Question 1

A solid right circular cylinder of radius 7 cm and height 21 cm is melted and recast into small identical bullets. Each bullet consists of a right circular cylinder surmounted by a hemisphere on one base. The total height of each bullet is 3.5 cm an…

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Introduction / Context: This question combines volume of a cylinder and hemisphere and the idea of melting and recasting a solid. The original solid is a large cylinder, and it is converted into smaller composite solids shaped like bullets. Each bullet consists of a cylindrical part plus a hemispherical cap. To find how many bullets can be formed, we compare the total volume of the original cylinder with the volume of one bullet and then take the integer number of complete bullets. Given Data / Assumptions: Original solid is a right circular cylinder with radius R = 7 cm and height H = 21 cm. Each bullet has radius r = 2.1 cm at the base. Each bullet has total height 3.5 cm and consists of a cylinder plus a hemisphere. The hemispherical part has radius 2.1 cm. Height of cylindrical part of each bullet = 3.5 cm minus 2.1 cm = 1.4 cm. Take pi = 22 / 7 for calculations. Concept / Approach: First, we find the volume of the large cylinder using V = pi * R^2 * H. Next, we compute the volume of one bullet as the sum of the volume of its cylindrical part and the volume of its hemispherical part. That means V_bullet = pi * r^2 * h_cyl + (2 / 3) * pi * r^3. Dividing the total original volume by the volume of one bullet gives the number of bullets. Because only complete bullets are counted, we take the integer part of this ratio. Step-by-Step Solution: Volume of original cylinder: V_big = pi * R^2 * H = (22 / 7) * 7^2 * 21. Compute V_big: (22 / 7) * 49 * 21 = (22 / 7) * 1029 = 22 * 147 = 3234 cubic centimetres. For each bullet, height of cylindrical part h_cyl = 3.5 cm - 2.1 cm = 1.4 cm. Volume of cylindrical part: V_cyl = pi * r^2 * h_cyl = (22 / 7) * 2.1^2 * 1.4. Volume of hemispherical part: V_hemi = (2 / 3) * pi * r^3 = (2 / 3) * (22 / 7) * 2.1^3. Performing the calculation gives V_bullet = V_cyl + V_hemi = approximately 38.808 cubic centimetres. Number of bullets = V_big / V_bullet = 3234 / 38.808 which is about 83.33. Therefore, the number of complete bullets that can be formed is 83. Verification / Alternative check: We can quickly check by multiplying 83 by the bullet volume: 83 * 38.808 is slightly less than 3234, so the material is sufficient for 83 complete bullets. If we tried 84 bullets, the required volume would exceed the original 3234 cubic centimetres, so 84 is not possible. This confirms that 83 is the maximum number of complete bullets that can be formed. Why Other Options Are Wrong: 89 and 92 would require more volume than available in the original cylinder and so are impossible. 74 and 79 are smaller than the correct integer count and would leave some usable volume unused, which contradicts the idea of forming as many complete bullets as possible. Common Pitfalls: A common mistake is to ignore the hemispherical part and treat the bullet as a simple cylinder, which underestimates its volume and leads to a larger incorrect count. Another error is to take the total height of the bullet as the height of the cylindrical part and forget to subtract the radius reserved for the hemisphere. Finally, rounding pi or intermediate values too early can slightly distort the final answer, so it is better to keep enough precision until the end and then choose the closest valid integer. Final Answer: Approximately 83 complete bullets can be made from the original cylinder.

Question 2

A solid cuboid of dimensions 50 cm × 40 cm × 30 cm is cut into 8 identical smaller cuboids by making 3 cuts, each cut parallel to one of its faces. What is the total surface area, in square centimetres, of all the 8 smaller cuboids together?

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Introduction / Context: This problem tests understanding of how cutting a solid affects total surface area. A large cuboid is divided into smaller identical cuboids by planes parallel to its faces. Although the total volume remains the same, new internal faces become external surfaces of the smaller pieces, which changes the sum of their surface areas. Such questions are common in aptitude exams because they involve both geometry and spatial reasoning. Given Data / Assumptions: Original cuboid dimensions: 50 cm, 40 cm and 30 cm. The cuboid is cut into 8 identical smaller cuboids. Three cuts are made, each parallel to a face, so that the three dimensions are each divided by 2. No material is lost in cutting. We must find the total surface area of all 8 smaller cuboids combined. Concept / Approach: To obtain 8 equal parts from a cuboid, one cut is made along each dimension, halving length, breadth and height. This yields smaller cuboids with dimensions equal to half of each original dimension. The surface area of a cuboid with dimensions l, b and h is given by 2 * (l * b + b * h + h * l). First, determine the dimensions of each smaller cuboid and compute its surface area. Then multiply by 8 for the total surface area of all parts. Step-by-Step Solution: Original dimensions: length L = 50 cm, breadth B = 40 cm, height H = 30 cm. To get 8 parts, each dimension is halved: L is cut into two segments of 25 cm, B into two segments of 20 cm, and H into two segments of 15 cm. So each small cuboid has dimensions: l = 25 cm, b = 20 cm, h = 15 cm. Surface area of one small cuboid: S_one = 2 * (l * b + b * h + h * l). Compute products: l * b = 25 * 20 = 500; b * h = 20 * 15 = 300; h * l = 15 * 25 = 375. Sum inside brackets: 500 + 300 + 375 = 1175. S_one = 2 * 1175 = 2350 square centimetres. There are 8 identical cuboids, so total surface area S_total = 8 * 2350 = 18800 square centimetres. Verification / Alternative check: We can compare with the surface area of the original cuboid. Original surface area S_original = 2 * (50 * 40 + 40 * 30 + 30 * 50) = 2 * (2000 + 1200 + 1500) = 2 * 4700 = 9400 square centimetres. After cutting, the total surface area 18800 is exactly double. This makes sense because new faces appear inside the original cuboid that become external surfaces of the smaller pieces, increasing the overall surface area. Why Other Options Are Wrong: 11750 and 14100 correspond to miscalculations where one or two face products are omitted or not doubled correctly. 23500 would arise if you mistakenly took the surface area of one small cuboid as the total or miscounted the number of parts. 16400 is another incorrect intermediate value that does not match any consistent application of the surface area formula and the number of pieces. Common Pitfalls: Learners sometimes assume the total surface area remains unchanged after cutting, which is true for volume but not for surface area. Another mistake is to divide the original surface area by 8 instead of recalculating for each smaller cuboid. Misapplying the formula for surface area and confusing l, b and h values also leads to errors. It is important to recompute based on the new dimensions and then multiply by the correct number of identical parts. Final Answer: The total surface area of all the 8 smaller cuboids is 18800 square centimetres.

Question 3

Ratio of the height of the cone and cylinder is $3 : 2$. Ratio of the radius of the cone and cylinder is $3 : 4$. Ratio of the volume of the cone and cylinder is $m : n$. The length of the cuboid is $(m + 11)$ and the breadth of the cuboid is $(n - …

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Concept & Volume Proportions: The key is establishing the relationship between the volumes of a cone and a cylinder using their given dimension ratios. The volume of a cone is V_{cone} = (1/3)\pi r^2 h and the volume of a cylinder is V_{cylinder} = \pi r^2 h. The ratio of their volumes is: \frac{V_{cone}}{V_{cylinder}} = (\frac{1/3)\pi r_1^2 h_1}{\pi r_2^2 h_2} Step-by-Step Solution: • Establish Ratios: Let the radii of the cone and cylinder be 3x and 4x respectively. Let the heights of the cone and cylinder be 3y and 2y respectively. • Calculate Volume Ratio (m : n): (m/n) = (\frac{1/3) \pi (3x)^2 (3y)}{\pi (4x)^2 (2y)} (m/n) = (1/3) × (9x^2/16x^2) × (3y/2y) (m/n) = (1/3) × (9/16) × (3/2) = (9/32) Assuming the simplest integer values for the ratio, m = 9 and n = 32. • Calculate Cuboid Dimensions: Length of cuboid L = m + 11 = 9 + 11 = 20. Breadth of cuboid B = n - 11 = 32 - 11 = 21. • Find Height of Cuboid (H): Total volume of cuboid = L × B × H = 11340. 20 × 21 × H = 11340 420 × H = 11340 \implies H = 27 • Evaluate Statements: I. 3m = 3(9) = 27 (Matches) II. (n/4) = (32/4) = 8 (Does not match) III. n - m - 15 = 32 - 9 - 15 = 8 (Does not match) Only statement I yields the correct height. Exam Strategy & Shortcut: When variables expressed directly from ratios (like m and n forming m+11) are used as absolute lengths, it strongly implies that the base simplified ratio values (9 and 32) are the actual intended absolute values. Plug k=1 in immediately to save time, and quickly verify if the resulting numbers divide cleanly (like 11340 \div 420 = 27). Common Pitfall: Forgetting the (1/3) factor in the cone's volume formula, which would drastically alter the m:n ratio and lead to incorrect cuboid dimensions. Final Answer: Therefore, the correct answer is Only I.

Question 4

Directions: Read the information carefully and answer the following questions.  A cube of side $C$ cm is placed inside the sphere of radius $R$ cm in such a way that sphere touches all the vertex of the cube. A cone of radius $\sqrt{3}R$ cm and heig…

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Concept & Mensuration of 3D Figures: When a cube is inscribed inside a sphere (touching all vertices), the body diagonal of the cube is exactly equal to the diameter of the sphere. Body Diagonal = \sqrt{3} × C Diameter = 2 × R Step-by-Step Solution: 1. Establish the relation between C and R: - Since the sphere touches all vertices of the cube, the longest distance inside the cube (body diagonal) spans across the sphere's center. - \sqrt{3}C = 2R - Squaring both sides: 3C^2 = 4R^2 \Rightarrow C^2 = (4/3)R^2 2. Calculate Total Surface Area (TSA) of the sphere: - TSA_{sphere} = 4\pi R^2 = 4 × (22/7) × R^2 = (88/7)R^2 3. Calculate Lateral Surface Area (LSA) of the cube: - The lateral surface area comprises 4 vertical faces. - LSA_{cube} = 4C^2 - Substituting C^2 = (4/3)R^2: LSA_{cube} = 4 × ((4/3)R^2) = (16/3)R^2 4. Find the Ratio: - Ratio = \frac{TSA_{sphere}}{LSA_{cube}} = (\frac{88/7)R^2}{(16/3)R^2} - Ratio = (88/7) × (3/16) = (11/7) × (3/2) = (33/14) Exam Strategy & Shortcut: Always leave \pi as 22/7 when ratios are involved because standard options will often perfectly cancel out common factors like 11 or 4. Also, write out formulas symbolically before plugging any values in to avoid redundant or nested calculations. Common Pitfall: A very common mistake is confusing Total Surface Area (6 faces) with Lateral Surface Area (4 faces) of the cube. The question explicitly asks for the lateral surface area (4C^2). Final Answer: Therefore, the correct answer is 33 : 14.

Question 5

Directions: Read the information carefully and answer the following questions.  A cube of side $C$ cm is placed inside the sphere of radius $R$ cm in such a way that sphere touches all the vertex of the cube. A cone of radius $\sqrt{3}R$ cm and heig…

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Concept & Algebraic Substitution in Geometry: By finding a common variable (in this case, the sphere's radius R) between the cube and the cone, we can link the given volume of the cone directly to the dimensional side of the cube. Volume of Cone = (1/3)\pi r^2 h Step-by-Step Solution: 1. Analyze the Cone's Volume: - Radius of cone = \sqrt{3}R - Height of cone = H - Volume = 3168 cm^3 - (1/3) × \pi × (\sqrt{3}R)^2 × H = 3168 2. Simplify the Cone Equation: - (1/3) × (22/7) × 3R^2 × H = 3168 - The 3s cancel out: (22/7) × R^2 × H = 3168 - R^2H = 3168 × (7/22) - 3168 / 22 = 144. So, R^2H = 144 × 7 = 1008. 3. Relate R back to C: - From the initial conditions (cube inscribed in sphere), the body diagonal equals the diameter: \sqrt{3}C = 2R. - Squaring both sides: 3C^2 = 4R^2 \Rightarrow R^2 = (3C^2/4). 4. Substitute and Solve for Relation: - Substitute R^2 into the simplified volume equation: ((3C^2/4)) × H = 1008 - 3C^2H = 1008 × 4 - 3C^2H = 4032 - C^2H = (4032/3) = 1344 - C^2 = (1344/H) Exam Strategy & Shortcut: Rather than calculating 144 × 7 immediately, keep the terms in factored form as 144 × 7 × (4/3) when bridging the two formulas. 144 / 3 = 48, and 48 × 28 = 1344. Delayed multiplication reduces arithmetic errors and speeds up the simplification process. Common Pitfall: Forgetting to properly square the \sqrt{3} coefficient when substituting the radius of the cone into the volume formula, which fundamentally alters the entire factor chain. Final Answer: Therefore, the correct answer is C^2 = (1344/H).

Question 6

The area of the four walls of a room is $120 	ext{ m}^2$ and the length is twice the breadth. If the height of the room is $4	ext{ m}$, then the area of the floor is

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Concept & Surface Area of Walls: The area of four walls of a room is given by the lateral surface area formula. Area of 4 walls = 2(l + b)h Step-by-Step Solution: • Given height h = 4 m and the area of the four walls = 120 m^2. • Let the breadth of the room be b. According to the problem, the length is l = 2b. • Substitute the known values into the area formula: 2(2b + b) × 4 = 120. • Simplify the equation: 2(3b) × 4 = 120 \Rightarrow 24b = 120. • Solving for breadth gives b = 5 m. • Since length is twice the breadth, l = 2 × 5 = 10 m. • The area of the floor is given by l × b. Thus, Floor Area = 10 × 5 = 50 m^2. Exam Strategy & Shortcut: Notice that the floor area is l × b = (2b) × b = 2b^2. By simplifying the wall area formula to 24b = 120, we immediately find b=5. We just need 2(5^2) = 50, avoiding the need to write down every individual dimension. Common Pitfall: A frequent error is confusing the total surface area of a cuboid with the area of its four walls. Remember that the four walls formula explicitly excludes the floor and ceiling. Final Answer: Therefore, the correct answer is 50 m^2.

Question 7

The dimensions of a room are $12.5	ext{ metres}$ by $9	ext{ metres}$ by $7	ext{ metres}$. There are $2$ doors and $4$ windows in the room; each door measures $2.5	ext{ metres}$ by $1.2	ext{ metres}$ and each window $1.5	ext{ metres}$ by $1	ex…

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Concept & Effective Paintable Area: When painting a room, doors and windows are typically excluded. To find the effective paintable area, subtract their total area from the total area of the four walls. Paintable Area = 2(l + b)h - (Area of Doors + Area of Windows) Step-by-Step Solution: • Identify the dimensions of the room: l = 12.5 m, b = 9 m, h = 7 m. • Calculate the total area of the four walls: 2(12.5 + 9) × 7 = 2(21.5) × 7 = 43 × 7 = 301 m^2. • Calculate the area of the 2 doors: 2 × (2.5 × 1.2) = 2 × 3 = 6 m^2. • Calculate the area of the 4 windows: 4 × (1.5 × 1) = 4 × 1.5 = 6 m^2. • Find the total unpaintable area: 6 + 6 = 12 m^2. • Determine the effective paintable area: 301 - 12 = 289 m^2. • Multiply by the rate to find total cost: 289 × 3.50. • Cost = 289 × (7/2) = 2023 / 2 = ₹ 1011.50. Exam Strategy & Shortcut: Using fractions for monetary rates can save time. Converting ₹ 3.50 to 7/2 makes multiplying with 289 cleaner, avoiding decimal multiplication errors when solving manually under time pressure. Common Pitfall: A common error is skipping the multiplication by the number of doors (2) or windows (4) and simply subtracting the area of a single door and window from the total wall area. Final Answer: Therefore, the correct answer is ₹ 1011.50.

Question 8

A hall, whose length is $16	ext{ m}$ and the breadth is twice its height, takes $168	ext{ m}$ of paper with $2	ext{ m}$ as its width to cover its four walls. The area of the floor is

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Concept & Surface Area Mapping: The area of the paper required to cover the walls is exactly equal to the surface area of the four walls of the hall. Area of Paper = Length of Paper × Width of Paper = 2(l + b)h Step-by-Step Solution: • Calculate the total area of the wallpaper: 168 m × 2 m = 336 m^2. • Let the height of the hall be h. The problem states breadth b = 2h. The length l = 16 m. • Set the area of the four walls equal to the paper area: 2(16 + 2h)h = 336. • Divide the equation by 2: (16 + 2h)h = 168. • Expand the terms: 16h + 2h^2 = 168. • Form a quadratic equation by dividing by 2 and moving all terms to one side: h^2 + 8h - 84 = 0. • Factor the quadratic equation: (h + 14)(h - 6) = 0. • Since height cannot be negative, h = 6 m. • Substitute h back to find breadth: b = 2(6) = 12 m. • Calculate the floor area: l × b = 16 × 12 = 192 m^2. Exam Strategy & Shortcut: Instead of factoring the quadratic conventionally, you can test values directly from (8 + h)h = 84. Since 8 × 10 is 80, h must be slightly less than 10. Testing h=6 immediately yields (8 + 6) × 6 = 14 × 6 = 84. Common Pitfall: A standard trap is solving for h (6 m) or b (12 m) and mistakenly selecting it as the final answer instead of completing the last step to calculate the area of the floor. Final Answer: Therefore, the correct answer is 192 m^2.

Question 9

The ratio of the height of a room to its semi-perimeter is $2 : 5$. It costs ₹ $260$ to paper the walls of the room with paper $50	ext{ cm}$ wide at ₹ $2$ per metre allowing an area of $15	ext{ sq. m}$ for doors and windows. The height of the room…

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Concept & Abstract Dimensions: The area covered by the wallpaper can be reverse-calculated from its total cost and cost per unit length. The semi-perimeter is (l + b). Total Wall Area = Area of Paper + Area of Doors/Windows Step-by-Step Solution: • Given the ratio of height to semi-perimeter (l+b) is 2:5. Let height h = 2x and semi-perimeter (l+b) = 5x. • The width of the paper is 50 cm = 0.5 m. The cost is ₹ 2 per running metre. • The area of a 1-metre length of paper is 1 m × 0.5 m = 0.5 m^2. Thus, the rate is ₹ 2 for every 0.5 m^2, which translates to a rate of ₹ 4 per 1 m^2. • Total cost for the paper is ₹ 260. The total area of paper used = 260 / 4 = 65 m^2. • The total area of the four walls includes the papered area plus the doors and windows. Total Wall Area = 65 + 15 = 80 m^2. • We know the formula for wall area is 2(l+b)h. Substitute our x variables: 2(5x)(2x) = 80. • Simplify the equation: 20x^2 = 80 \Rightarrow x^2 = 4 \Rightarrow x = 2. • Calculate the height: h = 2x = 2(2) = 4 m. Exam Strategy & Shortcut: Calculate the effective rate per square meter immediately. ₹2 for 0.5 m width means ₹4 per sq.m. The 260 cost gives 65 sq.m, plus 15 equals 80 sq.m. Knowing area is 2(5x)(2x) = 20x^2, 20x^2=80 provides x=2 in seconds. Common Pitfall: A tricky element is confusing the "₹2 per metre" rate as a rate per square meter. Because the paper is only 0.5 m wide, 1 running meter represents only 0.5 square meters of material. Final Answer: Therefore, the correct answer is 4 m.

Question 10

A closed aquarium of dimensions 30 cm × 25 cm × 20 cm is made up entirely of glass plates held together with tapes. The total length of tape required to hold the plates together (ignore the overlapping tapes) is (Hotel Management, 2009)

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Concept & Total Length of Edges: The tape is used along the joints of the glass plates. Since it's a cuboidal aquarium, the joints correspond exactly to the edges of the cuboid. Total Edge Length = 4(l + b + h) Step-by-Step Solution: • Given: Length (l) = 30 cm, Breadth (b) = 25 cm, Height (h) = 20 cm. • A cuboid has 12 edges in total: 4 lengths, 4 breadths, and 4 heights. • Step 1: Sum the dimensions Sum = l + b + h = 30 + 25 + 20 = 75 cm. • Step 2: Multiply by 4 Total Tape Length = 4 × 75 cm = 300 cm. Exam Strategy & Shortcut: Recognize that "length of tape" means "perimeter of all faces combined but counted without overlap", which is just the sum of all edges. Add the three given numbers (30+25+20 = 75) and multiply by 4. Common Pitfall: Many students calculate the surface area or volume by mistake, or they only sum l + b + h once (Option A) or multiply by 2 (Option C) instead of 4. Final Answer: Therefore, the correct answer is 300 cm.

Question 11

The dimensions of a room are 15 m, 10 m and 8 m. The volume of a bag is 2.25 m³. The maximum number of bags that can be accommodated in the room is

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Concept & Maximizing Space (Volume Division): To find the maximum number of identical items that can fit into a larger space, divide the total volume of the space by the volume of a single item. Number of items = \frac{Total Volume}{Volume of one item} Step-by-Step Solution: • Given: Dimensions of room = 15 m × 10 m × 8 m. Volume of one bag = 2.25 m^3. • Step 1: Calculate total volume of the room Volume = 15 × 10 × 8 = 1200 m^3. • Step 2: Divide by the volume of a single bag Number of bags = (1200/2.25) • Step 3: Simplify the fraction Multiply numerator and denominator by 100 to remove decimals: (120000/225) Divide by 25: (4800/9) Divide by 3: (1600/3) = 533.33... • Since we can only accommodate whole bags, we take the integer part (floor value). Maximum bags = 533. Exam Strategy & Shortcut: Instead of multiplying 15 × 10 × 8 fully, keep it as a product and write 2.25 as a fraction ((9/4)). Number of bags = (15 × 10 × 8/\frac{9){4}} = (15 × 10 × 8 × 4/9). Cancel out common factor 3: (5 × 10 × 8 × 4/3) = (1600/3) \approx 533. Common Pitfall: Rounding up to 534 is a common mistake. Even if the decimal was .9, you cannot fit a whole bag into a partial space. Always round down (take the floor) for physical packing problems. Final Answer: Therefore, the correct answer is 533.

Question 12

The length of a hall is 20 metres and the width is 16 metres. The sum of the areas of the floor and roof is equal to the sum of the areas of the four walls. Find the volume of the hall.

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Concept & Volume of a Cuboid: To find the volume of the hall, we first need to determine its height. The problem states a clear relationship between the areas of the top/bottom and the lateral surfaces (walls). Volume = Length × Breadth × Height Area\_of\_Floor\_and\_Roof = 2 × (Length × Breadth) Area\_of\_Four\_Walls = 2 × Height × (Length + Breadth) Step-by-Step Solution: 1. Given Data: - Length (L) = 20 m - Width / Breadth (B) = 16 m 2. Calculate Area of Floor and Roof: - Area = 2 × (20 × 16) = 2 × 320 = 640 m² 3. Set Up the Equation: - The area of the four walls is 2H(L + B) = 2H(20 + 16) = 2H(36) = 72H - According to the condition: 72H = 640 4. Find the Height (H): - H = 640 / 72 = 80 / 9 m 5. Calculate Volume: - Volume = L × B × H = 20 × 16 × (80 / 9) - Volume = 320 × (80 / 9) = 25600 / 9 - Volume ≈ 2844.44 m³ Exam Strategy & Shortcut: When you reach the fraction 25600 / 9, realize that 25 + 6 = 11, which isn't divisible by 9, meaning it will be a recurring decimal. 25600 / 9 = 2844.44... Knowing the fraction breakdown instantly guides you to the correct choice without exhaustive division. Common Pitfall: A frequent error is forgetting to multiply the floor area by 2 when equating it to the walls. "Floor and roof" implies two identical rectangular areas. Final Answer: Therefore, the correct answer is 2844.4 m³.

Question 13

It is required to construct a big rectangular hall to accommodate 500 persons, allowing 22.5 m³ space per person. The height of the hall is to be kept at 7.5 m, while the total inner surface area of the walls must be 1200 sq. m. Then the length and …

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Concept & System of Equations in Geometry: This problem creates a system of two equations based on volume capacity and lateral surface area. Total\_Volume = Number\_of\_Persons × Volume\_per\_Person Area\_of\_Walls = 2H(L + B) Step-by-Step Solution: 1. Find Total Required Volume: - Volume = 500 persons × 22.5 m³/person = 11250 m³ 2. Form Equation 1 (Volume): - Volume = L × B × H - 11250 = L × B × 7.5 - L × B = 11250 / 7.5 = 1500 3. Form Equation 2 (Area of Walls): - Area = 2 × 7.5 × (L + B) = 15(L + B) - We are given Area = 1200 sq. m. - 15(L + B) = 1200 - L + B = 80 4. Solve for L and B: - We need two numbers that add up to 80 and multiply to 1500. - The numbers are 50 and 30. Exam Strategy & Shortcut: Once you deduce that L + B = 80 and L × B = 1500, immediately scan the options. (a) 40+30=70 (Reject) (b) 45+35=80, but 45×35 = 1575 (Reject) (c) 50+30=80, and 50×30 = 1500 (Accept) (d) 60+20=80, but 60×20 = 1200 (Reject) Common Pitfall: Using the total surface area formula 2(LB + BH + HL) instead of the area of the four walls 2H(L + B). "Total inner surface area of the walls" strictly excludes the floor and ceiling. Final Answer: Therefore, the correct answer is 50 m and 30 m.

Question 14

The length of the longest rod that can be placed in a room of dimensions 10 m × 10 m × 5 m is

Accepted Answer

Concept & Diagonal of a Cuboid: The longest rod that can fit inside a rectangular room must span from one bottom corner to the opposite top corner. This is exactly the main diagonal of the cuboid. Diagonal = \sqrt{L^2 + B^2 + H^2} Step-by-Step Solution: 1. Identify the Dimensions: - Length (L) = 10 m - Breadth (B) = 10 m - Height (H) = 5 m 2. Apply the Diagonal Formula: - Diagonal = \sqrt{10^2 + 10^2 + 5^2} - Diagonal = \sqrt{100 + 100 + 25} - Diagonal = \sqrt{225} 3. Calculate Final Value: - \sqrt{225} = 15 m Exam Strategy & Shortcut: This is a standard Pythagorean quadruplet conceptually (10, 10, 5 \rightarrow 15). Notice that all terms share a common factor of 5. You can pull out the 5 to simplify mental math: 5 × \sqrt{2^2 + 2^2 + 1^2} = 5 × \sqrt{4 + 4 + 1} = 5 × \sqrt{9} = 5 × 3 = 15. Common Pitfall: Mistaking the "longest rod" for the diagonal of the floor (10\sqrt{2}), ignoring the height dimension completely. Final Answer: Therefore, the correct answer is 15.

Question 15

How many bricks, each measuring 25 cm × 11.25 cm × 6 cm, will be needed to build a wall 8 m × 6 m × 22.5 cm?

Accepted Answer

Concept & Volume Division for Object Counts: To find the number of standard units (bricks) needed to form a larger volume (wall), simply divide the total volume by the unit volume, provided all dimensions are in the same measurement unit. Number = (Total Volume/Volume of One Unit) Step-by-Step Solution: 1. Standardize the Units: - Convert all wall dimensions from meters to centimeters to match the brick dimensions. - Wall Length = 8 m = 800 cm - Wall Height = 6 m = 600 cm - Wall Thickness = 22.5 cm 2. Set Up the Ratio: - Number of bricks = (800 × 600 × 22.5/25 × 11.25 × 6) 3. Simplify and Cancel: - Notice that 22.5 is exactly double 11.25. So, (22.5/11.25) = 2. - The expression becomes: (800 × 600 × 2/25 × 6) - Cancel the 6 from 600: (800 × 100 × 2/25) - Cancel 25 from 100: 800 × 4 × 2 4. Final Calculation: - Number of bricks = 800 × 8 = 6400 Exam Strategy & Shortcut: Never multiply all dimensions to find total volumes before dividing. Instead, write them out as a single large fraction and cancel common factors directly. Spotting relationships like 22.5 = 2 × 11.25 instantly cuts the calculation time in half. Common Pitfall: A classic error is failing to convert the 8 m and 6 m dimensions into centimeters, leading to an answer that is orders of magnitude too small. Final Answer: Therefore, the correct answer is 6400.

Question 16

50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m³, then the rise in the water level in the tank will be :

Accepted Answer

Concept & Volume Displacement: According to Archimedes' principle (in a simplified geometric context), the total volume of water displaced directly correlates to the rise in the water level spread across the surface area of the tank. Rise in Level = (Total Volume Displaced/Surface Area of Tank Base) Step-by-Step Solution: 1. Calculate Total Volume Displaced: - Volume displaced by 1 man = 4 m³ - Volume displaced by 50 men = 50 × 4 = 200 m³ 2. Calculate Base Area of the Tank: - Area = Length × Breadth = 40 m × 20 m = 800 m² 3. Calculate the Rise in Water Level: - Rise (H) = Volume / Area - Rise = 200 / 800 = (1/4) meter = 0.25 m 4. Convert Units: - The options are in centimeters. - 0.25 m × 100 cm/m = 25 cm. Exam Strategy & Shortcut: Since total displacement is 200, and the area is 800, the fraction is obviously (1/4). Knowing instantly that (1/4) of a meter is 25 cm allows you to solve this entirely in your head within 10 seconds. Common Pitfall: Forgetting to convert the final answer from meters to centimeters. If you stop at 0.25, you won't find it in the options and might start over unnecessarily. Final Answer: Therefore, the correct answer is 25 cm.

Question 17

A swimming bath is 24 m long and 15 m broad. When a number of men dive into the bath, the height of the water rises by 1 cm. If the average amount of water displaced by one of the men be 0.1 cu. m, how many men are there in the bath?

Accepted Answer

Concept & Reverse Volume Displacement: This is the inverse of a standard displacement problem. Here, you use the rise in water level to calculate the total displaced volume, and then divide by the displacement per person to find the number of people. Total Displaced Volume = Base Area × Rise in Height Number of People = (Total Displaced Volume/Volume Displaced per Person) Step-by-Step Solution: 1. Standardize Units: - Base Length = 24 m - Base Breadth = 15 m - Rise in Height = 1 cm = 0.01 m 2. Calculate Total Displaced Volume: - Total Volume = 24 × 15 × 0.01 - Total Volume = 360 × 0.01 = 3.6 m³ 3. Find the Number of Men: - Average displacement per man = 0.1 m³ - Number of men = 3.6 / 0.1 = 36 Exam Strategy & Shortcut: Handle the multiplication quickly: 24 × 15 is just 24 × (30/2) = 12 × 30 = 360. Multiplying by 0.01 shifts the decimal to 3.6. Dividing by 0.1 shifts it back to 36. Common Pitfall: Using 1 for the height without converting cm to m. This would give a volume of 360 m³, resulting in 3600 men, which is physically impossible for that pool and missing from the options. Always check unit consistency. Final Answer: Therefore, the correct answer is 36.

Question 18

A school room is to be built to accommodate 70 children so as to allow 2.2 m² of floor and 11 m³ of space for each child. If the room be 14 metres long, what must be its breadth and height?

Accepted Answer

Concept & Intersecting Constraints: The room is governed by two constraints: floor area (which dictates breadth since length is fixed) and total volume (which dictates height once the floor area is known). Total Area = Number of Children × Area per Child Total Volume = Number of Children × Volume per Child Step-by-Step Solution: 1. Find Total Floor Area & Determine Breadth: - Total Area required = 70 × 2.2 = 154 m² - We know Area = Length × Breadth - 154 = 14 × Breadth - Breadth = (154/14) = 11 m 2. Find Total Volume & Determine Height: - Total Volume required = 70 × 11 = 770 m³ - We know Volume = Area × Height - 770 = 154 × Height - Height = (770/154) = 5 m Exam Strategy & Shortcut: There is a direct relationship between Volume, Area, and Height. Since Volume = Area × Height, the height of the room is simply the ratio of Volume per child to Area per child. Height = \frac{11 m^3}{2.2 m^2} = 5 m. Looking at the options, only (b) has a height of 5 m, so you can choose it instantly without even calculating the breadth! Common Pitfall: Unnecessarily calculating the total volume (770) and total area (154) just to find height, rather than realizing height applies uniformly per child. Final Answer: Therefore, the correct answer is 11 m, 5 m.

Question 19

A rectangular tank measuring $5 	ext{ m} 	imes 4.5 	ext{ m} 	imes 2.1 	ext{ m}$ is dug in the centre of the field measuring $13.5 	ext{ m}$ by $2.5 	ext{ m}$. The earth dug out is evenly spread over the remaining portion of the field. How muc…

Accepted Answer

Concept & Formula: To find the rise in the level of the field, we need to divide the volume of the earth dug out by the area of the remaining field over which it is spread. Rise in level = \frac{Volume of earth dug out}{Total area of field - Area of tank base} Step-by-Step Solution: 1. Find the volume of earth dug out: Volume = 5 m × 4.5 m × 2.1 m = 47.25 m^3 2. Find the total area of the field: Area = 13.5 m × 2.5 m = 33.75 m^2 3. Find the base area of the dug tank: Tank Area = 5 m × 4.5 m = 22.5 m^2 4. Calculate the remaining area where earth is spread: Remaining Area = 33.75 - 22.5 = 11.25 m^2 5. Calculate the rise in level: Rise = (47.25/11.25) = 4.2 m Exam Strategy & Shortcut: Work with fractions to simplify calculations: 47.25 / 11.25 = (189/4/45/4) = (189/45) = (21/5) = 4.2. Common Pitfall: Spreading the earth over the *entire* field area instead of subtracting the area of the dug tank first. Final Answer: Therefore, the correct answer is 4.2 m.

Question 20

A plot of land in the form of a rectangle has dimensions $240 	ext{ m} 	imes 180 	ext{ m}$. A drainlet $10 	ext{ m}$ wide is dug all around it (outside) and the earth dug out is evenly spread over the plot, increasing its surface level by $25 	…

Accepted Answer

Concept & Formula: The volume of the earth dug from the drainlet equals the volume of the earth spread over the plot. Volume of earth = Area of plot × Increase in height Volume of drainlet = Area of drainlet × Depth Step-by-Step Solution: 1. Find the area of the plot: Plot Area = 240 × 180 = 43200 m^2 2. Find the volume of earth spread: Rise = 25 cm = 0.25 m Volume spread = 43200 × 0.25 = 10800 m^3 3. Find the area of the outer drainlet: Outer dimensions = (240 + 10 + 10) × (180 + 10 + 10) = 260 m × 200 m Outer Area = 52000 m^2 Drainlet Area = Outer Area - Plot Area = 52000 - 43200 = 8800 m^2 4. Equate volumes to find the depth (d): 8800 × d = 10800 d = (10800/8800) = (108/88) = (27/22) \approx 1.2272... m Exam Strategy & Shortcut: Instead of calculating huge areas, calculate the drainlet area using the formula: 2w(L+B+2w), where w is width. 2(10)(240+180+20) = 20(440) = 8800. Equate 8800 × d = 43200 × 0.25 directly. Common Pitfall: Forgetting to add twice the width of the path (10+10=20) to both the length and the breadth when calculating the outer area. Final Answer: Therefore, the correct answer is 1.227 m.

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