Welcome to Kienthucykhoa.com, your ultimate destination for expert educational content. In this article, we will provide step-by-step solutions to exercise problems in the Grade 5 Mathematics notebook. These solutions include detailed explanations and answers for each exercise. By following our strategies, students will have the opportunity to review the concepts covered in their textbook. So, let’s get started!
Exercise 1 from Page 67 in Grade 5 Mathematics Notebook
Problem Statement: Calculate:
- 34.28 + 19.47
- 408.23 – 62.81
- 17.29 + 14.43 + 9.36
Solution:
- Arrange the numbers vertically, making sure that the digits are aligned in their respective columns.
- Perform addition/subtraction as you would with natural numbers.
- Place the decimal point directly beneath the decimal points in the numbers being added/subtracted.
Answer and Solution:
Calculate:
- 34.28 + 19.47 = 53.75
- 408.23 – 62.81 = 345.42
- 17.29 + 14.43 + 9.36 = 41.08
For a visual representation of the solution, refer to the following image:
Exercise 2 from Grade 5 Mathematics Notebook, Page 67
Problem Statement: Find the value of x.
- a) x – 3.5 = 2.4 + 1.5
- b) x + 6.4 = 27.8 – 8.6
Solution:
- Calculate the value on the right-hand side (RHS) of the equation.
- Determine the role of x in the equation and solve using the following rules:
- To find the subtrahend, add it to the difference.
- To find the unknown term, subtract the known term from the sum.
Answer and Solution:
Find the value of x.
- a) x – 3.5 = 2.4 + 1.5
- x – 3.5 = 3.9
- x = 3.9 + 3.5
- x = 7.4
- b) x + 6.4 = 27.8 – 8.6
- x + 6.4 = 19.2
- x = 19.2 – 6.4
- x = 12.8
Exercise 3 in Grade 5 Mathematics Notebook, Page 68
Problem Statement: Calculate using the most convenient method.
- a) 14.75 + 8.96 + 6.26 = …………………..
- b) 66.79 – 18.89 – 12.11 = ………………..
Solution:
Apply the following formulas:
- a + b + c = (a + c) + b
- a – b – c = a – (b + c)
Answer and Solution:
Calculate using the most convenient method.
- a) 14.75 + 8.96 + 6.26 = (14.75 + 6.25) + 8.96 = 21 + 8.96 = 29.96
- b) 66.79 – 18.89 – 12.11 = 66.79 – (18.89 + 12.11) = 66.79 – 31 = 35.79
Exercise 4 from Grade 5 Mathematics Notebook, Page 68
Problem Statement: The total area of three orchards is 5.4 hectares. The area of the first orchard is 2.6 hectares. The area of the second orchard is 0.8 hectares smaller than the first orchard. What is the area of the third orchard in square meters? (Solve using two methods)
Solution:
Method 1:
- Calculate the combined area of the second and third orchards: Combined Area = Total Area of Three Orchards – Area of the First Orchard.
- Calculate the area of the second orchard: Area of Second Orchard = Area of First Orchard – 0.8 hectares.
- Calculate the area of the third orchard: Area of Third Orchard = Combined Area of the Second and Third Orchards – Area of the Second Orchard.
Method 2:
- Calculate the area of the second orchard: Area of Second Orchard = Area of First Orchard – 0.8 hectares.
- Calculate the combined area of the first and second orchards.
- Calculate the area of the third orchard: Area of Third Orchard = Total Area of Three Orchards – Combined Area of the First and Second Orchards.
Answer and Solution:
Solution:
Method 1:
The areas of the second and third orchards are:
- 5.4 – 2.6 = 2.8 hectares
- The area of the second orchard is:
- 2.6 – 0.8 = 1.8 hectares
- The area of the third orchard is:
- 2.8 – 1.8 = 1 hectare = 10,000 square meters.
Method 2:
The area of the second orchard is:
- 2.6 – 0.8 = 1.8 hectares
- The combined area of the first and second orchards is:
- 2.6 + 1.8 = 4.4 hectares
- The area of the third orchard is:
- 5.4 – 4.4 = 1 hectare = 10,000 square meters.
Answer: 10,000 square meters
For more detailed solutions, including a downloadable Word and PDF file, click here to visit Kienthucykhoa.com. We offer these resources completely free of charge.
Remember, continuous practice is the key to mastering mathematics!
