Official Syllabus Item: 18, 47
Mensuration
3D Mensuration: Volume & Surface Area of Cube and Cuboid
Master volume, total surface area, lateral surface area (area of 4 walls), space diagonal, water reservoirs, melting/recasting, and brickwork capacity for RIMC Maths 2026.
Overview & Fundamental Concept
Three-dimensional solid mensuration of cubes and cuboids represents a foundational scoring pillar in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Serial Numbers 18 ('Volume of Cube and Cuboids') and 47 ('Volume and surface area of cube and cuboids, etc.') of Chapter 8 in the official brochure, solid geometry questions feature prominently in both Part A (5 Marks) and Part B (10 Marks). The RIMC descriptive evaluation standards demand spatial clarity and arithmetic rigor: drawing neat isometric/oblique sketches with dashed hidden edges, distinguishing precisely between Total Surface Area (TSA), Lateral Surface Area (LSA / Area of 4 walls), and 5-face Open Cistern Area, applying space diagonal formulas ($d = \sqrt{l^2 + b^2 + h^2}$), converting fluently between cubic metric volumes and liquid capacities ($1\text{ m}^3 = 1000\text{ litres}$), tracking volume conservation during phase-change recasting, and accounting for door/window deductions and mortar percentages in masonry brickwork.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 18 | Volume of Cube and Cuboids: Fundamental formulas V = a³ and V = lbh, space diagonals, and melting/recasting problems. | Standard 5-mark and 10-mark descriptive problems testing 3D spatial properties and conservation of matter. |
| Item 47 | Surface Area & Practical Mensuration: Total surface area, lateral surface area (4 walls), open boxes, water tanks/capacities, and brick masonry. | Directly specified in syllabus item 47. Tests multi-step real-world applications including wall painting and fluid dynamics. |
1. The Cuboid (Rectangular Parallelepiped)
A cuboid is a solid three-dimensional polyhedron bounded by six rectangular faces. Any two adjacent faces meet at a right angle ($90^\circ$). It has $6$ faces, $12$ edges, and $8$ vertices. If $l, b,$ and $h$ denote its length, breadth, and height respectively:
• Volume ($V$): The measure of space enclosed by the cuboid:
$$V = \text{Base Area} \times \text{Height} = l \times b \times h$$
• Total Surface Area (TSA): The sum of the areas of all six rectangular faces:
$$\text{TSA} = 2(l \times b + b \times h + l \times h) = 2(lb + bh + lh)$$
• Lateral Surface Area (LSA) / Area of Four Walls: The sum of the areas of the four vertical walls (excluding floor and ceiling):
$$\text{LSA} = 2(lh + bh) = 2(l + b)h = \text{Perimeter of Base} \times \text{Height}$$
• Surface Area of an Open Cistern / Box (5 Faces): When the top face is absent:
$$\text{Area}_{\text{open}} = lb + 2(l + b)h = \text{Base Area} + \text{Area of 4 Walls}$$
• Space Diagonal ($d$): The length of the longest straight rod that can fit inside the cuboid, joining two opposite vertices:
$$d = \sqrt{l^2 + b^2 + h^2}$$
2. The Cube (Regular Hexahedron)
A cube is a special cuboid whose length, breadth, and height are all equal ($l = b = h = a$). All six faces are congruent squares of side $a$:
• Volume ($V$):
$$V = a^3 \implies \text{Edge } a = \sqrt[3]{V}$$
• Total Surface Area (TSA):
$$\text{TSA} = 6a^2$$
• Lateral Surface Area (LSA) / 4 Vertical Faces:
$$\text{LSA} = 4a^2$$
• Space Diagonal ($d$):
$$d = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$$
Scaling and Ratio Properties:
If the side of a cube is scaled by a factor of $k$ (i.e. $a' = ka$):
• The edge length ratio is $1 : k$.
• The surface area ratio is $1 : k^2$.
• The volume ratio is $1 : k^3$.
3. Fluid Capacity, Rates of Flow & Submersion
Water reservoir and flow dynamics problems connect linear mensuration to fluid kinematics:
Standard Capacity Conversion Units:
$$1\text{ m}^3 = 1000\text{ Litres (L)} = 1\text{ Kilolitre (kL)}$$$$1\text{ Litre} = 1000\text{ cm}^3 = 1000\text{ mL} \implies 1\text{ cm}^3 = 1\text{ mL}$$$$1000\text{ Litres} = 1\text{ m}^3 \implies 1\text{ Litre} = \frac{1}{1000}\text{ m}^3 = 10^{-3}\text{ m}^3$$
Rate of Flow Through a Pipe:
When water flows through a pipe of uniform cross-section at a constant speed $v$ for time $t$, the water extruded forms a column of length $L = v \times t$:
$$\text{Volume of Water Delivered} = \text{Cross-Sectional Area of Pipe} \times (\text{Speed} \times \text{Time})$$
Rise in Liquid Level on Solid Submersion:
By Archimedes' Principle, when a solid insoluble object is completely submerged in a liquid-filled vessel, the volume of liquid displaced equals the volume of the submerged solid:
$$\text{Rise in Liquid Level } (\Delta h) = \frac{\text{Volume of Submerged Solid}}{\text{Base Area of Vessel}} = \frac{V_{\text{solid}}}{l_{\text{tank}} \times b_{\text{tank}}}$$
4. Applied Problems: Melting, Hollow Boxes & Brickwork
1. Conservation of Volume in Melting & Recasting:
When solids are melted and recast into a different shape, total mass and volume remain invariant (assuming zero foundry loss):
$$\sum V_{\text{initial solids}} = V_{\text{recast solid}}$$Example: If three solid cubes of edges $a_1, a_2, a_3$ are melted into one large cube of edge $A$, then $A^3 = a_1^3 + a_2^3 + a_3^3$.
2. Hollow Boxes with Uniform Wall Thickness ($t$):
For a closed rectangular box with external dimensions $L, B, H$ and wall thickness $t$:
$$\text{Internal Dimensions: } l = L - 2t, \quad b = B - 2t, \quad h = H - 2t$$$$\text{Volume of Material (Wood/Metal)} = V_{\text{external}} - V_{\text{internal}} = (L \times B \times H) - (l \times b \times h)$$Special Note for OPEN Box: If the box has no lid, thickness is subtracted only once from height: $h = H - t$.
3. Masonry Wall Construction with Mortar:
$$\text{Net Volume of Wall} = \text{Gross Wall Volume} - \text{Openings (doors/windows)} - \text{Mortar Volume}$$$$\text{Number of Bricks} = \frac{\text{Net Brickwork Volume}}{\text{Volume of One Single Brick}}$$
Illustrated Conceptual Diagrams & Visual Schematics
3D Cuboid Isometric Geometry & Space Diagonal Blueprint
3D Metrics: Volume = $l \times b \times h$. Total Surface Area = $2(lb + bh + hl)$. Lateral Surface Area (Area of 4 walls) = $2h(l + b)$. Longest Rod that can be placed inside (Space Diagonal) = $\sqrt{l^2 + b^2 + h^2}$.
Topper Strategy & Descriptive Step Breakdown
Mastering RIMC Mathematics Standard
In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.
Official Descriptive Marking Scheme & Step Allocation
In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:
Step 1: 3D Geometric Diagram & Given Variables
20% of marks
Draw an accurate 3D isometric or oblique sketch of the cube, cuboid, or tank. Mark length $l$, breadth $b$, height $h$, or edge $a$ clearly. Write explicit 'Given:' dimensions with standardized units.
Step 2: Formula Quoting & Model Formulation
25% of marks
State the appropriate formula algebraically before plugging in numbers (e.g., $d = \sqrt{l^2+b^2+h^2}$, $\text{LSA} = 2(l+b)h$, or $V = a_1^3 + a_2^3 + a_3^3$).
Step 3: Dimensional Harmonization & Arithmetic Steps
35% of marks
Ensure all linear dimensions are converted to the same unit (all metres or all centimetres). Show cubic root extraction or prime factorizations step by step.
Step 4: Unitary Conversion & Final Boxed Answer
20% of marks
Convert final cubic units to litres ($1\text{ m}^3 = 1000\text{ L}$) or cost (₹) as asked. State the final answer in a complete concluding sentence enclosed in a box.
Examiner Traps & Common Blunders (Marks Penalties)
Using TSA for an Open Cistern or Tank 2 to 3 marks deducted for computing 6 faces instead of 5 faces ($lb + 2(l+b)h$).
Examiner Trap: Using $2(lb + bh + lh)$ for an open tank or swimming pool, counting the top open surface as a physical wall.
Unit Mismatch in Capacity (m³ vs Litres vs cm³) Severe: factor-of-1000 error results in complete loss of calculation marks.
Examiner Trap: Multiplying tank dimensions in metres ($4 \times 3 \times 2 = 24\text{ m}^3$) and writing capacity as $24\text{ Litres}$ instead of $24,000\text{ Litres}$.
Confusing Space Diagonal with Face Diagonal Total deduction on diagonal questions; maximum 1 mark for diagram.
Examiner Trap: Calculating the longest rod that can fit in a room as $\sqrt{l^2 + b^2}$ instead of the 3D space diagonal $\sqrt{l^2 + b^2 + h^2}$.
Subtracting Thickness Once in Closed Hollow Boxes Compounding error invalidates internal volume; 3 to 4 marks lost.
Examiner Trap: Computing internal length as $L - t$ instead of $L - 2t$ (thickness exists on both opposite walls of a closed box).
Ignoring Mortar Deductions in Masonry Walls 2 marks deducted for failing to account for net brickwork volume.
Examiner Trap: Dividing total wall volume directly by brick volume when the question specifies that mortar occupies $10\%$ of the wall.
Cadet Topper Mental Verification Shortcuts
Space Diagonal Integer Triplets
Remember standard integer space diagonals: $(12, 9, 8) \implies \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m}$. $(2, 3, 6) \implies \sqrt{4 + 9 + 36} = \sqrt{49} = 7\text{ m}$. $(1, 2, 2) \implies \sqrt{1 + 4 + 4} = 3\text{ m}$.
The 3-Cube Recasting Invariant (3-4-5 to 6)
When cubes of edges $3, 4,$ and $5$ are melted together: $3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216 = 6^3$. The recast cube's edge is always $6$! For multiples like $(6, 8, 10)$, the edge is $12$.
Direct Capacity in Litres from Centimetres
If dimensions are in centimetres, capacity in litres is simply $\frac{l \times b \times h}{1000}$. If dimensions are in metres, capacity in litres is $l \times b \times h \times 1000$.
Area of 4 Walls from Perimeter
Area of four walls is simply $\text{Perimeter of floor} \times \text{Height}$. If given perimeter of room is $30\text{ m}$ and height is $4\text{ m}$, area of 4 walls is $30 \times 4 = 120\text{ m}^2$ in one step.
Key Formulas & Conversion Cheat Sheet
Cuboid Volume & Diagonals
$$V = lbh, \quad d = \sqrt{l^2 + b^2 + h^2}$$
$d$ is the longest straight line/rod fitting inside the cuboid.
Cuboid Surface Areas
$$\text{TSA} = 2(lb + bh + lh), \quad \text{LSA (4 Walls)} = 2(l + b)h, \quad \text{Open Box} = lb + 2(l + b)h$$
LSA excludes ceiling and floor; Open box has 5 faces.
Cube Metrics
$$V = a^3, \quad \text{TSA} = 6a^2, \quad \text{LSA} = 4a^2, \quad d = a\sqrt{3}$$
$a$ is edge length; ratio of volumes of two cubes is $(a_1/a_2)^3$.
Fluid Capacity Relations
$$1\text{ m}^3 = 1000\text{ L}, \quad 1\text{ L} = 1000\text{ cm}^3, \quad 1\text{ cm}^3 = 1\text{ mL}$$
Convert all lengths to metres before multiplying by 1000 for litres.
Pipe Flow Rate
$$\text{Volume} = \text{Cross-Section Area} \times \text{Velocity} \times \text{Time}$$
Ensure velocity (e.g. km/h) is converted to m/s or m/min matching cross-section units.
Hollow Closed Box Material Volume
$$V_{\text{material}} = LBH - (L - 2t)(B - 2t)(H - 2t)$$
$t$ is uniform wall thickness; subtract $t$ only once for height in open boxes.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Part A - Descriptive Space Diagonal
5 Marks
Find the length of the longest pole that can be placed in a room $12\text{ m}$ long, $9\text{ m}$ broad, and $8\text{ m}$ high.
Step-by-Step Model Solution:
Step 1: Identify the Geometric Entity
The longest straight pole that can fit inside a rectangular room is along its space diagonal, joining two diametrically opposite vertices.
Given dimensions of the room:
$$\text{Length } l = 12\text{ m}, \quad \text{Breadth } b = 9\text{ m}, \quad \text{Height } h = 8\text{ m}$$
Step 2: Formula for Space Diagonal of a Cuboid
The length of the space diagonal $d$ is given by:
$$d = \sqrt{l^2 + b^2 + h^2}$$
Step 3: Substitute Values and Calculate
$$d = \sqrt{12^2 + 9^2 + 8^2}$$ $$d = \sqrt{144 + 81 + 64}$$ $$d = \sqrt{289}$$
Since $17^2 = 289$:
$$d = 17\text{ m}$$
$$\mathbf{\text{Conclusion: The length of the longest pole that can be placed in the room is } 17\text{ metres.}}$$
The longest straight pole that can fit inside a rectangular room is along its space diagonal, joining two diametrically opposite vertices.
Given dimensions of the room:
$$\text{Length } l = 12\text{ m}, \quad \text{Breadth } b = 9\text{ m}, \quad \text{Height } h = 8\text{ m}$$
Step 2: Formula for Space Diagonal of a Cuboid
The length of the space diagonal $d$ is given by:
$$d = \sqrt{l^2 + b^2 + h^2}$$
Step 3: Substitute Values and Calculate
$$d = \sqrt{12^2 + 9^2 + 8^2}$$ $$d = \sqrt{144 + 81 + 64}$$ $$d = \sqrt{289}$$
Since $17^2 = 289$:
$$d = 17\text{ m}$$
$$\mathbf{\text{Conclusion: The length of the longest pole that can be placed in the room is } 17\text{ metres.}}$$
Rough Column Working:
12^2 = 144
9^2 = 81
8^2 = 64
Sum = 144 + 81 + 64 = 289
sqrt(289) = 17 m.
12^2 = 144
9^2 = 81
8^2 = 64
Sum = 144 + 81 + 64 = 289
sqrt(289) = 17 m.
Step Mark Allocation:
- Identifying longest pole as space diagonal with diagram: 1 Marks
- Stating formula d = sqrt(l^2 + b^2 + h^2): 1.5 Marks
- Substitution and calculation: sqrt(144 + 81 + 64) = sqrt(289): 1.5 Marks
- Solving d = 17 m and boxed concluding statement: 1 Marks
Examiner Scoring Key: Do not confuse the floor diagonal (sqrt(12^2 + 9^2) = 15 m) with the 3D space diagonal. The pole spans the full height of the room.
Part A - Descriptive Melting & Recasting
5 Marks
Three solid metal cubes of edges $3\text{ cm}, 4\text{ cm},$ and $5\text{ cm}$ respectively are melted together to form a single new solid cube. Find:
(i) The edge of the new cube.
(ii) The total surface area of the new cube.
(iii) The percentage change in total surface area.
(i) The edge of the new cube.
(ii) The total surface area of the new cube.
(iii) The percentage change in total surface area.
Step-by-Step Model Solution:
Step 1: Calculate Total Volume of the Three Cubes
Let the edge lengths of the three cubes be $a_1 = 3\text{ cm}, a_2 = 4\text{ cm},$ and $a_3 = 5\text{ cm}$.
$$\text{Volume of Cube 1 } V_1 = a_1^3 = 3^3 = 27\text{ cm}^3$$ $$\text{Volume of Cube 2 } V_2 = a_2^3 = 4^3 = 64\text{ cm}^3$$ $$\text{Volume of Cube 3 } V_3 = a_3^3 = 5^3 = 125\text{ cm}^3$$
$$\text{Total Combined Volume } V = V_1 + V_2 + V_3 = 27 + 64 + 125 = 216\text{ cm}^3$$
Step 2: (i) Find the edge of the new cube ($A$)
By conservation of volume during melting and recasting:
$$\text{Volume of new cube } = A^3 = 216\text{ cm}^3$$ $$A = \sqrt[3]{216} = 6\text{ cm}$$
$$\mathbf{\text{The edge of the new cube is } 6\text{ cm.}}$$
Step 3: (ii) Calculate Total Surface Area of the new cube
$$\text{TSA}_{\text{new}} = 6A^2 = 6 \times (6)^2 = 6 \times 36 = 216\text{ cm}^2$$
$$\mathbf{\text{The total surface area of the new cube is } 216\text{ cm}^2.}$$
Step 4: (iii) Calculate percentage change in surface area
Initial combined surface area of the three separate cubes:
$$\text{TSA}_{\text{initial}} = 6a_1^2 + 6a_2^2 + 6a_3^2 = 6(3^2 + 4^2 + 5^2) = 6(9 + 16 + 25) = 6(50) = 300\text{ cm}^2$$
$$\text{Decrease in Surface Area} = \text{TSA}_{\text{initial}} - \text{TSA}_{\text{new}} = 300 - 216 = 84\text{ cm}^2$$ $$\text{Percentage Decrease} = \frac{84}{300} \times 100\% = \frac{84}{3}\% = 28\%$$
$$\mathbf{\text{Conclusion: (i) Edge } = 6\text{ cm}; \; (ii) \; \text{TSA} = 216\text{ cm}^2; \; (iii) \; 28\% \text{ decrease in surface area.}}$$
Let the edge lengths of the three cubes be $a_1 = 3\text{ cm}, a_2 = 4\text{ cm},$ and $a_3 = 5\text{ cm}$.
$$\text{Volume of Cube 1 } V_1 = a_1^3 = 3^3 = 27\text{ cm}^3$$ $$\text{Volume of Cube 2 } V_2 = a_2^3 = 4^3 = 64\text{ cm}^3$$ $$\text{Volume of Cube 3 } V_3 = a_3^3 = 5^3 = 125\text{ cm}^3$$
$$\text{Total Combined Volume } V = V_1 + V_2 + V_3 = 27 + 64 + 125 = 216\text{ cm}^3$$
Step 2: (i) Find the edge of the new cube ($A$)
By conservation of volume during melting and recasting:
$$\text{Volume of new cube } = A^3 = 216\text{ cm}^3$$ $$A = \sqrt[3]{216} = 6\text{ cm}$$
$$\mathbf{\text{The edge of the new cube is } 6\text{ cm.}}$$
Step 3: (ii) Calculate Total Surface Area of the new cube
$$\text{TSA}_{\text{new}} = 6A^2 = 6 \times (6)^2 = 6 \times 36 = 216\text{ cm}^2$$
$$\mathbf{\text{The total surface area of the new cube is } 216\text{ cm}^2.}$$
Step 4: (iii) Calculate percentage change in surface area
Initial combined surface area of the three separate cubes:
$$\text{TSA}_{\text{initial}} = 6a_1^2 + 6a_2^2 + 6a_3^2 = 6(3^2 + 4^2 + 5^2) = 6(9 + 16 + 25) = 6(50) = 300\text{ cm}^2$$
$$\text{Decrease in Surface Area} = \text{TSA}_{\text{initial}} - \text{TSA}_{\text{new}} = 300 - 216 = 84\text{ cm}^2$$ $$\text{Percentage Decrease} = \frac{84}{300} \times 100\% = \frac{84}{3}\% = 28\%$$
$$\mathbf{\text{Conclusion: (i) Edge } = 6\text{ cm}; \; (ii) \; \text{TSA} = 216\text{ cm}^2; \; (iii) \; 28\% \text{ decrease in surface area.}}$$
Rough Column Working:
3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216
A = cbrt(216) = 6 cm
TSA new = 6 * 36 = 216 cm^2
TSA old = 6 * (9 + 16 + 25) = 300 cm^2
Decrease = 84 / 300 * 100 = 28%.
3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216
A = cbrt(216) = 6 cm
TSA new = 6 * 36 = 216 cm^2
TSA old = 6 * (9 + 16 + 25) = 300 cm^2
Decrease = 84 / 300 * 100 = 28%.
Step Mark Allocation:
- Calculating individual and total volumes = 216 cu cm: 1.5 Marks
- Solving cube root to get new edge A = 6 cm: 1 Marks
- Calculating new TSA = 216 sq cm: 1 Marks
- Calculating initial TSA = 300 sq cm and percentage decrease = 28%: 1.5 Marks
Examiner Scoring Key: Coincidentally, for an edge of 6 cm, Volume (216 cu cm) and TSA (216 sq cm) have identical numerical magnitudes, though their physical units differ.
Part B - Room Whitewashing & Deductions
10 Marks
A classroom is $8.5\text{ m}$ long, $6.5\text{ m}$ broad, and $3.4\text{ m}$ high. It has two doors each measuring $1.5\text{ m} \times 1\text{ m}$ and two windows each measuring $2\text{ m} \times 1.5\text{ m}$. Find:
(i) The total area of the four walls requiring whitewashing.
(ii) The area of the ceiling requiring whitewashing.
(iii) The total cost of whitewashing the walls and ceiling at the rate of ₹$18\text{ per m}^2$.
(i) The total area of the four walls requiring whitewashing.
(ii) The area of the ceiling requiring whitewashing.
(iii) The total cost of whitewashing the walls and ceiling at the rate of ₹$18\text{ per m}^2$.
Step-by-Step Model Solution:
Step 1: Given Dimensions of the Classroom
$$\text{Length } l = 8.5\text{ m}, \quad \text{Breadth } b = 6.5\text{ m}, \quad \text{Height } h = 3.4\text{ m}$$
Step 2: Calculate Gross Area of Four Walls
The gross lateral surface area of the four walls is:
$$\text{Gross Area of 4 Walls} = 2(l + b)h$$ $$\text{Gross Area} = 2(8.5\text{ m} + 6.5\text{ m}) \times 3.4\text{ m}$$ $$\text{Gross Area} = 2(15\text{ m}) \times 3.4\text{ m} = 30 \times 3.4 = 102\text{ m}^2$$
Step 3: Calculate Total Area of Openings (Doors and Windows)
• Area of 2 Doors:
$$\text{Area of 1 Door} = 1.5\text{ m} \times 1\text{ m} = 1.5\text{ m}^2$$ $$\text{Area of 2 Doors} = 2 \times 1.5 = 3\text{ m}^2$$
• Area of 2 Windows:
$$\text{Area of 1 Window} = 2\text{ m} \times 1.5\text{ m} = 3\text{ m}^2$$ $$\text{Area of 2 Windows} = 2 \times 3 = 6\text{ m}^2$$
$$\text{Total Area of Openings} = 3\text{ m}^2 + 6\text{ m}^2 = 9\text{ m}^2$$
Step 4: (i) Calculate Net Area of Four Walls for Whitewashing
$$\text{Net Wall Area} = \text{Gross Area of 4 Walls} - \text{Total Area of Openings}$$ $$\text{Net Wall Area} = 102\text{ m}^2 - 9\text{ m}^2 = 93\text{ m}^2$$
$$\mathbf{\text{Area of four walls for whitewashing } = 93\text{ m}^2.}$$
Step 5: (ii) Calculate Area of the Ceiling
The ceiling is a horizontal rectangle of length $l$ and breadth $b$:
$$\text{Area of Ceiling} = l \times b = 8.5\text{ m} \times 6.5\text{ m} = 55.25\text{ m}^2$$
$$\mathbf{\text{Area of ceiling for whitewashing } = 55.25\text{ m}^2.}$$
Step 6: (iii) Calculate Total Area and Total Expenditure
$$\text{Total Whitewashing Area} = \text{Net Wall Area} + \text{Ceiling Area}$$ $$\text{Total Whitewashing Area} = 93\text{ m}^2 + 55.25\text{ m}^2 = 148.25\text{ m}^2$$
Given rate of whitewashing $= ₹18\text{ per m}^2$:
$$\text{Total Cost} = 148.25\text{ m}^2 \times ₹18/\text{m}^2$$ $$\text{Total Cost} = 148.25 \times 18 = 2668.50$$
$$\mathbf{\text{Conclusion: (i) Net Wall Area } = 93\text{ m}^2; \; (ii) \text{ Ceiling } = 55.25\text{ m}^2; \; (iii) \text{ Total Cost } = ₹2,668.50.}$$
$$\text{Length } l = 8.5\text{ m}, \quad \text{Breadth } b = 6.5\text{ m}, \quad \text{Height } h = 3.4\text{ m}$$
Step 2: Calculate Gross Area of Four Walls
The gross lateral surface area of the four walls is:
$$\text{Gross Area of 4 Walls} = 2(l + b)h$$ $$\text{Gross Area} = 2(8.5\text{ m} + 6.5\text{ m}) \times 3.4\text{ m}$$ $$\text{Gross Area} = 2(15\text{ m}) \times 3.4\text{ m} = 30 \times 3.4 = 102\text{ m}^2$$
Step 3: Calculate Total Area of Openings (Doors and Windows)
• Area of 2 Doors:
$$\text{Area of 1 Door} = 1.5\text{ m} \times 1\text{ m} = 1.5\text{ m}^2$$ $$\text{Area of 2 Doors} = 2 \times 1.5 = 3\text{ m}^2$$
• Area of 2 Windows:
$$\text{Area of 1 Window} = 2\text{ m} \times 1.5\text{ m} = 3\text{ m}^2$$ $$\text{Area of 2 Windows} = 2 \times 3 = 6\text{ m}^2$$
$$\text{Total Area of Openings} = 3\text{ m}^2 + 6\text{ m}^2 = 9\text{ m}^2$$
Step 4: (i) Calculate Net Area of Four Walls for Whitewashing
$$\text{Net Wall Area} = \text{Gross Area of 4 Walls} - \text{Total Area of Openings}$$ $$\text{Net Wall Area} = 102\text{ m}^2 - 9\text{ m}^2 = 93\text{ m}^2$$
$$\mathbf{\text{Area of four walls for whitewashing } = 93\text{ m}^2.}$$
Step 5: (ii) Calculate Area of the Ceiling
The ceiling is a horizontal rectangle of length $l$ and breadth $b$:
$$\text{Area of Ceiling} = l \times b = 8.5\text{ m} \times 6.5\text{ m} = 55.25\text{ m}^2$$
$$\mathbf{\text{Area of ceiling for whitewashing } = 55.25\text{ m}^2.}$$
Step 6: (iii) Calculate Total Area and Total Expenditure
$$\text{Total Whitewashing Area} = \text{Net Wall Area} + \text{Ceiling Area}$$ $$\text{Total Whitewashing Area} = 93\text{ m}^2 + 55.25\text{ m}^2 = 148.25\text{ m}^2$$
Given rate of whitewashing $= ₹18\text{ per m}^2$:
$$\text{Total Cost} = 148.25\text{ m}^2 \times ₹18/\text{m}^2$$ $$\text{Total Cost} = 148.25 \times 18 = 2668.50$$
$$\mathbf{\text{Conclusion: (i) Net Wall Area } = 93\text{ m}^2; \; (ii) \text{ Ceiling } = 55.25\text{ m}^2; \; (iii) \text{ Total Cost } = ₹2,668.50.}$$
Rough Column Working:
Gross walls = 2*(8.5+6.5)*3.4 = 30 * 3.4 = 102 m^2
Doors = 2 * 1.5 = 3 m^2
Windows = 2 * 3 = 6 m^2
Total openings = 9 m^2
Net walls = 102 - 9 = 93 m^2
Ceiling = 8.5 * 6.5 = 55.25 m^2
Total Area = 93 + 55.25 = 148.25 m^2
Cost = 148.25 * 18 = ₹2668.50.
Gross walls = 2*(8.5+6.5)*3.4 = 30 * 3.4 = 102 m^2
Doors = 2 * 1.5 = 3 m^2
Windows = 2 * 3 = 6 m^2
Total openings = 9 m^2
Net walls = 102 - 9 = 93 m^2
Ceiling = 8.5 * 6.5 = 55.25 m^2
Total Area = 93 + 55.25 = 148.25 m^2
Cost = 148.25 * 18 = ₹2668.50.
Step Mark Allocation:
- Calculating gross area of 4 walls = 102 sq m: 2 Marks
- Calculating total opening area = 9 sq m and net wall area = 93 sq m: 2.5 Marks
- Calculating ceiling area = 55.25 sq m: 2 Marks
- Finding total whitewashing area = 148.25 sq m: 1.5 Marks
- Multiplying by ₹18/sq m to get ₹2,668.50 and boxed answer: 2 Marks
Examiner Scoring Key: Do not whitewash the floor! Only the four walls (minus openings) and ceiling are whitewashed.
Part B - Fluid Flow & Tank Inflow
10 Marks
Water flows into a rectangular water reservoir $12\text{ m}$ long and $9\text{ m}$ wide through a pipe of rectangular cross-section $20\text{ cm} \times 15\text{ cm}$ at the rate of $12\text{ km/h}$. Find:
(i) The volume of water delivered by the pipe in $45\text{ minutes}$ (in cubic metres and litres).
(ii) The rise in water level in the reservoir in $45\text{ minutes}$.
(i) The volume of water delivered by the pipe in $45\text{ minutes}$ (in cubic metres and litres).
(ii) The rise in water level in the reservoir in $45\text{ minutes}$.
Step-by-Step Model Solution:
Step 1: Harmonize All Units into Metres
Reservoir dimensions:
$$\text{Length } L = 12\text{ m}, \quad \text{Breadth } B = 9\text{ m}$$ $$\text{Base Area of Reservoir } = L \times B = 12 \times 9 = 108\text{ m}^2$$
Pipe cross-section dimensions:
$$\text{Internal Width } = 20\text{ cm} = \frac{20}{100} = 0.2\text{ m} = \frac{1}{5}\text{ m}$$ $$\text{Internal Depth } = 15\text{ cm} = \frac{15}{100} = 0.15\text{ m} = \frac{3}{20}\text{ m}$$
$$\text{Cross-Sectional Area of Pipe } (A_p) = \frac{1}{5} \times \frac{3}{20} = \frac{3}{100}\text{ m}^2 = 0.03\text{ m}^2$$
Step 2: Determine Length of Water Column Discharged in $45\text{ minutes}$
Speed of water flow $= 12\text{ km/h} = 12,000\text{ m/h}$.
Time elapsed $t = 45\text{ minutes} = \frac{45}{60}\text{ hours} = \frac{3}{4}\text{ hours}$.
Length of water column extruded $l_w$ is:
$$l_w = \text{Speed} \times \text{Time} = 12000\text{ m/h} \times \frac{3}{4}\text{ h} = 9000\text{ m}$$
Step 3: (i) Calculate Volume of Water Delivered
$$\text{Volume Delivered } (V) = A_p \times l_w = \frac{3}{100}\text{ m}^2 \times 9000\text{ m} = 3 \times 90 = 270\text{ m}^3$$
Convert to Litres using $1\text{ m}^3 = 1000\text{ Litres}$:
$$\text{Volume in Litres} = 270 \times 1000 = 270,000\text{ Litres}$$
$$\mathbf{\text{Volume delivered } = 270\text{ m}^3 = 270,000\text{ Litres.}}$$
Step 4: (ii) Calculate Rise in Water Level in the Reservoir
Let the rise in water level in the reservoir be $h\text{ metres}$.
$$\text{Volume in Reservoir} = \text{Base Area} \times \text{Rise in Height}$$ $$270\text{ m}^3 = 108\text{ m}^2 \times h$$ $$h = \frac{270}{108} = \frac{30}{12} = \frac{5}{2} = 2.5\text{ m}$$
$$\mathbf{\text{Conclusion: (i) Volume } = 270\text{ m}^3 \; (270,000\text{ Litres}); \; (ii) \text{ Rise in water level } = 2.5\text{ metres (or } 250\text{ cm).}}$$
Reservoir dimensions:
$$\text{Length } L = 12\text{ m}, \quad \text{Breadth } B = 9\text{ m}$$ $$\text{Base Area of Reservoir } = L \times B = 12 \times 9 = 108\text{ m}^2$$
Pipe cross-section dimensions:
$$\text{Internal Width } = 20\text{ cm} = \frac{20}{100} = 0.2\text{ m} = \frac{1}{5}\text{ m}$$ $$\text{Internal Depth } = 15\text{ cm} = \frac{15}{100} = 0.15\text{ m} = \frac{3}{20}\text{ m}$$
$$\text{Cross-Sectional Area of Pipe } (A_p) = \frac{1}{5} \times \frac{3}{20} = \frac{3}{100}\text{ m}^2 = 0.03\text{ m}^2$$
Step 2: Determine Length of Water Column Discharged in $45\text{ minutes}$
Speed of water flow $= 12\text{ km/h} = 12,000\text{ m/h}$.
Time elapsed $t = 45\text{ minutes} = \frac{45}{60}\text{ hours} = \frac{3}{4}\text{ hours}$.
Length of water column extruded $l_w$ is:
$$l_w = \text{Speed} \times \text{Time} = 12000\text{ m/h} \times \frac{3}{4}\text{ h} = 9000\text{ m}$$
Step 3: (i) Calculate Volume of Water Delivered
$$\text{Volume Delivered } (V) = A_p \times l_w = \frac{3}{100}\text{ m}^2 \times 9000\text{ m} = 3 \times 90 = 270\text{ m}^3$$
Convert to Litres using $1\text{ m}^3 = 1000\text{ Litres}$:
$$\text{Volume in Litres} = 270 \times 1000 = 270,000\text{ Litres}$$
$$\mathbf{\text{Volume delivered } = 270\text{ m}^3 = 270,000\text{ Litres.}}$$
Step 4: (ii) Calculate Rise in Water Level in the Reservoir
Let the rise in water level in the reservoir be $h\text{ metres}$.
$$\text{Volume in Reservoir} = \text{Base Area} \times \text{Rise in Height}$$ $$270\text{ m}^3 = 108\text{ m}^2 \times h$$ $$h = \frac{270}{108} = \frac{30}{12} = \frac{5}{2} = 2.5\text{ m}$$
$$\mathbf{\text{Conclusion: (i) Volume } = 270\text{ m}^3 \; (270,000\text{ Litres}); \; (ii) \text{ Rise in water level } = 2.5\text{ metres (or } 250\text{ cm).}}$$
Rough Column Working:
Pipe area = 0.2 * 0.15 = 0.03 m^2
Time = 45/60 = 0.75 h
Distance = 12,000 * 0.75 = 9,000 m
Volume = 0.03 * 9000 = 270 m^3 = 270,000 L
Height = 270 / (12 * 9) = 270 / 108 = 2.5 m.
Pipe area = 0.2 * 0.15 = 0.03 m^2
Time = 45/60 = 0.75 h
Distance = 12,000 * 0.75 = 9,000 m
Volume = 0.03 * 9000 = 270 m^3 = 270,000 L
Height = 270 / (12 * 9) = 270 / 108 = 2.5 m.
Step Mark Allocation:
- Converting pipe cross-section to sq metres (0.03 sq m): 2 Marks
- Converting time and flow speed to find column length = 9,000 m: 2.5 Marks
- Calculating volume delivered = 270 cu m and 270,000 Litres: 2.5 Marks
- Dividing by reservoir base area (108 sq m) to get rise = 2.5 m and boxed answer: 3 Marks
Examiner Scoring Key: Harmonizing all dimensions into metres right at Step 1 prevents catastrophic unit errors.
Part A - Descriptive Hollow Box
5 Marks
A closed wooden box has external dimensions $30\text{ cm} \times 25\text{ cm} \times 20\text{ cm}$. If the wood is $1.5\text{ cm}$ thick throughout, find:
(i) The volume of the wood used to make the box.
(ii) The mass of the box if $1\text{ cm}^3$ of wood weighs $0.8\text{ g}$.
(i) The volume of the wood used to make the box.
(ii) The mass of the box if $1\text{ cm}^3$ of wood weighs $0.8\text{ g}$.
Step-by-Step Model Solution:
Step 1: Calculate External Volume ($V_{\text{ext}}$)
External dimensions:
$$L = 30\text{ cm}, \quad B = 25\text{ cm}, \quad H = 20\text{ cm}$$ $$V_{\text{ext}} = L \times B \times H = 30 \times 25 \times 20 = 15,000\text{ cm}^3$$
Step 2: Calculate Internal Dimensions and Internal Volume ($V_{\text{int}}$)
Uniform thickness of wood: $t = 1.5\text{ cm}$.
Since the box is closed, thickness must be subtracted twice from each external dimension:
$$l = L - 2t = 30 - 2(1.5) = 30 - 3 = 27\text{ cm}$$ $$b = B - 2t = 25 - 2(1.5) = 25 - 3 = 22\text{ cm}$$ $$h = H - 2t = 20 - 2(1.5) = 20 - 3 = 17\text{ cm}$$
$$V_{\text{int}} = l \times b \times h = 27 \times 22 \times 17$$ $$27 \times 22 = 594$$ $$V_{\text{int}} = 594 \times 17 = 10,098\text{ cm}^3$$
Step 3: (i) Calculate Volume of the Wood
$$\text{Volume of Wood} = V_{\text{ext}} - V_{\text{int}} = 15,000\text{ cm}^3 - 10,098\text{ cm}^3 = 4,902\text{ cm}^3$$
$$\mathbf{\text{Volume of wood } = 4,902\text{ cm}^3.}$$
Step 4: (ii) Calculate Mass of the Box
$$\text{Mass} = \text{Volume of Wood} \times \text{Density}$$ $$\text{Mass} = 4902\text{ cm}^3 \times 0.8\text{ g/cm}^3 = 3921.6\text{ g} = 3.9216\text{ kg}$$
$$\mathbf{\text{Conclusion: (i) Volume of wood } = 4,902\text{ cm}^3; \; (ii) \text{ Mass of box } = 3921.6\text{ g (or } 3.92\text{ kg).}}$$
External dimensions:
$$L = 30\text{ cm}, \quad B = 25\text{ cm}, \quad H = 20\text{ cm}$$ $$V_{\text{ext}} = L \times B \times H = 30 \times 25 \times 20 = 15,000\text{ cm}^3$$
Step 2: Calculate Internal Dimensions and Internal Volume ($V_{\text{int}}$)
Uniform thickness of wood: $t = 1.5\text{ cm}$.
Since the box is closed, thickness must be subtracted twice from each external dimension:
$$l = L - 2t = 30 - 2(1.5) = 30 - 3 = 27\text{ cm}$$ $$b = B - 2t = 25 - 2(1.5) = 25 - 3 = 22\text{ cm}$$ $$h = H - 2t = 20 - 2(1.5) = 20 - 3 = 17\text{ cm}$$
$$V_{\text{int}} = l \times b \times h = 27 \times 22 \times 17$$ $$27 \times 22 = 594$$ $$V_{\text{int}} = 594 \times 17 = 10,098\text{ cm}^3$$
Step 3: (i) Calculate Volume of the Wood
$$\text{Volume of Wood} = V_{\text{ext}} - V_{\text{int}} = 15,000\text{ cm}^3 - 10,098\text{ cm}^3 = 4,902\text{ cm}^3$$
$$\mathbf{\text{Volume of wood } = 4,902\text{ cm}^3.}$$
Step 4: (ii) Calculate Mass of the Box
$$\text{Mass} = \text{Volume of Wood} \times \text{Density}$$ $$\text{Mass} = 4902\text{ cm}^3 \times 0.8\text{ g/cm}^3 = 3921.6\text{ g} = 3.9216\text{ kg}$$
$$\mathbf{\text{Conclusion: (i) Volume of wood } = 4,902\text{ cm}^3; \; (ii) \text{ Mass of box } = 3921.6\text{ g (or } 3.92\text{ kg).}}$$
Rough Column Working:
V_ext = 30 * 25 * 20 = 15,000
Internal: 27 * 22 * 17 = 10,098
V_wood = 15,000 - 10,098 = 4,902 cm^3
Mass = 4902 * 0.8 = 3921.6 g = 3.92 kg.
V_ext = 30 * 25 * 20 = 15,000
Internal: 27 * 22 * 17 = 10,098
V_wood = 15,000 - 10,098 = 4,902 cm^3
Mass = 4902 * 0.8 = 3921.6 g = 3.92 kg.
Step Mark Allocation:
- Calculating external volume = 15,000 cu cm: 1 Marks
- Calculating internal dimensions (27 x 22 x 17 cm) and internal volume = 10,098 cu cm: 1.5 Marks
- Subtracting to find volume of wood = 4,902 cu cm: 1.5 Marks
- Calculating mass = 3921.6 g (3.92 kg) and boxed answer: 1 Marks
Examiner Scoring Key: Subtracting 1.5 cm once instead of 2 * 1.5 = 3 cm is a common error in closed box problems.
Part B - Brickwork Masonry with Mortar
10 Marks
A solid brick wall is to be constructed with dimensions $15\text{ m}$ long, $4\text{ m}$ high, and $30\text{ cm}$ thick. The wall contains two identical rectangular door openings of $2\text{ m} \times 1.25\text{ m}$ each. Mortar (cement and sand mixture) occupies $10\%$ of the net volume of the wall. Find the number of bricks required if each brick measures $20\text{ cm} \times 10\text{ cm} \times 7.5\text{ cm}$.
Step-by-Step Model Solution:
Step 1: Harmonize Units into Centimetres
$$\text{Wall Length } L = 15\text{ m} = 1500\text{ cm}$$ $$\text{Wall Height } H = 4\text{ m} = 400\text{ cm}$$ $$\text{Wall Thickness } B = 30\text{ cm}$$
$$\text{Gross Volume of Wall} = L \times H \times B = 1500 \times 400 \times 30 = 18,000,000\text{ cm}^3$$
Step 2: Calculate Volume of the Two Door Openings
Dimensions of each door opening:
$$\text{Height} = 2\text{ m} = 200\text{ cm}, \quad \text{Width} = 1.25\text{ m} = 125\text{ cm}, \quad \text{Thickness} = 30\text{ cm}$$
$$\text{Volume of 1 Door} = 200 \times 125 \times 30 = 750,000\text{ cm}^3$$ $$\text{Volume of 2 Doors} = 2 \times 750,000 = 1,500,000\text{ cm}^3$$
Step 3: Calculate Net Solid Volume of the Wall
$$\text{Solid Wall Volume} = \text{Gross Volume} - \text{Volume of 2 Doors}$$ $$\text{Solid Wall Volume} = 18,000,000\text{ cm}^3 - 1,500,000\text{ cm}^3 = 16,500,000\text{ cm}^3$$
Step 4: Deduct Mortar Volume to find Net Brick Volume
Mortar occupies $10\%$ of the solid wall volume.
Therefore, bricks occupy the remaining $90\%$ ($100\% - 10\%$):
$$\text{Net Volume Occupied by Bricks} = 90\% \text{ of } 16,500,000$$ $$\text{Net Brick Volume} = \frac{90}{100} \times 16,500,000 = 9 \times 1,650,000 = 14,850,000\text{ cm}^3$$
Step 5: Calculate Volume of One Brick and Number of Bricks
Dimensions of one brick: $20\text{ cm} \times 10\text{ cm} \times 7.5\text{ cm}$.
$$\text{Volume of 1 Brick} = 20 \times 10 \times 7.5 = 1500\text{ cm}^3$$
The total number of bricks $N$ required is:
$$N = \frac{\text{Net Volume Occupied by Bricks}}{\text{Volume of 1 Brick}}$$ $$N = \frac{14,850,000}{1500} = \frac{148,500}{15} = 9,900\text{ bricks}$$
$$\mathbf{\text{Conclusion: The number of bricks required to construct the wall is } 9,900\text{ bricks.}}$$
$$\text{Wall Length } L = 15\text{ m} = 1500\text{ cm}$$ $$\text{Wall Height } H = 4\text{ m} = 400\text{ cm}$$ $$\text{Wall Thickness } B = 30\text{ cm}$$
$$\text{Gross Volume of Wall} = L \times H \times B = 1500 \times 400 \times 30 = 18,000,000\text{ cm}^3$$
Step 2: Calculate Volume of the Two Door Openings
Dimensions of each door opening:
$$\text{Height} = 2\text{ m} = 200\text{ cm}, \quad \text{Width} = 1.25\text{ m} = 125\text{ cm}, \quad \text{Thickness} = 30\text{ cm}$$
$$\text{Volume of 1 Door} = 200 \times 125 \times 30 = 750,000\text{ cm}^3$$ $$\text{Volume of 2 Doors} = 2 \times 750,000 = 1,500,000\text{ cm}^3$$
Step 3: Calculate Net Solid Volume of the Wall
$$\text{Solid Wall Volume} = \text{Gross Volume} - \text{Volume of 2 Doors}$$ $$\text{Solid Wall Volume} = 18,000,000\text{ cm}^3 - 1,500,000\text{ cm}^3 = 16,500,000\text{ cm}^3$$
Step 4: Deduct Mortar Volume to find Net Brick Volume
Mortar occupies $10\%$ of the solid wall volume.
Therefore, bricks occupy the remaining $90\%$ ($100\% - 10\%$):
$$\text{Net Volume Occupied by Bricks} = 90\% \text{ of } 16,500,000$$ $$\text{Net Brick Volume} = \frac{90}{100} \times 16,500,000 = 9 \times 1,650,000 = 14,850,000\text{ cm}^3$$
Step 5: Calculate Volume of One Brick and Number of Bricks
Dimensions of one brick: $20\text{ cm} \times 10\text{ cm} \times 7.5\text{ cm}$.
$$\text{Volume of 1 Brick} = 20 \times 10 \times 7.5 = 1500\text{ cm}^3$$
The total number of bricks $N$ required is:
$$N = \frac{\text{Net Volume Occupied by Bricks}}{\text{Volume of 1 Brick}}$$ $$N = \frac{14,850,000}{1500} = \frac{148,500}{15} = 9,900\text{ bricks}$$
$$\mathbf{\text{Conclusion: The number of bricks required to construct the wall is } 9,900\text{ bricks.}}$$
Rough Column Working:
Wall = 1500 * 400 * 30 = 18,000,000 cm^3
Doors = 2 * (200 * 125 * 30) = 1,500,000 cm^3
Net solid = 16,500,000 cm^3
Bricks 90% = 14,850,000 cm^3
1 brick = 20 * 10 * 7.5 = 1500 cm^3
N = 14,850,000 / 1500 = 9,900 bricks.
Wall = 1500 * 400 * 30 = 18,000,000 cm^3
Doors = 2 * (200 * 125 * 30) = 1,500,000 cm^3
Net solid = 16,500,000 cm^3
Bricks 90% = 14,850,000 cm^3
1 brick = 20 * 10 * 7.5 = 1500 cm^3
N = 14,850,000 / 1500 = 9,900 bricks.
Step Mark Allocation:
- Converting all dimensions to cm and finding gross volume = 18,000,000 cu cm: 2 Marks
- Calculating volume of 2 door openings = 1,500,000 cu cm: 2 Marks
- Deducting openings to get 16,500,000 cu cm and applying 90% for mortar deduction: 2.5 Marks
- Calculating volume of 1 brick = 1500 cu cm: 1.5 Marks
- Dividing to find 9,900 bricks and boxed conclusion: 2 Marks
Examiner Scoring Key: Accounting for the thickness of the door openings (matching wall thickness 30 cm) and applying the 10% mortar deduction correctly are crucial for full marks.
RIMC Practice & Drill Exercises
Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.
Question 1
5 Marks
The volume of a cube is $729\text{ cm}^3$. Find its total surface area and the length of its space diagonal.
Model Descriptive Solution:
Let edge be $a$. Volume $V = a^3 = 729 \implies a = \sqrt[3]{729} = 9\text{ cm}$. Total surface area $\text{TSA} = 6a^2 = 6(9)^2 = 6 \times 81 = 486\text{ cm}^2$. Space diagonal $d = a\sqrt{3} = 9\sqrt{3} \approx 9 \times 1.732 = 15.588\text{ cm}$.
Question 2
5 Marks
A rectangular water tank is $8\text{ m}$ long, $6\text{ m}$ wide, and $2.5\text{ m}$ deep. How many litres of water can it hold when completely full?
Model Descriptive Solution:
Volume of tank $V = l \times b \times h = 8 \times 6 \times 2.5 = 120\text{ m}^3$. Since $1\text{ m}^3 = 1000\text{ Litres}$, capacity in litres $= 120 \times 1000 = 120,000\text{ Litres}$.
Question 3
5 Marks
The dimensions of a cuboid are in the ratio $5 : 3 : 2$ and its total surface area is $558\text{ cm}^2$. Find its volume.
Model Descriptive Solution:
Let dimensions be $5x, 3x, 2x$. $\text{TSA} = 2(lb + bh + lh) = 2(15x^2 + 6x^2 + 10x^2) = 2(31x^2) = 62x^2$. Given $\text{TSA} = 558 \implies 62x^2 = 558 \implies x^2 = 9 \implies x = 3$. Dimensions are $15\text{ cm}, 9\text{ cm}, 6\text{ cm}$. Volume $= 15 \times 9 \times 6 = 810\text{ cm}^3$.
Question 4
5 Marks
How many cubes of side $3\text{ cm}$ can be cut from a wooden block measuring $18\text{ cm} \times 12\text{ cm} \times 9\text{ cm}$?
Model Descriptive Solution:
Number of cubes $= \frac{\text{Volume of block}}{\text{Volume of 1 cube}} = \frac{18 \times 12 \times 9}{3 \times 3 \times 3} = 6 \times 4 \times 3 = 72\text{ cubes}$.
Question 5
10 Marks
An open rectangular cistern is made of iron plate $2\text{ cm}$ thick. Its external dimensions are $1.04\text{ m} \times 84\text{ cm} \times 52\text{ cm}$. Find the capacity of the cistern in litres and the weight of the empty cistern if $1\text{ cm}^3$ of iron weighs $7.8\text{ g}$.
Model Descriptive Solution:
Convert external dimensions to cm: $L = 104\text{ cm}, B = 84\text{ cm}, H = 52\text{ cm}$. Thickness $t = 2\text{ cm}$. Since the cistern is open at the top, subtract $2t$ from length and breadth, but only $t$ from height: $l = 104 - 4 = 100\text{ cm}, b = 84 - 4 = 80\text{ cm}, h = 52 - 2 = 50\text{ cm}$. Internal capacity $= 100 \times 80 \times 50 = 400,000\text{ cm}^3 = \frac{400,000}{1000} = 400\text{ Litres}$. External volume $= 104 \times 84 \times 52 = 454,272\text{ cm}^3$. Volume of iron plate $= 454,272 - 400,000 = 54,272\text{ cm}^3$. Weight $= 54,272 \times 7.8 = 423,321.6\text{ g} = 423.32\text{ kg}$.
Question 6
10 Marks
A solid cube of edge $12\text{ cm}$ is submerged into a rectangular container $24\text{ cm}$ long and $16\text{ cm}$ wide containing water. Assuming the cube is completely immersed without any overflow, find the rise in the water level of the container.
Model Descriptive Solution:
Volume of submerged cube $= a^3 = 12^3 = 1728\text{ cm}^3$. Base area of container $= 24 \times 16 = 384\text{ cm}^2$. Rise in water level $h = \frac{\text{Volume of cube}}{\text{Base area of container}} = \frac{1728}{384} = 4.5\text{ cm}$.
Question 7
5 Marks
If each edge of a cube is increased by $50\%$, find the percentage increase in its surface area and its volume.
Model Descriptive Solution:
Let original edge be $a$. Increased edge $a' = 1.5a$. Original TSA $= 6a^2$; New TSA $= 6(1.5a)^2 = 6(2.25a^2) = 13.5a^2$. Increase in TSA $= 13.5a^2 - 6a^2 = 7.5a^2$. Percentage increase in TSA $= \frac{7.5a^2}{6a^2} \times 100\% = 125\%$. Original volume $= a^3$; New volume $= (1.5a)^3 = 3.375a^3$. Increase in volume $= 2.375a^3$. Percentage increase in volume $= 237.5\%$.
Question 8
5 Marks
The lateral surface area of a cuboid is $224\text{ cm}^2$. Its length is $18\text{ cm}$ and height is $4\text{ cm}$. Find its breadth and total surface area.
Model Descriptive Solution:
$\text{LSA} = 2(l + b)h \implies 224 = 2(18 + b) \times 4 \implies 224 = 8(18 + b) \implies 18 + b = 28 \implies b = 10\text{ cm}$. Total surface area $\text{TSA} = 2(lb + bh + lh) = 2(18 \times 10 + 10 \times 4 + 18 \times 4) = 2(180 + 40 + 72) = 2(292) = 584\text{ cm}^2$.
Question 9
10 Marks
A godown measures $40\text{ m} \times 25\text{ m} \times 15\text{ m}$. Find the maximum number of wooden crates each measuring $1.5\text{ m} \times 1.25\text{ m} \times 0.5\text{ m}$ that can be stored in the godown.
Model Descriptive Solution:
Volume of godown $= 40 \times 25 \times 15 = 15,000\text{ m}^3$. Volume of 1 wooden crate $= 1.5 \times 1.25 \times 0.5 = \frac{3}{2} \times \frac{5}{4} \times \frac{1}{2} = \frac{15}{16} = 0.9375\text{ m}^3$. Maximum number of crates $N = \frac{15000}{15/16} = \frac{15000 \times 16}{15} = 1000 \times 16 = 16,000\text{ crates}$.
Question 10
10 Marks
A swimming pool is $50\text{ m}$ long and $20\text{ m}$ wide. Its depth increases uniformly from $1\text{ m}$ at the shallow end to $4\text{ m}$ at the deep end. Find the volume of water in the pool when completely filled.
Model Descriptive Solution:
The longitudinal vertical cross-section of the pool is a trapezium of parallel depths $1\text{ m}$ and $4\text{ m}$ with length $50\text{ m}$ as the perpendicular distance between them. Area of cross-section $= \frac{1}{2}(a + b)h = \frac{1}{2}(1 + 4) \times 50 = \frac{5}{2} \times 50 = 125\text{ m}^2$. The volume of the pool is the area of this cross-section multiplied by the constant width ($20\text{ m}$): $V = 125\text{ m}^2 \times 20\text{ m} = 2500\text{ m}^3$. In litres: $2500 \times 1000 = 2,500,000\text{ Litres}$.
High-Yield FAQs & Revision Notes
What is the formula for the longest rod that can fit in a room?
The longest rod fits along the 3D space diagonal: $d = \sqrt{l^2 + b^2 + h^2}$. For a cube of edge $a$, the space diagonal is $d = a\sqrt{3}$.
How many faces does an open cistern or swimming pool have?
An open cistern has $5$ faces: $1$ bottom rectangular base plus $4$ vertical walls. Its surface area is $A = lb + 2(l + b)h$.
How many litres are in 1 cubic metre and 1 cubic centimetre?
$1\text{ m}^3 = 1000\text{ Litres}$. $1\text{ Litre} = 1000\text{ cm}^3$. $1\text{ cm}^3 = 1\text{ mL}$.
Why do we subtract thickness twice for closed boxes but once for open boxes?
A closed box has both top and bottom lids, so thickness $t$ is subtracted at both ends ($H - 2t$). An open box has no top lid, so thickness is only present at the bottom, meaning internal height is $H - t$.
How does scaling the edge of a cube affect its surface area and volume?
If the edge is multiplied by $k$, the total surface area increases by $k^2$ times, and the volume increases by $k^3$ times.
What is Lateral Surface Area (LSA) of a cuboid?
Lateral Surface Area is the area of the four vertical walls of the cuboid, excluding the top ceiling and the bottom floor: $\text{LSA} = 2(l + b)h$.
How to calculate the rise in water level when a solid is dropped into a tank?
The volume of liquid displaced equals the volume of the submerged solid. Therefore, $\text{Rise in level } (\Delta h) = \frac{\text{Volume of solid}}{\text{Base area of tank}}$.
What is the difference between a cuboid and a cube?
A cuboid has three generally unequal dimensions ($l \ne b \ne h$), with rectangular faces. A cube is a regular cuboid where length, breadth, and height are all equal ($l = b = h = a$), with square faces.
How do you calculate the number of bricks required for a wall?
Number of bricks $= \frac{\text{Net volume of wall (after deducting doors, windows, and mortar)}}{\text{Volume of one single brick}}$.
Can Volume and Total Surface Area of a cube have the same numerical value?
Yes! When edge $a = 6\text{ units}$, $V = 6^3 = 216\text{ cubic units}$ and $\text{TSA} = 6(6^2) = 216\text{ square units}$. Their numerical values match, though their dimensional units differ.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 18, 47 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| Checkpoint 1 | Always state given dimensions with units and draw a neat 3D isometric sketch. |
| Checkpoint 2 | Space diagonal formulas: Cuboid d = sqrt(l^2 + b^2 + h^2); Cube d = a * sqrt(3). |
| Checkpoint 3 | Open box / cistern has 5 faces: Area = lb + 2(l + b)h. |
| Checkpoint 4 | Capacity conversions: 1 cu m = 1000 Litres; 1 Litre = 1000 cu cm. |
| Checkpoint 5 | For closed boxes, internal dims = L - 2t; for open boxes, internal height = H - t. |
| Checkpoint 6 | Masonry walls: Net brick volume = Gross volume - Openings - Mortar volume. |