Fight Against School Fights In Lae (i): School's Culture Vs Students' Culture

STOP: school fights in Lae city schools


UPDATED 17th December 2018

This is first of five write-up about school fights among Lae city schools. I am going to look at what was wrong, how groups are formed, impacts on students' lives, what Morobe education officials and politicians can do, what can be done and what if what can be done is not done.

School, as an organisation, functions within a culture: school facilitates a way to do things and participants (students and teachers) adhere to it. This then creates vibrant learning atmosphere within which stems adjacent nomenclatures like students' ethos and staff's code of practice.

So he who is at the helm of any learning institution must come up with the ultimate solution. (This is a topic for my other post so keep in check)

It is important to ensure the ethos and codes remain healthy and functional. Apparently, there is more to be desired from schools in Lae - the lae city schools.

To begin with, every stakeholder involving and receiving this vital government service (Education) must first asks ‘What went wrong?!’. This question supersedes when, why, who or how. Only if this can can effectively analysed, then stakeholders and concerned citizens can use it as pointer to fight against school fights. 

So what happened, then? 

A culture (note: I call a culture and not cult) was created by students (students' culture) and it existed parallel to that of the affected schools (established schools' culture). Education psychology dictates that a culture can be negative or positive whether it is schools' or students'. In Lae city schools an unpopular and detrimental students' culture has existed right under the noses to school administrators and local education authorities which seem to have continued unobstructed over the years.

For whatever reason, students' way of doing things in and around schools went undetected (or ignored!). It flourished in schools like Bugandi Secondary School, Bumayong Lutheran Secondary School, Lae Secondary School and Busu Seconday School.

This culture has tentacles in primary schools, too. It has gone from bad to worse - uncontrollable!

Today the culture created by students evidently has prominence because it had an influence on negative behaviours about school. More-so, 'generational' structure into which every student is a subject makes it difficult for him/her to take an independent stand against it. Almost unavoidable!

How are students identified?

What happened among schools in Lae is contrary to good student culture. Students are distinctly identified on two obscure but effective conditions: where and why.

WHERE: This takes precedence. Students from same area mostly attend the same primary school. They know their seniors and their seniors know theirs. They, by default, easily identify themselves with whom they know, hence generation groups takes stronghold.

WHY: Being part of a group is survival instinct many animals display. Due to peer effect within school - to actually survive and thrive in a city school - students have to identify themselves with their peers.

What is bad is the fact that students are absorbed into these groups where the atmosphere is completely opposite to norms of every school. Instead, it promotes rebellion and disobedience. A grave concern when not only practising good character and personality at young ages are vital, but also good academic achievement.

A student by default joins a group or align themselves with one. Other reasons like smoke buddies, alcohol mates, class mates, etc are supplementary as they fall under 'where' and 'why'.

Lae city schools have a situation where negative students' culture exists parallel to schools' culture. It has got to a point when the Lae school administrators, school boards, Morobe Provincial Education board, Provincial Educations Adviser, Morobe Provincial Administrator, Morobe Governor and other Morobe Education officials must not let bad students' culture takes over schools' culture.

Four Impacts Within the Education System

The impacts of negative student's culture on students' lives are many both now or later and whether they are in the village or workplace. Those stated below are impacts on educational strand - what is happening within educational circles.

Rebellion: Students are influenced to develop negative ethos by doing something against school's prescribed norm. Such students' behaviour are to prove that they can do something against school's principles and get away! Call it the test of 'daring'. For example, deliberately disobeying teachers or causing injury to other students or taking drugs. This can lead to other serious incidences like having sex, taking alcohol, skelim boros, involve fights and killings, etc.

High Drop Out: There is likely to be a high rate of students failing their examinations. This is a concern for parents; it is also a concern for those running the school; and a concern for those at the helm of the Education System in Morobe.

Education Gap: The is a lap - a generational gap - where Morobe would have had a population of half educated individuals roaming the streets and villages.

Substandard Secondary Schools: Standard of a school is linked to students' performance. It has been a disgrace for Lae schools as far as students performances is concerned. It is shameful to class a school as 'low standard'. Substandard schools would be appropriate for schools in Lae City.

It is now time to think differently. 

So, what has been done to address this problem? Here is what the Morobe Provincial Education Chairman had to say.


In my next post I’ll explain the roles and responsibilities of School administrators, school board, Morobe Provincial Education board, Provincial Educations Adviser, Provincial Administrator, Morobe Governor; what they must do their school levels.

Algebra II - Higher Order Questions Sorted by Topic

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GCSE Foundation Maths - Sorted by Topic

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Addition Methods | Adding by Parts

Adding two or more numbers can be fun and easy when done in parts - by parts I mean unit, ten, hundred,....

For example

Work out the value of 5 + 15 + 34 + 58 + 72

add the tens:                      10 + 30 + 50 + 70 = 160  (you see, 70 + 30 give 100, 10 + 50 give 60 )
add the units:               5 +  5  +   4 +   8 +  2 =   24    (5 + 5 = 10, 8 + 2 = 10: that's 20, plus 4 give 24)

Now add 160 and 24

160 + 24 = 184

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Maths is Fun ^o^



4 Components of Primary and Secondary Algebra and Related Questions

1. Substituting
2. Expanding
3. Factorising
4. Solving
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Algebra is often regarded by both students and teachers as a tough topic to learn or teach. I think the important thing (and from experience) is to know the general outline forehand  makes it easier. Regardless, of primary (Yr 5, 6, 8) or secondary (Yr 9, 10, 11, 12), the layout is similar. Here, you will see sample questions (and solutions) I used in tuition groups at both primary and secondary levels to introduce Algebra. Questions adapted from UK GCSE maths exam papers, 2010 - 2012.

SEE MORE ON GRADE 10 EXAM QUESTIONS HERE

1. Substitution

a. Write down the value of abc when a = 10, b = 2 and c = 0
The answer is 0 (but, many students will write 20)
……………..1 mark
Maths knowledge: any number multiply by 0 is 0
b. Work out the value of 1/2x - 3y when x= 10 and y= 2

 5 - 6 = - 1 (many students write 1 instead)
Concept tested: Addition and subtraction of -ve and +ve numbers. (Reinforce that differences between 6 - 5 and 5 -6 or ask students to think about number line, starting at 5 and moving left 6 places)

2 marks
c. Find the value of 3x + 2y when x = 4 and y = 5

 12 - 10 = 2 ............. :)

2 marks

2. Expanding Brackets (note that expanding and factorising are opposites)

 Expand the following expressions

a. 3(2y – 5) =                                                       ……………..1 mark

 6y - 15 ( many student forget to do 3 x - 15)

b. 4(2m + 3n) =                                                   ……………..1 mark
8m + 12n

c. x(x – 10) =                                                       ……………..1 mark

 x^2 - 10x

3. Factorising

Factorise the following expressions  (note that expanding and factorising are opposites)

a. 2a + 10 =                                                   ……………..1 mark
Highest Common Factor (HCF) of 2a and 10 is 2

2(a + 10)

b. 4 + 6x =                                                      ……………..1 mark
HCF = 2

2(2 + 3x)

c. 3x – 9=                                                        ……………..1 mark
3 (x - 3)
d. 2x^2  + 4x

 HCF = 2x
2x (x + 2)

2 marks
4. Solving equation

Solve the following equations to find the value of x

a. 4x = 20                                                       ……………..1 mark
x = 5

b. 3x - 7 = 8                                                   ……………..1 mark

 3x = 8 + 7 ( it is important to get the order 8 + 7 right and not 7 + 8: even the answer is same, the answer may not be the same when - ing, see the example below)
3x = 15
  x = 5
c. 8(x + 12) = 100

 8x + 96 = 100 ..................Expand the brackets
8x = 100 - 96................... (subtract 96 on both sides (remember balancing equations?)
x = 4/8 (Why divide 8? In order to find the value of x, you must divide LHS and RHS by 8)
x = 1/2 (or 0.5)

2 marks

Solve the following to find the value of y

a. y/3 = 9                         ……………..1 mark

   y = 27 .............multiply 3 x 9 ( now, this is important as you can use this to solve complex equations that have a divisor)

b.  2y/5 = 4                               ……………..1 mark
2y = 20.......... ( 20 = 5 x 4)
y = 10

c.  2y + 3 / 2 = 5

 2y + 3 = 10.......... ( 10 = 2 x 5)
2y = 10 - 3 ...........( subtracting 3 on both sides of the equation)
2y = 7
y = 7/2
y = 3.5

2 marks

Division Methods | Long Division and Division by Chunking

These two methods can be very useful. I recommend introducing them together - give the students options to work with.
-------------------------

Method 1 - Long Division

Solve 351divided by 9
 Steps
1. 9 goes into 35, 3 times
2. 9 x 3 - 27
3. 35 - 27 = 8 ......bring the unit, 1, down. That makes 81
4. 9 goes into 81, 9 times.
5. Now, 81 - 81 = 0....no remainder, the answer is 39
 
Note: If there is remainder, your answer will be a mixed number (or a decimal number)
 

Method 2 - Division by Chunking - This is about calculating in parts

Solve 351divided by 9

351
  81        (9 x 9) .....you know that from x9 table
----
270        (9 x 30)

ANSWER 39 ......add 30 and 9 to give you the answer (easy aye :))


YOUTUBE Links
Long Division
Division by Chunking


Multiplication Methods | Long Multiplication, Chunking and Grid

Multiply big numbers by breaking them into parts is the easiest way, especially when working with non calculator papers.

Here, are illustrations on how to make calculations by using Long Multiplication, Multiplication by Chunking and Multiplication by Grid

Work out the value of 135 x 27

Method 1 - Long Multiplication by Chunking

             135
             x22
             -----
             270 .........multiply 2 x 135
         +2700..........put a 0 under 0, then multiply 2 x 135 and add the results
           ------
           2907
         ---------

Method 2 - Multiplication by Chunking

In 135, we have 100 + 30 + 5. Multiply each part by 22

           100 x 22 = 2200
             30 x 22 =   660...easy mental calculation, if 3 x 22 = 66, 30 x 22= 660  
               5 x 22      110
                              ------
                              2970     .......add 2200, 660 and 110 together       ...Happy now? :)
                            ---------

Method 3 - Multiplication by Grid

100   30   5
    0   20   2
---------------
      0      0      0               ....x 0
2000  600  100            ......x20
  200    60    10            ........x2
------------------
2200   660   110          .....add the number down
-------------------
    2970                           .....................Take the grand total (add across)       ......Get it? :( haha!
...............

Teachers' Note - it is important to use simple methods without diverting from traditional ones. However, introducing an optional method can help others students with difficulty in application of the other.

Students' Note - It is better to know two other methods that one.



How to Memorise Multiples of 9

This is the best and easiest technique I learnt in my first year of teaching. Could have been really helpful if I had known in back in Grade 2...

The idea is discovering the pattern in multiples of 9.

Source: PNG Teachers on Facebook

1 x 9 = 09
2 x 9 = 18
3 x 9 = 27
4 x 9 = 36
5 x 9 = 45
6 x 9 = 54

Do you see the pattern developing?
The tens increase from 0, 10, 20, 30, 40, 50....
Units decrease from 9, 8, 7, 6, 5, 4, ....

Now, you can smile :) and complete the rest...

7 x 9 =
8 x 9 =
9 x 9 =
10 x 9 =
11 x 9 =
12 x 9 =
13 x 9 =

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Hope this helps :)

Memorising Vs Collective Summation | 4x, x6, x7, x8, x12 & x 13

My Gr 2 headmaster in the village would let us out for recess if only we recite our multiplication tables. Those were the times. Here, is a strategy and if used properly can be very effective too.

I'd like to called it 'Collective Summation'. It uses the idea of patterns and simple multiplication & addition.

Students can easily recall the first 5 multiples of 4

4 x 1 = 4
4 x 2 = 8......I will us this to show the strategy
4 x 3 = 12
4 x 4 = 16
4 x 5 = 20

Strategy 1
Key: remember the last multiple and add 4.

4 x 2 = 8

then 4 x 3 = 12 = 4 + 8
        4 x 4 = 16 = 4 + 12
        4 x 5 = 20 = 4 + 16

Instead of responsive recall of multiples of 4, what you are doing is actually working with what you know and adding a 4 to the subsequent number to get the next.

This strategy can be applied to hard-to-recall multiples in 6, 7, 8 and 13 times tables.

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Strategy 2

Using square numbers:

4 = 2 x 2
9 = 3 x 3
16 = 4 x 4
25 = 5 x 5
36 = 6 x 6
49 = 7 x 7
64 = 8 x 8
81 = 9 x 9
100 = 10 x 10
121 = 11 x 11
144 = 12 x 12
169 = 13 x 13

If you know, you can easily work out the multiples, when these numbers are doubled.
2 x 2 = 4            2 x 4 = 8 = 4 + 4

3 x 3 = 9            3 x 6 = 18 = 9 + 9
-----------------------------------------
16 = 4 x 4          4 x 8 = 32 = 16 + 16
25 = 5 x 5          5 x 10 = 50 = 25+ 25

36 = 6 x 6           6 x 12  = 72 = 36 + 36
49 = 7 x 7           7 x 14 = 58= 49 + 49

Got the idea? Now, try these

64 = 8 x 8              8 x 16 =
81 = 9 x 9              9 x 18 =

100 = 10 x 10              10 x 20 =
121 = 11 x 11              11 x 11 =

144 = 12 x 12              12 x 24 =
169 = 13 x 13               13 x 26 =

Teacher's Note: You may have a technique you have used in class over time - this is what I personally find useful. It can give your students confidence in you, too. (You know what I mean when you cannot do 12 x 24 in front of your student :))

Students' Note: You will find this useful when doing calculation involving 12 x 12, 12 x 24, 13 x 13 and 13x 26. It can save you lots of exam time.

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Perimeter | Finding a missing length

The diagram shows a rectangle and a square.




The perimeter of the rectangle is the same as the perimeter of the square.
Work out the length of one side of the rectangle.

Solution.........perimeter, p, is the sum of lengths around a shape


rectangle, p = 8+8+2+2 = 20 cm

square, s = 20/4 = 5 cm ....4 is the number of equal sides in a square

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Students note
- showing a clear working out will result in awarding of full marks
- key to getting this question right is knowing how to find perimeter and 4 sides of a square are equal.

Teachers note
- students should know the formulae p = 2( l + w) and p = 4s, for rectangle and square respectively. But, I think the best strategy is to show students the concept, not formula, as the concept can be used in both regular and irregular shapes. Remembering to use formulae can create confusion in contextual exam setting if a student can not recall them easily.
 

Perimeter | Adding the lengths by collecting like terms

In the diagram, all measurements are given in centimetres.


All angles are right angles.

Show that the perimeter of the shape can be written as 2(3x + 5).
----------------------------
Solution
perimeter = side 1 + side 2 + side 3 + ...... (two sides are missing)
                = 2x + x + 3 + 2 + horizontal missing side + vertical missing side
                = 2x + x + 3 + 2 + 2x + 3 + x + 2
                = 6x + 10
                = 2(3 + 10) ......you've got to factorise to show

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Note: to get 4 full marks for this question, student must show (by calculation).

Prime Number, Square Number, Sequence

Here are 5 rows of numbers.

 

Row A

2

4

6

8

10

12

14

16

 

 

Row B

3

5

7

9

11

13

15

17

 

 

Row C

2

3

5

7

11

13

17

 

 

 

Row D

1

2

5

10

20

50

100

200

 

 

Row E

1

2

4

8

16

32

64

 

 

 

All the numbers are even in one of the rows.

 

(a)   Which row?

 

……………

 

(1)

 

(b)   The numbers in Row C are the first 7 prime numbers written in order of sizes.

 

 

 

Write the next prime number.

 

……………

(1)

 

(c)   Write down a square number from Row D.

 

 

 

……………

 

(1)

 

The numbers in Row E are the first 7 numbers of a sequence.

 

 

 

(d)   Work out the next number in the sequence.

 

……………

 

(1)

 

(Total 4 marks)

Answers

 

(a) Row A

(b) 19

(c) 100 (can also be 1)

(d) 128 ( each number doubles, nth term =  2n)

 

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Students note:

- knowing your square numbers (say the first 10) will help with finding answers to square roots, e.g. square root of 100 is 10

 

- there are two ways of finding nth term in a sequence: finding the right pattern and or working our the rule.

 

Teachers note:

It can be helpful to use the same test format when setting test