Singapore Primary Math Explained sc-math

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Updated: 25 October, 2007

Shop A has 156 kg of rice. Shop B has 72 kg of rice. After both shops sold an equal amount of rice, the ratio of rice that shop A has to shop B is 4:1. Find the amount of rice sold by each shop. (3 marks)

1 unit => 84 ÷ 3 = 28 kg

Shop B sold 72 − 28 = 44 kg

As you read my solutions, please note that for the sake of clarity, they are probably more "wordy" than necessary. In the answer above, instead of writing out the statements like I did, a pupil will probably work out the value of 1 unit on the side and put the figure 28 at the unsold unit for shop B. He thens mentally subtracts 28 from 72 and put the value 44 at the sold portion of the bar for shop B. He only writes the final answer statement.

6 |

14 |

6 |

14 |

7 |

14 |

7 |

14 |

6 |

14 |

1 |

14 |

Total chairs = 14 ×

1 |

14 |

note: to get 14 × 112 quickly without a calculator, do (14×100) + (14×10) + (14×2) to get 1400 + 140 + 28. Alternatively, do (112×10) + (112×4).

**Note (updated 25 Oct 07):** If you try to confirm the answer, you'll realize that you cannot arrange 672 (6/14 of 1568) chairs into rows of 13. Probably the person who memorized this question got the numbers wrong (change 112 to 117). Hopefully, its not a mistake in the original paper as pupils who tried to confirm their answer may get confused and loose time double-checking their work.

Tank A is filled with water to its brim. Water is then poured from Tank A to Tank B until both share the same height. What is this height? (4 marks)

Both Tank A and Tank B are 40 cm wide.

Therefore the ratio of the volume of water in each tank to the surface area of the water at it's front glass panel is the same for both tanks.

Since volume remains the same,

area of surface A = area of surface B + area of surface C

surface A = 50 cm × 60 cm = 3000 cm²

surface B + C = (50 cm + 70 cm) × height

height = 3000 ÷ 120 = 25 cm

As you can see, the solutions to the three questions are not impossibly difficult. I am sure there are other ways of solving the questions. For example, in question 3, some may prefer to calculate the total volume of tank A and then divide the total volume by the combined base area of tanks A and B to get the common height.

The solutions are not difficult once you know what to do. The point is that to be able to know what to do, especially under pressure during exams, pupils must have sufficient exposure and practise in analyzing and solving problems. For example, one of the recommended heuristics is "simply the problem". Pupils who had practised "simplifying the problem" will have a better chance spotting the possibility of simplifying question 3 by working on the area rather than the volume.

I believe that primary 6 pupils of average intelligence should be able to solve these challenging questions (except for the few questions aimed at the exceptionally bright pupils) if they

- - are keen to learn and are willing to put in the time and effort to practise properly;
- - had been practising properly since the lower primary (no accumulated deficit to hinder understanding of lessons in higher primary);
- - had received proper guidance on problem solving skills;
- - are attentive in class; and
- - have a reasonable command of the English language (able to read and understand the question).

(You may wish to read Primary math problem solving heuristics, More math practise = better math grades? and How to solve math problems if you haven't done so.)

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