RBSE Solutions Class 9 Maths Chapter 12 Surface Area and Volume of Cube and Cuboid Exercise 12.2

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 12 Surface Area and Volume of Cube and Cuboid Ex 12.2

Question 1. The dimension of a match box is 3 cm x 2 cm x 1 cm. Find the volume of such 12 packets.
Answer: The length (l) of a match box is 3 cm, its breadth (b) is 2 cm, and its height (h) is 1 cm.
The volume of one match box is calculated by multiplying its length, breadth, and height: \( l \times b \times h = 3 \text{ cm} \times 2 \text{ cm} \times 1 \text{ cm} = 6 \text{ cm}^3 \).
To find the volume of 12 such match boxes, we multiply the volume of one box by 12: \( 6 \text{ cm}^3 \times 12 = 72 \text{ cm}^3 \). Understanding the volume helps in efficiently packing items for transport.
In simple words: The size of one matchbox is 3 cm by 2 cm by 1 cm. Its volume is 6 cubic cm. For 12 such boxes, the total volume is 72 cubic cm.

🎯 Exam Tip: Remember to multiply the length, breadth, and height to find the volume of a cuboid. Then multiply by the number of identical items.

Question 2. The perimeter of one face of a cube is 24 cm. Find the volume of the cube.
Answer: Let one side of the cube be \( l \). A face of a cube is a square, and the perimeter of a square is \( 4 \times \text{side} \).
Given that the perimeter of one face is 24 cm, we have: \( 4l = 24 \)
\( \implies l = \frac{24}{4} = 6 \text{ cm} \).
The volume of a cube is calculated as \( (\text{side})^3 \).
So, the volume of this cube is \( (6)^3 = 6 \times 6 \times 6 = 216 \text{ cm}^3 \). The volume shows how much space the cube takes up.
In simple words: A cube has 6 faces, all squares. If the edge of one square face has a perimeter of 24 cm, then each side is 6 cm. To find the volume, you multiply the side by itself three times, which gives 216 cubic cm.

🎯 Exam Tip: Be careful not to confuse the perimeter of a face with the total surface area of the cube. Always write the formula before substituting values.

Question 3. Three cubes of metal whose edges are 3 cm, 4 cm and 5 cm respectively are melted and recast into a single cube. Find the edge of this new cube.
Answer: Let the edge of the new cube be \( x \).
First, we find the volume of each small cube:
Volume of the first cube \( = (3 \text{ cm})^3 = 27 \text{ cm}^3 \).
Volume of the second cube \( = (4 \text{ cm})^3 = 64 \text{ cm}^3 \).
Volume of the third cube \( = (5 \text{ cm})^3 = 125 \text{ cm}^3 \).
When the three cubes are melted and recast into a single new cube, their total volume remains the same. This principle of conservation of volume is key in many real-world metalworking and casting processes.
So, the volume of the new cube is the sum of the volumes of the three small cubes:
\( x^3 = 27 + 64 + 125 \)
\( x^3 = 216 \)
To find the edge \( x \) of the new cube, we take the cube root of 216:
\( x = \sqrt[3]{216} = 6 \text{ cm} \).
In simple words: Three small metal cubes are melted and turned into one big cube. First, find the volume of each small cube. Then, add these volumes to get the total volume of the new, bigger cube. Finally, find the side length of this new cube by taking the cube root of its total volume.

🎯 Exam Tip: Remember that melting and recasting changes the shape but keeps the total volume constant. Ensure you calculate cube roots correctly.

Question 4. The length and breadth of a water tank are 2.5 m and 2 m respectively. It can hold 1500 litres of water. Find the depth of water tank.
Answer: The length \( (l) \) of the water tank is 2.5 m and its breadth \( (b) \) is 2 m.
The capacity (volume) of the tank is 1500 litres. We know that \( 1 \text{ m}^3 = 1000 \text{ litres} \).
So, we convert the volume from litres to cubic meters: \( \text{Volume} = \frac{1500}{1000} = 1.5 \text{ m}^3 \).
Let the depth (height) of the water tank be \( h \). The volume of a cuboid is given by \( l \times b \times h \).
Therefore, \( 2.5 \times 2 \times h = 1.5 \)
\( 5 \times h = 1.5 \)
\( \implies h = \frac{1.5}{5} = 0.3 \text{ m} \).
To express the depth in centimeters, we multiply by 100: \( h = 0.3 \times 100 = 30 \text{ cm} \). Knowing the tank's depth helps in calculating how much water it can store for daily use.
In simple words: We know the length and width of a water tank and how many liters it can hold. First, change the liters to cubic meters. Then, use the formula for the volume of a tank (length x width x height) to find the missing height (depth).

🎯 Exam Tip: Always ensure all units are consistent (e.g., all meters or all centimeters) before performing calculations. Remember the conversion factor between litres and cubic meters.

Question 5. The length, breadth and height of a wall are 4 m, 15 cm and 3 m respectively. How many bricks of size 20 cm x 10 cm x 8 cm will be required to make it? If the cost of bricks is Rs. 120 per thousand, then find the total cost of bricks.
Answer: First, we convert all dimensions to centimeters for consistency.
For the wall:
Length \( (l_1) = 4 \text{ m} = 400 \text{ cm} \).
Breadth \( (b_1) = 15 \text{ cm} \).
Height \( (h_1) = 3 \text{ m} = 300 \text{ cm} \).
Volume of the wall \( (V_1) = l_1 \times b_1 \times h_1 = 400 \times 15 \times 300 = 1,800,000 \text{ cm}^3 \).
For one brick:
Length \( (l_2) = 20 \text{ cm} \).
Breadth \( (b_2) = 10 \text{ cm} \).
Height \( (h_2) = 8 \text{ cm} \).
Volume of one brick \( (V_2) = l_2 \times b_2 \times h_2 = 20 \times 10 \times 8 = 1600 \text{ cm}^3 \).
The number of bricks required is the total volume of the wall divided by the volume of one brick:
Number of bricks \( = \frac{\text{Volume of wall}}{\text{Volume of one brick}} = \frac{1800000}{1600} = 1125 \text{ bricks} \).
Now, we calculate the total cost.
Cost of 1000 bricks = Rs. 120.
Cost of 1 brick \( = \frac{\text{Rs. } 120}{1000} \).
Total cost of 1125 bricks \( = \frac{120}{1000} \times 1125 = \frac{135000}{1000} = \text{Rs. } 135 \). Practical construction projects always require careful calculation of material quantities to manage costs and avoid waste.
In simple words: First, find the total space of the wall and the space of one brick, making sure all measurements are in centimeters. Divide the wall's volume by the brick's volume to find out how many bricks are needed. Then, use the given cost per thousand bricks to calculate the total cost for all the bricks.

🎯 Exam Tip: Be careful with unit conversions, ensuring all dimensions are in the same unit before calculating volumes. Clearly show steps for both brick count and cost calculation.

Question 6. A village has a water tank of dimension 20 m x 15 m x 6 m. How many litres of water it can hold. If per day consumption of water is 1000 litres. For how many days will the water of this tank last?
Answer: The dimensions of the water tank are: Length \( (l) = 20 \text{ m} \), Breadth \( (b) = 15 \text{ m} \), Height \( (h) = 6 \text{ m} \).
The volume of the tank \( = l \times b \times h = 20 \text{ m} \times 15 \text{ m} \times 6 \text{ m} = 1800 \text{ m}^3 \).
To convert this volume to litres, we use the conversion factor \( 1 \text{ m}^3 = 1000 \text{ litres} \).
Capacity of the tank in litres \( = 1800 \times 1000 = 1,800,000 \text{ litres} \).
The water consumed per day is 1000 litres.
To find how many days the water will last, we divide the total capacity by the daily consumption:
Number of days \( = \frac{\text{Total capacity of tank}}{\text{Consumption per day}} = \frac{1800000}{1000} = 1800 \text{ days} \). Such calculations are important for village planning to ensure a sustainable water supply for the community.
In simple words: First, find the total volume of the water tank in cubic meters, then convert this volume into liters. Once you know the total liters and how many liters are used each day, divide the total by the daily usage to find out for how many days the water will last.

🎯 Exam Tip: Always convert cubic meters to litres correctly using the factor of 1000. Clearly show both the capacity calculation and the duration calculation.

Question 7. A wall is 8 m long, 4 m high and 35 cm thick. There is one door of size 2 m x 1 m and two windows of size 1.20 m x 1 m. Find the cost of construction at the rate Rs. 1500 per cubic metres.
Answer: First, we convert all dimensions to meters for consistency.
For the wall:
Length \( = 8 \text{ m} \).
Height \( = 4 \text{ m} \).
Thickness \( = 35 \text{ cm} = 0.35 \text{ m} \).
Volume of the entire wall (if there were no openings) \( = 8 \times 4 \times 0.35 = 11.2 \text{ m}^3 \).
For the door:
Dimensions of one door: \( 2 \text{ m} \times 1 \text{ m} \).
Volume of space for 1 door \( = 2 \times 1 \times 0.35 = 0.7 \text{ m}^3 \).
For the windows:
Dimensions of each window: \( 1.20 \text{ m} \times 1 \text{ m} \).
Volume of space for 1 window \( = 1.20 \times 1 \times 0.35 = 0.42 \text{ m}^3 \).
Volume of space for 2 windows \( = 2 \times 0.42 = 0.84 \text{ m}^3 \).
Total volume of all openings \( = 0.7 \text{ m}^3 + 0.84 \text{ m}^3 = 1.54 \text{ m}^3 \).
Actual volume of the wall to be constructed \( = \text{Volume of entire wall} - \text{Total volume of openings} \).
Actual volume \( = 11.2 \text{ m}^3 - 1.54 \text{ m}^3 = 9.66 \text{ m}^3 \).
The cost of construction is Rs. 1500 per cubic meter.
Total cost of construction \( = 9.66 \times 1500 = \text{Rs. } 14490 \). Subtracting the volumes of doors and windows ensures that construction costs are accurately estimated, preventing overspending.
In simple words: First, change all measurements to meters. Calculate the total volume of the wall as if it had no openings. Then, calculate the volume of the door and windows separately. Subtract the volume of the openings from the total wall volume to get the actual amount of wall material needed. Finally, multiply this actual volume by the cost per cubic meter to find the total construction cost.

🎯 Exam Tip: Remember to convert all units to a common base (meters are preferred here) before calculating volumes. Always subtract the volume of openings from the total volume of the structure.

Question 8. How many bricks of size 25 cm x 15 cm x 6 cm will be required for a wall 5 m long, 2 m high and 50 cm thick, while 10% of space occupied by cement?
Answer: First, we convert all dimensions to meters.
For the wall:
Length \( (l_w) = 5 \text{ m} \).
Height \( (h_w) = 2 \text{ m} \).
Thickness \( (t_w) = 50 \text{ cm} = 0.50 \text{ m} \).
Total volume of the wall \( = l_w \times h_w \times t_w = 5 \times 2 \times 0.50 = 5 \text{ m}^3 \).
Space occupied by cement \( = 10\% \) of the total volume.
Volume of cement \( = 0.10 \times 5 = 0.5 \text{ m}^3 \).
Actual volume to be filled by bricks \( = \text{Total volume of wall} - \text{Volume of cement} \).
Actual volume for bricks \( = 5 \text{ m}^3 - 0.5 \text{ m}^3 = 4.5 \text{ m}^3 \).
For one brick:
Dimensions of one brick: \( 25 \text{ cm} \times 15 \text{ cm} \times 6 \text{ cm} \).
Convert brick dimensions to meters: \( 0.25 \text{ m} \times 0.15 \text{ m} \times 0.06 \text{ m} \).
Volume of 1 brick \( = 0.25 \times 0.15 \times 0.06 = 0.002250 \text{ m}^3 \).
Total number of bricks required \( = \frac{\text{Actual volume for bricks}}{\text{Volume of 1 brick}} \).
Number of bricks \( = \frac{4.5}{0.002250} = 2000 \text{ bricks} \). Considering the space taken by mortar (cement) ensures that the right quantity of bricks is ordered, preventing shortages or excess material.
In simple words: First, calculate the total volume of the wall in cubic meters. Since 10% of this space is for cement, subtract that amount to find the volume that bricks will fill. Then, find the volume of one brick (also in cubic meters). Divide the brick-filled volume by the volume of one brick to get the total number of bricks needed.

🎯 Exam Tip: Remember to subtract the volume taken by cement or mortar from the total volume of the structure before calculating the number of bricks. Ensure all units are consistent.

Question 9. A field is 150 metres long and 100 metres wide. If the pit in the pond is 200 metres long and 50 metres wide is dug to a depth of 0.75 m and the earth is spread uniformly in the field. By how much the level of field is raised?
Answer: Dimensions of the field: Length \( = 150 \text{ m} \), Breadth \( = 100 \text{ m} \).
Dimensions of the pit: Length \( = 200 \text{ m} \), Breadth \( = 50 \text{ m} \), Depth \( = 0.75 \text{ m} \).
Volume of earth taken out from the pit \( = \text{Length} \times \text{Breadth} \times \text{Depth} = 200 \times 50 \times 0.75 \text{ m}^3 \).
Volume of earth \( = 7500 \text{ m}^3 \).
When this earth is spread uniformly over the field, it raises the field's level. Let the height by which the field level is raised be \( h \) meters.
The volume of the raised earth in the field will be equal to the volume of earth taken out from the pit.
So, \( \text{Length of field} \times \text{Breadth of field} \times h = \text{Volume of earth} \).
\( 150 \times 100 \times h = 7500 \)
\( 15000 \times h = 7500 \)
\( \implies h = \frac{7500}{15000} = \frac{1}{2} = 0.50 \text{ m} \).
To convert meters to centimeters, multiply by 100: \( h = 0.50 \times 100 = 50 \text{ cm} \). This method of spreading excavated earth to raise ground level is a common practice in land development and flood control.
In simple words: First, calculate how much earth is dug out from the pit by multiplying its length, width, and depth. This amount of earth is then spread evenly over the field. So, the volume of the dug-out earth is equal to the new volume added to the field. Use the field's length and width, along with this volume, to find out how much the field's level increases.

🎯 Exam Tip: The key here is the conservation of volume: the volume of earth removed equals the volume of earth added. Ensure you set up the equation correctly to find the unknown height.