## Teaching Approach ### Engage

Here, we engage our students and grab their attention before we start going through the topic by sharing with them:
• How the topic they are about to learn can be applied in real life
• History of Mathematics (Secondary)
• Discussions based on the students’ initial understanding of the topic to make connections between past and present learning experiences

### Evaluate

Finally, at the end of the lesson, we evaluate our students’ understanding via the following:
• Evaluation of students’ workings and correction of misconceptions
• Clarification of common misunderstanding of concepts
• Reinforcement of concepts
• Recapitulation based on the students’ unique needs

### Elaborate

Here we extend our students’ understanding by:
• Expanding on the concepts they have learned
• Making connections to other related concepts
• Applying their understanding to the world around them
This is done by exposing them to:
• Real world application questions
• Contextual questions
• Specially adapted questions based on concepts tested in schools and national exams.

### Explore

Next, we get them directly involved with the topic and materials and start on their exploratory journey by:
• Conducting Exploratory Activities
• Allowing students to interact with Flash applets or videos specially selected to invoke students’ interest and at the same time educate them

### Explain

Once we have gained our students’ attention, we start to:
• Explain the concepts involved in the everyday applications and the videos or flash applets used
• Determine levels of understanding and identify possible misconceptions
• This is done via the use of our specially designed Just Mathematics notes.

### JUSTEDU students are exposed to every single Maths concept ever tested in school

#### Higher Order Thinking PRIMARY 5/6 QUESTION

Gillian and Cadence packed 4000 presents for an orphanage. The time Gillian took to pack 7 presents  was similar to the time Cadence took to pack 9 presents. They earned 45₵ for every present packed. An additional \$1.50 was earned for every 30 presents packed. How much more did Cadence earn than Gillian?

#### CLICK FOR SOLUTIONS

4000 ÷ (7 + 9) = 250

250 × 7 = 1750

250 × 9 = 2250

1750 × 45₵ = \$787.50

1750 ÷ 30 = 58 R 10

\$787.50 + 58 × \$1.50 = \$874.50

2250 × 45₵ = \$1012.50

2250 ÷ 30 = 75

\$1012.50 + 75 × \$1.50 = \$1125

\$1125 − \$874.50 = \$250.50

Ans: \$250.50

#### CHALLENGING QUESTION (P3)

Maurice had some \$5-notes and twice as many \$2-notes.
The total value of all her notes is \$207.
Find the value of all her \$2-notes.

#### CLICK FOR SOLUTIONS

2 × \$2 = \$4
\$5 + \$4 = \$9 per group
\$207 ÷ \$9 = 23 groups
23 × 2 = 46 (Number of \$2 notes)
46 × \$2 = \$92

Ans: \$92

#### Higher Order Thinking PRIMARY 5/6 QUESTION

A container measuring 1.2 m by 65 cm by 40 cm was filled with water. At 3.30 p.m., water from a tap was turned on to fill the container at a rate of 3.25 l per minute. When the container was filled, the base of the container cracked and water leaked out of the container at a rate of 1250 ml per minute.

 (a) How many litres of water were there in the tank at first? (b) At what time will the tank be completely filled?

#### CLICK FOR SOLUTIONS

 (a) × 120 × 65 × 40 = 104 000 cm3= 104 000 ml = 104 l Ans: 104 l (b) － =  × 120 × 65 × 40 = 52 000 cm3 52 000 ÷ 3250 = 16 min 3250 – 1250 = 2000 cm3 × 120 × 65 × 40 = 156 000 cm3 156 000 ÷ 2000 = 78 min 78 + 16 = 94 min = 1 h 34 min 1 h 34 min after 3.30 p.m. is 5.04 p.m. Ans: 5.04 p.m.

#### PRIMARY 5 HEURISTICS

Shawn, Collin and Eddie bought some marbles to play against each other.
In the first round, Shawn lost half of his marbles to Eddie.
In the second round, Collin won some marbles from Eddie and his number of marbles doubled.

In the third round, Shawn lost of his marbles to Collin.
At the end of the game, the three boys had the same number of marbles.

If Colin had 360 marbles at first, how many marbles did the three boys have altogether?

#### CLICK FOR SOLUTIONS

When Shawn lost of his marbles to Collin, When Eddie lost (Collin won), Collin’s marbles were halved: When Shawn lost half of his marbles to Eddie: 3 units = 360 1 unit = 360 ÷ 3 = 120 24 units = 24 × 120 = 2880

Ans: 2880

### SECONDARY

How our students benefit from Just Mathematics?
– Holistic and Accelerated Approach to Learning Math Concepts
– Specially Designed In-house Curriculum Framework
– Ample Practices with Comprehensive Worked Examples

Appreciation of Maths
Through our special sections:
– History of Maths
– Investigative Maths

Customised curriculum to meet the needs of IP students

Exam-Oriented (School/’O’-Level Based) Revision Papers
– To provide intensive practice
– And for tutors to assess students’ understanding before their examinations

#### SECONDARY 2 MATHS

A contractor agrees to lay a road of length 3000 m in 20 days. He employs 80 men who work 8 hours a day. After 15 working days, he finds that only 2000 m of the road is completed. How many more men should he employ so as to finish the work on time if each man now works 10 hours a day?

#### CLICK FOR SOLUTIONS

 No. of hours worked in 15 days = 8 × 15 = 120 Remaining no. of hours to work = 5 × 10 = 50 Remaining length of road to lay = 2000 – 1000 = 1000 m

Let the number of workers be x and the total number of hours be y.
Since the number of workers is inversely proportional to the total number of hours, Let the number of workers be x and the total length of road be l.
Since the number of workers is directly proportional to the length of road, Therefore, 96 – 80 = 16 more men are required.

#### SECONDARY 3 E. MATHS QUESTION

In a bridge structure, the mast above each pillar is 87 metres high. The gradient at which the longest steel cable is connected to the centre of the bridge is such that the ratio of vertical distance : horizontal distance is 29 : 57.

a) Show that the angle, α , is 63.03° correct to 2 decimal places.

b) Calculate the length of 2 pieces of the longest steel cable attached from the top of the mast to the centre of the bridge.

#### SECONDARY 3 E. MATHS QUESTION

In the diagram, ΔFCE is similar to ΔABE. Given that ΔABC has an area of 24 cm2,
3BC = 2CE and 4DE = DA,

 a) find the area of ΔACE; b) find the area of ΔFCE; c) show that area of ΔCDE = area of ΔABC; d) find the value of #### CLICK FOR SOLUTIONS

 a) b) c) d) #### SECONDARY 4 A. MATHS QUESTION

Crime Scene Investigation
A dead body was discovered in Hotel California, Room 104 at 10.00 a.m. The pathologist measured the temperature of the body to be 34°C and the room temperature at that instant is 20°C.

• Deriving from Newton’s law of cooling, the rate at which the temperature of a body changes is given by the formula , where T is the temperature of the body after cooling for n minutes.
• Assuming that the room temperature is fixed at 20°C throughout and the deceased at the time of death has a normal body temperature of 37°C,

What is the time of death?

#### CLICK FOR SOLUTIONS 