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All exercises 1, 2, 3... of Concentrated Practice - Exercise 88 will be meticulously explained by Mytour in the document Solving 5th-grade math problems on pages 108, 109, 110 of VBT Volume 2, Concentrated Practice, Exercise 88. Please refer to it and compare it with your work to better understand the lesson content.
Solving 5th-grade math problems on pages 108, 109, 110 of VBT Volume 2, Concentrated Practice, Exercise 88.
Part 1 - 1. Solving problem 1 - Workbook for 5th-grade math exercises Volume 2, Page 108.
Problem Statement:
Given the decimal number 54.172
The value of digit 7 is:
A. 7
B. 7/10
C. 7/100
D. 7/1000
Solution Method
Identify the position of digit 7 then determine the value of digit 7 in the given number.
Answer
Digit 7 in the decimal number 54.172 is in the hundredths place, so its value is 7/100.
Select C
Part 1 - 2. Solve problem 2 - Workbook for 5th-grade math exercises Volume 2, Page 108
Problem Statement:
The amount deposited in savings is 1,000,000 dong. One month later, the total amount of deposit and interest is 1,005,000 dong. How much is the interest as a percentage of the deposit?
A. 105%
B. 10.05%
C. 100.05%
D. 0.5%
Solution Method
- Calculate the interest amount = total amount - deposit amount.
- To find the percentage ratio of the interest amount to the deposit amount, divide the interest amount by the deposit amount, then multiply the quotient by 100 and add the % symbol to the right of the obtained product.
Answer
The interest amount is:
1,005,000 - 1,000,000 = 5000 (dong)
The percentage ratio of the interest amount to the deposit amount is:
5000 : 1,000,000 = 0.005 = 0.5%
Select D.
Part 1 - 3. Solve problem 3 - Workbook for 5th-grade math exercises Volume 2, Page 109
Problem Statement:
4200 meters is equal to how many kilometers?
A. 420km
B. 42km
C. 4.2km
D. 0.42km
Solution Method
We have: 1km = 1000m.
To convert a number from meters to kilometers, simply divide the number by 1000.
Answer
Convert 4200m = 4200 : 1000 = 4.2km
Select C
Part 2 - 1. Solve problem 1 - Workbook for 5th-grade math exercises Volume 2, Page 109
Problem Statement:
Set up and calculate
356.37 + 542.81
416.3 - 252.17
25.14 x 3.6
78.24 ÷ 1.2
Solution Method
Set up and solve according to the rules learned for addition, subtraction, multiplication, and division of decimal numbers.
Answer
Part 2 - 2. Solve problem 2 - Workbook for 5th-grade math exercises Volume 2, Page 109
Problem Statement:
a) 5m 5cm = ........m;
b) 5m2 5dm2 = ........m2
Solution Method
Based on the relationship between the measurement units to write the given measurements in mixed form, then write them in decimal form.
Answer
a) 5m 5cm = 5.05 m;
b) 5m2 5dm2 = 5.05 m2
Part 2 - 3. Solve problem 3 - Workbook for 5th-grade math exercises Volume 2, Page 110
Problem Statement:
Given rectangle ABCD and parallelogram AMCN with dimensions indicated on the diagram. Calculate the area of rectangle ABCD and parallelogram AMCN using two different methods.
Solution Method
Method 1:
- Area of parallelogram AMCN = Area of rectangle ABCD + area of triangle AND + area of triangle BCM.
- Apply the formulas to calculate the areas of shapes:
+ Area of rectangle = length x width.
+ Area of triangle = base length x height : 2.
Method 2:
- Join A with C to obtain two triangles ACN and ACM with equal areas (because they have equal base lengths and equal heights).
- Area of parallelogram AMCN = area of triangle ACN + area of triangle ACM.
Answer
Method 1:
Observing the diagram, we have: AD = BC = 8cm; BM = ND = 4cm.
So the area of triangle AND = the area of triangle BMC.
Area of triangle AND is:
4 x 8 : 2 = 16 (cm2)
Area of rectangle ABCD is:
10 x 8 = 80 (cm2)
We have: Area of parallelogram AMCN = Area of rectangle ABCD + Area of triangle AND + Area of triangle BCM.
Area of parallelogram AMCN is:
80 + 16 + 16 = 112 (cm2)
Answer: 112cm2.
Method 2:
Joining points AC, we get two triangles ACN and ACM with equal areas (because they have equal base lengths AM = CN = 14cm and equal heights AD = BC = 8cm).
Area of triangle ACN is:
14 x 8 : 2 = 56 (cm2)
We have: Area of parallelogram AMCN = area of triangle ACN + area of triangle ACM.
Area of parallelogram AMCN is:
56 x 2 = 112 (cm2)
Answer: 112cm2
Part 2 - 4. Solve problem 4 - Math workbook for grade 5 exercise book 2 page 110
Question:
Solution:
Based on comparing two decimal numbers:
- Compare the integer parts of the two numbers like comparing two natural numbers, the decimal number with the larger integer part is greater.
- If the integer parts of the two numbers are equal, then compare the decimal parts, from the tenths place, hundredths place, thousandths place, etc., until a corresponding place is reached, the decimal number with the larger digit in the corresponding place is greater.
- If the integer part and the decimal part of the two numbers are equal, then the two numbers are equal.
Answer
Above is the guidance section for solving Exercise Book Mathematics Grade 5 Workbook 2 pages 108, 109, 110 Concentrated Practice, Exercise 88. Please continue to read the instructions for Solving Grade 5 Math on pages 111, 112 Workbook 2, Self-assessment, Exercise 89, or review the previous exercise Solving Grade 5 Math on pages 106, 107, 108 Workbook 2, Practice, Exercise 87 to reinforce and solidify your knowledge.
