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Solving Math for grade 3 on pages 52, 53, 54, 55 of Workbook Volume 2
Solving exercises on pages 52, 53, 54, 55 of Math Workbook Grade 3 Volume 2, titled Connecting Knowledge with Life:
Concentration Practice
Practice Set 1
1. Solve Exercise 1 Page 52 Math Textbook Grade 3
Exercise: Set up and calculate.
Solution Guide:
Step 1: Set up the multiplication and division operations.
Step 2:
- For multiplication: Perform multiplication from right to left.
- For division: Perform sequentially from left to right.
Answer:
2. Solve Exercise 2 Page 52 Math Textbook Grade 3
Exercise: Plane A is flying at an altitude of 6,504 meters. Plane A is flying at twice the altitude of plane B. Plane B is flying at three times the altitude of plane C. What is the altitude of plane C?
Solution Guide:
Step 1: Calculate the altitude of plane B while flying = altitude of plane A : 2.
Step 2: Calculate the altitude of plane C while flying = altitude of plane B : 3.
Answer:
The altitude of plane B while flying is:
6,504 : 2 = 3,252 (m)
The altitude of plane C while flying is:
3,252 : 3 = 1,084 (m)
Answer: 1,084 m.
3. Solve Exercise 3 Page 52 Math Textbook Grade 3
Exercise: Number?
a) ? x 4 = 1,668 b) ? : 3 = 819
Solution Guide:
a) To find a factor, divide the product by the other factor.
b) To find the divisor, multiply the quotient by the divisor.
Answer:
a) ? x 4 = 1,668
1,668 : 4 = 417
b) ? : 3 = 819
819 x 3 = 2,457
4. Solve Exercise 4 Page 52 Math Textbook Grade 3
Exercise: a) Two water bugs A and B swim to a cluster of algae (as in the picture). Water bug A swims along a zigzag path consisting of 4 equal segments, water bug B swims along a zigzag path consisting of 3 equal segments. Which water bug swims a shorter distance?
b) Number?
The distance the shrimp swims is a zigzag path consisting of 5 equal segments. Knowing that the distance the shrimp swims is equal to the distance water bug A swims. Each segment of that zigzag path is ? cm long.
Solution Guide:
a) Distance of each water bug = Length of each segment of the zigzag path x number of segments of each path.
Compare the swimming distances of the two water bugs and answer the question.
b) Length of each segment = Distance the shrimp swims : 5.
Answer:
a) The swimming distance of water bug A is x centimeters long:
515 x 4 = 2,060 (cm)
The swimming distance of water bug B is x centimeters long:
928 x 3 = 2,784 (cm)
Because 2,060 cm is less than 2,784 cm, the swimming distance of water bug A is shorter.
b) The swimming distance of the shrimp is equal to the swimming distance of water bug A and is 2,575 (cm)
So each segment of the zigzag path is:
2,060 : 5 = 412 (cm)
Fill in the number 412.
Answer: a) The swimming distance of water bug A is shorter
b) 412 cm
5. Solve Exercise 5 Page 53 Math Textbook Grade 3
Problem: Knowing that 8 identical batteries weigh 1,680 g. Each robot without batteries weighs 2,000 g.
a) How much does each battery weigh?
b) After installing the number of batteries as shown, which robot is the lightest and how much does it weigh?
Solution Guide:
a) The weight of each battery = Total weight of 8 batteries : 8.
b) Step 1: Count the number of batteries in each figure and calculate the weight of the batteries installed in each robot.
Step 2: The weight of each robot = Weight of the robot without batteries + weight of the batteries installed.
Step 3: Compare the weights of the robots and draw conclusions.
Answer:
a) The weight of each battery is:
1,680 : 8 = 210 (g)
b) The weight of 5 batteries installed in robot A is
210 x 5 = 1,050 (g)
Robot A weighs a total of:
1,050 + 2,000 = 3,050 (g)
The weight of 6 batteries installed in robot B is
210 x 6 = 1,260 (g)
Robot B weighs a total of:
1,260 + 2,000 = 3,260 (g)
The weight of robot figure C is:
1,680 + 2,000 = 3,680 (g)
Practice 2
1. Solve Exercise 1 Page 53 Mathematics Textbook Grade 3
Problem: Set up and solve.
9 362 : 9 1 214 x 6 2 790 : 3 912 x 7
Solution Guide:
Step 1: Set up the calculation.
Step 2:
- For multiplication: Perform multiplication from right to left.
- For division: Perform division from left to right.
Answer:
2. Solve Exercise 2 Page 53 Mathematics Textbook Grade 3
Problem: Number?
In the amusement park, Uncle Nam wants to attach lights along each edge of the cubic-shaped house, except for the edges adjacent to the ground. Each edge needs to attach a light string that is 450 cm long.
a) Uncle Nam needs to attach a total of ? light strings.
b) The total length of those light strings is ? centimeters.
Solution Guide:
a) Number of light strings needed = Number of edges of the cubic shape - Number of edges adjacent to the ground.
b) Length of the light strings = length of one light string x number of light strings.
Answer:
a) We have a cubic shape consisting of 12 edges.
Observing the picture, we see that this house has 4 edges adjacent to the ground that do not need to attach light strings.
So Uncle Nam needs to attach a total of ? light strings:
12 - 4 = 8 (light strings)
We fill in the number 8.
b) The total length of those light strings is:
450 x 8 = 3 600 (cm)
We fill in the number 3600.
- See more: Solve exercises on pages 57, 58, 59, 60 Mathematics Textbook Grade 3 Volume 2, the book Connect knowledge with life
3. Solve Exercise 3 Page 54 Mathematics Textbook Grade 3
Problem: Which path will each caterpillar take to reach its leaf house? Knowing that each caterpillar only crawls along a path that matches its color and also crawls only to the leaf with the result of the calculation on itself that caterpillar.
Solution Guide:
Step 1: Calculate the result of the calculation on itself of the caterpillar.
Step 2: Draw the appropriate path for each caterpillar.
Answer:
We have:
721 x 6 = 4 326
4 328 : 6 = 721 (remainder 2)
So the path of each caterpillar is as follows:
4. Solve Exercise 4 Page 54 Mathematics Textbook Grade 3
Problem: Which giant can lift the most kilograms?
Solution Guide: Calculate the weight of the object that each giant lifts and then compare the weight results to answer the question.
Answer:
- Giant A lifted 3 horses, each weighing 450 kg.
So, giant A lifted a total of 450 x 3 = 1,350 (kg).
- Giant B lifted 1 elephant weighing 1,245 kg and 1 dog weighing 25 kg.
So, giant B lifted: 1,245 + 25 = 1,270 (kg)
- Giant C lifted 1 rock weighing 2,612 kg.
Practice 3
1. Solve Exercise 1 Page 55 Mathematics Textbook Grade 3
Problem: Calculate the value of the expression
a) (2,000 + 7,015) : 3 b) (102 + 901) x 7
c) 2,515 : (1 + 4) d) 705 x (8 - 2)
Solution: Perform operations inside parentheses first for expressions containing brackets.
Answer:
a) (2,000 + 7,015) : 3 = 9,015 : 3
= 3,005
b) (102 + 901) x 7 = 1,003 x 7
= 7,021
c) 2,515 : (1 + 4) = 2,515 : 5
= 503
d) 705 x (8 - 2) = 705 x 6
= 4,230
2. Solve Exercise 2 Page 55 Mathematics Textbook Grade 3
Problem: A ship carries 7,863 crates. Some crates are unloaded, reducing the remaining number of crates to one-third of the original amount. How many crates are left on the ship?
Solution: The number of crates left on the ship = Number of crates initially : 3.
Answer:
The number of crates left on the ship is:
7,863 : 3 = 2,621 (crates)
Answer: 2,621 crates
3. Solve Exercise 3 Page 55 Mathematics Textbook Grade 3
Problem: Number?
A supervisor uses identical rectangular prism blocks to pave a road that is 4,555 m long. The rectangle on the top face of each block has a length of 5 m.
How many blocks did the supervisor use?
Solution: The number of stone blocks to pave the road = Length of the road : length of the rectangular prism.
Answer:
The number of stone blocks the supervisor used to pave the road is:
4,555 : 5 = 911 (blocks)
Answer: 911 stone blocks.
4. Solve Exercise 4 Page 55 Mathematics Textbook Grade 3
Problem: Number?
A wall is built around a square-shaped plot of land ABCD. Each side measures 2,324 steps.
a) The wall is ? steps long.
b) A guardhouse is built at the midpoint I of side AB.
The segment AI is ? steps long.
Solution:
a) The length of the wall equals the perimeter of the square plot ABCD.
b) The length of segment AI = length of segment AB : 2.
Answer:
a) The perimeter of the wall is 2,324 x 4 = 9,296 (steps)
So, the wall is 9,296 steps long.
b) Since I is the midpoint of side AB, the length of segment AI is 2,324 : 2 = 1,162 steps.
Thus, segment AI is 1,162 steps long.
We hope that with the solution to exercises on pages 52, 53, 54, 55 of Mathematics Grade 3 Workbook 2, Connecting Knowledge with Life, students and parents have found useful materials and can visualize how to solve the exercises.
