Question 1

A right angled triangle has sides of length 3 cm, 4 cm and 5 cm. The triangle is rotated about the side of length 3 cm to generate a right circular cone. What is the volume of the cone formed, in cubic centimetres?

Accepted Answer

Introduction / Context: This problem connects coordinate geometry of a right angled triangle with the mensuration formula for the volume of a right circular cone. When a right triangle is rotated about one of its sides, the resulting solid is a cone. Identifying which side becomes the height and which side becomes the radius is critical. Once that is clear, we apply the standard cone volume formula. Problems like this check both visualization skills and formula application, and they are common in competitive exams involving quantitative aptitude. Given Data / Assumptions: Right triangle sides are 3 cm, 4 cm and 5 cm. The triangle is right angled, and 3 cm and 4 cm are the perpendicular sides, 5 cm is the hypotenuse. The triangle is rotated about the side of length 3 cm. We assume a full rotation of 360 degrees. We must find the volume of the resulting cone. Concept / Approach: In a right triangle with perpendicular sides a and b, and hypotenuse c, if the triangle is rotated about one perpendicular side, say a, then: The axis of rotation (side a) becomes the height h of the cone. The other perpendicular side b traces a circle and becomes the radius r of the cone base. The hypotenuse becomes the slant height, which is not needed for volume. The volume V of a right circular cone with radius r and height h is: V = (1 / 3) * pi * r^2 * h Step-by-Step Solution: Step 1: Identify height and radius. Triangle sides: 3 cm, 4 cm, 5 cm, with 3 and 4 as perpendicular sides. Rotation is about the side of 3 cm, so height h = 3 cm. Other perpendicular side 4 cm becomes radius r = 4 cm. Step 2: Apply cone volume formula. V = (1 / 3) * pi * r^2 * h V = (1 / 3) * pi * 4^2 * 3 V = (1 / 3) * pi * 16 * 3 V = 16 * pi cubic centimetres. Thus, the volume of the cone formed is 16 pi cubic centimetres. Verification / Alternative check: We can check the choice of height and radius logically. The axis of rotation is fixed, so the distance from any point on the triangle to this axis describes a circle. When rotating around the 3 cm side, the 3 cm side itself stays fixed and forms the axis, so it must be the height. The side opposite the axis, which is 4 cm, moves in a circle at constant distance 4 cm from the axis, so that distance is the radius. This validates our selection of h and r, and hence the computed volume. Why Other Options Are Wrong: Option A (12 pi) and Option B (15 pi) correspond to incorrect combinations or mistaken identification of height or radius. Option D (20 pi) assumes larger dimensions than actually used. Option E (10 pi) is also not consistent with the formula. Only 16 pi cubic centimetres is consistent with the correct substitution of r = 4 cm and h = 3 cm. Common Pitfalls: Many students mistakenly take the hypotenuse as the radius or height. Others mix up which side becomes the axis of rotation. Another common error is to forget the factor 1 over 3 in the cone volume formula. Always clearly identify the rotating side, mark it as the height, and take the perpendicular side as the radius when the triangle is right angled. Final Answer: The volume of the cone formed is 16 pi cubic centimetres.

Question 2

A solid rectangular block has dimensions 6 cm by 12 cm by 15 cm. It is to be cut exactly into a certain number of equal small cubes, all of the same size, with no waste of material. What is the smallest possible number of such cubes?

Accepted Answer

Introduction / Context: This is a standard problem on cutting a rectangular solid into the maximum size identical cubes while using the entire volume. The key idea is that the edge of each small cube must be a common measure of all three dimensions of the rectangular block, so that an integer number of cubes fits exactly along each side. The smallest number of cubes will occur when each cube has the largest possible edge length, which is equal to the greatest common divisor of the three dimensions. Given Data / Assumptions: Dimensions of the rectangular block are 6 cm, 12 cm and 15 cm. The block is solid and cut completely into equal cubes. There is no wastage of material. We seek the least possible number of cubes. Concept / Approach: Let the side of each small cube be s cm. For the cutting to be exact, s must divide each of 6, 12 and 15 exactly. Therefore s must be a common divisor of the three dimensions. To minimize the number of cubes, we want to maximize s. Hence s equals the greatest common divisor (gcd) of 6, 12 and 15. Once s is found, the number of cubes N is: N = (6 * 12 * 15) / s^3 Step-by-Step Solution: Step 1: Compute gcd of 6, 12 and 15. gcd(6, 12) = 6. gcd(6, 15) = 3. So overall gcd is 3. Step 2: Take cube side s = 3 cm. Step 3: Count how many cubes along each dimension. Along 6 cm side: 6 / 3 = 2 cubes. Along 12 cm side: 12 / 3 = 4 cubes. Along 15 cm side: 15 / 3 = 5 cubes. Step 4: Total number of cubes. N = 2 * 4 * 5 = 40. Thus, the least possible number of equal cubes is 40. Verification / Alternative check: Using the volume method: Volume of block = 6 * 12 * 15 = 1080 cubic centimetres. Volume of each cube with side 3 cm = 3^3 = 27 cubic centimetres. Number of cubes N = 1080 / 27 = 40. The result matches the earlier computation, confirming correctness. Why Other Options Are Wrong: Option A (30) suggests a smaller number of cubes, which would require a larger cube size than 3 cm that still divides all dimensions, but no such divisor exists. Option C (10) and Option D (20) correspond to still larger cube sizes and are not compatible with all three dimensions. Option E (60) corresponds to a smaller cube size, which would increase, not decrease, the number of cubes. Common Pitfalls: Some students mistakenly choose the smallest dimension as the cube edge, or use the least common multiple instead of the greatest common divisor. Others forget that the cube size must divide all three dimensions exactly. Always check divisibility and use the gcd to achieve the largest cube size and smallest count. Final Answer: The smallest possible number of equal cubes is 40.

Question 3

A hall is 15 m long and 12 m broad. The sum of the areas of the floor and the ceiling is equal to the sum of the areas of the four walls. Based on this condition, what is the volume of the hall, in cubic metres?

Accepted Answer

Introduction / Context: This is a mensuration problem involving a rectangular hall where we are given a relationship between the areas of the floor, ceiling and walls. The hall is assumed to be a rectangular cuboid. The statement that the sum of the floor area and the ceiling area equals the sum of the wall areas allows us to find the height. Once the height is known, we can calculate the volume. Such problems test conceptual understanding of surface area formulas for cuboids and the ability to set up and solve simple equations. Given Data / Assumptions: Length of the hall, l = 15 m. Breadth of the hall, b = 12 m. Height of the hall, h = unknown. Floor area = l * b. Ceiling area = l * b (same as floor). Wall areas: two walls of size l by h and two walls of size b by h. Condition: floor area + ceiling area = total area of four walls. Concept / Approach: For a rectangular hall: Floor area = l * b. Ceiling area = l * b. Total area of the four walls = 2 * l * h + 2 * b * h = 2 * h * (l + b). The given condition is: 2 * l * b = 2 * h * (l + b) From this we can solve for h. After finding h, volume V is: V = l * b * h Step-by-Step Solution: Step 1: Write the floor and ceiling area. Floor area = 15 * 12 = 180 square metres. Ceiling area = 180 square metres. Sum of floor and ceiling areas = 2 * 180 = 360 square metres. Step 2: Express wall areas. Total wall area = 2 * l * h + 2 * b * h = 2 * h * (15 + 12) = 54 * h. Step 3: Use given condition. 360 = 54 * h h = 360 / 54 = 20 / 3 metres. Step 4: Compute volume of the hall. V = l * b * h = 15 * 12 * 20 / 3. V = 180 * 20 / 3 = 60 * 20 = 1200 cubic metres. So the volume of the hall is 1200 cubic metres. Verification / Alternative check: We can recheck the condition using h = 20 / 3 metres. Wall area = 2 * 15 * 20 / 3 + 2 * 12 * 20 / 3 = 2 * 100 + 2 * 80 = 200 + 160 = 360 square metres. Floor plus ceiling area is 360 square metres as computed earlier. Hence the condition is satisfied, so the derived height and volume are correct. Why Other Options Are Wrong: Option A (720) corresponds to a smaller height that would not satisfy the wall condition. Option B (900) and Option D (1800) also lead to inconsistent wall area sums. Option E (600) assumes a height that is half of the correct height. Only 1200 cubic metres satisfies both the area relation and the volume formula. Common Pitfalls: Some learners mistakenly take only one of floor or ceiling areas in the equation instead of both. Others forget that there are two pairs of opposite walls leading to the factor 2 in the total wall area. Careful setup of the equality and correct arithmetic division are crucial to avoid errors. Final Answer: The volume of the hall is 1200 cubic metres.

Question 4

A cistern is 6 m long and 4 m wide and contains water up to a depth of 1 m 25 cm. Considering the bottom and the submerged walls, what is the total area of the wet surface of the cistern, in square metres?

Accepted Answer

Introduction / Context: This is a classic cistern or tank problem from mensuration. The wet surface area means the total area of the surfaces that are in contact with water. For an open rectangular cistern partly filled with water, the wet surface usually includes the bottom and the vertical walls up to the water level. Understanding which surfaces are wet and calculating their areas correctly is essential for solving this type of question. Given Data / Assumptions: Length of cistern, L = 6 m. Breadth of cistern, B = 4 m. Water depth, h = 1 m 25 cm = 1.25 m. The cistern is open at the top. Wet surfaces are the bottom and the four vertical walls up to height h. Concept / Approach: The wet surface area A wet is the sum of: Area of bottom = L * B. Area of two longer side walls = 2 * (L * h). Area of two shorter side walls = 2 * (B * h). Therefore: A wet = L * B + 2 * L * h + 2 * B * h We simply substitute the given values and compute. Step-by-Step Solution: Step 1: Compute area of bottom. Bottom area = L * B = 6 * 4 = 24 square metres. Step 2: Compute area of two longer side walls. One long wall area = L * h = 6 * 1.25 = 7.5 square metres. Two long walls area = 2 * 7.5 = 15 square metres. Step 3: Compute area of two shorter side walls. One short wall area = B * h = 4 * 1.25 = 5 square metres. Two short walls area = 2 * 5 = 10 square metres. Step 4: Sum all wet surfaces. Total wet area = bottom + long walls + short walls Total wet area = 24 + 15 + 10 = 49 square metres. Thus, the total wet surface area is 49 square metres. Verification / Alternative check: We can rewrite the formula in a compact form: A wet = L * B + 2 * h * (L + B). Substituting L = 6, B = 4, h = 1.25: L + B = 10, 2 * h * (L + B) = 2 * 1.25 * 10 = 25, and bottom area is 24. Therefore total area = 24 + 25 = 49 square metres, confirming our earlier calculation. Why Other Options Are Wrong: Option B (50) and Option C (53.5) are close but incorrect, likely stemming from rounding or arithmetic mistakes when multiplying or adding wall areas. Option D (55) arises if the depth is taken incorrectly or if an extra surface is mistakenly included. Option E (45) indicates that one or more surfaces, such as either pair of walls or the bottom, were omitted. Common Pitfalls: Students often forget to convert mixed units like 1 m 25 cm fully into metres. Another common error is to include the top surface, which is not in contact with water, or to miss one pair of walls. Being systematic about which surfaces are actually in contact with water avoids these mistakes. Final Answer: The total area of the wet surface of the cistern is 49 square metres.

Question 5

Each brick measures 25 cm by 11.25 cm by 6 cm. A wall has dimensions 8 m by 6 m by 22.5 cm (that is, thickness 22.5 cm). Assuming that there is no space for mortar, how many such bricks will be needed to build the wall completely?

Accepted Answer

Introduction / Context: This is a typical volume based problem involving bricks and walls. The question gives the dimensions of a single brick and the dimensions of a wall and asks how many bricks are required to construct the wall. The assumption is that there is no space for mortar and that the wall is solid. The key skill tested is unit conversion between metres and centimetres and division of volumes to count identical bricks. Given Data / Assumptions: Brick dimensions: 25 cm by 11.25 cm by 6 cm. Wall dimensions: 8 m by 6 m by 22.5 cm. Since 1 m = 100 cm, wall length = 800 cm, wall height = 600 cm, wall thickness = 22.5 cm. Wall is solid and there is no volume for mortar. All bricks are placed without gaps. Concept / Approach: The number of bricks required is given by: Number of bricks = Volume of wall / Volume of one brick We must ensure that all dimensions are in the same units before computing volumes. Since brick dimensions are in centimetres, we convert the wall dimensions to centimetres as well. Step-by-Step Solution: Step 1: Convert wall dimensions to centimetres. Wall length L = 8 m = 800 cm. Wall height H = 6 m = 600 cm. Wall thickness T = 22.5 cm. Step 2: Compute volume of the wall. Volume wall = L * H * T Volume wall = 800 * 600 * 22.5 cubic centimetres. Volume wall = 10,800,000 cubic centimetres. Step 3: Compute volume of one brick. Volume brick = 25 * 11.25 * 6 cubic centimetres. Volume brick = 1687.5 cubic centimetres. Step 4: Compute number of bricks. Number of bricks = 10,800,000 / 1687.5 Number of bricks = 6400. Thus, 6400 bricks are required to build the wall. Verification / Alternative check: We can simplify the division by factoring: 1687.5 = 16875 / 10. Then 10,800,000 / 1687.5 = (10,800,000 * 10) / 16875. Both numerator and denominator are divisible by 375, giving 288,000 / 750, which simplifies to 6400. This algebraic simplification confirms the numerical result. Why Other Options Are Wrong: Option A (5600) and Option D (7200) represent miscalculations usually caused by rounding the thickness or brick measurements improperly. Option B (600) and Option E (4800) are far too small and would correspond to an unrealistic wall volume relative to brick size. Only 6400 matches the precise volume ratio. Common Pitfalls: The main source of error is forgetting to convert metres to centimetres, leading to inconsistent units. Another frequent mistake is rounding 11.25 or 22.5 too early, which slightly changes the computed number of bricks. Always maintain exact decimals until the final division and ensure all three wall dimensions are included when computing volume. Final Answer: The number of bricks needed to build the wall is 6400.

Question 6

A cubical box has an edge of 1 metre. Small solid cubes each having an edge of 10 cm are to be placed inside the box so that it is completely filled without gaps. How many such small cubes can be put inside the box?

Accepted Answer

Introduction / Context: This question checks the understanding of relationships between units as well as volumes of cubes. The task is to determine how many smaller cubes of a given edge length can exactly fill a larger cubical box. It tests familiarity with converting metres to centimetres and the idea that the number of small cubes is the cube of the ratio of edge lengths, provided the packing is perfect and there are no gaps. Given Data / Assumptions: Edge of large cube (box) = 1 m. Edge of each small cube = 10 cm. 1 m = 100 cm. The box is completely filled with no gaps. Concept / Approach: First, express both edge lengths in the same unit. After that, the number of small cubes along one edge of the large cube equals the ratio of the large edge to the small edge. Since the solid is three dimensional, the total number of cubes is that ratio raised to the power 3. Equivalently, we can compute the volume of the large cube and divide by the volume of one small cube. Step-by-Step Solution: Step 1: Convert the large cube edge from metres to centimetres. Edge of large cube = 1 m = 100 cm. Edge of small cube = 10 cm. Step 2: Compute how many small cubes fit along one edge. Number along one edge = 100 / 10 = 10. Step 3: Use three dimensional count. Total small cubes = 10 * 10 * 10 = 10^3 = 1000. Thus, 1000 small cubes can be fitted into the box. Verification / Alternative check: Volume method: Volume of large cube = (1 m)^3 = 1 cubic metre. One cubic metre equals 100 cm * 100 cm * 100 cm = 1,000,000 cubic centimetres. Volume of each small cube = 10^3 = 1000 cubic centimetres. Number of cubes = 1,000,000 / 1000 = 1000. This matches the earlier result. Why Other Options Are Wrong: Option A (10) and Option B (100) ignore one or two of the three dimensions respectively. Option D (10000) corresponds to 10^4 and would be correct only if there were an extra dimension, which is not possible. Option E (500) is an arbitrary wrong value. Only 1000 is consistent with the geometry and unit conversion. Common Pitfalls: Some candidates forget to convert metres to centimetres and directly divide 1 by 10, giving 0.1 as a misleading ratio. Others compute surface area instead of volume or count along only one dimension. Always use consistent units and remember that a cube is three dimensional, so the number of small cubes is the cube of the linear ratio. Final Answer: A total of 1000 cubes of edge 10 cm can be placed inside the cubical box of edge 1 metre.

Question 7

The total surface area of a cube is 1734 square centimetres. Find the volume of the cube in cubic centimetres.

Accepted Answer

Introduction / Context: Here we have a cube whose total surface area is given and we are required to find its volume. This connects two standard formulas of a cube: the formula for surface area and the formula for volume. The problem tests the ability to move from area to side length and then from side length to volume, which is a common sequence in aptitude questions on three dimensional geometry. Given Data / Assumptions: Total surface area of the cube, S = 1734 square centimetres. Let side of the cube be a centimetres. Cube is solid and all edges are equal. Concept / Approach: For a cube with side length a: Total surface area S = 6 * a^2. Volume V = a^3. We are given S and need V. So we first solve 6 * a^2 = 1734 to find a, and then compute a^3. Step-by-Step Solution: Step 1: Use surface area formula. S = 6 * a^2 = 1734. Step 2: Solve for a^2. a^2 = 1734 / 6. a^2 = 289. Step 3: Take square root. a = square root of 289 = 17 centimetres. Step 4: Compute volume of the cube. V = a^3 = 17^3. V = 17 * 17 * 17 = 289 * 17 = 4913 cubic centimetres. Thus, the volume of the cube is 4913 cubic centimetres. Verification / Alternative check: We can verify by reversing the steps. If side length a = 17 cm, then surface area S = 6 * 17^2 = 6 * 289 = 1734 square centimetres, which matches the given value. This confirms that a has been calculated correctly and thus the volume is correct as well. Why Other Options Are Wrong: Option A (2334 cubic cm) and Option B (3356 cubic cm) do not correspond to any cube with an integer side that satisfies the given surface area. Option D (3478 cubic cm) is not equal to 17^3 and is not consistent with the surface area. Option E (4096 cubic cm) is 16^3, which would imply a cube with side 16 cm and surface area 6 * 256 = 1536 square centimetres, not 1734. Common Pitfalls: A common error is to forget that the surface area formula includes a factor of 6 and to equate a^2 directly with 1734, which is incorrect. Others may compute a^2 correctly but make arithmetic mistakes when cubing 17. It is also important to keep units consistent and recognize that we are moving from area units to volume units, which explains the change from squared to cubed centimetres. Final Answer: The volume of the cube is 4913 cubic centimetres.

Question 8

A wall has a volume such that it is 5 times as high as it is broad and 8 times as long as it is high. If the volume of the wall is 12.8 cubic metres, what is the breadth of the wall, in centimetres?

Accepted Answer

Introduction / Context: This problem tests the understanding of proportional dimensions and volume in a three dimensional object. The wall has its height expressed in terms of breadth and its length expressed in terms of height. With the ratios between the three dimensions and the total volume, we must deduce the actual breadth. Such problems are common in aptitude exams where relationships between dimensions are expressed in words instead of directly giving all numerical values. Given Data / Assumptions: Let breadth of the wall be b metres. Height is 5 times breadth, so height = 5b metres. Length is 8 times height, so length = 8 * (5b) = 40b metres. Volume of the wall V = 12.8 cubic metres. Wall is rectangular in shape. Concept / Approach: Volume of a rectangular solid is given by: V = length * breadth * height Here each dimension is in terms of b, so we substitute and form an equation in b. After solving for b, we convert the result from metres to centimetres because options are in centimetres. Step-by-Step Solution: Step 1: Express all dimensions in terms of b. Breadth = b metres. Height = 5b metres. Length = 40b metres. Step 2: Write volume equation. V = length * breadth * height = 40b * b * 5b. V = 200 * b^3. Given volume V = 12.8 cubic metres. So, 200 * b^3 = 12.8. Step 3: Solve for b^3. b^3 = 12.8 / 200 = 0.064. Step 4: Take cube root. b = cube root of 0.064. Since 0.4^3 = 0.064, b = 0.4 metres. Step 5: Convert to centimetres. b = 0.4 metres = 0.4 * 100 = 40 centimetres. Thus, the breadth of the wall is 40 cm. Verification / Alternative check: We can recompute the volume with b = 0.4 metres. Height = 5 * 0.4 = 2 metres. Length = 8 * 2 = 16 metres. Volume V = 16 * 0.4 * 2 = 12.8 cubic metres, which matches the given volume exactly, confirming that the breadth is indeed 0.4 metres or 40 centimetres. Why Other Options Are Wrong: Option B (30 cm), Option C (20 cm), and Option D (10 cm) would all produce different volumes when substituted back into the dimensional relationships and thus would not equal 12.8 cubic metres. Option E (50 cm) similarly leads to an incorrect volume. Only 40 cm yields the correct given volume. Common Pitfalls: A common error is to misinterpret the phrase 8 times as long as it is high and mistakenly multiply directly by 8 twice, or confuse which dimension is which. Another pitfall is forgetting to convert metres to centimetres at the end, which would result in choosing an option expressed in metres rather than centimetres. Careful reading and unit conversion are crucial. Final Answer: The breadth of the wall is 40 cm.

Question 9

The diagonal of a rectangle is equal to square root of 41 centimetres and its area is 20 square centimetres. What is the perimeter of the rectangle, in centimetres?

Accepted Answer

Introduction / Context: This problem involves a rectangle for which the diagonal and area are given. We are required to find the perimeter. This connects three basic relationships: the Pythagoras relation between sides and diagonal, the formula for area, and the formula for perimeter. Problems of this type test the ability to translate given data into equations and solve them systematically, often using algebraic identities. Given Data / Assumptions: Let the sides of the rectangle be a centimetres and b centimetres. Diagonal = square root of 41 centimetres, so diagonal squared = 41. Area of rectangle = 20 square centimetres, so a * b = 20. Need perimeter P = 2 * (a + b). Concept / Approach: For a rectangle: By Pythagoras theorem: a^2 + b^2 = diagonal^2. Given diagonal^2 = 41, so a^2 + b^2 = 41. Given area: a * b = 20. We recall identity: (a + b)^2 = a^2 + b^2 + 2ab. Using this, we can directly find a + b once we know a^2 + b^2 and ab. Then we compute the perimeter as 2 * (a + b). Step-by-Step Solution: Step 1: Use diagonal information. a^2 + b^2 = 41. Step 2: Use area information. ab = 20. Step 3: Use identity for (a + b)^2. (a + b)^2 = a^2 + b^2 + 2ab. (a + b)^2 = 41 + 2 * 20 = 41 + 40 = 81. Step 4: Find a + b. a + b = square root of 81 = 9. Step 5: Compute perimeter. Perimeter P = 2 * (a + b) = 2 * 9 = 18 centimetres. So, the perimeter of the rectangle is 18 centimetres. Verification / Alternative check: For completeness, we can find actual values of a and b by solving quadratic equations, but that is not necessary. For instance, suppose a and b are roots of the quadratic x^2 - (a + b)x + ab = 0. That is x^2 - 9x + 20 = 0. The roots are x = 4 and x = 5. So sides are 4 and 5. Check: area = 4 * 5 = 20, diagonal^2 = 4^2 + 5^2 = 16 + 25 = 41, diagonal = square root of 41. Perimeter = 2 * (4 + 5) = 18, which matches our answer. Why Other Options Are Wrong: Option A (9 cm) is just a + b, not twice that value. Option C (20 cm) is the area, not the perimeter. Option D (41 cm) is diagonal squared, and Option E (16 cm) might come from confusing a side with the perimeter. None of these satisfy all given conditions together. Common Pitfalls: A common mistake is to compute a and b separately without using the identity, which increases the chance of algebra errors. Others may confuse diagonal with its square, leading to wrong equations. Remember the useful shortcut that when both a^2 + b^2 and ab are known, the sum a + b is found easily using (a + b)^2 = a^2 + b^2 + 2ab. Final Answer: The perimeter of the rectangle is 18 centimetres.

Question 10

Three solid cubes with edge lengths 1 cm, 6 cm and 8 cm are melted together to form a single new solid cube. What is the total surface area of the cube so formed, in square centimetres?

Accepted Answer

Introduction / Context: This question combines the ideas of volume conservation and surface area of a cube. When several solid cubes are melted and recast into a new cube, the total volume of the metal remains the same but the surface area changes. We must compute the combined volume of the original cubes, then find the side length of the new cube, and finally compute its surface area. Such problems are common in aptitude exams to test spatial reasoning and the use of formulas in multiple steps. Given Data / Assumptions: Edge length of first cube = 1 cm. Edge length of second cube = 6 cm. Edge length of third cube = 8 cm. All three cubes are melted and recast into one new cube. No loss of material is assumed. Concept / Approach: For a cube with side a: Volume V = a^3. Total surface area S = 6 * a^2. Since volume is conserved when the cubes are melted and recast, we first find the sum of the volumes of the three cubes. Let that sum be V total. If the side of the new cube is A, then: A^3 = V total Then the new surface area S new is: S new = 6 * A^2 Step-by-Step Solution: Step 1: Compute volumes of the three cubes. V1 = 1^3 = 1 cubic centimetre. V2 = 6^3 = 216 cubic centimetres. V3 = 8^3 = 512 cubic centimetres. Step 2: Compute total volume. V total = 1 + 216 + 512 = 729 cubic centimetres. Step 3: Find side of new cube. Let side of new cube be A. A^3 = 729. Since 9^3 = 729, A = 9 centimetres. Step 4: Compute surface area of new cube. S new = 6 * A^2 = 6 * 9^2 = 6 * 81 = 486 square centimetres. So, the surface area of the new cube is 486 square centimetres. Verification / Alternative check: We can cross check by factoring 729 as 9 * 81 and recognizing that 81 is 9^2, confirming that 729 is indeed 9^3. Also, note that if we had chosen a different side, such as 8 or 10, the volume would have been 512 or 1000 respectively, which are not equal to 729. Thus, 9 cm is the unique correct side length. Why Other Options Are Wrong: Option B (586), Option C (686) and Option D (786) all correspond to incorrect side lengths and would not give a cube volume of 729 cubic centimetres. Option E (400) would arise if someone used side 8 or another incorrect value and misapplied the surface area formula. Only 486 corresponds to the correctly computed side length of 9 cm. Common Pitfalls: A common mistake is to sum the edges and not the volumes, or to average the sides and treat that as the new side length. Another error is to forget that the new cube side must be the cube root of total volume, not just the square root. Finally, some students incorrectly assume that surface area is conserved, which is not true when shapes change while volume is preserved. Final Answer: The surface area of the cube formed is 486 square centimetres.

Question 11

A town has a population of 4000 people, and each person requires 140 litres of water per day. There is a rectangular water tank measuring 18 m by 12 m by 8 m, which is completely full. For how many whole days will the water in this tank be sufficien…

Accepted Answer

Introduction / Context: This question is from the application of volume to real life situations involving water storage and consumption. We are given the capacity of a tank and the daily water requirement of a population. The main steps are to convert the volume of the tank into litres and then divide by the daily consumption to find the number of days for which the supply will last. Such problems are common for practising unit conversions between cubic metres and litres. Given Data / Assumptions: Population of the town = 4000 people. Per head daily water requirement = 140 litres. Tank dimensions: 18 m by 12 m by 8 m. Tank is completely full initially. Water usage is uniform and only for this population. Concept / Approach: First, compute the volume of the tank in cubic metres. Then convert cubic metres to litres using: 1 cubic metre = 1000 litres Next, compute the total daily water consumption by multiplying population and per head requirement. Finally, divide the tank capacity in litres by daily consumption to obtain the number of days the water will last. Step-by-Step Solution: Step 1: Compute volume of tank in cubic metres. Volume tank = length * breadth * height Volume tank = 18 * 12 * 8 cubic metres. Volume tank = 1728 cubic metres. Step 2: Convert volume to litres. Capacity in litres = 1728 * 1000 = 1,728,000 litres. Step 3: Compute daily water requirement. Daily requirement per person = 140 litres. Total daily requirement = 4000 * 140 litres. Total daily requirement = 560,000 litres. Step 4: Compute number of days. Number of days = 1,728,000 / 560,000. Number of days = 3.0857 approximately. Considering whole days of full requirement, water will suffice for 3 full days. Verification / Alternative check: If we supply water for 3 days, total water consumed = 3 * 560,000 = 1,680,000 litres. Remaining water = 1,728,000 - 1,680,000 = 48,000 litres, which is less than one full day of requirement. Hence three days are possible, but a full fourth day is not. This confirms that the correct answer in whole days is 3. Why Other Options Are Wrong: Option B (4 days) would require 4 * 560,000 = 2,240,000 litres, which exceeds the tank capacity. Option C (5 days) and Option E (6 days) are even larger overestimates. Option D (2 days) underestimates the supply duration because two days require only 1,120,000 litres and there will still be enough water left for more than one additional day. Common Pitfalls: Students may forget the conversion factor between cubic metres and litres or may miscalculate the total daily requirement by missing a zero. Some also round the days incorrectly or interpret 3.08 as 4 instead of taking only complete days of supply. Always convert units carefully and think about whether the question asks for exact days or whole days. Final Answer: The water in the tank will be sufficient for 3 days for the town population.

Question 12

On a certain religious day, 50 men take a dip in a rectangular water tank that is 40 m long and 20 m broad. If each man displaces an average of 4 cubic metres of water, by how much will the water level in the tank rise, in centimetres?

Accepted Answer

Introduction / Context: This question involves the principle that when bodies are immersed in water, they displace water equal to their own volume. The displaced water causes the water level in the container to rise. Here, a group of men are taking a dip in a rectangular tank, and each man displaces a known volume of water. We must compute how much the water level rises. This problem is a straightforward application of volume conservation and the formula for the volume of a rectangular prism. Given Data / Assumptions: Number of men = 50. Average volume of water displaced by each man = 4 cubic metres. Tank length L = 40 m. Tank breadth B = 20 m. Rise in water level is uniform across the tank. Concept / Approach: Total volume of water displaced by all men equals the volume corresponding to the rise in water level in the tank. For a rectangular tank, volume increase due to rise in level h is: Volume rise = L * B * h Total displaced volume V disp is: V disp = number of men * volume displaced per man Equating these volumes gives L * B * h = V disp, from which we solve for h. Finally, we convert h from metres to centimetres. Step-by-Step Solution: Step 1: Compute total displaced volume. V disp = 50 * 4 = 200 cubic metres. Step 2: Express volume rise in terms of h. Volume rise = L * B * h = 40 * 20 * h. Volume rise = 800 * h cubic metres. Step 3: Equate displaced volume with volume rise. 800 * h = 200. h = 200 / 800 = 1 / 4 metre. Step 4: Convert h to centimetres. h = 1 / 4 metre = 0.25 metre. 0.25 metre = 0.25 * 100 = 25 centimetres. So, the water level rises by 25 centimetres. Verification / Alternative check: We can check that 25 cm is reasonable. The base area of the tank is 40 * 20 = 800 square metres. Multiplying this by a rise of 0.25 metre gives a volume of 800 * 0.25 = 200 cubic metres, which equals the total displaced volume. Thus the calculation is consistent. Why Other Options Are Wrong: Option A (20 cm) would correspond to a rise of 0.20 metre, which would only add 160 cubic metres of water, less than the displaced 200 cubic metres. Option C (35 cm) and Option D (50 cm) correspond to rises that would yield volumes larger than 200 cubic metres. Option E (10 cm) is much too small and would increase the volume by only 80 cubic metres. Common Pitfalls: The most common mistakes are forgetting to convert metres to centimetres or miscalculating the total displaced volume by mistakenly multiplying by 10 or missing a zero. Another error is to assume the tank is square instead of rectangular, using the wrong base area. Always compute the total displaced volume first, then divide by the correct base area to find the height of the water rise. Final Answer: The water level in the tank will rise by 25 centimetres.

Question 13

A solid wooden block has dimensions 5 cm by 10 cm by 20 cm. Identical blocks are to be arranged to form a solid wooden cube of the smallest possible size, with no gaps. How many such blocks are required to construct this cube?

Accepted Answer

Introduction / Context: This problem involves assembling a three dimensional cube from identical rectangular blocks. Each block has dimensions 5 cm by 10 cm by 20 cm. Our goal is to construct the smallest possible cube whose dimensions are compatible with these blocks. The key idea is that the cube side must be a common multiple of the three given dimensions, and the arrangement of blocks must fill the cube exactly with whole blocks, no cutting and no gaps. This is a typical puzzle style question that tests spatial reasoning and understanding of least common multiples in three dimensions. Given Data / Assumptions: Block dimensions: 5 cm, 10 cm and 20 cm. We can orient blocks in any direction, but they must fit exactly. The final solid is a cube, so all three sides of the final shape must be equal. We want the cube of minimum possible size. Concept / Approach: Let the side of the cube be L centimetres. For the cube to be filled exactly, along each of its three dimensions, we must be able to fit an integer number of block edges, possibly matching 5, 10 or 20 along each direction depending on orientation. The smallest such cube will have side length equal to the least common multiple of 5, 10 and 20. Once L is known, we compute the volume of the cube, divide by the volume of one block, and obtain the number of blocks required. Step-by-Step Solution: Step 1: Find the least common multiple (LCM) of the block edges. Edges are 5, 10 and 20 centimetres. LCM of 5, 10 and 20 is 20 centimetres. Thus the smallest possible cube side length L = 20 centimetres. Step 2: Arrange blocks to fit the cube. We can orient each block so that its longest side (20 cm) lies along one cube edge. Then along that edge, only one block fits: 20 / 20 = 1. The 10 cm side can align with another cube edge, giving 20 / 10 = 2 blocks along that direction. The 5 cm side can align with the remaining cube edge, giving 20 / 5 = 4 blocks along that direction. Step 3: Compute number of blocks. Number of blocks along three dimensions: 1, 2 and 4. Total number of blocks = 1 * 2 * 4 = 8. Thus 8 blocks are required to form the smallest solid cube. Verification / Alternative check: Volume of each block = 5 * 10 * 20 = 1000 cubic centimetres. Volume of final cube with side 20 cm = 20^3 = 8000 cubic centimetres. Number of blocks = 8000 / 1000 = 8, which confirms our earlier count. The cube is completely filled without gaps, and the side length is indeed the smallest multiple that accommodates all block dimensions. Why Other Options Are Wrong: Option A (6), Option C (12), Option D (16), and Option E (24) correspond to numbers of blocks that would either not make a perfect cube or require a cube with a side larger than 20 cm. None of these counts match the exact volume ratio or a consistent integer arrangement in three dimensions for the smallest cube. Common Pitfalls: Some learners attempt to use the greatest common divisor instead of the least common multiple or forget that the final shape must be a cube. Others might try to force a smaller cube, which is impossible given the dimensions. The safest method is to use the least common multiple to determine the cube side and then verify by volume division. Final Answer: The minimum number of such blocks required to form a solid cube is 8.

Question 14

A cistern has a capacity of 8000 litres and its external dimensions are 3.3 m × 2.6 m × 1.1 m. The thickness of each of the four vertical walls is 5 cm. Assuming the top is open, what is the thickness (in centimetres) of the bottom slab of the ciste…

Accepted Answer

Introduction / Context: In this aptitude question on volume and surface area, we are dealing with a rectangular cistern (tank) whose external dimensions and wall thickness are known. We are also given its internal capacity in litres. The goal is to determine the thickness of the bottom slab. This type of question tests understanding of the relationship between internal and external dimensions, unit conversion between litres and cubic centimetres, and basic volume formulas for cuboids. Given Data / Assumptions: Capacity of the cistern = 8000 litres. External dimensions: length = 3.3 m, breadth = 2.6 m, height = 1.1 m. Thickness of each vertical wall = 5 cm. Cistern is open at the top, so only the bottom slab thickness reduces the internal height. 1 litre = 1000 cubic centimetres. Concept / Approach: To find the bottom thickness, first convert everything into centimetres. Then compute the internal length and internal breadth by subtracting twice the wall thickness from the external dimensions. Let the bottom thickness be t cm. The internal height will then be (external height − t). The volume of the inner hollow region must be equal to the given capacity, converted into cubic centimetres. We then solve the resulting linear equation for t. Step-by-Step Solution: Convert external dimensions to cm: 3.3 m = 330 cm, 2.6 m = 260 cm, 1.1 m = 110 cm. Internal length = 330 − 2 * 5 = 320 cm. Internal breadth = 260 − 2 * 5 = 250 cm. Let bottom thickness be t cm, so internal height = 110 − t cm. Capacity = 8000 litres = 8000 * 1000 = 8000000 cm^3. Volume of cistern interior = 320 * 250 * (110 − t). So 320 * 250 * (110 − t) = 8000000. 320 * 250 = 80000, so 80000 * (110 − t) = 8000000. Divide both sides by 80000: 110 − t = 100. Therefore t = 110 − 100 = 10 cm. Verification / Alternative check: Substitute t = 10 cm back: internal height = 100 cm. Then interior volume = 320 * 250 * 100 = 8000000 cm^3, which equals 8000 litres. This confirms that the computed bottom thickness is consistent with the given capacity of the cistern. Why Other Options Are Wrong: 5 cm would give internal height 105 cm and a larger volume than 8000 litres. 15 cm and 20 cm would give smaller internal heights and capacities less than 8000 litres. 90 cm is far too large as a bottom thickness and is unrealistic for this geometry, so only 10 cm satisfies the volume condition exactly. Common Pitfalls: A common mistake is to forget to convert litres into cubic centimetres or to subtract the wall thickness from both sides when finding internal length and breadth. Another frequent error is to subtract the wall thickness from the height twice, even though there is only one bottom slab and the top is open. Some students also try to use metres for some dimensions and centimetres for others, which leads to unit inconsistency and wrong answers. Final Answer: Thus, the thickness of the bottom slab of the cistern is 10 cm.

Question 15

In a rectangular room measuring 12 m in length, 8 m in breadth and 9 m in height, what is the length of the longest pole that can be placed inside the room from one corner to the opposite corner?

Accepted Answer

Introduction / Context: This problem is a classic application of three-dimensional Pythagoras theorem. You are asked to find the longest pole that can fit inside a rectangular room. That longest pole will lie along the space diagonal of the cuboid formed by the room, connecting one vertex to the opposite vertex. The question tests your understanding of distances in 3D geometry and how to extend the Pythagorean theorem from two to three dimensions. Given Data / Assumptions: Length of the room, l = 12 m. Breadth (width) of the room, b = 8 m. Height of the room, h = 9 m. The room is rectangular in all three dimensions. The longest pole lies along the space diagonal of this rectangular box. Concept / Approach: In a cuboid (rectangular box) with sides l, b and h, the length of the space diagonal d is given by the formula: d = sqrt(l^2 + b^2 + h^2) This comes from applying Pythagoras theorem twice: first to find the diagonal of the base rectangle, then to combine that with the height. Once we compute d using the given dimensions, we choose the matching option in metres. Step-by-Step Solution: Compute l^2 = 12^2 = 144. Compute b^2 = 8^2 = 64. Compute h^2 = 9^2 = 81. Sum of squares = 144 + 64 + 81 = 289. Space diagonal d = sqrt(289). Since 289 = 17^2, d = 17 m. Verification / Alternative check: We can cross-check by remembering that 17 is part of the Pythagorean triple 8–15–17. Here the numbers are different, but the square sum 289 exactly equals 17^2, so there is no rounding. Also, among the options, 17 m is the only value consistent with the exact square root of 289. Therefore, the calculation is reliable and the pole length of 17 m is correct. Why Other Options Are Wrong: Values 14 m, 15 m, and 16 m all correspond to smaller squared lengths: 196, 225 and 256 respectively, which do not match 289. A value like 18 m would give a squared length of 324, which is greater than 289 and does not describe the space diagonal for these dimensions. Hence none of these other options fits the formula for the diagonal of this room. Common Pitfalls: A common error is to compute only the diagonal of the floor (using 12 m and 8 m) and forget to include the height, which gives a shorter length than the true space diagonal. Another mistake is mixing up units, but here everything is in metres, so consistent usage avoids that issue. Students should also remember to square and add correctly before taking the square root. Final Answer: The length of the longest pole that can be placed inside the room is 17 m.

Question 16

During a uniform rain shower, 5 cm of rain falls on level ground. What is the total volume of water, in cubic metres, that falls on a field of area 1.5 hectares?

Accepted Answer

Introduction / Context: This question links basic geometry with units used in real-life measurement of rainfall. When a certain depth of rain falls uniformly over a region, the total volume of water can be found by multiplying the area of the region by the rainfall depth, provided both are expressed in consistent units. The problem tests familiarity with hectares, conversion between centimetres and metres, and the simple formula for volume involving area and height (or depth). Given Data / Assumptions: Rainfall depth = 5 cm. Area of the field = 1.5 hectares. 1 hectare = 10,000 square metres. We assume the rain falls uniformly over the entire area. We must express the final answer in cubic metres. Concept / Approach: Volume is given by: Volume = Area * Depth To use this formula correctly, both area and depth must be in compatible units (for example, square metres and metres). Therefore, we convert hectares to square metres and centimetres to metres. Then we multiply the area by the converted depth to obtain the volume in cubic metres. Step-by-Step Solution: Convert area: 1 hectare = 10000 m^2, so 1.5 hectares = 1.5 * 10000 = 15000 m^2. Convert depth: 5 cm = 5 / 100 = 0.05 m. Use Volume = Area * Depth. Volume = 15000 m^2 * 0.05 m. Volume = 750 m^3. Verification / Alternative check: We can estimate to check reasonableness. A depth of 0.05 m is quite small compared to 1 m, and the area is 15000 m^2. If a depth of 1 m fell, the volume would be 15000 m^3. Because the actual depth is 1/20 of a metre, the volume should be roughly 15000 / 20 = 750 m^3. This confirms the exact calculation. Why Other Options Are Wrong: 75 m^3 is ten times too small and would correspond to incorrectly using 0.005 m as the depth. 7500 m^3 and 75000 m^3 are too large and would result from misusing the area or depth conversion. 500 m^3 is also not consistent with the proportional reasoning above and would correspond to an incorrect intermediate step in the calculation. Common Pitfalls: Students often forget to convert centimetres to metres and directly multiply by 5 instead of 0.05. Some also confuse hectares with square kilometres or square metres, leading to major errors. Another common mistake is to convert the area correctly but still report the final answer in cubic centimetres instead of cubic metres. Keeping unit conversions clear at each step helps avoid these errors. Final Answer: The total volume of water that falls on the field is 750 cubic metres.

Question 17

Two right circular cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. What is the ratio of their volumes?

Accepted Answer

Introduction / Context: This problem checks understanding of how the volume of a right circular cone depends on its dimensions. Instead of giving specific numerical values for height and radius, the question provides ratios. We must use the formula for the volume of a cone and manipulate these ratios to find the ratio of the volumes. This kind of question is common in aptitude exams, because it tests formula knowledge and algebraic reasoning without heavy computation. Given Data / Assumptions: Height ratio of the two cones: h1 : h2 = 1 : 3. Radius ratio of the two bases: r1 : r2 = 3 : 1. We are asked for the ratio V1 : V2 of their volumes. Both cones are right circular cones. Concept / Approach: The volume V of a right circular cone is given by: V = (1/3) * pi * r^2 * h For two cones, we can write their volumes using proportional radii and heights. Because only the ratio is asked, we do not need actual numeric values of pi or the constant 1/3, as these factors will cancel when forming the ratio V1 : V2. The key idea is that volume is directly proportional to r^2 and to h. Step-by-Step Solution: Let h1 = 1k and h2 = 3k for some positive constant k. Let r1 = 3m and r2 = 1m for some positive constant m. Then V1 is proportional to r1^2 * h1 = (3m)^2 * (1k) = 9m^2 * k. V2 is proportional to r2^2 * h2 = (1m)^2 * (3k) = 3m^2 * k. Thus, V1 : V2 = 9m^2*k : 3m^2*k = 9 : 3 = 3 : 1. Verification / Alternative check: We can choose simple numeric values: take h1 = 1, h2 = 3, r1 = 3, r2 = 1. Then V1 = (1/3)*pi*3^2*1 = 3pi. V2 = (1/3)*pi*1^2*3 = pi. So V1 : V2 = 3pi : pi = 3 : 1, which matches the derived ratio. This confirms that the algebraic reasoning is correct. Why Other Options Are Wrong: Ratios such as 1 : 3 or 2 : 1 confuse the roles of height and radius in the volume formula. Ratios 4 : 1 and 5 : 1 correspond to incorrect combinations of the squared radius and height. Because both dimensions change and the radius is squared, small mistakes in combining the ratios quickly give the wrong answer. Common Pitfalls: A common error is to treat volume as directly proportional to r * h instead of r^2 * h, forgetting the square on the radius. Another pitfall is misreading the ratios and reversing them, leading to an inverted answer. Students should carefully write intermediate expressions and remember that constants like (1/3)*pi cancel when forming a ratio. Final Answer: The ratio of the volumes of the two cones is 3 : 1.

Question 18

A hollow iron pipe is 21 cm long and has an external diameter of 8 cm. The thickness of the pipe is 1 cm and the density of iron is 8 g per cubic centimetre. What is the weight of the pipe in kilograms?

Accepted Answer

Introduction / Context: This question combines geometry (volume of a hollow cylinder) with basic physics (mass = density * volume). The pipe is modelled as a hollow circular cylinder, where we know the external diameter, the thickness, the length, and the density of iron. We must first compute the volume of iron used and then convert that volume into mass in kilograms. This type of problem is common in aptitude tests, engineering entrance exams and technical interviews. Given Data / Assumptions: Length of the pipe, h = 21 cm. External diameter = 8 cm, so external radius R = 4 cm. Thickness of pipe = 1 cm, so internal radius r = 4 − 1 = 3 cm. Density of iron = 8 g per cubic centimetre. The pipe is hollow only along its axis; the ends are open. Concept / Approach: The iron occupies the region between the outer cylinder and the inner cylinder. So the volume of iron is the difference between the volumes of two cylinders with the same height but different radii. The volume V of a cylinder is: V = pi * radius^2 * height Hence, for the pipe: V = pi * (R^2 − r^2) * h. Once we know V in cubic centimetres, mass in grams is density multiplied by volume, and then we convert grams to kilograms. Step-by-Step Solution: External radius R = 8 / 2 = 4 cm. Internal radius r = 4 − 1 = 3 cm. Height h = 21 cm. Volume of iron V = pi * (R^2 − r^2) * h. Compute R^2 − r^2 = 4^2 − 3^2 = 16 − 9 = 7. So V = pi * 7 * 21 = 147pi cubic centimetres. Mass in grams = density * volume = 8 * 147pi = 1176pi g. Using pi ≈ 3.14, mass ≈ 1176 * 3.14 ≈ 3696 g. Convert to kilograms: 3696 g = 3.696 kg. Verification / Alternative check: We can quickly approximate: R^2 − r^2 = 7, V ≈ 147 * 3.14 ≈ 462 cm^3. Multiplying by density 8 gives about 3696 g, which fits well within typical rounding. Among the options, 3.696 kg matches this computed value exactly, so the answer is consistent and precise. Why Other Options Are Wrong: 3.60 kg and 3.20 kg are too low and would result from approximating pi very roughly or miscalculating the difference of squares. Values like 36.0 kg and 36.9 kg are unrealistically large given the small dimensions of the pipe, and they would result from errors such as using centimetres as metres without proper conversion or mistakenly multiplying by 10 instead of converting to kilograms correctly. Common Pitfalls: A frequent mistake is to use the external radius alone, ignoring the hollow inner part, thereby overestimating the volume. Another common error is forgetting to convert the final mass into kilograms, leaving the answer in grams. Some students also calculate R − r instead of R^2 − r^2, which is incorrect for volume differences. Final Answer: The weight of the hollow iron pipe is 3.696 kg (approximately).

Question 19

A right circular cone has slant height 10 m and vertical height 8 m. Find the curved surface area of the cone in square metres.

Accepted Answer

Introduction / Context: This question deals with the geometry of a right circular cone. You are given the slant height and the vertical height, and you are asked to find the curved surface area. The curved surface area formula uses the radius and slant height, so we must first determine the radius using Pythagoras theorem applied to the right triangle formed by the radius, height and slant height. This is a standard problem that checks both geometric understanding and formula usage. Given Data / Assumptions: Slant height of the cone, l = 10 m. Vertical (perpendicular) height of the cone, h = 8 m. The cone is a right circular cone. We must find curved surface area (CSA) in square metres. Concept / Approach: For a right circular cone, we use two main formulas: l^2 = r^2 + h^2 (from Pythagoras theorem) Curved surface area = pi * r * l First, we compute the radius r from the relation between slant height, height and radius. Once r is known, we substitute into the curved surface area formula and simplify to get the final expression in terms of pi. Step-by-Step Solution: Given l = 10 m, h = 8 m. From Pythagoras theorem: l^2 = r^2 + h^2. 10^2 = r^2 + 8^2, so 100 = r^2 + 64. Therefore r^2 = 100 − 64 = 36. Thus r = sqrt(36) = 6 m. Curved surface area = pi * r * l = pi * 6 * 10. Curved surface area = 60 pi square metres. Verification / Alternative check: We can verify by approximate numeric value. Using pi ≈ 3.14, CSA ≈ 60 * 3.14 ≈ 188.4 m^2. A cone with radius 6 m and slant height 10 m should indeed have an area of around 190 m^2, which feels reasonable for these dimensions. None of the other options, when multiplied by pi, matches the correct product of 6 and 10. Why Other Options Are Wrong: 40 pi, 50 pi, 70 pi and 80 pi sq.m correspond to either using incorrect values of radius or slant height, or miscalculating the Pythagoras relation. For example, 40 pi could arise from mistakenly using r = 4 m, and 80 pi might come from incorrectly multiplying 8 and 10 without first verifying the radius. Only 60 pi comes from the correct combination of r = 6 m and l = 10 m. Common Pitfalls: Students sometimes confuse height and slant height, using the height directly in the CSA formula instead of the slant height. Another error is to forget to take the square root when computing the radius from r^2. It is also common to mistakenly use the total surface area formula, which includes the base area, instead of the curved surface area formula required here. Final Answer: The curved surface area of the cone is 60 pi sq.m.

Question 20

For a right circular cylinder, the radius is increased by 20% and the height is increased by 15%. By what percentage does the curved surface area of the cylinder increase?

Accepted Answer

Introduction / Context: This question tests percentage change concepts applied to geometry. The curved surface area of a cylinder depends on both its radius and height. Here, both radius and height are increased by given percentages, and we are asked to find the resulting overall percentage increase in curved surface area. This requires understanding of how multiplicative changes combine and avoiding the mistake of simply adding percentage increases directly. Given Data / Assumptions: Initial radius of the cylinder = r. Initial height of the cylinder = h. Radius increases by 20%, so new radius = 1.20r. Height increases by 15%, so new height = 1.15h. We are concerned with the curved surface area (CSA). Concept / Approach: The curved surface area of a cylinder is given by: CSA = 2 * pi * r * h If r and h change to r′ and h′, then new CSA is: CSA′ = 2 * pi * r′ * h′ The ratio CSA′ / CSA equals (r′ / r) * (h′ / h). After finding this ratio, we convert it into percentage increase by subtracting 1 and multiplying by 100%. Step-by-Step Solution: New radius r′ = 1.20r. New height h′ = 1.15h. Original CSA = 2 * pi * r * h. New CSA = 2 * pi * (1.20r) * (1.15h). CSA′ = 2 * pi * r * h * (1.20 * 1.15). Compute factor: 1.20 * 1.15 = 1.38. Thus, CSA′ / CSA = 1.38. Percentage increase = (1.38 − 1) * 100% = 0.38 * 100% = 38%. Verification / Alternative check: We can check with simple values: let r = 10 and h = 10. Then original CSA = 2 * pi * 10 * 10 = 200pi. New r = 12, new h = 11.5, so new CSA = 2 * pi * 12 * 11.5 = 276pi. The ratio 276 / 200 = 1.38, again giving 38% increase. This numerical check confirms our percentage calculation. Why Other Options Are Wrong: 33% or 35% would result from averaging or incorrectly adding the individual percentage changes instead of multiplying the scale factors. 41% and 44% are too high and would imply a larger combined factor than 1.38. The correct method uses multiplication of (1 + 0.20) and (1 + 0.15), not addition of 20% and 15% alone. Common Pitfalls: A frequent mistake is to simply add 20% and 15% to get 35%, ignoring the fact that the changes are multiplicative. Some students also mistakenly apply the formula for total surface area instead of curved surface area, which could change the result. It is crucial to work with ratios and scale factors when dealing with percentage changes in products like r * h. Final Answer: The curved surface area of the cylinder increases by 38%.

Volume and Surface Area Questions