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Adrian Ng Principal Trainer. Proudly Presents Heuristics Approach Solving Challenging Primary Mathematical Problems. Guess and Check ( 3 guesses + Look for pattern ).

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proudly presents heuristics approach solving challenging primary mathematical problems

Adrian Ng

Principal Trainer

Proudly Presents

Heuristics Approach

Solving Challenging Primary Mathematical Problems

slide3
String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question

Big (90/2=45cm)

slide4
String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question

Big (90/2=45cm)

Small (120/5=24cm)

slide5
String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question

Big (90/2=45cm)

Small (120/5=24cm)

Diff ( 0 )

slide6
String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question

Big (90/2=45cm)

Small (120/5=24cm)

Diff ( 0 )

no.

length

no.

length

2520-0= 2520

0

45x0=0

105

105x24=2520

slide7
String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question

Big (90/2=45cm)

Small (120/5=24cm)

Diff ( 0 )

no.

length

no.

length

2520-0= 2520

0

45x0=0

105

105x24=2520

2520-2499= 21 (pattern)

+120

2544-45= 2499

1

45x1=45

106

106x24=2544

Gap/pattern

2520/21=120

2520-0= 2520 (gap)

0+120= 120

120x45=5400

120+105= 225

225x24=5400

( 0 )

slide8
String of 2 big balloons is 90cm. String of 5 small balloons is 1.2m. If both strings are of the same length, there would be 105 more small balloons than the big balloons. How many balloons are there altogether? 2009 PSLE question

Big (90/2=45cm)

Small (120/5=24cm)

Diff ( 0 )

no.

length

no.

length

2520-0= 2520

0

45x0=0

105

105x24=2520

2520-2499= 21 (pattern)

+120

2544-45= 2499

1

45x1=45

106

106x24=2544

Gap/pattern

2520/21=120

2520-0= 2520 (gap)

0+120= 120

120x45=5400

120+105= 225

225x24=5400

( 0 )

120+225=345

Ans: 345 balloons

slide9

Note: This question was provided by students who have sat for the 2009 examination. It may vary from the actual examination question.

slide11
Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question

Friends

Group

1

1

1

1

1

1

1

2

1

1

3

1

1

1

1

1

1

1

4

1

1

1

1

5

1

6

1

1

1

1

4

4

4

4

4

4

Total (each)

slide12
Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question

Friends

Group

1

1

1

1

1

1

1

2

1

1

3

1

1

1

1

1

1

1

4

1

1

1

1

5

1

6

1

1

1

1

4

4

4

4

4

4

Total (each)

2.00pm to 4.30pm  150min

slide13
Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question

Friends

Group

1

1

1

1

1

1

1

2

1

1

3

1

1

1

1

1

1

1

4

1

1

1

1

5

1

6

1

1

1

1

4

4

4

4

4

4

Total (each)

2.00pm to 4.30pm  150min

150min/6 = 25 min per group

slide14
Six friends decided to rent computers from 2.00pm to 4.30pm. Four of them were playing while the other two would watch. If the cycle continues, and each of them played for equal number of minutes, how many minutes will each person get to play? 2009 PSLE question

Friends

Group

1

1

1

1

1

1

1

2

1

1

3

1

1

1

1

1

1

1

4

1

1

1

1

5

1

6

1

1

1

1

4

4

4

4

4

4

Total (each)

2.00pm to 4.30pm  150min

150min/6 = 25 min per group

25x4 = 100 min

Ans: 100 min

slide15

Note: This question was provided by students who have sat for the 2009 examination. It may vary from the actual examination question.