EIJP ISSN: 2751-000X
VOLUME04 ISSUE11
75
WORKING ON COLLABORATIVE PROBLEMS IN ELEMENTARY MATH LESSONS
Kasimova M.M.
Associate Professor of Bukhara State Pedagogical Institute, Uzbekistan
Ikromova Sevinch
2nd year student of Bukhara State Pedagogical Institute, Uzbekistan
AB O U T ART I CL E
Key words:
Problem, math problem, arithmetic
problem, arithmetic operations, addition,
subtraction, multiplication, division, finding the
sum, generalization, comparison, conclusion,
complex problem.
Received:
02.11.2024
Accepted
: 07.11.2024
Published
: 12.11.2024
Abstract:
Problem is a complex category, which
does not have a single definition accepted by
everyone, but representatives of different fields
have interpreted the concept of problem in
different ways depending on their orientations.
This article describes the characteristics of
arithmetical problems, the methods and means of
teaching to solve problems related to working
together in primary school. Specific cases of
developing students' problem-solving skills by
discussing and solving any arithmetical problems
are highlighted.
INTRODUCTION
It is not difficult to distinguish a series of problems that are performed in the same
sequence and solved by the same actions from complex text problems. Such issues can be said to be one
type of issues. But complex problems with some important features were accepted as typical
arithmetical problems in the methodology course.
A characteristic feature of typical problems is that they are much more difficult than non-typical problems
and it is necessary to use special reasoning methods to solve them.
We will discuss below the methods of solving typical arithmetic problems of joint solution.
The content of the issues related to this type is different.
a) when a number of people, crews or work tools (tractor, excavator, bulldozer) perform a certain work
separately, they are told how long it will take to complete the work, and when they work together, they
are asked to find the deadline for the completion of the work;
b) when several people, crews or work tools work together, a deadline is given for the completion of a
certain job, and one of them is asked to find the time spent on this job when working alone. );
VOLUME04 ISSUE11
DOI:
https://doi.org/10.55640/eijp-04-11-17
Pages:75-79
EUROPEAN INTERNATIONAL JOURNAL OF PEDAGOGICS
ISSN: 2751-000X
VOLUME04 ISSUE11
76
c) when a number of workers or crews start a certain job at the same time, and after a few days, some
of them are transferred to another job, it is asked how many days the entire job or the rest of the job will
be completed;
g) when a number of workers or crews start a certain work at the same time, and after a few days, when
new assistants join them, they are asked to complete the whole work or the remaining part of the work
in how many days, etc.
To solve such problems, it is necessary to start by finding the time limit for execution alone, if it is given
in hours, it is one hour, if it is given in days, it is one day. Issues related to joint work can be found in
elementary school textbooks. But in the process of solving problems of this type, students can often be
expected to make some mistakes.
Consider the following issue.
Problem: 150 bicycles need to be repaired. The master does this work in 10 days, and the apprentice in 15
days. If they work together, how many days will it take them to do it?
A brief statement of the problem:
Master - in 10 days
150 bikes -? in the day
Apprentice - 15 days
When this problem is given for solving, the following solution options are shown.
Option 1: 1) 10+15=25 (per day)
2) 150:25=6 (days)
Option 2: 150: (10+15)=150:25=6(days)
Option 3:
1) 150:10=15 (items)
2) 150:15=10(s)
3) 15+10=25 (each)
4) 150:25=6 (days)
Answer: If the master and the apprentice work together, they will complete the work in 6 days.
When analyzing these solution options, every time the result is 6 in all three solution options, solution
options 1 and 2 are error solutions.
If you ask the student after each action, "What did you find?" when asked the question, they cannot
explain the results of the action. Indeed, "What did you get by adding 10 to 15?" ” or, “What did you find
by divid
ing 150 by 25 days?” We cannot find answers to such questions as the solution to the problem.
Therefore, option 3 is the correct solution to this problem.
We divide 150 by 10 and find out how many bicycles the master repaired in one day. We divide 150 by 15
and find out how many bicycles the student repaired in one day. By adding 10 to 15, we find how many
bicycles the master and apprentice repaired in one day, and divide 150 by 25. It is found that the master
and apprentice together repaired 150 bicycles in how many days.
Another great feature of this problem is that if you put 300, 450, 1800 and other numbers instead of 150,
the problem does not change. The problem can be stated as a problem in the following adequate form.
Matter. A master can do a job in 10 days, an apprentice in 15 days. If they work together, how many days
will it take them to do it?
This problem is a problem to be solved by students of the 4th grade.
Let's look at the following issues. There are 3 trucks with different carrying capacity. When each car
transports the cargo in the base by itself: the first car can transport it in 10 hours, the second car in 12
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hours and the third car in 15 hours. How many hours will it take to carry this load when the machine works
together? Let's state the problem:
1 - avt alone - at 10 o'clock
2 - avt alone - how many hours together in 12 hours?
3 - avt alone - at 15 hours
1) What part of the load does the first car transport alone in one hour?
1 : 10 = (part)
2) What part of the load can the second car alone transport in one hour?
1 : 12 = (part)
3) What part of the load does the third car transport alone in one hour?
1 : 15 = (part)
4) What fraction of the load will the three cars transport in one hour?
+ + = (part)
5) In how many hours will the three cars together transport the entire load?
1 : = 4 (hours)
Answer: 4 hours.
Issue 2. One working assignment had to be completed in 12 days. 4 days after he started work, another
worker came to help him and the whole work was completed in 8 days. When the second worker worked
alone, he could finish the whole job in a few days
After discussing the issue, we present the solution in the following sequence of questions.
1) What part of the work can the first worker do in a day?
1 : 12 =
12
1
(part)
2) What part of the work did the first worker do in 4
3
1
days?
4
3
1
*
12
1
=
36
13
(part)
3) How much work is left?
1 -
36
13
=
36
23
(part)
4) How many days did both workers work together?
8 - 4
3
1
= 3
3
2
(days)
5) What part of the work did the first worker do in 3
3
2
days?
3
3
2
*
12
1
=
36
11
(part)
6) What part of the work did the second worker do in 3
3
2
days?
36
23
-
36
11
=
3
1
(part)
7) How many days can the second worker do the work alone?
3
3
2
:
3
1
= 11(day)
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Answer: 11 days.
Problem: Bikbayeva N, Grifanova K 4th grade Mathematics. Page 112 of 2020. Issue 4
Two master bricklayers worked together to collect 119,920 bricks. The first master worked for 17 days
and collected 2,860 bricks every day. The second master picked the remaining bricks in 23 days. How
many bricks did the second master pick every day?
When the reader analyzes this issue, it is known that two masters worked together to pick 119920 bricks,
the first master worked for 17 days picking 2860 bricks every day, and the second master picked 23 bricks
for the remaining bricks. should know that the master is required to find the number of bricks he picks in
one day. The problem is first analyzed and a solution is sought based on a short condition. Although the
problem is about working together, the problem also focuses on performing arithmetic operations on
multi-digit numbers.
The period of preparation for this type of problems begins in 2-3 grades, and in the 4th grade textbook,
we often meet problems of this type. By teaching them to solve problems of this type, it is considered as
a basis for elementary school students to master high-grade mathematics in depth.
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