Hi, appreciate someone can help to solve for this question. Thanks in advance!

Source: Anglo-Chinese School Primary

Hi, appreciate someone can help to solve for this question. Thanks in advance!

Source: Anglo-Chinese School Primary

BigDevil

For this kind of question, I always taught my daughter to think logically, rather than to memorize any formula/method.

At first, there were

45 45 45 …. 45 in unknown number of containers, with excess of 33.

Next, to place 53 beads into the containers, Kate needed to place 8 more beads into each container, first using the 33 beads she had left.

8 | 8 | 8 | 8 | 1 | ||||

7 | 8 | 8 | 8 | 8 |

The first row is the 33 beads left, second row is the 39 beads she needed.

We can then see there were a total of 9 containers.

Acestarling

BigDevil,

Thank you for sharing this.

Normally at one look at the question and the “standard conventional method” will pops up in mind and seldom will give it much further second thought. Didn’t realised this problem can be approached in this “unique” way.

BigDevil

Glad you found it interesting.

Just wanted to train my children to think and analyse, rather than spoon feeding them formulas. After analysing the problem and solution, I would then guide them to come up with the formula themselves.

Although their end formula was exactly the same as the one given by sky minecrafter, they had the satisfaction of knowing that they themselves came up with the formula, and that made it easier for them to remember the formula for future use.

elitemathstudio

39 + 33 = 72

53 – 45 = 8

72/8 = 9

DoAppliedLearning

Total Container = 1 Unit

EQ1: 45 units = Total beads – 33

EQ2: 53 Units = Total Beads + 39

EQ2 – EQ1: 8 Units = 72

1 Unit = 9 (containers)

alfretztay

The above problem could be solved as follows :

45 x number of containers + 33 = 53 x number of containers – 39

number of containers = (33 + 39)/(53 – 45) = 9

Ans : 9 containers.

Khong Pek Mao

Hi Alfredztay,

After reading your long comment, I agreed with you said about different type of methods.

So in my P4 pupils, they can understand the use of units and numbers. In fact, for P4 pupils, I have talked to their parents there may be a need to talk to their Maths teacher during the Meeting the Parents session.

Fortunately, none of them need to. This is because the pupils can perform in their problem solving.

We are always aware that different kid learn in different ways.

Those can use the school method will always perform well, because they know the tricks. I believe the tutors here are experts in this, and your children will always top in class.

Those who half half will always get borderline marks, because no matter how hard they try, the problem always got something to trick them. Their parent will always ask them to “work hard”. In fact both parents and kid do not know how?

Those who do not know how to use the school method will just sit down and relax, because they know that no matter what they do, they are always wrong.

We should be more concerned on the second and third group of pupils. Maths are very flexible. There are many ways to solve them, not just one school method.

So long that a kid can perform more than 10 questions, there should not be any reason to stop him to use a new skill.

alfretztay

“Hi Alfredztay,

After reading your long comment, I agreed with you said about different type of methods. …

The above problem could be solved as follows :

45 x number of containers + 33 = 53 x number of containers – 39

number of containers = (33 + 39)/(53 – 45) = 9

Ans : 9 containers.”

I just provided my solution, as above, to the question. Where and what is the long comment?

Ender

Number of containers be u

Number of beads = 45u + 33

Number of beads is also = 53u – 39

53u – 39 = 45u + 33

8u = 72

u = 9 ###

Acestarling

I notice a few of the solutions provided here use various methods such as assigning a variable to the “number of containers” ( algebra ? ) , matrix , excess and shortage. It is good to see everyone sharing the various ways to solve a problem, but do be careful whether the method you used is “suitable” for the level.

I believe at P4 level, a student is expected to understand the “excess and shortage” concept, but I am not too sure if the other methods are also being covered / taught at P4 level in every school ? I have heard cases where a student use algebra to solve a P4 problem but the solution was “rejected” by the teacher on the pretext that the method was not taught to the student and advice the student to use the “Before / After” concept. ( Perhaps sometimes the rational of setting a particular type of question is to test the student’s understanding and application of a certain mathematical concept ? ) Of course this situation may be subjective depending on how strictly that teacher follows the MOE padegogy. But I think it is good to be more careful if your child’s teacher is “strict” on the “approved” techniques / methods to use, especially for lower primary level. Perhaps some school teacher here may give us some advice ?

If this is a question appears in a P5 or P6 papers, I would have no qualms about which method to use.

Khong Pek Mao

This is a Matrix question.

Container | Left | Total | |

Ratio 1 | 45u | 33 | 45u + 33 |

Ratio 2 | 53u | -39 | 53u – 39 |

45u + 33 = 53u – 39

33 + 39 = 53u – 45u

8u = 72

1u = 72 ÷ 8 = 9