : B : Total =5:4:9 = 10 : 8 : 18 26 www.jimmymaths.com [email protected] After: A : B : Total =1:1:2 = 9 : 9 : 18 Step 2: Find the difference between Ali’s starting amount and ending amount 10u – 9u = 1u Step 3: Find 1 unit 1 unit = $20 Step 4: Find Billy’s amount in the end 9 units = $180 (Ans) 27 www.jimmymaths.com [email protected] Constant Difference Concept Constant Difference means the difference remained the same. Usually, this concept applies for question related to “Age”. As years go by, the age difference between 2 people will always be the same, because both will grow old together. Example 3 The ages of Ali and Billy are in the ratio of 4 : 7. In 3 years’ time, their ages will be in the ratio of 3 : 5. How old is Billy now? Important Thought: Step 1: Make the difference for Ali and Billy the same Before: A : B : Difference =4:7:3 = 8 : 14 : 6 28 www.jimmymaths.com [email protected] After: A : B : Difference =3:5:2 = 9 : 15 : 6 Step 2: Find the difference between Ali’s starting age and final age 9u – 8u = 1u Step 3: Find 1 unit 1 unit = 3 years Step 4: Find Billy’s age now 14 units = 42 years old (Ans) 29 www.jimmymaths.com [email protected] Everything Changed Concept As the name says, everything changed! Every part changed, the difference changed, the total changed… Nothing remained the same. This type of questions are usually the last few 5 marks questions in the paper. So make sure you know how to do this type of questions. There are a few methods to solve this type of question. I will use an example below to show a method which you can use. Example 4 The ratio of Ali’s money to Billy’s money was 2 : 1. After Ali saved another $60 and Billy spent $150, the ratio became 4 : 1. How much money did Ali have at first? Solutions: Step 1: Write down the starting ratio and apply the changes. A B Before 2u 1u Change +60 - 150 After 2u + 60 1u – 150 Step 2: Compare the final units with the final ratio. A : B = 2u + 60 : 1u – 150 =4 : 1 Step 3: Cross multiply the final units with the final ratio Important Thought: 30 www.jimmymaths.com [email protected] 1 × (2u + 60) = 4 × (1u – 150) 2u + 60 = 4u – 600 Step 4: Solve for 1 unit 2 units = 60 + 600 = $660 (Ans) 31 www.jimmymaths.com [email protected] Excess and Shortage Concept This is another type of common question which you are given 2 conditions. One condition leads to an excess (extra) while the other leads to a shortage (lack). It can also be both conditions lead to excess or both conditions lead to shortage. Example 1 Tom packed 5 balls into each bag and found that he had 8 balls left over. If he packed 7 balls into each bag, he would need another 4 more balls. a) How many bags did he have? b) How many balls did he have altogether? Step 1: Find the difference in the number of balls in each bag 7–5=2 Step 2: Find the number of bags by distributing the 8 extra balls 8÷2=4 Step 3: Find the number of bags by distributing the shortage of 4 balls 4÷2=2 Step 4: Find the total number of bags 4 + 2 = 6 bags (Ans for a) 32 www.jimmymaths.com [email protected] Step 5: Find the total number of balls 6 × 5 + 8 = 38 Or 6 × 7 – 4 = 38 balls (Ans for b) Example 2 When a stack of books were packed into bags of 4, there would be 3 books left over. When the same number of books were packed into bags of 6, there would still be 3 books left over. What could be the least number of books in the stack? Important: You are not able to use the earlier method because the number of bags is not the same in both cases. Step 1: List out the possible number of books for the first case 7, 11, 15, 19, 23… Step 2: List out the possible number of books for the second case 9, 15, 21, 27… Step 3: Find the lowest common number in both cases 15 (Ans) 33 www.jimmymaths.com [email protected] Example 3 Mary had some money to buy some pens. If she bought 12 pens, she would need $8 more. If she bought 9 pens, she would be left with $4. How much money did Mary have? 12 pens Amount of money $8 9 pens $4 Step 1: Find the difference in the pens 12 – 9 = 3 Step 2: Add up the excess and the shortage 8 + 4 = 12 Step 3: Find the cost of 1 pen 12 ÷ 3 = $4 Step 4: Find the total amount of money 4 × 12 – 8 = $40 Or 4 × 9 + 4 = $40 (Ans) 34 www.jimmymaths.com [email protected] Difference Concept Similar to the previous concept, here are some common types of problem sums which require you to find the difference to solve the question. Example 1 John packed some marbles equally into 10 bags. 4 bags were torn, so he transferred the marbles from the 4 bags equally into each of the remaining 6 bags. The remaining bags had 20 more marbles as a result. How many marbles were there altogether? Step 1: Find the total difference in marbles for the 6 bags 6 × 20 = 120 Step 2: Find the number of marbles in each bag at first 120 ÷ 4 = 30 Step 3: Find the total number of marbles 30 × 10 = 300 marbles (Answer) 35 www.jimmymaths.com [email protected] Example 2 Alice and Betty were given an equal amount of money each. Alice spent $20 each day and Betty spent $25 each day. When Alice had $157 left, Betty had $82 left. How much did each girl receive? Step 1: Find the difference in spending for each day 25 – 20 = 5 Step 2: Find the total difference 157 – 82 = 75 Step 3: Find the total days 75 ÷ 5 = 15 days Step 4: Find the amount for each girl 15 × 20 + 157 = $457 (Answer) Or 15 × 25 + 82 = $457 (Answer) 36 www.jimmymaths.com [email protected] Grouping Concept Below are some common questions which require you to group items together followed by finding the number of groups. Example 1 Mark bought an equal number of shorts and shirts for $100. A shirt cost $8 and each pair of shorts cost $12. How much did he spend on the shirts? Step 1: Group 1 shirt and 1 pair of shorts 8 + 12 = 20 Step 2: Find the number of groups 100 ÷ 20 = 5 Step 3: Find the amount spent on the shirts 5 × 8 = $40 (Ans) Example 2 John bought 2 more pairs of shorts than shirts and paid a total of $124. A shirt cost $8 and each pair of shorts cost $12. (a) How many shirts did he buy? (b) How many pairs of shorts did he buy? Step 1: Find the cost of 2 pairs of shorts 12 × 2 = $24 37 www.jimmymaths.com [email protected] Step 2: Find the total cost of equal number of shorts and shirts 124 – 24 = $100 Step 3: Group 1 shirt and 1 pair of shorts 8 + 12 = 20 Step 4: Find the number of groups 100 ÷ 20 = 5 (Ans for a) Step 5: Find the number of pairs of shorts 5 + 2 = 7 (Ans for b) Example 3 Mary makes $2.50 for every bottle of pills she sells. She will be paid a bonus of $10 for every 50 bottles she sell. How many bottles must she sell in order to make a total of $450? Step 1: Find the total amount for 1 group of 50 bottles 50 × 2.50 + 10 = $135 Step 2: Find the number of groups 450 ÷ 135 = 3 R 45 Step 3: Find the number of bottles she sell to earn the remainder of $45 45 ÷ 2.5 = 18 Step 4: Find the total number of bottles 3 × 50 + 18 = 168 (Ans) 38 www.jimmymaths.com [email protected] Example 4 A class consists of an equal number of boys and girls. Mr Tan gave 2 lollipops to every 3 girls and 5 lollipops to every 7 boys. If the total number of lollipops the boys received is 4 more than the total number of lollipops the girls received, how many students are there in the class? Solutions: Step 1: Find lowest common multiple of 3 and 7 3 7 21 Step 2: Find the number of lollipops in 1 group. Girls : 2 7 14 Boys : 5 3 15 Step 3: Find the difference between the number of lollipops each group of girls and boys receive. 15 – 14 = 1 Step 4: Find the number of groups. 4 1 4 Step 5: Find the number of students. 2 4 21 168 (Answer) 39 www.jimmymaths.com [email protected] Number × Value Concept This is a very useful concept. You need to multiply the number of objects by its value to get the total amount. Example 1 The ratio of the number of 50 cents coins to 1 dollar coin is 3 : 1. The total value of the coins is $12.50. How many coins are there in total? Step 1: Write down the ratio of 50 cents : $1 3:1 Step 2: Group three 50 cents coins and one $1 coin into 1 group 3 × 0.5 = $1.50 1 × 1 = $1 Step 3: Find the total value of 1 group $1.50 + $1 = $2.50 Step 4: Find the number of groups 12.50 ÷ 2.5 = 5 Step 5: Find the total number of coins 5 × 4 = 20 coins (Ans) 40 www.jimmymaths.com [email protected] Example 2 James has 44 more 20 cents than 50 cents coins. The total value of the 50 cents coins is $2 more than the total value of the 20 cents coins. How much money does James have? Step 1: Write down the number of 20 cents coins and 50 cents coins in terms of units 50 cents u 20 cents u + 44 Step 2: Find the total value of the 20 cents coins and 50 cents coins 50 cents u × 0.5 = $0.5u 20 cents (u + 44) × 0.2 = $ (0.2u + 8.8) Step 3: Find the difference in their value 0.5u – (0.2u + 8.8) = 0.3u – 8.8 Step 4: Equate the difference to $2 0.3u – 8.8 = 2 Step 5: Find 1 unit 0.3u = 8.8 + 2 0.3u = 10.8 1u = 36 Step 6: Find the total amount of money 0.5 × 36 + 0.2 × 36 + 8.8 = $34 (Ans) 41 www.jimmymaths.com [email protected] Example 3 1 kg of cherries cost $0.50 more than 1 kg of grapes. Mr Tan collected 3 $312 for selling some cherries and grapes. He sold as much cherries as 4 grapes. The total amount collected from the sale of grapes was $24 more than the amount collected from the sale of cherries. a) Find the mass of grapes sold. b) Find the cost of 1kg of cherries. Step 1: Write down the ratio of Cherries : Grapes Cherries : Grapes = 3 : 4 Step 2: Find the total amount spent on cherries and grapes Cherries = (312 – 24) ÷ 2 = $144 Grapes = 144 + 24 = $168 Step 3: Find the cost of 1 unit and cherries and 1 unit of grapes 1 unit of Cherries = $144 ÷ 3 = $48 1 unit of Grapes = $168 ÷ 4 = $42 Step 4: Find the difference between 1 unit of cherries and 1 unit of grapes $48 – $42 = $6 Step 5: Find 1 unit $6 ÷ $0.5 = 12 Step 6: Find the mass of grapes sold 12 × 4 = 48 kg (Ans for a) Step 7: Find the cost of 1kg of cherries $168 ÷ $48 = $3.50 (Cost of 1 kg of grapes) $3.50 + $0.50 = $4 (Ans for b) 42 www.jimmymaths.com [email protected] Guess and Check/ Assumption/ Supposition Concept Guess and Check method can be pretty time consuming. There is a faster and more systematic way which require you to make assumption or supposition. From your assumption, you then tweak it slightly and see the pattern. Example 1 Miss Lee bought some pencils for her class of 8 students. Each girl received 5 pencils and each boy received 2 pencils. She bought a total of 22 pencils. How many boys were there in the class? Step 1: Start with an assumption (You can start with girls or boys). Suppose there are 8 girls Step 2: Find the total number of pencils 8 × 5 = 40 Step 3: Change your assumption Suppose there are 7 girls, 1 boy. Step 4: Find the total number of pencils 7 × 5 + 1 × 2 = 37 Step 5: Spot the pattern 40 – 37 = 3 (When the boys increase by 1, the total pencils decrease by 3) Step 6: Find the total difference 40 – 22 = 18 Step 7: Find the number of boys 18 ÷ 3 = 6 boys (Ans) Check your answer! Number of girls = 8 – 6 = 2 Total pencils = 2 × 5 + 6 × 2 = 22 (Correct) 43 www.jimmymaths.com [email protected] Example 2 Miss Tan has a total of 100 two-dollar and five-dollar notes. The total value of the five-dollar notes was $290 more than that of the two-dollar notes. How much money does she have? Step 1: Start with an assumption Suppose there are 100 five-dollar notes. Step 2: Find the difference in the value of five-dollar notes and two-dollar notes 100 × $5 = $500 Step 3: Change your assumption Suppose there are 99 five-dollar notes, 1 two-dollar note Step 4: Find the difference in the value of five-dollar notes and two-dollar notes 99 × $5 – 1 × $2 = $493 Step 5: Spot the pattern $500 – $493 = $7 (When the two-dollar notes increase by 1, the difference decreases by $7) Step 6: Find the total difference $500 – $290 = $210 Step 7: Find the number of two dollar notes 210 ÷ 7 = 30 Step 8: Find the number of five-dollar notes 100 – 30 = 70 Step 9: Find the total amount of money 30 × $2 + 70 × $5 = $410 (Ans) 44 www.jimmymaths.com [email protected] Simultaneous Equations/ Units and Parts Concept Solving Simultaneous Equations is similar to solving Units and Parts. You need to form 2 equations to solve for 2 unknowns. Example 1 Amy and Billy had a total of $400. Amy spent 1 of her sum and Billy 4 2 of his. They then had a total of $255 left. How much did Amy 5 spend? spent Step 1: Let Amy’s money be 4 units, Billy’s money be 5 parts A 4u B 5p Step 2: Form a first equation using their total amount of money at first 4u + 5p = 400 Step 3: Find the amount of money Amy and Billy have left A 4u – u = 3u B 5p – 2p = 3p Step 4: Form a second equation using their total amount of money left 3u + 3p = 255 45 www.jimmymaths.com [email protected] Step 5: Simplify the second equation to make the number of units the same as the first equation 3u + 3p = 255 u + p = 85 (Divide every term by 3) 4u + 4p = 340 (Multiply every term by 4) Step 6: Use the first equation minus the second equation to find 1 part 4u 5 p 400 (4u 4 p 340) p 60 Step 7: Find 1 unit 85 – 60 = $25 (Ans) Example 2 Ahmad and Bernard had a total of $240. Ahmad spent 2 of his money and 3 3 of his money. As a result, Bernard had $8 more than 5 Ahmad. How much did they spend altogether? Bernard spent Step 1: Let Ahmad’s money be 3 units, Bernard’s money be 5 parts A 3u B 5p Step 2: Form a first equation using their total amount of money at first 3u + 5p = 240 Step 3: Find the amount of money Ahmad and Bernard have left A 3u – 2u = u B 5p – 3p = 2p 46 www.jimmymaths.com [email protected] Step 4: Form a second equation using the $8 which Bernard have more than Ahmad 2p – u = 8 Step 5: Simplify the second equation to make the number of units the same as the first equation 2p – u = 8 6p – 3u = 24 (Multiply every term by 3) Step 6: Use the first equation plus the second equation to find 1 part 3u 5 p 240 (6 p 3u 24) 11p 264 p 24 Step 7: Find 1 unit 5p = 120 3u = 240 – 120 1u = 40 Step 8: Find the amount which they spend altogether 3p + 2u = 3 × 24 + 2 × 40 = $152 (answer) 47 www.jimmymaths.com [email protected] Working Backwards Concept This is another common concept to use when you are given a final value, and you need to work backwards to find the initial value. Example 1 A bus left an interchange carrying some passengers with it. 1 At the first stop, of the people in it alighted and 5 people boarded it. 4 1 At the 2nd stop, of the people in it alighted and 20 people boarded the 2 bus. When it left the 2nd stop, there were 60 passengers in it. How many passengers were there in the bus when it left the interchange? Step 1: Find the number of people before the 2nd stop 60 – 20 = 40 40 × 2 = 80 Step 2: Find the number of people before the 1st stop 80 – 5 = 75 75 ÷ 3 × 4 = 100 people (Ans) Example 2 Amy and Betty have a total of 48 sweets. Amy gave Betty. Betty then gave 1 of her sweets to 4 1 of her total sweets to Amy. In the end, each of 3 them had the same number of sweets. How many sweets did each of them have at first? Step 1: Find the number of sweets each of them had in the end. 48 ÷ 2 = 24 48 www.jimmymaths.com [email protected] Step 2: Draw a table and fill in the sweets in the end. A B 24 24 Start AB BA End Step 3: Find the number of sweets before Betty gave to Amy. B 24 ÷ 2 × 3 = 36 A 48 – 36 = 12 A B 12 36 24 24 Start AB BA End Step 4: Find the number of sweets at first. A 12 ÷ 3 × 4 = 16 sweets (Ans) B 48 – 16 = 32 sweets (Ans) Start AB BA End A 16 B 32 12 36 24 24 49 www.jimmymaths.com [email protected] Double If Concept Here are 2 types of very common “double if” questions. Example 1 A farmer has some chickens and ducks. If he sells 2 chickens and 3 ducks every day, there will be 50 chickens left when all the ducks have been sold. If he sells 3 chickens and 2 ducks every day, there will be 25 chickens left when all the ducks have been sold. a) how many ducks are there? b) how many chickens are there? Step 1: Write down the selling ratio for both cases Case 1: Chicken : Duck = 2u : 3u Case 2: Chicken : Duck = 3u : 2u Step 2: Make the ratio of ducks to be the same as all the ducks are sold out in both cases. Case 1 (Times 2 to both sides) Chicken : Duck = 2u : 3u = 4u : 6u Case 2 (Times 3 to both sides) Chicken : Duck = 3u : 2u = 9u : 6u Step 3: Form an equation using the chickens left 4u + 50 = 9u + 25 50 www.jimmymaths.com [email protected] Step 4: Find 1 unit 5u = 50 – 25 1u = 5 Step 5: Find the total ducks 5 × 6 = 30 ducks Step 6: Find the total chickens 5 × 4 + 50 = 70 chickens Or 5 × 9 + 25 = 70 chickens Example 2 If Sam gives Terry $350, he will have 4 times as much money as Terry. If Sam gives Terry $550, he will have thrice as much money as Terry. How much money does Sam have? Step 1: Using the first ‘if’, write down the final ratio S:T = 4u :1u Step 2: Find the starting amount S:T = 4u + 350 : 1u – 350 Step 3: Using the second ‘if’, find the final amount S:T = 4u + 350 – 550 : 1u – 350 + 550 = 4u – 200 : 1u + 200 Step 4: Write down the final ratio S:T = 4u – 200 : 1u + 200 =3:1 51 www.jimmymaths.com [email protected] Step 5: Form an equation 3 × (1u + 200) = 1 × (4u – 200) Step 6: Find 1 unit 3u + 600 = 4u – 200 4u – 3u = 600 + 200 1u = 800 Step 7: Find the amount for Sam 4 × 800 + 350 = $3550 52 www.jimmymaths.com [email protected] Conclusion We have come to the end of this guide. As you are learning the concepts which I just shared, keep in mind that your child may have other methods of doing the same question. Keep an open mind to more methods and ideas. In Math, sometimes, there are more than 1 way to get the same answers. And it will be beneficial for your child to learn more methods. Knowing more methods will ensure your child have more tools to use in exams. The concepts which I have shared with you are a small part of all the concepts which your child needs to learn for PSLE. The rest of the concepts are covered in more details in our PSLE Math Online Course and tuition classes. If you need more help, check out more details in the next page. As the saying goes, “Give a man a fish and you feed him for a day. Teach a man how to fish and you can feed him for a lifetime.” Impart the essential problem solving skills to your child. With the right skills and attitude, your child will certainly score the grades he deserve for PSLE. Most importantly, Have fun learning together with your child! 53 www.jimmymaths.com [email protected] Online Courses and Tuition Classes Online Courses Find out more by clicking below: https://jimmymaths.com/our-courses/ - Learn from recorded videos - Get access to lots of common exam questions to ensure sufficient practice - Get unlimited support and homework help If you have any questions, you can email us at [email protected] Group Tuition Classes Together with a team of tutors, we conduct Math tuition at tuition centre, Grade Solution Learning Centre, located near Kovan MRT. We also have a new branch near Bugis MRT. You can find out more details below. Contact Us Call: 6280 6841 Website: https://gradesolution.com.sg 54