Thinking skills are skills used in solving mathematical problems, such as classifying, comparing, sequencing, analysing parts and wholes, identifying patterns and relationships, induction, deduction, and spatial visualisation. Examples of heuristics are drawing a diagram, making a list, using equations, using guess-and-check, looking for patterns, making suppositions, acting it out, working backwards, working before-after, re- stating the problems, simplifying the problem and solving part of the problem.

1.Making Suppositions – Mr Zanizam 2.Working Backwards – Mr Long 3.Working Before and After – Mr Joseph Lim 4.Re-stating the problem/Simplifying the problem – Mrs Elizabeth Simon

In a coop, there are 6 spiders and chickens. If there are 36 legs in all, find the number of chickens in the coop. Step 1 – Assume all animals are chickens. 6 chickens’ legs  2 × 6 = 12 Why do we need to find the difference? Since spiders have more legs than chickens, it is easier to “hide” their legs and assume all animals are chickens. Therefore, each spider is hiding 6 legs. Step 2 – Find the difference in the number of legs between each pair of animals. 8 – 2 = 6

In a coop, there are 6 spiders and chickens. If there are 36 legs in all, find the number of chickens in the coop. Step 3 – Find the difference between the number of chickens’ legs and the total number of legs of the two animals. 36 – 12 = 24  (hidden spiders’ legs) Step 4 – Find the number of spiders. 24 ÷ 6 = 4 Step 5 – Find the number of chickens. 6 – 4 = 2 Checking: 4 × 8 = 32 (spiders’ legs) 2 × 2 = 4 (chickens’ legs) = 36

There were 15 Math Olympiad questions. 5 marks were awarded for every correct answer and for each wrong answer, 1 mark was deducted. Sam scored 33 marks. How many questions did he answer correctly? Assuming all questions  5 x 15 answered correctly= 75 Difference in marks  75 – 33 = 42 Marks deducted for each  wrong answer= 6 No. of wrong answers  42 ÷ 6 = 7 No. of correct answers  15 – 7 = 8

Patrick had a sum of money. On the first day, he spent of the money and donated $30 to charity. On the second day, he spent of the money he still had and donated $20 to charity. On the third day, he spent of the money he still had and donated $10 to charity. In the end, he had $10 left. How much did he have at first?

Patrick had a sum of money. On the first day, he spent of the money and donated $30 to charity. spent On the second day, he spent of the money he still had and donated $20 to charity. 1 st Day 2 nd Day spent On the third day, he spent of the money he still had and donated $10 to charity. 3 rd Day 20 d spent 10 d left = 20 2 × 20 = = 6060 ÷ 2 = = = ÷ 3 = × 4 = 160 He had $160 at first. In the end, he had $10 left. How much did he have at first? 30 d Checking: 160 ÷ 4 = 40 (spent on 1 st day) 160 – 40 – 30 = 90 (left after 1 st day) 90 ÷ 3 = 30 (spent on 2 nd day) 90 – 30 – 20 = 40 (left after 1 st day) 40 ÷ 2 = 20 (spent on 3 rd day) 20 – 10 = 10 (left after 3 rd day)

John had some books. He gave May half of the books plus 1 book. He gave Mike half of the remaining books plus 2 books. If he had 5 books left after this, how many books had John at first?

John had some books. He gave May half of the books plus 1 book. May 1 He gave Mike half of the remaining books plus 2 books. Mike 2 If he had 5 books left after this, how many books had John at first? = 7 7 × 2 = = × 2 = 30 He had 30 books at first. Checking: 30 ÷ = 16 (given to May) 30 – 16 = 14 (left after giving to May) 14 ÷ = 9 (given to Mike) 14 – 9 = 5 (left after giving to Mike)

Working Before and After

Before and After Example 1 Clement had thrice as many apples as pears at his stall at first. After he sold 129 apples and threw away 8 rotten pears, he had only half as many apples as pears left. How many pears had he left?

Before Apples Pears 88 1 part Clement had thrice as many apples as pears at his stall at first. Apples Pears 129 sold apples 8 pears thrown away left After After he sold 129 apples and threw away 8 rotten pears, he had only half as many apples as pears left. How many pears had he left?

8 × 3 = 24 5 units  129 – 24 = unit  105 ÷ 5 = 21 2 units  21 × 2 = 42 He had 42 pears left. Question: How many pears had he left? Before Apples Pears 8881u thrown away 1u Apples Pears 129 sold 8 thrown away left After 1u

Before and After Example 2 James had twice as many pens as erasers at first. After he gave away 170 pens and 10 erasers, he had thrice as many erasers as pens left. How many pens had he at first?

Before Pens Erasers 1 part James had twice as many pens as erasers at first. After he gave away 170 pens and 10 erasers, he had thrice as many erasers as pens left. How many pens had he at first? Pens Erasers 170 given away 10 lost away left After

10 × 2 = 20 5 units  170 – 20 = unit  150 ÷ 5 = 30 6 units  30 × 6 = = 200 He had 200 pens at first. Question: How many pens had he at first? Pens Erasers 170 given away 10 lost away left After Before Pens Erasers 8 10 lost away 10 1u

RE-STATING AND SIMPLIFYING THE PROBLEM

Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have? PROBLEM 1

Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have? (A)10¢ (C)20¢ No. of coins Value of coins $1.20 (A) 10¢ (C) 20¢

10¢ 20¢ What is the difference between them? 20¢ – 10¢ = 10¢ Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have?

Value of coins 10¢ 20¢ $ ¢ – 10¢ = 10¢ 120¢ ÷ 10¢ = 12 (sets of 10¢ and 20¢) Each of them has 12 coins. Aric has some 10-cent coins. Cathy has some 20-cent coins. Both have the same number of coins. Cathy has $1.20 more than Aric. How many coins does each of them have?

Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi. Find the amount of money that Devi had. PROBLEM 2

(D)10¢ (B)20¢ No. of coins 2 Value of coins (D)10¢ (B)20¢ $0.40 Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi. Find the amount of money that Devi had.

The 2 10-cent coins will give a value of 2 × $0.10 = $ ¢ 20¢ No. of coins 2 Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi. Find the amount of money that Devi had.

Value of coins (D)10¢ (B) 20¢ $0.40 The value of the extra 2 coins can be found in the model below. $0.20 Value of coins (D)10¢ (B)20¢ $0.40$0.20 Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi. Find the amount of money that Devi had. To make the no. of coins the same, we remove the 2 coins from 10¢ model.

$ $0.40 = $ ¢ – 10¢ = 10¢ 60¢ ÷ 10¢ = 6 (sets of 10¢ and 20¢) Value of coins (D) 10¢ (B) 20¢ $0.40$0.20 Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi. Find the amount of money that Devi had.

2 × $0.10 = $0.20 $ $0.40 = $ ¢ – 10¢ = 10¢ 60¢ ÷ 10¢ = 6 (sets of 10¢ and 20¢) = 8 (No. of coins Devi has) 8 × $0.10 = $0.80 (Devi) Devi had a number of 10-cent coins and Bala had a number of 20-cent coins. Devi had 2 more coins than Bala but Bala had $0.40 more than Devi. Find the amount of money that Devi had.