Question 1

The curved surface area of a cylindrical pillar is 264 square metres and its volume is 924 cubic metres. What is the ratio of the diameter of the base to the height of the pillar?

Accepted Answer

Introduction / Context: Here we have a cylindrical pillar for which both the curved surface area and the volume are given. The task is to find the ratio of the diameter of the base to the height. This problem tests the ability to use two standard formulas of a cylinder together and eliminate variables to find a ratio, rather than any single value. It is a good example of using algebra with geometric formulas. Given Data / Assumptions: Curved surface area (CSA) = 264 m^2. Volume (V) = 924 m^3. Radius of the base = r metres. Height of the cylinder = h metres. We must find the ratio (diameter : height) = (2r : h). Concept / Approach: For a cylinder, we use: CSA = 2 * pi * r * h Volume V = pi * r^2 * h We are given both CSA and V, so we can form a ratio CSA / V to eliminate pi and h. This will give a relationship involving only r. From that, we solve for r and then find h from either formula. Finally, we compute the ratio 2r : h and simplify it to the smallest whole-number ratio. Step-by-Step Solution: CSA = 2 * pi * r * h = 264. V = pi * r^2 * h = 924. Compute CSA / V: (2 * pi * r * h) / (pi * r^2 * h) = 264 / 924. Left side simplifies to 2 / r. So 2 / r = 264 / 924. Simplify 264 / 924: divide both by 12 gives 22 / 77 = 2 / 7. Thus 2 / r = 2 / 7, so r = 7 m. Now use volume: 924 = pi * 7^2 * h = pi * 49 * h. So h = 924 / (49 * pi). Using CSA instead is easier: 264 = 2 * pi * 7 * h = 14 * pi * h. Hence h = 264 / (14 * pi). With pi ≈ 22 / 7, h ≈ 6 m. Diameter = 2r = 14 m, height ≈ 6 m, so ratio 14 : 6 = 7 : 3. Verification / Alternative check: Take r = 7 and h = 6. Then CSA = 2 * pi * 7 * 6 = 84pi, which with pi ≈ 22 / 7 gives 84 * 22 / 7 = 264 m^2. Volume = pi * 49 * 6 = 294pi, which is 294 * 22 / 7 = 924 m^3. Both values match the given data, confirming r = 7 and h = 6 are correct and, therefore, the ratio is valid. Why Other Options Are Wrong: Ratios like 3 : 7 or 6 : 7 either invert or misrepresent the relationship between diameter and height. For instance, 3 : 7 would imply a very slender cylinder, which does not satisfy the given CSA and volume simultaneously. Ratios 7 : 6 or 5 : 3 correspond to other inconsistent combinations of r and h that fail to satisfy both formulas at once. Common Pitfalls: Students sometimes forget that the question asks for diameter, not radius, and incorrectly give r : h instead of 2r : h. Another typical error is not simplifying the fraction 264 / 924 correctly, which leads to an incorrect r. It is also easy to make algebraic mistakes when forming the ratio CSA / V, so careful simplification is important. Final Answer: The required ratio of diameter to height of the cylindrical pillar is 7 : 3.

Question 2

How many bricks, each measuring 25 cm × 11.25 cm × 6 cm, are required to build a solid wall of dimensions 8 m × 6 m × 22.5 cm, assuming the wall is completely solid without gaps?

Accepted Answer

Introduction / Context: This question is a classic volume-and-counting problem related to construction. We are given the dimensions of a wall and an individual brick, and we must determine how many such bricks are required to build the wall if it is completely solid. The problem checks understanding of unit conversion between metres and centimetres, volume calculation for cuboids, and the idea that the number of bricks equals total wall volume divided by volume of one brick. Given Data / Assumptions: Dimensions of the wall: 8 m by 6 m by 22.5 cm. Dimensions of each brick: 25 cm by 11.25 cm by 6 cm. The wall is solid, with no gaps or mortar considered. Bricks are placed without cutting and fill the wall volume exactly. Concept / Approach: Both the wall and each brick are rectangular solids (cuboids). The number of bricks needed is: Number of bricks = Volume of wall / Volume of one brick We must ensure all dimensions are in the same unit, preferably centimetres. So we convert the metre dimensions of the wall to centimetres before calculating volumes. Step-by-Step Solution: Convert wall dimensions: 8 m = 800 cm, 6 m = 600 cm, thickness = 22.5 cm. Volume of wall = 800 cm * 600 cm * 22.5 cm. Volume of wall = 800 * 600 * 22.5 = 10,800,000 cubic centimetres. Volume of one brick = 25 cm * 11.25 cm * 6 cm. Volume of one brick = 25 * 11.25 * 6 = 1687.5 cubic centimetres. Number of bricks = 10,800,000 / 1687.5. Compute: 10,800,000 / 1687.5 = 6400. Verification / Alternative check: We can simplify the fraction by factoring. Note that 1687.5 * 100 = 168750, and 10,800,000 * 100 = 1,080,000,000, so the ratio remains the same. Performing long division or using proportional reasoning confirms 6400 exactly. This matches one of the given options, which supports our calculation. Why Other Options Are Wrong: 5600, 6000, 6800 and 7200 would imply either underestimating or overestimating the wall volume or miscalculating the brick volume. They usually arise from mistakes such as using 22 cm instead of 22.5 cm, or rounding 11.25 cm incorrectly. Only 6400 gives an exact integer count consistent with the precise geometric volumes. Common Pitfalls: A frequent error is mixing metres and centimetres in a single calculation, which gives incorrect volumes. Some students also round off 11.25 cm to 11 or 12, leading to wrong brick volumes. Others mis-handle the decimal thickness 22.5 cm or treat it as 22 cm. Always convert units carefully and handle decimals correctly before dividing to find the number of bricks. Final Answer: The number of bricks required to build the wall is 6400.

Question 3

A rectangular brick has dimensions 10 cm × 4 cm × 3 cm. What is the total surface area of the brick in square centimetres?

Accepted Answer

Introduction / Context: This straightforward geometry question asks for the total surface area of a rectangular brick. A brick is essentially a cuboid, and its surface area can be found using the standard formula that involves the sum of products of its side lengths. The problem checks whether you remember the surface area formula for a cuboid and can apply it correctly to given dimensions in centimetres. Given Data / Assumptions: Length of the brick, l = 10 cm. Breadth (width) of the brick, b = 4 cm. Height (thickness) of the brick, h = 3 cm. The brick has six rectangular faces, as a cuboid. Concept / Approach: The total surface area (TSA) of a cuboid with dimensions l, b and h is given by: TSA = 2 * (l * b + b * h + l * h) This formula accounts for each pair of opposite faces: two faces of size l × b, two of size b × h, and two of size l × h. We substitute the given values into this formula to obtain the surface area in square centimetres. Step-by-Step Solution: Compute l * b = 10 * 4 = 40. Compute b * h = 4 * 3 = 12. Compute l * h = 10 * 3 = 30. Sum of products = 40 + 12 + 30 = 82. TSA = 2 * (l * b + b * h + l * h) = 2 * 82. TSA = 164 square centimetres. Verification / Alternative check: We can verify by explicitly accounting for each face: two faces of 40 cm^2 (total 80 cm^2), two faces of 12 cm^2 (total 24 cm^2), and two faces of 30 cm^2 (total 60 cm^2). Adding these gives 80 + 24 + 60 = 164 cm^2, which matches the formula-based calculation, confirming that the total surface area is correct. Why Other Options Are Wrong: 84 sq.cm would result from taking only one of each face instead of both. 124 sq.cm and 180 sq.cm correspond to incorrect sums or misapplied formulas. 200 sq.cm would require larger face products than those actually present in this brick. None of these values match both the formula and the detailed face-by-face calculation. Common Pitfalls: A common mistake is to forget that each face occurs twice and therefore forget to multiply by 2 at the end. Another error is to confuse volume and surface area, mistakenly multiplying all three dimensions instead of using the surface area formula. Finally, incorrect arithmetic when adding 40, 12 and 30 also leads to wrong results. Final Answer: The total surface area of the brick is 164 sq.cm.

Question 4

The circumference of the circular base of a solid cylinder is 44 cm and its height is 15 cm. What is the volume of the cylinder in cubic centimetres?

Accepted Answer

Introduction / Context: This problem asks for the volume of a cylinder when the circumference of its base and its height are given. Instead of giving the radius directly, the question provides the circumference, so we must first relate circumference to radius and then use the volume formula. This tests understanding of multiple circle formulas and how to combine them correctly. Given Data / Assumptions: Circumference of the base, C = 44 cm. Height of the cylinder, h = 15 cm. The base is a circle of radius r. We must find the volume in cubic centimetres. Concept / Approach: For a circle, the circumference is related to the radius by: C = 2 * pi * r For a cylinder, the volume V is given by: V = pi * r^2 * h Here, we first express r in terms of C and pi, then substitute into the volume formula. Alternatively, we can use a derived formula expressing volume directly in terms of circumference and height: V = (C^2 * h) / (4 * pi). Both approaches give the same result when done carefully. Step-by-Step Solution: Given C = 44 cm. From C = 2 * pi * r, r = C / (2 * pi) = 44 / (2 * pi) = 22 / pi. Volume V = pi * r^2 * h. Substitute r = 22 / pi and h = 15: V = pi * (22 / pi)^2 * 15. V = pi * (484 / pi^2) * 15 = (484 * 15) / pi. Compute numerator: 484 * 15 = 7260. So V = 7260 / pi cubic centimetres. Using pi ≈ 3.14159, V ≈ 7260 / 3.14159 ≈ 2311 cubic centimetres. Verification / Alternative check: Alternatively, we can approximate r using pi ≈ 22 / 7: r = 44 / (2 * 22 / 7) = 44 * 7 / 44 = 7 cm. Then V ≈ pi * 7^2 * 15 ≈ 3.14 * 49 * 15 ≈ 3.14 * 735 ≈ 2309.9 cm^3, very close to 2311 cm^3. The small difference is due to different approximations of pi, and 2311 cubic cm is a reasonable rounded value. Why Other Options Are Wrong: The other options 2759, 2247 and 2614 cubic cm do not correspond to V = 7260 / pi with any standard approximation of pi. They reflect miscalculations such as using an incorrect value of radius or applying the area formula incorrectly. 2000 cubic cm is clearly too small for a cylinder of this circumference and height. Common Pitfalls: Students may incorrectly treat the given circumference as the diameter or directly plug it in as the radius. Others forget to square r in the volume formula or mishandle the division by pi. It is also easy to make arithmetic mistakes when multiplying or dividing large numbers, so careful calculation is important. Final Answer: The volume of the cylinder is approximately 2311 cubic cm.

Question 5

A conical tent is erected by the army at a base camp. The tent has height 3 m and base diameter 10 m. If each person requires 3.92 cubic metres of air space, approximately how many persons can be accommodated inside the tent?

Accepted Answer

Introduction / Context: This question uses the volume of a right circular cone to model the air space inside a conical tent. The tent dimensions are given, along with the air space required per person. By finding the total internal volume and dividing by the volume needed per person, we can estimate how many people can be seated inside the tent. This is a practical application of the cone volume formula in a real-world style scenario. Given Data / Assumptions: Height of the conical tent, h = 3 m. Base diameter = 10 m, so radius r = 5 m. Volume required per person = 3.92 m^3. The tent is completely filled up to its full volume by air and people. We are asked for an approximate number of persons. Concept / Approach: The volume V of a right circular cone is given by: V = (1/3) * pi * r^2 * h First compute the total internal volume of the tent using this formula. Then, to find how many persons can be accommodated, divide the total volume by the volume required per person. Because the answer must be a whole number of people, we pick the nearest suitable integer based on the calculation and the given options. Step-by-Step Solution: Radius r = 10 / 2 = 5 m. Height h = 3 m. Volume of tent V = (1/3) * pi * r^2 * h. V = (1/3) * pi * 5^2 * 3 = (1/3) * pi * 25 * 3. The factor 3 cancels with 1/3, so V = 25 * pi cubic metres. Using pi ≈ 3.14, V ≈ 25 * 3.14 = 78.5 m^3 (approximately). Volume per person = 3.92 m^3. Number of persons ≈ 78.5 / 3.92 ≈ 20.02. Thus, approximately 20 persons can be seated inside the tent. Verification / Alternative check: We can perform a quick check by multiplying 20 * 3.92 = 78.4 m^3, which is extremely close to the total tent volume of about 78.5 m^3. If we try 21 persons, 21 * 3.92 ≈ 82.32 m^3, which exceeds the tent volume. This confirms that 20 is the most reasonable maximum number of persons that fit given the stated air requirement per person. Why Other Options Are Wrong: Options 17, 18 and 19 are all too small and would leave unused volume given the assumed space requirement. Option 22 would demand a total air volume of about 86.24 m^3, which the tent does not have. Therefore, those options do not match the physical constraint of the tent's volume compared to per-person requirement. Common Pitfalls: Students sometimes confuse diameter with radius and incorrectly use r = 10 m in the cone volume formula, which doubles the radius and quadruples the area, leading to huge errors in volume. Another pitfall is using an incorrect approximation for pi or miscalculating 25 * pi. Finally, some may incorrectly round the number of persons up rather than choosing the nearest feasible integer that does not exceed the tent volume. Final Answer: The tent can accommodate approximately 20 persons.

Question 6

A wall is 24 m long, 8 m high and 60 cm thick. It is constructed using bricks each of size 24 cm × 12 cm × 8 cm. If 10% of the wall volume is occupied by mortar, how many bricks are required to build the wall?

Accepted Answer

Introduction / Context: This construction-based aptitude problem extends the brick-counting idea by adding the realistic detail that some portion of the wall volume is filled with mortar rather than bricks. We must calculate the total wall volume, subtract the part occupied by mortar, and then divide the remaining brick volume by the volume of a single brick. The question tests volume calculations, unit conversions and proportional reasoning. Given Data / Assumptions: Wall dimensions: length = 24 m, height = 8 m, thickness = 60 cm. Brick dimensions: 24 cm × 12 cm × 8 cm. 10% of the wall volume is mortar, hence 90% is brick volume. The wall is assumed solid, apart from mortar gaps. Concept / Approach: We treat both the wall and each brick as cuboids. The number of bricks required is the ratio of the volume actually occupied by bricks to the volume of one brick. Because mortar occupies 10% of the wall volume, only 90% of the wall volume is taken up by bricks. Therefore: Number of bricks = (0.9 * Volume of wall) / Volume of one brick All dimensions must be converted into the same unit, typically centimetres, before computing volumes. Step-by-Step Solution: Convert wall dimensions to cm: 24 m = 2400 cm, 8 m = 800 cm, thickness = 60 cm. Volume of wall = 2400 * 800 * 60 cubic centimetres. Volume of wall = 115,200,000 cubic cm. Volume of one brick = 24 * 12 * 8 cubic cm. Volume of one brick = 2304 cubic cm. If there were no mortar, number of bricks = 115,200,000 / 2304 = 50,000. But only 90% of wall volume is brick: brick volume = 0.9 * 115,200,000 = 103,680,000 cubic cm. Number of bricks = 103,680,000 / 2304 = 45,000. Verification / Alternative check: We can cross-check quickly: since 10% of the space is mortar, the brick count should be 90% of the no-mortar case. Without mortar, we needed 50,000 bricks, and 90% of 50,000 is 0.9 * 50,000 = 45,000, which matches our detailed calculation. This consistency confirms that our approach and arithmetic are correct. Why Other Options Are Wrong: 35,000 bricks would correspond to too small a brick volume, implying more mortar than stated. 55,000 and 65,000 are larger than the 50,000 bricks required without mortar, which is impossible. 50,000 bricks would be correct only if mortar were ignored; it does not reflect the 10% mortar content specified in the question. Common Pitfalls: One common mistake is forgetting to account for mortar and simply dividing total wall volume by the brick volume. Another is to subtract 10% from the number of bricks instead of 10% from the volume, which in this problem happens to give the same numeric count but may mislead in more complex cases. Students also often mix metres and centimetres, leading to incorrect wall volumes. Final Answer: The number of bricks required to build the wall is 45000.

Question 7

A solid sphere of radius 10 cm is melted and recast into 8 solid spherical balls of equal radius. What is the surface area of each small spherical ball in square centimetres?

Accepted Answer

Introduction / Context: This question involves conservation of volume when a solid object is melted and recast into several smaller objects. A large solid sphere is melted and formed into 8 smaller solid spheres of equal radius. Because no material is lost or gained, the total volume remains the same. We use this idea to find the radius of each smaller sphere, and then compute its surface area. This problem tests understanding of volume–surface area relationships for spheres. Given Data / Assumptions: Radius of original sphere, R = 10 cm. Number of smaller spheres = 8. Each smaller sphere is solid and has equal radius r. All material is conserved; no loss in melting or recasting. Concept / Approach: The volume of a sphere of radius x is: V = (4/3) * pi * x^3 Since the large sphere is recast into 8 equal small spheres, total volume is conserved: (4/3) * pi * R^3 = 8 * (4/3) * pi * r^3 This leads to R^3 = 8 * r^3, so r^3 = R^3 / 8. After finding r, the surface area of one small sphere is: Surface area = 4 * pi * r^2 Step-by-Step Solution: Original radius R = 10 cm. Volume of original sphere = (4/3) * pi * R^3 = (4/3) * pi * 10^3. Total volume of 8 small spheres = 8 * (4/3) * pi * r^3. Equate volumes: (4/3) * pi * 10^3 = 8 * (4/3) * pi * r^3. Cancel common factors (4/3 * pi) to get 10^3 = 8 * r^3. So r^3 = 1000 / 8 = 125. Therefore r = cube root of 125 = 5 cm. Surface area of one small sphere = 4 * pi * r^2 = 4 * pi * 5^2. Surface area = 4 * pi * 25 = 100 pi square centimetres. Verification / Alternative check: We can check total surface area increase. Original surface area = 4 * pi * 10^2 = 400 pi. Each small sphere has surface area 100 pi, and there are 8 of them, so combined surface area is 8 * 100 pi = 800 pi. The total surface area has doubled, which is expected when splitting a solid into multiple pieces of smaller size, confirming the internal logic. Why Other Options Are Wrong: Values like 80 pi, 50 pi, 64 pi and 75 pi sq.cm correspond to guessing a smaller radius than 5 cm or miscomputing r^3. For example, 80 pi would occur if r were wrongly taken as 4 cm, giving 4 * pi * 16 = 64 pi, still not listed. The only value consistent with r = 5 cm is 100 pi sq.cm. Common Pitfalls: A common mistake is to assume that the radius scales linearly with the number of spheres, which is incorrect because volume depends on the cube of the radius. Another error is to forget that volume is conserved and instead try to directly relate surface areas. Correct reasoning requires working through the volume equation first, then computing surface area from the derived radius. Final Answer: The surface area of each small spherical ball is 100 pi sq.cm.

Question 8

The volume of a cube is 729 cubic centimetres. What is the total surface area of the cube in square centimetres?

Accepted Answer

Introduction / Context: This problem links the volume of a cube to its surface area. Once the edge length of a cube is known, its volume and surface area can both be calculated easily. In this question, the volume is given and we must reverse the volume formula to find the edge length first, then apply the surface area formula. Such questions test your familiarity with powers and roots, particularly cube roots, and formula application. Given Data / Assumptions: Volume of the cube, V = 729 cm^3. Edge length (side) of the cube = a cm. We must find the total surface area in square centimetres. Concept / Approach: For a cube of side a, the volume is: V = a^3 The total surface area (TSA) is: TSA = 6 * a^2 First, we find a by taking the cube root of the given volume. Once a is known, we substitute it into the surface area formula to obtain the result. Step-by-Step Solution: Given V = 729 cm^3. For a cube, V = a^3, so a^3 = 729. We recognize 729 = 9^3 (since 9 * 9 * 9 = 729). Therefore, a = 9 cm. Total surface area TSA = 6 * a^2. Compute a^2 = 9^2 = 81. TSA = 6 * 81 = 486 square centimetres. Verification / Alternative check: We can reverse the calculation: if the cube has side 9 cm, its volume would be 9^3 = 729 cm^3, which matches the given volume. This confirms that a = 9 cm is correct. Then TSA = 6 * 81 = 486 cm^2 fits perfectly with the formula and the computed side length, so our answer is consistent in both directions. Why Other Options Are Wrong: The values 456, 446, 476 and 500 sq.cm all correspond to incorrectly squaring the side or using an approximate side length that is not the true cube root of 729. For example, 500 sq.cm would require a non-integer side length. Only 486 sq.cm comes from the correct side length of 9 cm and the standard surface area formula for a cube. Common Pitfalls: Some students mistakenly take the square root instead of the cube root of the volume, leading to a side length of sqrt(729) rather than cube root. Others confuse the formulas for volume and surface area or forget that a cube has 6 identical faces. Careful recall of definitions and formulas helps avoid these errors. Final Answer: The total surface area of the cube is 486 sq.cm.

Question 9

The circumferences of two circles are 264 metres and 352 metres respectively. What is the difference between the area of the larger circle and the area of the smaller circle, in square metres?

Accepted Answer

Introduction / Context: This question compares two circles using their circumferences instead of their radii. We are asked to find the difference in their areas. Rather than computing radii separately and then areas, it is often convenient to express area directly in terms of circumference, which simplifies the calculations. This problem tests algebraic manipulation of circle formulas and careful handling of pi in denominators. Given Data / Assumptions: Circumference of smaller circle, C1 = 264 m. Circumference of larger circle, C2 = 352 m. We must find A2 − A1, where A1 and A2 are the areas of the smaller and larger circles respectively. Concept / Approach: For a circle of radius r, the circumference C and area A are: C = 2 * pi * r A = pi * r^2 We can express area directly in terms of circumference using r = C / (2 * pi): A = pi * (C / (2 * pi))^2 = C^2 / (4 * pi) Thus, the difference in areas is: A2 − A1 = (C2^2 − C1^2) / (4 * pi) Step-by-Step Solution: Given C1 = 264 m, C2 = 352 m. Use A = C^2 / (4 * pi). Difference in areas ΔA = (C2^2 − C1^2) / (4 * pi). Compute squares: C2^2 = 352^2 = 123904, C1^2 = 264^2 = 69696. C2^2 − C1^2 = 123904 − 69696 = 54208. So ΔA = 54208 / (4 * pi) = 54208 / (4 * pi). Simplify denominator: 4 * pi ≈ 4 * 3.14159 ≈ 12.56636. Compute ΔA ≈ 54208 / 12.56636 ≈ 4313.7 square metres. To match given options, we round to the nearest whole number: approximately 4312 sq.m. Verification / Alternative check: We can approximate radii and areas separately. For C1 = 264, r1 ≈ 264 / (2 * 3.14) ≈ 42.1 m, so A1 ≈ 3.14 * 42.1^2 ≈ 5570 m^2. For C2 = 352, r2 ≈ 352 / (2 * 3.14) ≈ 56.1 m, so A2 ≈ 3.14 * 56.1^2 ≈ 9880 m^2. The difference A2 − A1 is around 4310 m^2, close to the more accurate computation, confirming 4312 sq.m is reasonable. Why Other Options Are Wrong: Values such as 2413, 1234 and 2143 sq.m are far too small compared to the approximate areas of the circles and would imply much closer circumferences. 3000 sq.m is also not consistent with the precise calculation from the formula. Only 4312 sq.m aligns with both the exact algebra and approximate reasoning. Common Pitfalls: Students may try to find radii and areas independently for each circle and make arithmetic mistakes along the way. Another error is to forget the squared term and treat area as directly proportional to circumference instead of the square of the radius. It is also easy to mishandle divisions involving pi, so using the area-in-terms-of-circumference formula helps streamline the work. Final Answer: The difference between the areas of the larger and smaller circles is approximately 4312 sq.m.

Question 10

The heights of two right circular cylinders are in the ratio 3 : 5 and their volumes are in the ratio 27 : 80. What is the ratio of their radii?

Accepted Answer

Introduction / Context: This problem involves comparing two right circular cylinders using ratios of their heights and volumes. The goal is to deduce the ratio of their radii. Because volume of a cylinder depends on both radius and height, ratios of volume are influenced by ratios of both dimensions. The question tests algebraic manipulation and understanding of how geometric quantities scale. Given Data / Assumptions: Heights of cylinders: h1 : h2 = 3 : 5. Volumes: V1 : V2 = 27 : 80. Radii of cylinders: r1 and r2, ratio required r1 : r2. Both shapes are right circular cylinders. Concept / Approach: Volume of a cylinder is given by: V = pi * r^2 * h Thus: V1 / V2 = (pi * r1^2 * h1) / (pi * r2^2 * h2) = (r1^2 * h1) / (r2^2 * h2) We substitute the given ratios for V1 : V2 and h1 : h2, and then solve for r1^2 / r2^2. Finally, we take square roots to obtain the ratio r1 : r2. Step-by-Step Solution: Given V1 / V2 = 27 / 80. Given h1 / h2 = 3 / 5. Using V proportionality: V1 / V2 = (r1^2 * h1) / (r2^2 * h2). So 27 / 80 = (r1^2 * h1) / (r2^2 * h2). Substitute h1 / h2 = 3 / 5: 27 / 80 = (r1^2 / r2^2) * (3 / 5). Rearrange: r1^2 / r2^2 = (27 / 80) * (5 / 3). Compute: (27 * 5) / (80 * 3) = 135 / 240 = 9 / 16. Thus r1^2 / r2^2 = 9 / 16. Taking square roots: r1 / r2 = 3 / 4. Verification / Alternative check: We can choose convenient values based on the ratio. Let h1 = 3k, h2 = 5k, r1 = 3m, r2 = 4m. Then V1 = pi * 3^2 * 3k = 27k * pi and V2 = pi * 4^2 * 5k = 80k * pi. Therefore V1 : V2 = 27k * pi : 80k * pi = 27 : 80, matching the given ratio. This confirms that r1 : r2 = 3 : 4 is consistent with all the information. Why Other Options Are Wrong: Options 1 : 3 and 2 : 1 ignore the squared relationship between radius and volume. Ratios like 4 : 7 or 5 : 4 do not produce the given volume ratio when combined with the height ratio 3 : 5. Only 3 : 4 satisfies the volume and height relationships simultaneously. Common Pitfalls: Students often miss the square on the radius in the volume formula and treat volume as proportional to r * h instead of r^2 * h. Another common mistake is to invert one of the ratios accidentally when cross-multiplying, leading to an incorrect radius ratio. Writing all steps clearly and keeping track of numerators and denominators helps avoid such errors. Final Answer: The ratio of the radii of the two cylinders is 3 : 4.

Question 11

A cuboidal block with dimensions 6 cm × 9 cm × 12 cm is cut up into an exact number of equal smaller cubes. What is the least possible number of equal cubes obtained?

Accepted Answer

Introduction / Context: This problem involves partitioning a cuboid into smaller cubes of equal size. To obtain the least number of equal cubes, we must use the largest possible edge length for the small cubes that still divides each of the three dimensions of the original block exactly. This approach relies on the concept of the greatest common divisor (GCD) for the dimensions and basic volume reasoning. Given Data / Assumptions: Dimensions of cuboidal block: 6 cm × 9 cm × 12 cm. The block is cut into identical cubes, each with side length s cm. The number of cubes must be an integer and as small as possible. Concept / Approach: For cubes to fit exactly, the side length s must divide each dimension 6, 9 and 12 without remainder. The largest such s is the greatest common divisor of these three numbers. Once s is known, the number of cubes equals the ratio of the volume of the cuboid to the volume of a cube: Number of cubes = (6 * 9 * 12) / s^3 Step-by-Step Solution: Compute the greatest common divisor of 6, 9 and 12. GCD(6, 9) = 3, GCD(3, 12) = 3, so s = 3 cm. Thus, each small cube has side 3 cm. Volume of cuboid = 6 * 9 * 12 cubic centimetres. Volume of cuboid = 648 cm^3. Volume of one cube = 3^3 = 27 cm^3. Number of cubes = 648 / 27 = 24. Verification / Alternative check: We can verify by counting cubes along each dimension. Along length 12 cm, we fit 12 / 3 = 4 cubes. Along breadth 9 cm, we fit 9 / 3 = 3 cubes. Along height 6 cm, we fit 6 / 3 = 2 cubes. Therefore, the total number of cubes is 4 * 3 * 2 = 24, matching the volume-based calculation. Why Other Options Are Wrong: Values such as 6, 9, 18 and 30 would correspond to using a smaller cube size or miscounting the cubes along one or more dimensions. For example, 18 would require incorrect division along at least one dimension. Because 3 cm is the largest edge length that divides all three dimensions exactly, any smaller cube would give more than 24 pieces, not fewer. Common Pitfalls: A typical mistake is to pick the smallest dimension (6 cm) as the cube side length without checking divisibility with 9 and 12. Others might use the least common multiple instead of the greatest common divisor by mistake. Students also sometimes forget that the number of cubes is a three-dimensional count (product along length, breadth and height) rather than just dividing by one dimension. Final Answer: The least possible number of equal cubes obtained is 24.

Question 12

Three solid cubes with edge lengths 3 cm, 4 cm and 5 cm are melted together and recast into one large solid cube. What is the ratio of the total surface area of the three original cubes to the surface area of the new large cube?

Accepted Answer

Introduction / Context: This question combines volume conservation with surface area comparisons. Three small cubes are melted and formed into a single larger cube. Because melting does not change the total volume of material, the sum of the volumes of the small cubes equals the volume of the large cube. We can use this fact to determine the edge length of the large cube, then find its surface area and compare it to the total surface area of the original cubes. This kind of question is common in aptitude exams involving solids. Given Data / Assumptions: Edge lengths of the three small cubes: 3 cm, 4 cm and 5 cm. The cubes are melted with no loss or gain of material. The resulting solid is a cube with edge length L. We need the ratio (total surface area of original cubes) : (surface area of new cube). Concept / Approach: For a cube of side a, the volume and surface area are: Volume = a^3 Surface area = 6 * a^2 First, we compute the total volume of the three small cubes and set it equal to L^3 to find L. Then we compute the total surface area of the three small cubes and the surface area of the large cube, and finally form their ratio. Step-by-Step Solution: Volume of cube with side 3 cm = 3^3 = 27 cm^3. Volume of cube with side 4 cm = 4^3 = 64 cm^3. Volume of cube with side 5 cm = 5^3 = 125 cm^3. Total volume = 27 + 64 + 125 = 216 cm^3. Let L be the edge of the new cube, so L^3 = 216. Since 216 = 6^3, L = 6 cm. Surface area of cube with side 3 cm = 6 * 3^2 = 6 * 9 = 54 cm^2. Surface area of cube with side 4 cm = 6 * 4^2 = 6 * 16 = 96 cm^2. Surface area of cube with side 5 cm = 6 * 5^2 = 6 * 25 = 150 cm^2. Total surface area of small cubes = 54 + 96 + 150 = 300 cm^2. Surface area of large cube (side 6 cm) = 6 * 6^2 = 6 * 36 = 216 cm^2. Required ratio = 300 : 216. Divide numerator and denominator by 12: 300 / 12 = 25, 216 / 12 = 18. So ratio = 25 : 18. Verification / Alternative check: The volume check is straightforward: 6^3 is indeed 216, matching the combined volume of the small cubes, so the side of the large cube is correctly found as 6 cm. The arithmetic for the surface areas is simple, and the ratio simplifies neatly, suggesting that the computation is correct. No inconsistencies appear when rechecking steps. Why Other Options Are Wrong: Ratios like 2 : 1 or 3 : 2 do not reflect the specific side lengths and would produce incorrect total surface area values. Ratios 27 : 20 and 27 : 16 have numerators and denominators inconsistent with the specific surface areas computed above. Only 25 : 18 matches the exact numeric values 300 and 216. Common Pitfalls: One common mistake is to add edge lengths instead of volumes, leading to an incorrect new cube side. Others miscompute 5^3 or 6^3. Students may also neglect to use the factor 6 in the surface area formula for cubes, or incorrectly reduce the final ratio. Careful step-by-step work avoids these issues. Final Answer: The ratio of the total surface area of the three original cubes to the surface area of the new large cube is 25 : 18.

Question 13

Find the total surface area of a cuboid that is 16 m long, 14 m broad and 7 m high.

Accepted Answer

Introduction / Context: Questions on the surface area of cuboids are common in aptitude tests and basic geometry. They help learners revise how the dimensions of a three dimensional object relate to the total area of its outer surfaces. Here, we are asked to find the total surface area of a cuboid whose length, breadth and height are given in metres. Given Data / Assumptions: The solid is a cuboid. Length l = 16 m. Breadth b = 14 m. Height h = 7 m. We are asked for the total surface area in square metres. Concept / Approach: For a cuboid, the total surface area is the sum of the areas of all six rectangular faces. The formula for total surface area (TSA) is: TSA = 2 * (l * b + b * h + l * h) This works because opposite faces are equal in area and there are three distinct pairs: top and bottom, front and back, left and right. Step-by-Step Solution: Step 1: Compute l * b = 16 * 14 = 224. Step 2: Compute b * h = 14 * 7 = 98. Step 3: Compute l * h = 16 * 7 = 112. Step 4: Add these three products: 224 + 98 + 112 = 434. Step 5: Multiply by 2 for the total surface area: TSA = 2 * 434 = 868 square metres. Verification / Alternative check: We can quickly check the reasonableness. Each face has area between about 100 and 250 square metres, and we have six faces, so a value near 6 * 150 = 900 square metres is expected. The calculated 868 square metres is close to this estimate and is internally consistent with the formula, so the value is reliable. Why Other Options Are Wrong: 796 square metres is less than the correct total surface area and would result from undercounting one of the faces or misadding the products. 812 square metres is also smaller than the correct value and indicates an error in adding 224, 98 and 112 or in doubling the sum. 902 square metres is larger than the correct value and could come from incorrectly rounding or adding an extra area term. 840 square metres is another nearby but incorrect approximation and does not match the exact formula based calculation. Common Pitfalls: Students often confuse lateral surface area with total surface area and forget to include the top and bottom faces. Another common mistake is to change units or to misplace a dimension. It is important to plug the length, breadth and height carefully into the formula and to perform the arithmetic accurately. Forgetting to multiply the sum by 2 is another frequent source of error. Final Answer: The total surface area of the cuboid is 868 square metres.

Question 14

A right circular cylinder circumscribes a hemisphere so that they have the same circular base. Find the ratio of the volume of the cylinder to the volume of the hemisphere.

Accepted Answer

Introduction / Context: This question compares the volumes of two common solids in mensuration: a right circular cylinder and a hemisphere. The cylinder is said to circumscribe the hemisphere with a common base, which means the hemisphere fits exactly inside the cylinder with the same radius. Understanding such relationships helps in visualizing three dimensional shapes and in quickly working with volume ratios in aptitude exams. Given Data / Assumptions: The hemisphere and the cylinder share the same circular base. Let the common radius be r. The height of the hemisphere is r. The cylinder circumscribes the hemisphere, so its height equals the height of the hemisphere. We want the ratio: volume of cylinder : volume of hemisphere. Concept / Approach: We recall the volume formulas: Volume of cylinder Vc = π * r^2 * h Volume of hemisphere Vh = (2 / 3) * π * r^3 Because the cylinder circumscribes the hemisphere with the same base, its height h equals r. Substituting h = r allows a direct comparison of the two volumes in terms of r and π, both of which will cancel in the ratio. Step-by-Step Solution: Step 1: Take height of cylinder h = r. Step 2: Volume of cylinder Vc = π * r^2 * r = π * r^3. Step 3: Volume of hemisphere Vh = (2 / 3) * π * r^3. Step 4: Form the ratio Vc : Vh = π * r^3 : (2 / 3) * π * r^3. Step 5: Cancel π * r^3 from numerator and denominator, giving 1 : (2 / 3). Step 6: Convert 1 : (2 / 3) to a ratio of whole numbers by multiplying both terms by 3, giving 3 : 2. Verification / Alternative check: We can also insert a specific value, such as r = 1 unit. Then volume of cylinder is π * 1^3 = π, and volume of hemisphere is (2 / 3) * π. The ratio π : (2 / 3)π again simplifies to 3 : 2. This numeric check confirms that the algebraic ratio is correct and independent of the chosen radius, because the radius cancels out. Why Other Options Are Wrong: 5 : 2 would mean the cylinder volume is much larger than it really is compared to the hemisphere, which does not follow from the standard formulas. 7 : 2 and 9 : 2 exaggerate the cylinder volume even more and have no basis in the geometry of the solids. 1 : 3 would suggest that the hemisphere is three times as large as the cylinder by volume, which is clearly impossible because the hemisphere fits inside the cylinder. Common Pitfalls: A common mistake is to confuse height with diameter and to take the cylinder height as 2r instead of r. That would correspond to a cylinder that encloses a sphere, not a hemisphere. Another error is forgetting the correct volume formula for a hemisphere and using 4 / 3 π r^3, which is the formula for a full sphere. Careful reading of the problem and recalling the correct formulas prevent such errors. Final Answer: The ratio of the volume of the cylinder to the volume of the hemisphere is 3 : 2.

Question 15

The capacity in litres of a rectangular water tank with internal dimensions 8 m by 6 m by 2.5 m is:

Accepted Answer

Introduction / Context: Capacity problems connect geometry with real life storage questions. Here, we deal with a rectangular water tank, which is just a cuboid. Once the internal dimensions are known, we can compute its volume in cubic metres and then convert that volume to litres, because capacity in practical contexts such as water storage is usually expressed in litres. Given Data / Assumptions: Length l = 8 m. Breadth b = 6 m. Height h = 2.5 m. The tank is assumed to be fully filled with water. 1 cubic metre corresponds to 1000 litres. Concept / Approach: The internal capacity of a rectangular tank is equal to its internal volume. For a cuboid, the volume is given by: Volume = l * b * h Once we compute the volume in cubic metres, we multiply by 1000 to convert to litres. This is because 1 m^3 of water occupies 1000 litres by definition of the litre unit. Step-by-Step Solution: Step 1: Compute the volume in cubic metres. Volume = 8 * 6 * 2.5. Step 2: First multiply 8 * 6 = 48. Step 3: Now multiply 48 * 2.5 = 120. So, volume = 120 m^3. Step 4: Convert cubic metres to litres. Capacity in litres = 120 * 1000 = 120000 litres. Verification / Alternative check: We can check the reasonableness. A 1 m by 1 m by 1 m cube holds 1000 litres. Our tank is 8 m by 6 m by 2.5 m, which has 8 * 6 * 2.5 = 120 such cubic metre units. Each contributes 1000 litres, so 120000 litres is consistent with this interpretation. The magnitude also makes sense for a fairly large storage tank. Why Other Options Are Wrong: 120 litres corresponds to only 0.12 m^3, which is far too small for a tank of this size. 1200 litres equals 1.2 m^3 and is still extremely small compared to the actual geometric volume. 12000 litres equals 12 m^3, again much smaller than the computed 120 m^3. 60000 litres equals 60 m^3, which is exactly half the correct volume and would result from mistakenly halving one dimension or misreading 2.5 m as 1.25 m. Common Pitfalls: The major pitfall is mishandling the unit conversion between cubic metres and litres. Some learners mistakenly think 1 m^3 equals 100 litres instead of 1000 litres. Another error comes from incorrect multiplication of the three dimensions, especially when one dimension is a decimal like 2.5. It is always safer to multiply step by step and, if necessary, first convert 2.5 to a fraction such as 5 / 2 to simplify mental arithmetic. Final Answer: The capacity of the tank is 120000 litres.

Question 16

How many solid iron rods, each of length 7 m and diameter 2 cm, can be manufactured from 0.88 cubic metre of iron?

Accepted Answer

Introduction / Context: This problem combines the concept of volume of a cylinder with unit conversion between cubic metres and cubic centimetres. Each iron rod is cylindrical in shape, and many such rods are made from a fixed total volume of molten iron. The task is to determine how many identical rods can be formed without any wastage of material. Given Data / Assumptions: Total volume of iron available = 0.88 m^3. Each rod is a solid cylinder. Length of each rod L = 7 m. Diameter of each rod = 2 cm, so radius r = 1 cm. The rods are assumed to be perfectly formed with no loss of volume. Concept / Approach: The volume of a cylinder is given by: Volume of one rod = π * r^2 * h Here r is in centimetres and h must also be in centimetres if we want the volume in cubic centimetres. Since the total volume is given in cubic metres, we convert that to cubic centimetres using: 1 m = 100 cm, so 1 m^3 = 100^3 = 1000000 cm^3. After obtaining both volumes in the same unit, we divide the total volume by the volume of one rod to get the number of rods. Step-by-Step Solution: Step 1: Convert the total volume to cubic centimetres. Total volume = 0.88 m^3 = 0.88 * 1000000 = 880000 cm^3. Step 2: Convert the length of one rod to centimetres. Length L = 7 m = 700 cm. Step 3: Radius r = 1 cm, so volume of one rod = π * 1^2 * 700 = 700 * π cm^3. Step 4: Number of rods = total volume / volume of one rod. Number of rods = 880000 / (700 * π). Step 5: Using π = 22 / 7, volume of one rod = 700 * 22 / 7 = 2200 cm^3. Step 6: Number of rods = 880000 / 2200 = 400. Verification / Alternative check: We can simplify 880000 / 2200 by cancelling zeros and factors. Dividing numerator and denominator by 100 gives 8800 / 22, which equals 400. This confirms the calculation and shows that the result is an exact integer, so the given data are consistent and there is no fractional rod. Why Other Options Are Wrong: 300 rods would use only 300 * 2200 = 660000 cm^3 of iron, leaving a large unused volume. 350 rods would correspond to 770000 cm^3, still less than the given total volume. 450 rods would require 990000 cm^3, which is more iron than available. 500 rods would need 1100000 cm^3, again exceeding the given 880000 cm^3 of iron. Common Pitfalls: Learners often forget to convert all dimensions to the same unit system, mixing metres with centimetres. Another frequent error is using the diameter instead of the radius in the cylinder formula or misapplying the factor of π. Rounding π too early can also lead to slight discrepancies, but here with π = 22 / 7 the answer comes out exactly as an integer. Carefully tracking units and using the correct formulas avoids these mistakes. Final Answer: The number of solid iron rods that can be made is 400 rods.

Question 17

The areas of three mutually perpendicular faces of a cuboid are 12 cm^2, 20 cm^2 and 15 cm^2. Find the volume of the cuboid in cubic centimetres.

Accepted Answer

Introduction / Context: This question tests the relationship between the face areas and the volume of a cuboid. Instead of giving the three edge lengths directly, we are given the areas of three faces that meet at a corner. Such problems are popular in aptitude exams because they require a small but important algebraic idea rather than straightforward substitution. Given Data / Assumptions: Let the edges of the cuboid be a, b and c centimetres. Area of face with sides a and b is ab = 12 cm^2. Area of face with sides b and c is bc = 20 cm^2. Area of face with sides c and a is ca = 15 cm^2. We want the volume V = a * b * c. Concept / Approach: The key idea is to multiply the three given face areas together: (ab) * (bc) * (ca) = a^2 * b^2 * c^2 = (abc)^2. From this, we can get the volume by taking the square root: abc = sqrt(ab * bc * ca). Therefore, we multiply the three given areas and then take the positive square root to find the volume of the cuboid. Step-by-Step Solution: Step 1: Compute the product of the three face areas. ab * bc * ca = 12 * 20 * 15. Step 2: First multiply 12 * 20 = 240. Step 3: Then multiply 240 * 15 = 3600. Step 4: We have (abc)^2 = 3600. Step 5: Take the positive square root: abc = sqrt(3600) = 60. So, the volume of the cuboid is 60 cm^3. Verification / Alternative check: We can attempt to find a, b and c explicitly. Suppose ab = 12 and bc = 20. Then (ab) * (bc) / (ca) should give b^2. That is: b^2 = (12 * 20) / 15 = 240 / 15 = 16. So b = 4 cm. Then a = 12 / b = 12 / 4 = 3 cm, and c = 20 / b = 20 / 4 = 5 cm. Now the volume is a * b * c = 3 * 4 * 5 = 60 cm^3, which matches the earlier result and confirms the correctness. Why Other Options Are Wrong: 55 cm^3, 45 cm^3 and 40 cm^3 do not match the computed product of the edges and result from incorrect algebra or arithmetic. 75 cm^3 would require a different combination of edges inconsistent with the given face areas, so it cannot be correct. Common Pitfalls: Students sometimes try to guess individual edge lengths from the given areas and may choose inconsistent values. Others may forget to take the square root after multiplying the three given areas, and mistakenly use 3600 cm^3 as the volume. Remembering the identity (ab) * (bc) * (ca) = (abc)^2 is very helpful and leads quickly to the correct volume without overcomplicating the algebra. Final Answer: The volume of the cuboid is 60 cubic centimetres.

Question 18

A flat-bottomed boat that is 3 m long and 2 m wide is floating on a lake. The boat sinks by 1 cm when a man steps into it. Find the mass of the man.

Accepted Answer

Introduction / Context: This problem combines ideas from mensuration with the principle of flotation. When a man steps into a boat, the boat sinks a little more into the water because it has to displace additional water to balance the extra weight. By measuring the change in water level, we can determine the mass of the man using the relationship between volume, density and weight. Given Data / Assumptions: Length of boat L = 3 m. Breadth of boat B = 2 m. Additional sinking depth when man steps in = 1 cm = 0.01 m. The bottom of the boat is flat, so the displaced volume is L * B * additional depth. Density of water is taken as 1000 kg per m^3. The system is in equilibrium, so weight of man equals weight of extra water displaced. Concept / Approach: Archimedes principle states that the upward buoyant force on an immersed body equals the weight of the fluid displaced. When the man gets into the boat, the extra volume of water displaced corresponds exactly to the weight of the man. The steps are: Volume of extra water displaced = L * B * extra depth. Mass of displaced water = volume * density. This mass equals the mass of the man because their weights balance. Step-by-Step Solution: Step 1: Compute the volume of extra water displaced. Volume = 3 * 2 * 0.01 = 0.06 m^3. Step 2: Use density of water, 1000 kg per m^3. Mass of displaced water = 0.06 * 1000 = 60 kg. Step 3: By Archimedes principle, mass of man = mass of additional water displaced = 60 kg. Verification / Alternative check: We can reason qualitatively: 0.1 m^3 of water would have a mass of 100 kg. Our displaced volume of 0.06 m^3 is a little more than half of 0.1 m^3, so the mass should be slightly more than 50 kg. The computed value of 60 kg fits this expectation, giving us confidence in the calculation. Why Other Options Are Wrong: 12 kg would correspond to a much smaller displaced volume and is unrealistic for an adult human mass. 72 kg and 96 kg are larger than the computed mass and would require the boat to sink more than the observed 1 cm. 48 kg is smaller than 60 kg and could only occur if either the dimensions or the sinking depth were less than given. Common Pitfalls: A common mistake is to forget to convert the sinking depth from centimetres to metres, which would lead to a tenfold error in the volume calculation. Another pitfall is to confuse mass with weight and insert an extra factor of gravitational acceleration when it is not needed, because we are working directly in mass units using density. Careful handling of units and straightforward application of Archimedes principle ensures the correct answer. Final Answer: The mass of the man is 60 kg.

Question 19

Water flows into a rectangular tank of plan dimensions 200 m by 160 m through a rectangular pipe of cross section 1.5 m by 1.25 m at a speed of 20 km per hour. In how many minutes will the water level in the tank rise by 2 m?

Accepted Answer

Introduction / Context: This problem applies volume flow rate concepts to a practical situation of filling a large rectangular tank through a pipe. The idea is that the volume of water entering per unit time must equal the increase in volume of water in the tank per unit time. Once the required volume corresponding to a 2 m rise is known, we can divide that by the flow rate to find the required time. Given Data / Assumptions: Tank length = 200 m. Tank breadth = 160 m. Required rise in water level = 2 m. Pipe cross section: 1.5 m by 1.25 m. Flow speed through pipe = 20 km per hour. No leakage or overflow is assumed. Concept / Approach: The key steps are: 1) Compute the volume of water needed to raise the tank level by 2 m: base area of tank * rise in height. 2) Compute the volumetric flow rate through the pipe: cross sectional area of pipe * flow speed. 3) Time required = required volume / flow rate. We must also convert the flow speed from km per hour to metres per minute so that the final time is in minutes. Step-by-Step Solution: Step 1: Compute the volume needed in the tank. Base area of tank = 200 * 160 = 32000 m^2. Required rise = 2 m. Required volume = 32000 * 2 = 64000 m^3. Step 2: Compute the pipe cross sectional area. Area of pipe = 1.5 * 1.25 = 1.875 m^2. Step 3: Convert flow speed to metres per minute. 20 km per hour = 20000 m per hour. Per minute = 20000 / 60 ≈ 333.33 m per minute. Step 4: Compute flow rate. Flow rate = 1.875 * 333.33 ≈ 625 m^3 per minute. Step 5: Compute time. Time = required volume / flow rate = 64000 / 625 = 102.4 minutes. Verification / Alternative check: We can check the magnitude: in about 100 minutes the water must fill 64000 m^3. A flow rate of 625 m^3 per minute for 100 minutes gives 62500 m^3, which is slightly less than 64000 m^3, so slightly more than 100 minutes is needed. The exact value 102.4 minutes is therefore reasonable and consistent with this quick mental estimate. Why Other Options Are Wrong: 80 minutes and 96 minutes are too small, since they would not supply enough volume at 625 m^3 per minute to reach 64000 m^3. 120 minutes and 140 minutes are larger than necessary and would fill the tank beyond the 2 m mark if the flow continued for that long. Common Pitfalls: A frequent error is incorrect unit conversion for speed, such as using 20 m per minute instead of converting 20 km per hour correctly. Another mistake is forgetting that the tank is very large, which leads some learners to underestimate the time. Some students also miscalculate the area of the pipe or the tank base. Carefully tracking units and performing each multiplication step separately reduces these errors. Final Answer: The water level in the tank will rise by 2 m in approximately 102.4 minutes.

Question 20

A solid cylindrical iron rod has height equal to its radius. It is melted and recast into solid spherical balls, each of radius equal to half the radius of the original rod. How many such balls are formed?

Accepted Answer

Introduction / Context: This question is about conservation of volume when a solid is melted and recast into new shapes. The mass and hence the volume of metal remain the same before and after recasting, assuming no loss. Here a cylindrical rod of iron is transformed into several spherical balls. We must relate the volume of the cylinder to the total volume of the spheres to determine how many spheres are formed. Given Data / Assumptions: Original solid is a cylinder. Height of cylinder h equals its radius R. New solids are spheres. Radius of each sphere r = R / 2. No loss of material during melting and recasting. Concept / Approach: We use the standard volume formulas: Volume of cylinder Vc = π * R^2 * h. Volume of sphere Vs = (4 / 3) * π * r^3. Since h = R, the cylinder volume simplifies. Because total volume is conserved, we set: Vc = n * Vs, where n is the number of spheres. Solving for n gives the required count. Step-by-Step Solution: Step 1: Use h = R for the cylinder. Vc = π * R^2 * R = π * R^3. Step 2: Radius of each sphere r = R / 2. Vs = (4 / 3) * π * (R / 2)^3. Step 3: Compute (R / 2)^3 = R^3 / 8. Vs = (4 / 3) * π * R^3 / 8 = (4 * π * R^3) / 24 = (1 / 6) * π * R^3. Step 4: Set total volume of spheres equal to cylinder volume. π * R^3 = n * (1 / 6) * π * R^3. Step 5: Cancel π * R^3 from both sides. 1 = n * (1 / 6) ⇒ n = 6. Verification / Alternative check: We can pick a convenient value, such as R = 2 units. Then cylinder volume Vc = π * 2^3 = 8π. Each sphere has radius 1 unit, so Vs = (4 / 3) * π * 1^3 = (4 / 3)π. The number of spheres is Vc / Vs = 8π / ((4 / 3)π) = 8 / (4 / 3) = 8 * 3 / 4 = 6. This numeric example confirms the symbolic computation. Why Other Options Are Wrong: 3 balls, 4 balls and 5 balls would together have less volume than the original cylinder, which would require some of the metal to vanish, contradicting the conservation of volume assumption. 8 balls would require more volume of iron than available from the original cylinder, which is not possible without adding extra material. Common Pitfalls: The main error is forgetting that the radius of each sphere is half the radius of the cylinder, which dramatically reduces the volume of each sphere because volume scales with the cube of the radius. Another pitfall is miscomputing the exponent when evaluating (R / 2)^3. Keeping track of the cube and simplifying carefully is essential in such problems. Final Answer: The number of spherical balls formed is 6 balls.

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