Samacheer Kalvi Class 5 Maths Solutions Term 3 Chapter 3 Measurements Exercise 3.2

Question 1. Volume of regular solids such as cube and cuboid can be found by multiplying the dimensions. Complete the given table by finding the volume of the given objects.

Answer: To find the volume of a solid, we multiply its length (l), breadth (b), and height (h). The formula for volume is \( \text{Volume} = \text{l} \times \text{b} \times \text{h} \). Below is the completed table with the volume for each object.

🎯 Exam Tip: Remember to always include the correct cubic units (e.g., cubic cm) when stating the volume of an object to get full marks.

Question 2. Complete the given table by finding the volume of the given objects.

Answer: We use the formula \( \text{Volume} = \text{l} \times \text{b} \times \text{h} \). If any dimension is missing, we can find it by dividing the given volume by the product of the other two dimensions. The completed table is shown below with all calculations.

🎯 Exam Tip: Always pay attention to the units (cm or m) and ensure consistency in your calculations, as mixing them up will lead to incorrect answers.

Question 3. Find the number of bricks of dimension 20 cm × 5 cm × 10 cm required to construct a wall of dimension 300 cm x 200 cm x 20 cm.
Answer: To find out how many bricks are needed for a wall, we divide the total volume of the wall by the volume of one single brick. This tells us how many small bricks fit into the large wall space.
\( \text{Volume of one brick} = 20 \, \text{cm} \times 5 \, \text{cm} \times 10 \, \text{cm} = 1000 \, \text{cubic cm} \)
\( \text{Volume of the wall} = 300 \, \text{cm} \times 200 \, \text{cm} \times 20 \, \text{cm} = 1,200,000 \, \text{cubic cm} \)
\( \text{Number of bricks} = \frac{\text{Volume of the wall}}{\text{Volume of one brick}} \)
\( \implies \text{Number of bricks} = \frac{1,200,000 \, \text{cubic cm}}{1000 \, \text{cubic cm}} \)
\( \implies \text{Number of bricks} = 1200 \)
Therefore, 1200 bricks are required.In simple words: We calculate the space each brick takes and the total space the wall needs. Then, we divide the wall's space by the brick's space to find how many bricks are needed.

🎯 Exam Tip: Always ensure that all dimensions (length, breadth, height) are in the same unit before performing volume calculations to avoid errors.

Question 4. How many sack of dimension 15 cm × 45 cm × 90 cm filled with rice can be kept in a room of dimension 3 m × 18 m × 9 m.
Answer: First, we need to make sure all units are the same. We will convert the room's dimensions from meters to centimeters (since 1 meter = 100 centimeters).
Room dimensions:
\( \text{Length} = 3 \, \text{m} = 3 \times 100 \, \text{cm} = 300 \, \text{cm} \)
\( \text{Breadth} = 18 \, \text{m} = 18 \times 100 \, \text{cm} = 1800 \, \text{cm} \)
\( \text{Height} = 9 \, \text{m} = 9 \times 100 \, \text{cm} = 900 \, \text{cm} \)
Sack dimensions:
\( \text{Length} = 15 \, \text{cm} \)
\( \text{Breadth} = 45 \, \text{cm} \)
\( \text{Height} = 90 \, \text{cm} \)
To find the number of sacks that can fit, we divide the volume of the room by the volume of one sack.
\( \text{Number of sacks} = \frac{\text{Volume of room}}{\text{Volume of one sack}} \)
\( \implies \text{Number of sacks} = \frac{300 \, \text{cm} \times 1800 \, \text{cm} \times 900 \, \text{cm}}{15 \, \text{cm} \times 45 \, \text{cm} \times 90 \, \text{cm}} \)
Now, we cancel out common factors:
\( \implies \text{Number of sacks} = \frac{300}{15} \times \frac{1800}{45} \times \frac{900}{90} \)
\( \implies \text{Number of sacks} = 20 \times 40 \times 10 \)
\( \implies \text{Number of sacks} = 8000 \)
Therefore, 8000 sacks of rice can be kept in the room.In simple words: First, we change all sizes to the same unit (centimeters). Then, we find the total space of the room and the space of one sack. Finally, we divide the room's space by the sack's space to find how many sacks fit.

🎯 Exam Tip: Always convert all units to be consistent (e.g., all centimeters or all meters) before starting any calculation, especially when comparing volumes of different objects.