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your-complete-guide-to-heuristic-math

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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
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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?
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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!
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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…
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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
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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
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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.
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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.
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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
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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.
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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.
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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?
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- 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.
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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.
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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.
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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.
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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
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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
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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
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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)
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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
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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)
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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
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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)
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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
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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)
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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
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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)
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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:
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1 × (2u + 60) = 4 × (1u – 150)
2u + 60 = 4u – 600
Step 4: Solve for 1 unit
2 units = 60 + 600 = $660 (Ans)
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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)
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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)
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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)
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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)
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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)
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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
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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)
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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)
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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)
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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)
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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)
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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)
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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)
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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
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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
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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)
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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
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Step 2: Draw a table and fill in the sweets in the end.
A
B
24
24
Start
AB
BA
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
AB
BA
End
Step 4: Find the number of sweets at first.
A  12 ÷ 3 × 4 = 16 sweets (Ans)
B  48 – 16 = 32 sweets (Ans)
Start
AB
BA
End
A
16
B
32
12
36
24
24
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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
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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
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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
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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!
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