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Guidance on solving exercises on page 50 of Math 5 textbook, Practice - Including solving methods
=> Observation: The commutative property of addition for decimal numbers:
When swapping two addends in a sum, the sum remains unchanged: a + b = b + a.
Solution Method:
Students need to remember the commutative property of addition: When swapping addends in a sum, the value of the sum remains the same. Therefore, students only need to perform the operation a + b to find the result of b + a.
Solve problem 2 on page 50 of Math 5 practice textbook
Problem:
Perform addition and then use the commutative property to retry:
a) 9.46 + 3.8;
b) 45.08 + 24.97;
c) 0.07 + 0.09.
Solution Method:
- First, perform addition of two decimal numbers as usual, you can draft the calculation and then fill in the given expression
- Then, swap the two addends and perform addition to compare the results.
- If the result of the operation a + b matches the result of the operation b + a => The operation is correct
- Otherwise, if the results do not match, perform the addition again.
Answers:
a) 9.46 + 3.8 = 13.26 retry 3.8 + 9.46 = 13.26
b) 45.08 + 24.97 = 70.05 retry 24.97 + 45.08 = 70.05
c) 0.07 + 0.09 = 0.16 retry 0.09 + 0.07 = 0.16.
Solve problem 3 on page 50 of Math 5 practice textbook
Problem:
A rectangle has a width of 16.34m, a length greater than the width of 8.32m. Calculate the perimeter of that rectangle.
Solution Method:
- The problem states:
+ Width: 16.34 m
+ Length greater than width: 8.32 m
- Problem requirement: Calculate the perimeter of the rectangle.
- Solution:
+ Find the length of the rectangle by adding the width to the length more than the width
+ Find the perimeter of the rectangle by adding the width (given) to the length (found), whatever the result is, multiply by 2.
Answers:
The length of the rectangle is:
16.34 + 8.32 = 24.66 (m)
The perimeter of the rectangle is:
(24.66 + 16.34) x 2 = 82 (m)
Answer: 82 m.
Solve problem 4 on page 50 of Math 5 practice textbook
Problem:
A shop sold 314.78m of fabric in the first week, and 525.22m in the second week. Knowing that the shop sold every day of the week, find the average number of meters of fabric sold per day by the shop?
Solution Method:
* Problem Summary:
First week: 314.78 meters of fabric
Second week: 525.22 meters of fabric
The shop sold every day of the week
Average per day: ? meters of fabric.
* Solution:
- First, find the total meters of fabric sold by the shop in two weeks by adding the meters sold in the first week to the meters sold in the second week
- Next, find the number of days in two weeks by multiplying 7 days (in 1 week) by 2.
- Finally, find the average meters of fabric sold per day by dividing the total meters of fabric sold by the shop in 2 weeks by the number of days in 2 weeks (found).
Answers:
The total meters of fabric sold by the shop in two weeks are:
314.78 + 525.22 = 840 (m)
The number of days in two weeks is:
7 x 2 =14 (days)
The average meters of fabric sold per day is:
840 : 14 = 60 (m)
Answer: 60 m of fabric.
Guidance on solving exercises on page 50 of Math 5 practice textbook - Concise
Exercise 1, Page 50, Mathematics Grade 5 Textbook
Exercise 2, Page 50, Mathematics Grade 5 Textbook
Exercise 3, Page 50, Mathematics Grade 5 Textbook
Exercise 4, Page 50, Mathematics Grade 5 Textbook
Above are suggestions for Solving exercises on page 50 Math 5 Textbook with detailed instructions. Prepare for the content of Sum of Decimal Numbers through Solving exercises on pages 51, 52 Math 5 Textbook and Practice on page 52 Math 5 Textbook through Solving exercises on page 52 Math 5 Textbook to excel in Grade 5 Mathematics.
Moreover, Solving exercises on pages 58, 59 Math 5 Textbook is an important lesson in the Grade 5 Mathematics curriculum that students need to pay special attention to.
In addition to the above content, students can also explore Solving exercises on page 60 Math 5 Textbook to enhance their Grade 5 Mathematics knowledge.
