Divide a natural number by a natural number whose quotient is a decimal

### 1.1. Knowledge to remember

**a)** __Example 1__: A square yard has a perimeter of 27m. How many meters long is the side of the yard?

We have to do the division: 27 : 4 = ?

Usually we set the calculation and then do the following:

- 27 divided by 4 get 6, write 6;

6 times 4 gets 24, 27 minus 24 gets 3, write 3;

- To divide tieeos, we write a comma to the right of 6 and add a 0 to the right of 3 to get 30.

30 divided by 4 get 7, write 7.

7 times 4 equals 28; 30 minus 28 equals 2, write 2.

- Add a 0 to the right of 2 to get 20; 20 divided by 4 get 5 write 5;

5 times 4 equals 20; 20 minus 20 equals 0 write 0

So 27 : 4 = 6.75 (m)

**b)** __Example 2__: 43: 52 = ?

This division has a divisor 43 less than the divisor 52, we can do the following:

- Convert 43 to 43.0
- Set the calculation and then calculate as division 43.0 : 52 (divide decimals by natural numbers).

***** **When dividing a natural number by a natural number and leaving a remainder, we continue to divide as follows:

- Write a comma to the right of the quotient.
- Add a zero to the right of the remainder and then divide.
- If there’s any more left over, then we write another zero to the right of the new remainder, then continue dividing, and so on, and so on.

### 1.2. Solve textbook exercises on page 68

**Lesson 1 **

Calculate then calculate:

a) 12 : 5 b) 15 : 8

23 : 44 75 : 12

882 : 36 81 : 4

__Solution guide:__

Sentence a

\(\begin{array}{l}

\left. \begin{array}{l}

twelfth\\

20

\end{array} \right|\begin{array}{*{20}{c}}

5\\

\hline

{2,4}

\end{array}\\

\,\,\,0

\end{array}\)

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

{23}&{}

\end{array}\\

\,\,30

\end{array} \right|\begin{array}{*{20}{c}}

4\\

\hline

{5,75}

\end{array}}\\

\begin{array}{l}

\,\,20\\

\,\,\,\,\,0

\end{array}

\end{array}\)

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

{822}&{}

\end{array}\\

162

\end{array} \right|\begin{array}{*{20}{c}}

{36}\\

\hline

{24,5}

\end{array}}\\

\begin{array}{l}

\,\,180\\

\begin{array}{*{20}{c}}

{}&0

\end{array}

\end{array}

\end{array}\)

sentence b

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

{15}&{}

\end{array}\\

\,\,70

\end{array} \right|\begin{array}{*{20}{c}}

8\\

\hline

{1,875}

\end{array}}\\

\begin{array}{l}

\,\,\,\,\,\,60\\

\begin{array}{*{20}{c}}

{}&{40}

\end{array}\\

\begin{array}{*{20}{c}}

{}&0

\end{array}

\end{array}

\end{array}\)

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

{75}&{}

\end{array}\\

\,\,30

\end{array} \right|\begin{array}{*{20}{c}}

{twelfth}\\

\hline

{6,25}

\end{array}}\\

\begin{array}{l}

\,\,\,\,\,\,60\\

\begin{array}{*{20}{c}}

{}&0

\end{array}

\end{array}

\end{array}\)

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

{81}&{}

\end{array}\\

\,\,\,ten

\end{array} \right|\begin{array}{*{20}{c}}

4\\

\hline

{20,25}

\end{array}}\\

\begin{array}{l}

\,\,\,\,\,\,20\\

\begin{array}{*{20}{c}}

{}&0

\end{array}

\end{array}

\end{array}\)

Lesson 2

Sew 25 sets of the same clothes with 70m of fabric. How many meters of fabric to sew 66 such clothes?

__Solution guide:__

Sewing a set of clothes that run out of fabric meters is

70:25=2.8(m)

Sewing 6 clothes out of fabric meters is:

2.8×6=16.8(m)

Answer: 16.8m.

lesson 3

Write the following fractions as decimals:

\(\frac{2}{5};\frac{3}{4};\frac{{18}}{5}\)

__Solution guide:__

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

2&{}

\end{array}\\

20

\end{array} \right|\begin{array}{*{20}{c}}

5\\

\hline

{0.4}

\end{array}}\\

{\,\,\,0}

\end{array}\)

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

3&{}

\end{array}\\

30

\end{array} \right|\begin{array}{*{20}{c}}

4\\

\hline

{0.75}

\end{array}}\\

\begin{array}{l}

\,\,\,20\\

\,\,\,\,\,\,0

\end{array}

\end{array}\)

\(\begin{array}{*{20}{l}}

{\left. \begin{array}{l}

\begin{array}{*{20}{c}}

{18}&{}

\end{array}\\

\,\,\,30

\end{array} \right|\begin{array}{*{20}{c}}

5\\

\hline

{3,6}

\end{array}}\\

{\,\,\,\,\,0}

\end{array}\)

So \(\frac{2}{5} = 0.4;\frac{3}{4} = 0.75;\frac{{18}}{5} = 3.6\)

### 1.3. Solving exercises in the textbook Practice page 68

Lesson 1

Calculate

a)5,9:2+13.06 b)35.04:4−6.87

c)167:25:4 d)8.76×4:8

__Solution guide:__

\(\begin{array}{l}

a)5,9:2 + 13.06\\

= 2.95 + 13.06\\

= 16.01\\

b)35.04:4 – 6.87\\

= 8.76 – 6.87\\

= 1.89\\

c)167:25:4\\

= 6.68:4\\

= 1.67\\

d)8.76 \times 4:8\\

= 35.04:8\\

= 4.38

\end{array}\)

Lesson 2

Calculate and compare the results:

a) 8.3×0.4 and 8.3×10:25

b) 4.2×1.25 and 4.2×10:8

c) 0.24×2.5 and 0.24×10:4

__Solution guide:__

a) 8.3×0.4=3.32; 8.3×10:25=83:25=3.32

So: 8.3×0.4=8.3×10:25

b) 4.2×1.25=5.25; 4.2×10:8=42:8=5,25;

So: 4.2×1.25=4.2×10:8

c) 0.24×2.5=0.6 0.24×10:4=2,4:4=0.6

So: 0.24×2,5=0.24×10:4.

lesson 3

A rectangular garden is 24m long and 2/5th the width. Calculate the perimeter and area of the garden.

__Solution guide:__

The width of the garden is:

\(24 \times \frac{2}{5} = 9.6\:(m)\)

The perimeter of the garden is:

(24+9.6)×2=67.2(m)

The area of the garden is:

24×9.6=230.4(m^{2})

Answer: Circumference: 67.2;

Area: 230.4m^{2}

**Lesson 4**

In 3 hours, the motorbike travels 93 km. In 2 hours, the car travels 103 km. How many kilometers more kilometers does the car travel every hour than the motorbike?

*Solution guide:*

The number of kilometers traveled by the motorbike per hour is:

93 : 3 = 31 (km)

The number of kilometers traveled by the car per hour is:

103 : 2 = 51.5 (km)

The number of kilometers per hour that the car travels more than the motorbike is:

51.5 – 31 = 20.5 (km)

Answer: 20.5km.

.

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