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Heuristics ApproachHeuristics ApproachSolving Challenging Primary Mathematical Solving Challenging Primary Mathematical

ProblemsProblems

Adrian NgAdrian NgPrincipal TrainerPrincipal Trainer

GUESS AND CHECK

(3 GUESSES + LOOK FOR

PATTERN)

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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)

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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)

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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 )

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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

0 45x0=0 105 105x24=2520

2520-0= 2520

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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

0 45x0=0 105 105x24=2520

2520-0= 2520

1 45x1=45 106 106x24=2544

2544-45= 2499

2520-2499= 21 (pattern)

( 0 )

2520-0= 2520 (gap)

Gap/pattern2520/21=120

+120

0+120= 120

120x45=5400

120+105= 225

225x24=5400

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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

0 45x0=0 105 105x24=2520

2520-0= 2520

1 45x1=45 106 106x24=2544

2544-45= 2499

2520-2499= 21 (pattern)

( 0 )

2520-0= 2520 (gap)

Gap/pattern2520/21=120

+120

0+120= 120

120x45=5400

120+105= 225

225x24=5400

120+225=345 Ans: 345 balloons

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

SYSTEMATIC LISTING

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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

1st 2nd 3rd 4th 5th 6th

1

2 11113 1111

4 1111

5 1111

6 1111

GroupFriends

444 444Total (each)

1111

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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

1st 2nd 3rd 4th 5th 6th

1 1111

2 11113 1111

4 1111

5 1111

6 1111

GroupFriends

444 444Total (each)

2.00pm to 4.30pm 150min

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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

1st 2nd 3rd 4th 5th 6th

1 1111

2 11113 1111

4 1111

5 1111

6 1111

GroupFriends

444 444Total (each)

2.00pm to 4.30pm 150min

150min/6 = 25 min per group

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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

1st 2nd 3rd 4th 5th 6th

1 1111

2 11113 1111

4 1111

5 1111

6 1111

GroupFriends

444 444Total (each)

2.00pm to 4.30pm 150min

150min/6 = 25 min per group25x4 = 100 minAns: 100 min

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

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