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.

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)32 + 4 = 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 15answered correctly = 75

Difference in marks 75 – 33 = 42

Marks deducted for each 5 + 1wrong 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?

2

1

3

14

1

Patrick had a sum of money.

On the first day, he spent of the money and donated $30 to charity.4

1

spent

On the second day, he spent of the money he still had and donated $20 to charity.3

1

1st Day

2nd Day

spent

On the third day, he spent of the money he still had and donated $10 to charity.2

1

3rd Day

20

d

spent

10

d

10

left

10 + 10 = 20

2 × 20 = 40

40

40 + 20 = 6060 ÷ 2 = 30

30 + 60 = 90

90 + 30 = 120120 ÷ 3 = 40

40 × 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 1st day)160 – 40 – 30 = 90 (left after 1st day)90 ÷ 3 = 30 (spent on 2nd day)90 – 30 – 20 = 40 (left after 1st day)40 ÷ 2 = 20 (spent on 3rd day)20 – 10 = 10 (left after 3rd 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?

5

2 + 5 = 7

7 × 2 = 14

14 + 1 = 15

15 × 2 = 30He had 30 books at first.

Checking:30 ÷ 2 + 1 = 16 (given to May)30 – 16 = 14 (left after giving to May)14 ÷ 2 + 2 = 9 (given to Mike)14 – 9 = 5 (left after giving to Mike)

Working Before and After

Before and AfterExample 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

8 8

1 part 1 part 1 part

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 = 105

1 unit 105 ÷ 5 = 21

2 units 21 × 2 = 42

He had 42 pears left.

Question:

How many pears had he left?

Before

Apples

Pears

8 8 81u 1u 1u 1u 1u1u 88 8

8 thrown away1u 1u

Apples

Pears

129 sold

8 thrown away

left

After

1u

1u 1u

Before and AfterExample 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 1 part

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 = 150

1 unit 150 ÷ 5 = 30

6 units 30 × 6 = 180

180 + 10 + 10 = 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 101u 1u 1u 1u 1u1u

1u1u1u

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 1PROBLEM 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¢ $1.20

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 2PROBLEM 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 = $0.20

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

$0.20

Value of coins

(D)10¢

(B)20¢ $0.40$0.20

$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.

$0.20 + $0.40 = $0.60

20¢ – 10¢ = 10¢

60¢ ÷ 10¢ = 6 (sets of 10¢ and 20¢)

Value of coins

(D) 10¢

(B) 20¢ $0.40$0.20

$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

$0.20 + $0.40 = $0.60

20¢ – 10¢ = 10¢

60¢ ÷ 10¢ = 6 (sets of 10¢ and 20¢)

6 + 2 = 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.