Review Table of units of mass measurement

### 1.1. Solve textbook exercises on pages 23 and 24

**Lesson 1 of the textbook page 23:**

a) Complete the following table of units of mass measurement:

Greater than kilogramki |
Kg |
Smaller kg |
||||

ton |
weights |
oats |
kg |
hg |
dag |
g |

1kg =10hg = \(\frac{1}{{10}}\) nest |

b) Comment: Two contiguous length units:

– Larger units are 10 times smaller than smaller units;

b) Comment: Two consecutive units of mass measurement:

– Larger units are 10 times smaller than smaller units;

– Small units are equal to \(\frac{1}{{10}}\) large units.

__Solution guide:__

Bigger kg |
Kg |
Smaller kg |
||||

ton |
weights |
oats |
kg |
hg |
dag |
g |

1 ton = 10 quintals |
1 quintal =10 yen =\(\frac{1}{{10}}\) tons |
1 nest = 10 kg =\(\frac{1}{{10}}\) |
1 kg =10 hg =\(\frac{1}{{10}}\) oats |
1 hg = 10 dag =\(\frac{1}{{10}}\) kg |
1 dag =10 g =\(\frac{1}{{10}}\) hg |
1g =\(\frac{1}{{10}}\) dag |

Lesson 2 Textbook page 24:

Write the correct number in the dot:

a) 18 oats = … kg

200 quintals = .. kg

35 tons = … kg

b) 430 kg = … oats

2500kg = … weights

16000kg = … tons

c) 2kg 326g = … g

6kg 3g = … g

d) 4008g = … kg … g

9050 kg = … tons … kg

__Solution guide:__

a) 18 oats = 180 kg

200 quintals = 20 000 kg

35 tons = 35 000 kg

b) 430 kg = 43 yen

2500kg = 25 quintals

16 000kg = 16 tons

c) 2kg 326g = 2kg + 326g = 2000g + 326g = 2326g

6kg 3g = 6kg + 3g = 6000g + 3g = 6003 g

d) 4008g = 4000g + 8g = 4 kg 8 g

9050 kg = 9000kg + 50kg = 9 tons 50kg.

Lesson 3 Textbook page 24:

Put the appropriate sign (>, <, =) in the dot:

2 kg 50g … 2500g 6090kg … 6 tons 8kg

13kg 85g … 13kg 805g \(\frac{1}{4}\) tons … 250 kg

__Solution guide:__

– 2 kg 50g = 2050g . But 2050g < 2500g.

So: 2 kg 50g < 2500g.

– 6 tons 8kg = 6008kg. Which 6090kg > 6008kg.

So: 6090kg > 6 tons 8kg.

– 13kg 85g = 13085g ; 13kg 805g = 13805g.

Which 13085g < 13805g.

So 13kg 85g < 13kg 805g.

– We have: 1 ton = 1000kg so \(\frac{1}{4}\) ton =1000 : 4 × 1 = 250 kg.

So \(\frac{1}{4}\) tons = 250 kg.

Lesson 4 Textbook page 24:

A shop in 3 days sells 1 ton of sugar. The first day sold 300kg. The second day sold twice as much as the first day. How many kilograms of sugar did the store sell on the third day?

__Solution guide:__

Exchange: 1 ton = 1000kg.

On the second day, the number of kilograms of sugar sold by the store is:

300 × 2 = 600(kg)

The first two days the store sold kilograms of sugar was:

300 + 600 = 900(kg)

On the third day, the number of kilograms of sugar sold by the store is:

1000 – 900 = 100(kg)

Answer: 100kg of sugar.

### 1.2. Solving exercises in the textbook Practice pages 24, 25

**Lesson 1 of the textbook page 24:**

The Hoa Binh School Union collected 1 ton 300 kg of waste paper. Hoang Dieu School Union collected 2 tons 700kg of waste paper. Know that for every 2 tons of scrap paper, 50,000 school notebooks can be produced. From the amount of scrap paper collected by both schools, how many student notebooks can be produced?

**Solution guide:**

Exchange 1 ton 300kg = 1300kg; 2 tons 700kg = 2700kg.

The number of kilograms of waste paper collected by both schools is:

1300kg + 2700kg = 4000 (kg)

4000kg = 4 tons

The number of notebooks produced per ton of scrap paper is:

50 000 : 2 = 25 000 (notebook)

4 tons of waste paper can produce the number of notebooks:

25 000 X 4 = 100 000 (notebook)

Answer 100 000 notebooks.

Lesson 2 Textbook page 24:

A worm weighs 60g. An ostrich weighs 120 kg. How many times heavier is the ostrich than the worm?

*Solution guide:*

Exchange: 120kg = 120000g

The ostrich weighs twice as much as the worm.

120000 : 60 = 2000 (times)

Answer: 2000 times.

Lesson 3 Textbook page 24:

Calculate the area of the plot of land with dimensions according to the figure below (created by rectangle ABCD and square CEMN).

*Solution guide:*

Area of rectangle ABCD is :

14 × 6 = 84 (m^{2})

The area of square CEMN is :

7 × 7 = 49 (m^{2})

The area of that land is:

84 + 49 = 133 (m^{2})

Answer: 133 m^{2}

Lesson 4 Textbook page 25:

Draw a rectangle that has the same area as rectangle ABCD but has different dimensions than that of rectangle ABCD.

*Solution guide:*

The area of rectangle ABCD is:

4 × 3 = 12 (cm^{2})

Write 12 products of two numbers (other than 4 and 3), for example:

12 = 6 × 2

So we can draw a rectangle MNPQ with length 6cm and width 2cm.

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