Solving 6th-grade math problems on pages 32 and 33: Sequencing mathematical operations.

=> Follow more 6th-grade math problem-solving here: solving 6th-grade math
- Solving 6th-grade math on page 27, Volume 1 of the Knowledge Connection book - Focused exercises on page 27
- Solving 6th-grade math on pages 39 and 40, Volume 1 of the Kite book - Exercise 9: Signs of divisibility by 3 and 9
- Solving 6th-grade math on page 30, Volume 1 of the Horizon of Creativity book - Exercise 9: Factors and multiples

Guide to solving 6th-grade math problems on pages 32 and 33 (Concise)
 

1. Solve 6th-grade math exercise on pages 32 and 33, question 73

a) 5 . 4² – 18 : 3²;
b) 3³ . 18 – 3³ . 12;
c) 39 . 213 + 87 . 39;
d) 80 – [130 – (12 – 4)²].
Solution:
a) 5 . 4² – 18 : 3² = 78;
b) 3³ . 18 – 3³ . 12 = 162;
c) 39 . 213 + 87 . 39 = 11700;
d) 80 – [130 – (12 – 4)²] = 14.

2. Solve 6th-grade math exercise on pages 32 and 33, question 74

Find the natural number x, given:
a) 541 + (218 – x) = 735
b) 5(x + 35) = 515
c) 96 – 3(x + 1) = 42
d) 12x – 33 = 3² . 3³
Solution:
a) x = 24
b) x = 68
c) x = 17
d) x = 23

3. Solve 6th-grade math exercise on pages 32 and 33, question 75

Insert the appropriate numbers into the squares.
Solution:
a) The value in the second square is 60 : 4 = 15 => the value in the first square is 15 - 3 = 12
b) The value in the second square is 11 + 4 = 15 => the value in the first square is 15 : 3 = 5
We have the following diagram:

Solved:
a) The value in the second box is 60 : 4 = 15 => the value in the first box is 15 - 3 = 12
b) The value in the second box is 11 + 4 = 15 => the value in the first box is 15 : 3 = 5
We have the following figure:

4. Solve question 76 on page 32 of Math Grade 6 textbook

Riddle: Nga uses four number 2s along with parentheses (if necessary) to write a series of calculations with results sequentially equal to 0, 1, 2, 3, 4. Help Nga achieve this.
Solution:
Calculations with results of 1, 2, 3, 4 using the number 2 and parentheses are:
2 - 2 + 2 - 2 = 0
2 . 2 :(2 + 2) = 1
2 : 2 + 2 : 2 = 2
(2 . 2 + 2) : 2 = 3
2 + 2 + 2 - 2 = 4

5. Solve 6th-grade math exercise on pages 32 and 33, question 77
 

Perform the following calculations:
a) 27 . 75 + 25 . 27 – 150
b) 12 : {390 : [500 – (125 + 35 . 7)]}
Solution:
a) 27 . 75 + 25 . 27 – 150 = 2550
b) 12 : {390 : [500 – (125 + 35 . 7)]} = 4

6. Solve 6th-grade math exercise on pages 32 and 33, question 78

Calculate the value of the expression:
12,000 – (1,500 . 2 + 1,800 . 3 + 1,800 . 2 : 3)
Solution:
12,000 – (1,500 . 2 + 1,800 . 3 + 1,800 . 2 : 3)
= 12,000 – (3,000 + 5,400 + 1,200)
= 12,000 – 9,600
= 2,400.

7. Solve question 79 on page 33 of Math Grade 6 textbook

Fill in the blanks in the following problem so that to solve it, we have to calculate the value of the expression mentioned in question 78.
An buys two pens priced at … dong each, buys three notebooks priced at … dong each, buys one book and one envelope. Knowing that the amount spent on three books is equal to the amount spent on two notebooks, the total amount to be paid is 12,000 dong. Calculate the price of an envelope.
Solution:
We have the expression:
12,000 - (1,500 . 2 + 1,800 . 3 + 1,800 . 2 : 3) = 2,400 dong
An buys two pens priced at 1,500 dong each, buys three notebooks priced at 1,800 dong each, buys one book and one envelope. Knowing that the amount spent on three books is equal to the amount spent on two notebooks, the total amount to be paid is 12,000 dong. Calculate the price of an envelope.
=> The price of an envelope is 2,400 dong.

8. Solve 6th-grade math exercise on pages 32 and 33, question 80

Solution:

9. Solve 6th-grade math exercise on page 33, question 81

Use a pocket calculator to compute
(274 + 318) . 6
34 . 29 + 14 . 35
49 . 62 - 32 . 51
Solution:
(274 + 318) . 6 = 3552
34 . 29 + 14 . 35 = 1476
49 . 62 - 32 . 51 = 1406

10. Solve question 82 on page 33 of Math Grade 6 textbook