of Process
by Instructor
unit
from pupil
for individuals
what were learnt in the previous lessons
the problem
3"
Understand the subject.
integer,but decimal integer."
problem
solving
3
"1.3"
"Between 1 and 2"
the plan
Write the way in the working paper.
Think out other ways, while time permits.
vidual
0.1 multiplied by 39 devided 3 is 0.1 multiplied by (39
3), which is 0.1 multiplied by 13, or 1.3.
Thus, 1.3 litter is the solution.
(2) 3.9 divided by 3 is 3 divided by 3 plus 0.9 divided by 3., which is 1 plus 0.3, or 1.3.
Thus, 1.3 litter is the solution.
(3) Using the algorithm of division.
Thus, 1.3 litter is the solution.
(4) 3.9 divided by 3 is 3.9 multiplied by 10, divided by 3, and divided by 10..
3.9 x 10 = 39, 39
3 = 13, 13 10 = 1.3
Thus, 1.3 litter is the solution.
Pupil who cannot get a clue of solution are given hints.
Give pictorial illustration in the working paper.
If the algorithm of division is one of those which pupil answer, it is suspended as a subject discussed in the next stage.
Whether pupil work enthusiastically.
solutions
Discuss on each solution.
Choose the solution of type (1) as the conclusion :
"3.9 is 0.1 multiplied by 39.
0.1 multiplied by 39 devided 3 is 0.1 multiplied by (39
3), which is 0.1 multiplied by 13, or 1.3.
Thus, 1.3 litter is the solution."
Let pupil show their solutions in their own expression.
Let each pupil think to which solution her/his solution is similar.
Explain by using pictorial illustration.
Let pupil dispose (2) by introducing 4.5
3.
If (1) does not come from pupil, determine (2) to be the solution, and then develop it to (1).
"decimal
integer is solved in the same way as integerinteger by taking 0.1 as base."
integer is solved in the same way as integerinteger by taking 0.1 as base.
and
Development
2.6
2
4.8
4
9.6
3
vidual
3.
the lesson