METHODOLOGY OF MATHEMATICAL SOLUTIONS OF SOME CHEMICAL PROBLEMS IN PRACTICAL LESSONS IN MATHEMATICS

Volume 04 Issue 12-2024

296

International Journal of Advance Scientific Research
(ISSN

–

2750-1396)

VOLUME

04

ISSUE

12

Pages:

296-301

OCLC

–

1368736135

A

BSTRACT

In this article, problems related to solutions and mixtures of different contents and the processes of solving
them in a mathematical way were seen in practical exercises in mathematics. This can be used in the
creation of new mixtures and the formation of colorful paints, which are important in many areas of
industry.

K

EYWORDS

Higher education, mathematics, chemistry, study, integration, example and problem, solution, mixture,
gold, silver, alcohol, copper, element, temperature, volume, concentration, application.

I

NTRODUCTION

It is an important task of today to pay attention to
the fact that the future specialists studying in
higher educational institutions have thorough
professional training based on the requirements

of the present time and become skilled masters of
their profession. How knowledgeable are the staff
and if he is skilled, he can greatly contribute to the
development of the country [1, 2].

Journal

Website:

http://sciencebring.co
m/index.php/ijasr

Copyright:

Original

content from this work
may be used under the
terms of the creative
commons

attributes

4.0 licence.

Research Article

METHODOLOGY OF MATHEMATICAL SOLUTIONS OF SOME
CHEMICAL PROBLEMS IN PRACTICAL LESSONS IN
MATHEMATICS

Submission Date:

December 15,

2024,

Accepted Date:

December 20, 2024,

Published Date:

December 30, 2024

Crossref doi:

https://doi.org/10.37547/ijasr-04-12-46

Zulfikaharov Ilhomjon Makhmudovich

Andijan Institute of Mechanical Engineering, Associate Professor of the "Information Technologies"
Department, Uzbekistan

Atajonova Saidakhon Baratalievna

Andijan Institute of Mechanical Engineering, Associate Professor (PhD) of the "Information Technologies"
Department, Uzbekistan

Volume 04 Issue 12-2024

297

International Journal of Advance Scientific Research
(ISSN

–

2750-1396)

VOLUME

04

ISSUE

12

Pages:

296-301

OCLC

–

1368736135

For this, first of all, professors and teachers who
teach mathematics, geometry, physics and
chemistry in schools, lyceums and higher
educational institutions should pay great
attention to improving the ability of students and
students to think mathematically.

It is important to teach pupils and students to
apply it to life issues, providing the integration of
mathematics with sciences such as geometry,
physics,

chemistry,

biology,

technology,

economics, in order to develop their logical
thinking in the process of teaching mathematics,
especially in practical training [1, 2] .

In order to implement the integration of
mathematics with the field of chemistry in
educational activities, we will consider several
such problems, paying attention to the process of
solving problems related to solutions and
mixtures in a mathematical way.

Issue 1. By mixing 10% and 25% salt solution, 3
kilograms of 20% solution was made. How many
kilograms were taken from each solution [7] .

Solving. If x kilograms are obtained from a 10%
solution, then 3-x kilograms are obtained from a
25% solution according to the condition of the
problem.

By average value

10 ∙ 𝑥 + 25(3 − 𝑥)

3

= 20

we get the equation and find the unknown

𝑥

.

10 ∙ 𝑥 + 75 − 25 ∙ 𝑥 = 60

𝑥 = 1

So, 1 kg of 10% li, 2 kg of 25% li was obtained.

Issue 2.

In one of the two mixtures of gold and

silver, the ratio of gold to silver is 2:3, and in the
second, it is 3:7. How much of each should be
taken to make 8 kilograms of a new 5:11
solution[7].

Solving.

If

𝑥

kilograms are taken

from solution 1 ,

it is necessary to take

8 − 𝑥

kilograms

from

solution 2 .

It is enough to solve the problem only with
respect to gold:

2
5

𝑥 +

3

10

(8 − 𝑥) = 2,5

4𝑥 + 24 − 3𝑥 = 25

𝑥 = 1

So, 1 kg of gold and 7 kg of silver are taken to form
a new solution.

Issue 3.

There are two containers of water at

different temperatures. If the ratio of the volumes
of water taken from the 1st and 2nd container is
1:2, a mixture of 35

0

C will be formed. If taken in a

ratio of 3:4, it is equal to 33

0

C.

Find the

temperature of the water in each container
(density and specific heat capacity of water are
unchanged) [7].

Solving.

If we take

the temperature of the water

in the 1st container

𝑡

1

and the temperature of the

water in the 2nd container

𝑡

2

:

Volume 04 Issue 12-2024

298

International Journal of Advance Scientific Research
(ISSN

–

2750-1396)

VOLUME

04

ISSUE

12

Pages:

296-301

OCLC

–

1368736135

{

1
3

𝑡

1

+

2
3

𝑡

2

= 35

3
7

𝑡

1

+

4
7

𝑡

2

= 33

we get a system of equations, from this system:

𝑡

1

= 21

and

𝑡

2

= 42

it turns out to be.

So, the temperature of the water in the 1st

container

𝑡

1

= 21

0

𝐶

will be the temperature of

the water in the 2nd container

𝑡

2

= 42

0

𝐶

.

Issue 4.

Two containers contain 4 and 6

kilograms of acid solutions of different
concentrations. If they are mixed, a 35% solution
is formed. If equal amounts of solution are taken
from the containers and mixed, a 36% acid
solution is formed. How many kilograms of acid
are in each container[7].

Solving.

If we find the acid concentration in each

container, the problem is solved. Concentration,
respectively,

𝑝

and

Let's say

𝑞

.

In that case

{

4 ∙ 𝑝 + 6 ∙ 𝑞

10

= 35

𝑝 + 𝑞

2

= 36

we create a system of equations, from which it
turns out that

𝑝 = 41%

and

𝑞 = 31%

.

It follows

that the concentration of acid in each container is
in kilograms

𝑚

2

= 6 ∙ 0,31 = 1,86

,

𝑚

1

= 4 ∙

0,41 = 1,64

Issue 5.

Gram with different contents in two

containers

𝑚

and

𝑛

grams of alcohol solution. A

homogenous amount of solution was taken from
each container and mixed by pouring into the 2nd
container. As a result, the same percentage
solution was formed in both containers. Find how
much solution was taken from each container[7].

Solving.

We denote

the initial content of alcohol

in the bottles by

𝑝

and

𝑞

. And from each

container

𝑥

gram of solution.

Then (Figure 1)

Figure 1.

Volume 04 Issue 12-2024

299

International Journal of Advance Scientific Research
(ISSN

–

2750-1396)

VOLUME

04

ISSUE

12

Pages:

296-301

OCLC

–

1368736135

as a result,

(𝑚 − 𝑥)𝑝 + 𝑥𝑞

𝑚

=

(𝑛 − 𝑥)𝑞 + 𝑝𝑥

𝑛

the equation is formed.

From this equation

𝑥 =

𝑚 ∙ 𝑛

𝑚 − 𝑛

we form the equation.

So, a gram of solution was taken from each
container

𝑚∙𝑛

𝑚−𝑛

and added to the other.

Issue 6.

There are 6 kilograms in 1 of two

containers, and 8 kilograms in 2, with alcohol
solutions of different concentrations. A certain
amount of solution was taken from the 1st
container, and twice as much solution was taken
from the 2nd container, and when it was poured
into another container and mixed, the
concentration of the solutions in the containers
remained equal. How much solution was taken
from each container[7].

Solving.

We denote

the concentration of alcohol

in the containers by

𝑝

and

𝑞

(Fig. 2).

Figure 2.

According to the condition of the issue:

𝑝(6 − 𝑥) + 2𝑞𝑥

6 + 𝑥

=

(8 − 2𝑥)𝑞 + 𝑝𝑥

8 − 𝑥

we form the equation

So, if we solve this equation, we will get 2.4 kg of
solution from container 1 and 4.8 kg from
container 2.

Issue 7.

There are 3 mixtures of elements

𝐴

,

𝐵

and

𝐶

,

1- consisting of only

𝐴

and

𝐵

, 2- consisting

of only

𝐵

and

𝐶

, 3- consisting of only

𝐴

and

𝐶

elements:

In the 1st mixture

𝐴: 𝐵 = 1: 2

, in the 2nd mixture

𝐵: 𝐶 = 1: 3

, in the 3rd mixture

𝐴: 𝐶 = 2: 1

. When

taken from the three mixtures in what

Volume 04 Issue 12-2024

300

International Journal of Advance Scientific Research
(ISSN

–

2750-1396)

VOLUME

04

ISSUE

12

Pages:

296-301

OCLC

–

1368736135

proportion, the new mixture will have

𝐴: 𝐵: 𝐶 =

11: 3: 8

ratio [7] .

Solving.

From each, respectively

𝑥

,

𝑦

and

we get

from

𝑧

kilograms.

In that case

𝐴:

1
3

𝑥 +

2
3

𝑧 =

11
22

(𝑥 + 𝑦 + 𝑧)

𝐵:

2
3

𝑥 +

1
4

𝑦 =

3

22

(𝑥 + 𝑦 + 𝑧)

𝐶:

3
4

𝑦 +

1
3

𝑧 =

4

11

(𝑥 + 𝑦 + 𝑧)

we form the equation

We make 1

𝑥 + 3𝑦 = 𝑧

and 3 of these equations.

48𝑥 − 51𝑦 = −47

from this

52𝑥 − 39𝑦 = 0

or

𝑥

𝑦

=

3

4

is formed.

𝑥 + 3𝑦 = 𝑧

dividing

𝑦

𝑧

=

4

15

ni by

𝑦

we get the

equality.

From this

𝑥: 𝑦 = 3: 4

𝑦: 𝑧 = 4: 15

it will be known. So,

3: 4: 15

was obtained from

all three mixtures.

Issue 8.

The price of copper is proportional to the

square of its mass. When 12 kilograms of copper
was divided into two parts, the price decreased by
1.6 times. In what proportion is copper
divided[7].

Solving.

Let

𝑦

be

the price,

𝑥

be

the mass, and let

𝑘

be the coefficient.

In that case

𝑦 = 𝑘𝑥

2

, the previous price

𝑦 = 144𝑘

.

144𝑘

𝑘𝑥

2

+ 𝑘(12 − 𝑥)

2

= 1,6

𝑥

2

− 12𝑥 + 27 = 0

𝑥

1

= 3, 𝑥

2

= 9

So it is divided in the ratio

1: 3

or

3: 1

.

With this, we will provide students studying in
higher educational institutions with the
integration of mathematics with subjects such as
physics,

chemistry,

biology,

technology,

economics in mathematics training. This
integration means that the processes of students'
ability to apply the acquired mathematical
knowledge in practice are contributed.

It is necessary to form mathematics lessons in a
creative way, to harmonize various learning
activities of students, and to provide quick
feedback between participants.

It is desirable to organize mathematics trainings
on a scientific and methodological basis, to form
the skills of applying students in practice in
accordance with their specializations, and to use
an effective incentive management mechanism in
trainings.

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301

International Journal of Advance Scientific Research
(ISSN

–

2750-1396)

VOLUME

04

ISSUE

12

Pages:

296-301

OCLC

–

1368736135

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