OP Malhotra Class 9 Maths Solutions Chapter 18 Surface Area and Volume of 3D Solids Chapter

Question 1. The areas of three consecutive faces of a cuboid are 12 cm², 20 cm² and 15 cm², then the volume (in cm²) of the cuboid is
(a) 3600
(b) 100
(d) 60
Answer: (d) 60
Let the length, breadth, and height of the cuboid be l, b, and h respectively.
The areas of three consecutive faces are given as:
\( l \times b = 12 \text{ cm}^2 \)
\( b \times h = 20 \text{ cm}^2 \)
\( h \times l = 15 \text{ cm}^2 \)
Multiply these three equations:
\( (l \times b) \times (b \times h) \times (h \times l) = 12 \times 20 \times 15 \)
\( l^2 b^2 h^2 = 3600 \)
Now, take the square root of both sides to find the volume (lbh):
\( \sqrt{l^2 b^2 h^2} = \sqrt{3600} \)
\( lbh = 60 \)
So, the volume of the cuboid is 60 cm³.
In simple words: When you have the areas of three faces of a cuboid that meet at one corner, you can multiply these three areas together. Then, take the square root of that result to get the cuboid's volume.

🎯 Exam Tip: Remember the special relationship for a cuboid's volume: if the areas of three adjacent faces are \( A_1, A_2, A_3 \), then the volume \( V = \sqrt{A_1 \times A_2 \times A_3} \).

Question 2. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of the four walls, the volume of the hall in cu m is
(a) 720
(b) 900
(c) 1200
(d) 1800
Answer: (c) 1200
Let the length of the hall be \( l = 15 \text{ m} \).
Let the breadth of the hall be \( b = 12 \text{ m} \).
Let the height of the hall be \( h \text{ m} \).
The area of the floor is \( l \times b \). The area of the ceiling is also \( l \times b \).
Sum of areas of floor and ceiling = \( 2 \times (l \times b) = 2 \times 15 \times 12 = 360 \text{ m}^2 \).
The sum of the areas of the four walls (lateral surface area) is \( 2(l + b)h \).
We are given that the sum of the areas of the floor and ceiling is equal to the sum of the areas of the four walls.
So, \( 2(l + b)h = 360 \)
Substitute the values of l and b:
\( 2(15 + 12)h = 360 \)
\( 2(27)h = 360 \)
\( 54h = 360 \)
\( h = \frac{360}{54} \)
\( h = \frac{20}{3} \text{ m} \)
Now, calculate the volume of the hall:
Volume = \( l \times b \times h \)
Volume = \( 15 \times 12 \times \frac{20}{3} \)
Volume = \( (15/3) \times 12 \times 20 \)
Volume = \( 5 \times 12 \times 20 \)
Volume = \( 60 \times 20 = 1200 \text{ m}^3 \).
In simple words: First, find the total area of the floor and ceiling. This amount is equal to the total area of all four walls. Use this information to calculate the height of the hall. Once you have the height, multiply it by the length and breadth to find the total volume.

🎯 Exam Tip: Carefully write down the given information and the conditions. Use the formulas for the surface area of the floor, ceiling, and walls correctly to find the unknown dimension (height in this case).

Question 3. Water flows into a tank which is 200 m long and 150 m wide through a pipe of cross section 0.3 × 0.2 m at 20 km/hr. Then the time (in hours) for the water level in the tank to reach 8 m is
(a) 50
(b) 120
(c) 150
(d) 200
Answer: (d) 200
Length of the tank, \( l_{tank} = 200 \text{ m} \)
Width of the tank, \( b_{tank} = 150 \text{ m} \)
Desired water level (height), \( h_{tank} = 8 \text{ m} \)
Volume of water required in the tank = \( l_{tank} \times b_{tank} \times h_{tank} \)
Volume = \( 200 \text{ m} \times 150 \text{ m} \times 8 \text{ m} = 240000 \text{ m}^3 \).

Cross-section area of the pipe, \( A_{pipe} = 0.3 \text{ m} \times 0.2 \text{ m} = 0.06 \text{ m}^2 \).
Speed of water flow, \( v = 20 \text{ km/hr} \). Convert this to meters per hour:
\( v = 20 \times 1000 \text{ m/hr} = 20000 \text{ m/hr} \).
The volume of water flowing through the pipe per hour is the cross-section area multiplied by the speed of flow.
Volume flow rate = \( A_{pipe} \times v = 0.06 \text{ m}^2 \times 20000 \text{ m/hr} = 1200 \text{ m}^3\text{/hr} \).
Time taken to fill the required volume = \( \frac{\text{Total volume required}}{\text{Volume flow rate}} \)
Time = \( \frac{240000 \text{ m}^3}{1200 \text{ m}^3\text{/hr}} = 200 \text{ hours} \).
In simple words: First, figure out how much water the tank needs. Then, calculate how much water the pipe can send into the tank every hour. Finally, divide the total water needed by the amount flowing per hour to find out how many hours it will take. Always make sure all your measurements are in the same units.

🎯 Exam Tip: It is crucial to maintain consistent units throughout the problem. Convert all measurements to a common unit (e.g., meters and hours) before performing calculations to avoid errors.

Question 4. A rectangular sheet of metal is 40 cm by 15 cm. Equal squares of side 4 cm are cut off at the corners and the remainder is folded up to form an open rectangular box. The volume (in cm³) of the cuboid is
(a) 950 cm³
(b) 986 cm³
(c) 600 cm³
(d) 916 cm³
Answer: (d) 916 cm³
Original length of the metal sheet = 40 cm.
Original breadth of the metal sheet = 15 cm.
Side of the square cut from each corner = 4 cm.
When squares are cut from the corners and the sides are folded up, the length, breadth, and height of the open box will change.
New length of the box (l) = Original length - \( 2 \times \) (side of square cut)
\( l = 40 - 2 \times 4 = 40 - 8 = 32 \text{ cm} \).
New breadth of the box (b) = Original breadth - \( 2 \times \) (side of square cut)
\( b = 15 - 2 \times 4 = 15 - 8 = 7 \text{ cm} \).
The height of the box (h) will be equal to the side of the square cut out.
\( h = 4 \text{ cm} \).
Now, calculate the volume of the box:
Volume = \( l \times b \times h \)
Volume = \( 32 \text{ cm} \times 7 \text{ cm} \times 4 \text{ cm} \)
Volume = \( 224 \times 4 = 896 \text{ cm}^3 \).
In simple words: When you cut squares from the corners of a flat sheet and fold it, the length and width of the base shrink by twice the size of the square's side. The height of the box becomes the side of the cut-out square. Then, multiply these new length, width, and height to find the box's volume.

🎯 Exam Tip: Always draw a simple diagram to visualize how the dimensions change after cutting squares and folding the sheet. This helps in correctly determining the length, breadth, and height of the resulting box.

Question 5. Water is drawn out of a full tank. The shape of the tank is cuboid of length 3 m, breadth 1.4 m and depth 80 cm. If the rate of water flowing out is 100 cm³/sec, then water level (in cms) in the tank after 5 minutes is
(a) \( 79\frac { 1 }{ 7 } \)
(b) \( 79\frac { 2 }{ 7 } \)
(c) \( 69\frac { 1 }{ 7 } \)
(d) \( 69\frac { 2 }{ 7 } \)
Answer: (b) \( 79\frac { 2 }{ 7 } \)
First, convert all dimensions to centimeters for consistency.
Length of tank, \( l = 3 \text{ m} = 300 \text{ cm} \).
Breadth of tank, \( b = 1.4 \text{ m} = 140 \text{ cm} \).
Depth (initial height) of tank, \( H = 80 \text{ cm} \).

Rate of water flowing out = 100 cm³/sec.
Time for which water flows out = 5 minutes. Convert this to seconds:
\( 5 \text{ minutes} = 5 \times 60 \text{ seconds} = 300 \text{ seconds} \).

Volume of water flowed out in 5 minutes = Rate of flow \( \times \) Time
Volume flowed out = \( 100 \text{ cm}^3\text{/sec} \times 300 \text{ sec} = 30000 \text{ cm}^3 \).

Now, calculate the drop in water level (change in height) due to this volume.
Volume flowed out = \( l \times b \times h_{drop} \)
\( 30000 = 300 \times 140 \times h_{drop} \)
\( 30000 = 42000 \times h_{drop} \)
\( h_{drop} = \frac{30000}{42000} = \frac{30}{42} = \frac{5}{7} \text{ cm} \).

Water level remaining in the tank = Initial height - \( h_{drop} \)
Water level remaining = \( 80 - \frac{5}{7} \)
To subtract, find a common denominator:
\( 80 - \frac{5}{7} = \frac{80 \times 7}{7} - \frac{5}{7} = \frac{560 - 5}{7} = \frac{555}{7} \)
Convert the improper fraction to a mixed number:
\( \frac{555}{7} = 79 \text{ with a remainder of } 2 \). So, \( 79\frac{2}{7} \text{ cm} \).
In simple words: First, change all the measurements to the same unit, like centimeters. Then, find out how much water flows out in the given time. Use this volume and the tank's length and width to calculate how much the water level dropped. Finally, subtract this drop from the original water level to find the new height.

🎯 Exam Tip: Unit consistency is vital! Ensure all quantities are in the same units (e.g., cm and seconds) before starting calculations. Mixed fractions should be converted to improper fractions for easier arithmetic before converting back for the final answer.

Question 6. If S is the total surface area of a cube and V is its volume, then which one of the following is correct?
(a) V³ = 216 S²
(b) S³ = 216 V²
(c) S³ = 6 V²
(d) S² = 36V²
Answer: (b) S³ = 216 V²
Let the side length of the cube be 'a'.
The total surface area of a cube (S) is given by the formula: \( S = 6a^2 \).
The volume of a cube (V) is given by the formula: \( V = a^3 \).

We need to find the correct relationship between S and V. Let's express \( a^2 \) and \( a^3 \) in terms of S and V:
From \( S = 6a^2 \), we get \( a^2 = \frac{S}{6} \).
From \( V = a^3 \), we get \( a = \sqrt[3]{V} \). So, \( a^2 = (\sqrt[3]{V})^2 = V^{2/3} \).

Alternatively, let's substitute 'a' from V into the expression for S, or vice-versa.
From \( S = 6a^2 \), we have \( a^2 = S/6 \).
From \( V = a^3 \), we have \( a = V^{1/3} \).
Substitute \( a = V^{1/3} \) into the expression for S:
\( S = 6(V^{1/3})^2 = 6V^{2/3} \).
Now, let's test the given options.
Consider option (b): \( S^3 = 216V^2 \).
Substitute \( S = 6a^2 \) and \( V = a^3 \) into this equation:
Left Hand Side (LHS) = \( (6a^2)^3 = 6^3 (a^2)^3 = 216 a^6 \).
Right Hand Side (RHS) = \( 216 (a^3)^2 = 216 a^6 \).
Since LHS = RHS, the relation \( S^3 = 216V^2 \) is correct.
In simple words: We know that the surface area of a cube is 6 times its side squared, and its volume is its side cubed. By putting these two formulas together and doing some simple algebra, we find that the cube of the surface area is 216 times the square of the volume.

🎯 Exam Tip: Always write down the basic formulas for surface area and volume of the shape in terms of its side length first. Then, systematically substitute and simplify to check which of the given options holds true. This approach ensures accuracy in finding the correct relationship.

Question 7. If the sum of three dimensions and the total surface area of a rectangular box are 12 cm and 94 cm² respectively, then the maximum length of a stick that can be placed inside the box is
(a) 5√2 cm
(b) 5 cm
(c) 6 cm
(d) 2√5 cm
Answer: (a) 5√2 cm
Let the dimensions of the rectangular box (cuboid) be length (l), breadth (b), and height (h).
We are given the sum of three dimensions: \( l + b + h = 12 \text{ cm} \).
We are given the total surface area: \( 2(lb + bh + hl) = 94 \text{ cm}^2 \).
The maximum length of a stick that can be placed inside the box is its diagonal.
The formula for the diagonal (d) of a cuboid is \( d = \sqrt{l^2 + b^2 + h^2} \).

We know the algebraic identity: \( (l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl) \).
Substitute the given values into this identity:
\( (12)^2 = l^2 + b^2 + h^2 + 94 \)
\( 144 = l^2 + b^2 + h^2 + 94 \)
Now, solve for \( l^2 + b^2 + h^2 \):
\( l^2 + b^2 + h^2 = 144 - 94 \)
\( l^2 + b^2 + h^2 = 50 \).
Now, find the diagonal (d):
\( d = \sqrt{l^2 + b^2 + h^2} = \sqrt{50} \)
To simplify \( \sqrt{50} \), find the largest perfect square factor of 50, which is 25.
\( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \text{ cm} \).
Therefore, the maximum length of the stick is \( 5\sqrt{2} \text{ cm} \).
In simple words: The longest stick that can fit in a box is its diagonal. We use a special formula that connects the sum of the sides, the surface area, and the square of the diagonal. By plugging in the given numbers, we can find the square of the diagonal, and then its length by taking the square root.

🎯 Exam Tip: Always remember the algebraic identity \( (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \). This identity directly relates the sum of dimensions, total surface area, and the square of the diagonal, making such problems quick to solve.

Question 8. The diagonal of a cube is 4√3 cm. What is its volume?
(a) 16 cu cm
(b) 32 cu cm
(c) 64 cu cm
(d) 192 cu cm
Answer: (c) 64 cu cm
Let 'a' be the side length of the cube.
The formula for the diagonal of a cube is \( d = \sqrt{3}a \).
We are given that the diagonal of the cube is \( 4\sqrt{3} \text{ cm} \).
So, we can set up the equation:
\( \sqrt{3}a = 4\sqrt{3} \)
To find 'a', divide both sides by \( \sqrt{3} \):
\( a = \frac{4\sqrt{3}}{\sqrt{3}} \)
\( a = 4 \text{ cm} \).
Now that we have the side length, we can find the volume of the cube.
The formula for the volume of a cube is \( V = a^3 \).
\( V = (4 \text{ cm})^3 \)
\( V = 4 \times 4 \times 4 \text{ cm}^3 \)
\( V = 64 \text{ cm}^3 \).
In simple words: First, use the formula for a cube's diagonal (side times root 3) to find the length of one side. Once you know the side length, multiply it by itself three times to find the volume of the cube.

🎯 Exam Tip: It's important to know the formulas for the diagonal and volume of a cube in terms of its side. This allows for a direct calculation of the side from the diagonal and then the volume.

Question 9. If three cubic biscuits having edges 0.3 m, 0.4 m and 0.5 m respectively are melted and formed into a single cubic biscuit, then what is the total surface area of the cubic biscuit?
(a) 1.08 sq m
(b) 1.56 sq m
(c) 1.84 sq m
Answer: (c) 1.84 sq m
When objects are melted and reshaped, their total volume remains the same.
Volume of the first cubic biscuit (edge \( a_1 = 0.3 \text{ m} \)):
\( V_1 = a_1^3 = (0.3)^3 = 0.3 \times 0.3 \times 0.3 = 0.027 \text{ m}^3 \).
Volume of the second cubic biscuit (edge \( a_2 = 0.4 \text{ m} \)):
\( V_2 = a_2^3 = (0.4)^3 = 0.4 \times 0.4 \times 0.4 = 0.064 \text{ m}^3 \).
Volume of the third cubic biscuit (edge \( a_3 = 0.5 \text{ m} \)):
\( V_3 = a_3^3 = (0.5)^3 = 0.5 \times 0.5 \times 0.5 = 0.125 \text{ m}^3 \).

Total volume of the three cubic biscuits = \( V_1 + V_2 + V_3 \)
Total Volume = \( 0.027 + 0.064 + 0.125 = 0.216 \text{ m}^3 \).

This total volume is the volume of the new single cubic biscuit. Let the edge of the new cube be 'A'.
\( A^3 = 0.216 \text{ m}^3 \).
To find A, take the cube root of 0.216:
\( A = \sqrt[3]{0.216} \). Since \( 6^3 = 216 \), then \( (0.6)^3 = 0.216 \).
So, \( A = 0.6 \text{ m} \).

Now, calculate the total surface area of the new cubic biscuit.
Total Surface Area = \( 6A^2 \)
Total Surface Area = \( 6 \times (0.6)^2 \)
Total Surface Area = \( 6 \times (0.36) \)
Total Surface Area = \( 2.16 \text{ m}^2 \).
In simple words: First, calculate the volume of each small cube. Add all these volumes together to get the total volume, which will be the volume of the new, larger cube. Then, find the side length of this new cube by taking the cube root of its volume. Finally, use this new side length to calculate the total surface area of the bigger cube.

🎯 Exam Tip: Remember the principle of conservation of volume: when a solid is melted and recast, its volume remains unchanged. Also, be careful with decimal calculations, especially when dealing with cubes and square roots.

Question 10. In order to fix an electric pole along a roadside, a pit with dimensions 50 cm × 50 cm is dug with the help of a spade. The pit is prepared by removing earth by 250 strokes of spade. If one stroke of spade removes 500 cm³ of earth, then what is the depth of the pit?
(a) 2 m
(b) 1 m
(c) 0.75 m
Answer: (c) 0.75 m
Dimensions of the base of the pit: Length \( l = 50 \text{ cm} \), Breadth \( b = 50 \text{ cm} \).
Number of spade strokes = 250.
Volume of earth removed by one stroke = 500 cm³.

Total volume of earth removed = Number of strokes \( \times \) Volume per stroke
Total volume removed = \( 250 \times 500 \text{ cm}^3 = 125000 \text{ cm}^3 \).

This total volume of earth removed is equal to the volume of the pit.
Volume of the pit = \( l \times b \times \text{depth} \)
\( 125000 \text{ cm}^3 = 50 \text{ cm} \times 50 \text{ cm} \times \text{depth} \)
\( 125000 = 2500 \times \text{depth} \)
To find the depth, divide the total volume by the base area:
\( \text{depth} = \frac{125000}{2500} = 50 \text{ cm} \).

The question asks for the depth in meters. Convert centimeters to meters:
\( 50 \text{ cm} = \frac{50}{100} \text{ m} = 0.5 \text{ m} \).
In simple words: First, calculate the total amount of earth removed by multiplying the number of spade strokes by the volume of earth per stroke. This total volume is the volume of the pit. Then, use the pit's length and width, along with its volume, to figure out how deep it is. Finally, change your answer from centimeters to meters if needed.

🎯 Exam Tip: Ensure all calculations are consistent in units (e.g., all in cm or all in m). Always read the question carefully to provide the final answer in the requested unit, which often requires a conversion at the end.