GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.5

Question 1. A matchbox measures 4 cm x 2.5 cm x 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Answer: The volume of one matchbox is calculated by multiplying its length, breadth, and height. So, \( 4 \times 2.5 \times 1.5 \text{ cm}^3 = 15 \text{ cm}^3 \). A packet containing 12 such boxes will have a total volume found by multiplying the volume of one matchbox by 12. Therefore, \( 15 \times 12 \text{ cm}^3 = 180 \text{ cm}^3 \).
In simple words: First, find the space one matchbox takes up. Then, multiply that amount by 12 to find the total space 12 matchboxes would take up.

Exam Tip: Remember to always state the units correctly for your final answer, especially for volume (cubic units).

Question 2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m³ = 1000 L)
Answer: The capacity of the tank is determined by multiplying its length, width, and depth. So, \( 6 \times 5 \times 4.5 \text{ m}^3 = 135 \text{ m}^3 \). To convert this volume from cubic meters to litres, we use the conversion factor \( 1 \text{ m}^3 = 1000 \text{ L} \). Therefore, the tank can hold \( 135 \times 1000 \text{ L} = 135000 \text{ L} \) of water.
In simple words: First, work out the tank's volume in cubic meters. Then, change that number into litres using the given conversion rate.

Exam Tip: Pay close attention to unit conversions, especially between cubic meters and litres, as this is a common step in such problems.

Question 3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Answer: Let the height of the cuboidal vessel be \( h \) meters. We are given the length \( l = 10 \text{ m} \), the width \( b = 8 \text{ m} \), and the desired capacity (volume) as \( 380 \text{ m}^3 \). The formula for the volume of a cuboid is \( l \times b \times h \).
So, \( l \times b \times h = 380 \)
\( \implies 10 \times 8 \times h = 380 \)
\( \implies 80h = 380 \)
\( \implies h = \frac{380}{80} \)
\( \implies h = \frac{38}{8} \)
\( \implies h = 4.75 \text{ m} \)
Therefore, the cuboidal vessel must be made 4.75 meters high to hold 380 cubic meters of liquid.
In simple words: We know the vessel's length, width, and the total volume it needs to hold. By dividing the total volume by the known length and width, we can find out how tall the vessel must be.

Exam Tip: Always make sure to isolate the unknown variable in your equation to solve for it. Remember to show your calculation steps clearly.

Question 4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m³?
Answer: First, we need to find the volume of the cuboidal pit. Given the length \( l = 8 \text{ m} \), breadth \( b = 6 \text{ m} \), and height (depth) \( h = 3 \text{ m} \). The volume of the pit is \( l \times b \times h = 8 \times 6 \times 3 \text{ m}^3 = 144 \text{ m}^3 \). The cost of digging is Rs. 30 per cubic meter. So, the total cost for digging the pit will be \( 144 \times \text{Rs. } 30 = \text{Rs. } 4320 \).
In simple words: Calculate the total space the pit takes up. Then, multiply that total space by the cost for each cubic meter to get the final digging cost.

Exam Tip: When dealing with costs, always ensure you calculate the total quantity first (in this case, volume) before applying the per-unit rate.

Question 5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Answer: Let the breadth of the cuboidal tank be \( b \) meters. We are given the length \( l = 2.5 \text{ m} \) and the depth \( h = 10 \text{ m} \). The capacity of the tank is 50000 litres. First, convert the capacity from litres to cubic meters: \( 50000 \text{ litres} = \frac{50000}{1000} \text{ m}^3 = 50 \text{ m}^3 \).
The formula for the volume of a cuboid is \( l \times b \times h \).
So, \( l \times b \times h = 50 \)
\( \implies 2.5 \times b \times 10 = 50 \)
\( \implies 25b = 50 \)
\( \implies b = \frac{50}{25} \)
\( \implies b = 2 \text{ m} \)
Hence, the breadth of the cuboidal tank is 2 meters.
In simple words: Convert the tank's capacity to cubic meters. Then, use the volume formula with the given length and depth to solve for the missing breadth.

Exam Tip: Always convert all units to be consistent (e.g., all to meters and cubic meters) before performing calculations to avoid errors.

Question 6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m x 15 m x 6 m. For how many days will the water of this tank last?
Answer: First, calculate the daily water requirement for the entire village. Each person needs 150 litres, and there are 4000 people, so the total daily requirement is \( 150 \times 4000 \text{ litres} = 600000 \text{ litres} \). Convert this to cubic meters: \( \frac{600000}{1000} \text{ m}^3 = 600 \text{ m}^3 \) per day.
Next, calculate the capacity of the tank. The tank measures 20 m long, 15 m wide, and 6 m high. Its volume is \( 20 \times 15 \times 6 \text{ m}^3 = 1800 \text{ m}^3 \).
Finally, to find how many days the water will last, divide the tank's capacity by the daily water requirement:
Number of days \( = \frac{\text{Capacity of the tank}}{\text{Requirement of water for the total population per day}} \)
\( = \frac{1800 \text{ m}^3}{600 \text{ m}^3/\text{day}} \)
\( = 3 \text{ days} \)
Therefore, the water in this tank will last for 3 days.
In simple words: Find out how much water the village uses each day. Then, find out how much water the tank holds. Divide the total water in the tank by the daily usage to see how many days it will last.

Exam Tip: Break down complex problems into smaller, manageable steps like calculating total requirement, then total capacity, and finally the duration. Ensure all units are consistent.

Question 7. A godown measures 40 m x 25 m x 15 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown.
Answer: First, calculate the capacity of the godown. Given its length \( l = 40 \text{ m} \), breadth \( b = 25 \text{ m} \), and height \( h = 15 \text{ m} \). The volume of the godown is \( l \times b \times h = 40 \times 25 \times 15 \text{ m}^3 = 15000 \text{ m}^3 \).
Next, calculate the capacity of one wooden crate. Given its length \( l = 1.5 \text{ m} \), breadth \( b = 1.25 \text{ m} \), and height \( h = 0.5 \text{ m} \). The volume of one crate is \( l \times b \times h = 1.5 \times 1.25 \times 0.5 \text{ m}^3 = 0.9375 \text{ m}^3 \).
To find the maximum number of wooden crates that can be stored, divide the total capacity of the godown by the capacity of a single crate:
Number of crates \( = \frac{\text{Volume of godown}}{\text{Volume of one crate}} \)
\( = \frac{15000}{0.9375} \)
\( = 16000 \)
Therefore, a maximum of 16000 wooden crates can be stored in the godown.
In simple words: Find the total space inside the godown. Then, find the space one wooden crate takes up. Divide the godown's space by the crate's space to find out how many crates can fit in.

Exam Tip: Ensure that all dimensions are in the same units before calculating volumes. This prevents errors when dividing total volume by individual object volume.

Question 8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Answer: The side of the original solid cube \( a = 12 \text{ cm} \).
The volume of the original solid cube is \( a^3 = (12)^3 = 1728 \text{ cm}^3 \).
When this cube is cut into eight cubes of equal volume, the volume of each new cube will be \( \frac{1728}{8} \text{ cm}^3 = 216 \text{ cm}^3 \).
Let the side of the new cube be \( x \) cm. Then its volume is \( x^3 \text{ cm}^3 \).
According to the question, \( x^3 = 216 \).
\( \implies x = \sqrt[3]{216} \)
\( \implies x = 6 \text{ cm} \)
So, the side of each new cube is 6 cm.
Now, let's find the ratio of their surface areas.
Surface area of the original cube \( = 6a^2 = 6(12)^2 \text{ cm}^2 \).
Surface area of the new cube \( = 6x^2 = 6(6)^2 \text{ cm}^2 \).
The ratio between their surface areas is \( \frac{\text{Surface area of the original cube}}{\text{Surface area of the new cube}} = \frac{6(12)^2}{6(6)^2} \).
\( = \frac{12^2}{6^2} = \frac{144}{36} = \frac{4}{1} = 4:1 \).
Hence, the side of the new cube is 6 cm, and the ratio between their surface areas is 4:1.
In simple words: First, find the volume of the big cube, then divide that by 8 to get the volume of each small cube. Use this volume to find the side length of the small cubes. After that, calculate the surface area for both the big cube and a small cube, and simplify their ratio.

Exam Tip: Remember that for similar solids, if the linear dimensions are in ratio \( k:1 \), then volumes are in ratio \( k^3:1 \) and surface areas are in ratio \( k^2:1 \). This can be a useful check for your calculations.

Question 9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Answer: First, convert the flow rate to meters per hour: \( 2 \text{ km} = 2 \times 1000 \text{ m} = 2000 \text{ m} \). So the river flows at 2000 m per hour.
In one hour, the volume of water falling into the sea can be considered as a cuboid with length \( l = 2000 \text{ m} \) (distance covered in one hour), breadth \( b = 40 \text{ m} \), and height (depth) \( h = 3 \text{ m} \).
Volume of water in one hour \( = l \times b \times h = 2000 \times 40 \times 3 \text{ m}^3 = 240000 \text{ m}^3 \).
To find the volume of water falling into the sea in one minute, divide the hourly volume by 60 (since there are 60 minutes in an hour):
Volume of water in one minute \( = \frac{240000}{60} \text{ m}^3 = 4000 \text{ m}^3 \).
Hence, 4000 cubic meters of water will fall into the sea in one minute.
In simple words: First, change the river's speed from kilometers per hour to meters per hour. Then, calculate how much water flows out in one hour by multiplying the length it travels by its width and depth. Finally, divide that total by 60 to find out how much water flows out in just one minute.

Exam Tip: Be careful with units and time conversions. Ensure all dimensions are in meters and convert the time frame (hours to minutes) appropriately before the final calculation.