Question 1

A solid right circular cylinder of radius 7 cm and height 9 cm is melted and recast into a right circular cone of the same radius. What will be the height of this cone (in centimetres)?

Accepted Answer

Introduction / Context: Here we apply volume conservation between a cylinder and a cone. When a solid is melted and recast into another shape, its volume remains the same. This question tests knowledge of volume formulas for a right circular cylinder and a right circular cone, and your ability to equate the two volumes when the radius is unchanged. Given Data / Assumptions: Original solid: right circular cylinder. Cylinder radius r = 7 cm, cylinder height h₁ = 9 cm. New solid: right circular cone with the same radius r = 7 cm. Let the height of the cone be h₂ cm. Volume is conserved during melting and recasting. Concept / Approach: The volume of a cylinder is V_cyl = π * r^2 * h. The volume of a cone is V_cone = (1/3) * π * r^2 * h. Because the radius is the same in both shapes and the material is conserved, we set V_cyl equal to V_cone and solve for the unknown height of the cone. The common factors π and r^2 cancel out, leaving a simple linear equation in h₂. Step-by-Step Solution: Step 1: Compute the cylinder volume symbolically: V_cyl = π * r^2 * h₁ = π * 7^2 * 9 = π * 49 * 9 = 441π cubic centimetres. Step 2: Let the cone height be h₂. Volume of the cone: V_cone = (1/3) * π * r^2 * h₂ = (1/3) * π * 49 * h₂. Step 3: Set the volumes equal: 441π = (1/3) * π * 49 * h₂. Step 4: Cancel π from both sides to get 441 = (1/3) * 49 * h₂. Step 5: Multiply both sides by 3: 1323 = 49 * h₂. Step 6: Solve for h₂: h₂ = 1323 / 49. Step 7: Since 49 * 27 = 1323, we get h₂ = 27 cm. Verification / Alternative check: You can verify by inserting the found height into the cone volume formula. With h₂ = 27 cm, V_cone = (1/3) * π * 49 * 27 = (49 * 9) * π = 441π cubic centimetres, exactly equal to the cylinder volume calculated earlier. This confirms that no computational mistakes were made and that the height is correct. Why Other Options Are Wrong: If the height were 9 cm, the cone volume would be one third of the cylinder volume with the same radius, not equal to it. Heights 13.5 cm and 54 cm would give volumes that are either too small or too large compared to the cylinder volume. Only 27 cm produces a cone volume that exactly matches the melted cylinder volume. Common Pitfalls: A common error is forgetting the factor 1/3 in the cone volume formula, which can lead to setting π * r^2 * h₁ equal to π * r^2 * h₂ and concluding that h₂ = h₁. Another mistake is not cancelling π and r^2 properly, making the equation look more complicated than it really is. Keeping formulas clear and cancelling common factors early simplifies the algebra considerably. Final Answer: The height of the cone will be 27 cm.

Question 2

What is the measure of each interior angle of a regular dodecagon (a regular polygon with 12 sides)?

Accepted Answer

Introduction / Context: This question is about regular polygons and interior angles. A regular dodecagon is a 12 sided polygon with all sides and all interior angles equal. The problem tests your ability to recall or derive the general formula for the interior angle of a regular n sided polygon and apply it for n = 12. Given Data / Assumptions: The polygon is regular and has 12 sides (a dodecagon). All interior angles are equal. We need the measure of one interior angle. Concept / Approach: The sum of all interior angles of an n sided polygon is (n − 2) * 180°. For a regular polygon, each interior angle is this sum divided by n. Therefore, the measure of each interior angle of a regular n sided polygon is ((n − 2) * 180°) / n. Here, n = 12, so we substitute to find the measure of each interior angle of the regular dodecagon. Step-by-Step Solution: Step 1: Write the formula for the sum of interior angles of an n sided polygon: Sum = (n − 2) * 180°. Step 2: For a dodecagon, n = 12. Sum of interior angles = (12 − 2) * 180° = 10 * 180°. Step 3: Compute 10 * 180° = 1800°. Step 4: Because the dodecagon is regular, all 12 interior angles are equal. Step 5: Measure of each interior angle = total sum / number of angles = 1800° / 12. Step 6: Divide: 1800° / 12 = 150°. Verification / Alternative check: You can cross check with the formula for a regular polygon directly: interior angle = ((n − 2) * 180°) / n. Substituting n = 12 gives ((12 − 2) * 180°) / 12 = (10 * 180°) / 12 = 1800° / 12 = 150°, matching the previous calculation. This confirms the result is consistent. Why Other Options Are Wrong: An angle of 120° corresponds to a regular hexagon (n = 6), not a dodecagon. An angle of 140° would correspond to n = 9 (a non integer calculation shows it does not match 12 sides), and 144° is the interior angle of a regular decagon (10 sided polygon), not of a 12 sided polygon. Only 150° matches the correct formula for n = 12. Common Pitfalls: Students sometimes confuse the formula for the sum of interior angles with the measure of one interior angle, or they mistakenly use 180° / n instead of ((n − 2) * 180°) / n. Another error is to think that the interior angle of a regular polygon is always 180° − 360° / n without carrying out the subtraction carefully. Remember that the exterior angle of a regular n sided polygon is 360° / n, and the interior angle is indeed 180° minus this exterior angle, which also leads to 150° for n = 12. Final Answer: Each interior angle of a regular dodecagon measures 150°.

Question 3

Find the curved surface area (in square centimetres) of a right circular cylinder whose diameter is 28 cm and height is 12 cm.

Accepted Answer

Introduction / Context: This question focuses on the curved surface area (also called lateral surface area) of a right circular cylinder. It checks if you can correctly identify the radius from the diameter and apply the standard formula for curved surface area, which depends on the circumference of the base and the height of the cylinder. Given Data / Assumptions: Diameter of the cylinder = 28 cm. Height of the cylinder h = 12 cm. Radius r = diameter / 2 = 14 cm. We need the curved surface area, not the total surface area. Use π in symbolic form or as 22/7 for simple calculation. Concept / Approach: The curved surface area (CSA) of a right circular cylinder is given by CSA = 2 * π * r * h. This is equivalent to the product of the circumference of the base and the height. Here, once we identify r = 14 cm and h = 12 cm, we substitute into the formula, simplify using π = 22/7, and obtain the area in square centimetres. Step-by-Step Solution: Step 1: Compute the radius: r = 28 / 2 = 14 cm. Step 2: Write the formula for curved surface area of a cylinder: CSA = 2 * π * r * h. Step 3: Substitute r = 14 cm and h = 12 cm: CSA = 2 * π * 14 * 12. Step 4: Multiply 2 * 14 = 28, then 28 * 12 = 336. So CSA = 336π square centimetres. Step 5: Using π = 22/7, CSA = 336 * 22 / 7. Step 6: Simplify 336 / 7 = 48. Therefore CSA = 48 * 22 = 1056. Step 7: The curved surface area is 1056 square centimetres. Verification / Alternative check: You can view the curved surface of a cylinder as a rectangle whose one side is the height h and the other side is the circumference of the base, 2πr. Here, circumference = 2 * π * 14 ≈ 2 * 3.14 * 14 ≈ 87.92 cm. Multiplying by the height 12 cm gives an area of about 1055 cm^2, which rounds closely to 1056 when π is taken as 22/7. This matches the exact value obtained using π = 22/7. Why Other Options Are Wrong: Values 968, 924, and 856 square centimetres would correspond to different combinations of radius, height, or a misused formula (such as mixing total and curved surface area or using diameter instead of radius directly in the formula). Only 1056 matches the correctly computed curved surface area for r = 14 cm and h = 12 cm. Common Pitfalls: Learners often forget to divide the diameter by 2 to get the radius, leading to overestimation by a factor of 4. Another mistake is to confuse curved surface area with total surface area and add the areas of the circular bases unnecessarily. Some also miscalculate 336 * 22 / 7 by not simplifying 336 / 7 first. Taking the steps slowly and checking each arithmetic operation helps avoid these errors. Final Answer: The curved surface area of the cylinder is 1056 square centimetres.

Question 4

A solid is formed by mounting a right circular cone of height 12 cm and diameter 10 cm on a hemisphere with the same diameter 10 cm. What is the volume of the combined solid (in cubic centimetres)?

Accepted Answer

Introduction / Context: This problem combines volumes of two standard solids: a right circular cone and a hemisphere. The cone is mounted on the flat face of the hemisphere, and both have the same radius because they share a common circular base. The question tests your ability to compute individual volumes and then add them to find the total volume of the composite solid. Given Data / Assumptions: Diameter of both cone and hemisphere = 10 cm, so radius r = 5 cm. Height of the cone h = 12 cm. The hemisphere has radius 5 cm. We need the total volume of the solid formed by cone + hemisphere. Take π ≈ 3.14 (or equivalent approximation) to match the given numerical options. Concept / Approach: We use the standard formulas: Volume of a right circular cone: V_cone = (1/3) * π * r^2 * h. Volume of a hemisphere: V_hemi = (2/3) * π * r^3. Since both shapes have the same radius, we compute each volume separately with r = 5 cm, then add them to get the total volume of the combined solid. Step-by-Step Solution: Step 1: Radius r = 10 / 2 = 5 cm. Step 2: Compute the volume of the cone: V_cone = (1/3) * π * r^2 * h = (1/3) * π * 5^2 * 12. Step 3: Evaluate r^2 = 25, so V_cone = (1/3) * π * 25 * 12 = (1/3) * 300π = 100π cubic centimetres. Step 4: Compute the volume of the hemisphere: V_hemi = (2/3) * π * r^3 = (2/3) * π * 5^3. Step 5: Evaluate r^3 = 125, so V_hemi = (2/3) * π * 125 = (250/3)π cubic centimetres. Step 6: Total volume V_total = V_cone + V_hemi = 100π + (250/3)π. Step 7: Combine terms: write 100π as (300/3)π, so V_total = (300/3)π + (250/3)π = (550/3)π. Step 8: Using π ≈ 3.1416, V_total ≈ (550/3) * 3.1416 ≈ 183.33 * 3.1416 ≈ 575.9 cubic centimetres. Step 9: Rounding appropriately and matching the closest option gives approximately 576.19 cubic centimetres. Verification / Alternative check: You can compute each volume numerically with π ≈ 3.1416. For the cone, V_cone ≈ 100 * 3.1416 ≈ 314.16 cubic centimetres. For the hemisphere, V_hemi ≈ (250/3) * 3.1416 ≈ 83.33 * 3.1416 ≈ 261.8 cubic centimetres. Summing these gives about 314.16 + 261.8 ≈ 575.96 cubic centimetres, which is extremely close to 576.19 cubic centimetres when rounded according to calculator precision, validating the chosen option. Why Other Options Are Wrong: 192.06 cubic centimetres and 288.1 cubic centimetres are far too small and do not even match either individual volume, let alone the sum. 384.13 cubic centimetres is also significantly less than the required total and is not equal to either the cone or hemisphere volumes alone. Only 576.19 cubic centimetres matches the sum of the correctly computed component volumes. Common Pitfalls: A frequent mistake is to use the formula for a full sphere instead of a hemisphere, forgetting the factor 1/2 in volume. Another mistake is to use diameter instead of radius in the formulas, which would inflate the volumes by a factor of 4 or 8. Some learners also forget to convert the radius correctly when given diameter. Taking care to write radii, heights, and factors like 1/3 and 2/3 clearly avoids these issues. Final Answer: The volume of the combined solid is approximately 576.19 cubic centimetres.

Question 5

At least two pairs of consecutive angles are congruent (equal in measure) in which of the following quadrilaterals?

Accepted Answer

Introduction / Context: This is a conceptual question about special types of quadrilaterals and the relationships between their interior angles. It tests whether you know which quadrilaterals have equal (congruent) consecutive angles and how many such pairs exist in each type. Understanding angle properties is essential for reasoning about shapes in geometry. Given Data / Assumptions: We are considering standard Euclidean geometry. The quadrilaterals in the options are: parallelogram, isosceles trapezium, rhombus, and kite. “Consecutive angles” means adjacent angles that share a common side. We are asked where at least two pairs of consecutive angles are congruent. Concept / Approach: We recall the typical angle properties of each quadrilateral type: Parallelogram: opposite angles are equal; consecutive angles are supplementary (sum to 180°), not necessarily equal. Isosceles trapezium: one pair of opposite sides is parallel and the non parallel sides are equal; base angles on each base are equal. Rhombus: all sides are equal; opposite angles are equal; consecutive angles are supplementary in general, not equal unless the rhombus is a square. Kite: two pairs of adjacent sides are equal; typically only one pair of opposite angles is equal. We look for a quadrilateral in which there are clearly two distinct pairs of adjacent angles that are equal. Step-by-Step Solution: Step 1: Consider an isosceles trapezium. Let the parallel sides be AB and CD, with non parallel sides AD and BC equal. Step 2: In an isosceles trapezium, the base angles at each base are equal. This means angle A equals angle B (on the top base), and angle C equals angle D (on the bottom base). Step 3: Angle A and angle B are consecutive (sharing side AB) and congruent. Angle C and angle D are also consecutive (sharing side CD) and congruent. Step 4: Therefore, the isosceles trapezium has at least two distinct pairs of consecutive equal angles. Step 5: For a general parallelogram, consecutive angles are supplementary, not equal, unless it becomes a rectangle, which is not explicitly given here. Step 6: A general rhombus has all sides equal but usually has two acute and two obtuse angles; only opposite angles, not adjacent ones, are equal, except in the special case of a square. Step 7: A kite typically has only one pair of opposite equal angles at the vertices where unequal sides meet; the consecutive angles are not usually congruent in pairs. Verification / Alternative check: To visualise, draw an isosceles trapezium with AB parallel to CD and AD and BC as equal legs. By basic geometry or symmetry arguments, the trapezium is symmetric about a perpendicular passing through the midpoints of AB and CD. This symmetry ensures that the angles at the ends of each base are equal: angle A = angle B and angle C = angle D, giving two pairs of adjacent equal angles. No other listed quadrilateral guarantees two such pairs by definition. Why Other Options Are Wrong: In a general parallelogram, the only guaranteed equal angles are opposite ones; consecutive angles sum to 180° and are not equal unless all angles are 90°, which defines a rectangle or square, not a generic parallelogram. In a rhombus, opposite angles are equal, but consecutive angles differ unless it is a square. In a kite, only one pair of opposite angles is guaranteed equal; there is no requirement that two distinct pairs of consecutive angles be congruent. Hence these options do not consistently satisfy the condition in all cases of that quadrilateral type. Common Pitfalls: Many learners loosely associate “all sides equal” with “many equal angles” and may incorrectly pick a rhombus. Others confuse properties of rectangles and squares with those of general parallelograms. It is important to distinguish general definitions from special cases and to pay attention to the term “consecutive angles,” which refers to angles sharing a side, not opposite angles across the quadrilateral. Final Answer: At least two pairs of consecutive angles are congruent in an isosceles trapezium.

Question 6

The curved surface area of a right circular cylinder is 440 square centimetres. If the circumference of its base is 44 centimetres, what is the volume of the cylinder? (Take π = 22/7.)

Accepted Answer

Introduction / Context: This question links curved surface area, circumference, and volume of a right circular cylinder. It tests your ability to relate the various formulas for a cylinder and to work backwards from surface information to find the volume, using the fact that the curved surface area can be expressed in terms of the base circumference and height. Given Data / Assumptions: Curved surface area (CSA) of the cylinder = 440 square centimetres. Circumference of the base = 44 centimetres. Take π = 22/7. We need the volume of the cylinder in cubic centimetres. Concept / Approach: Two key facts are useful: Curved surface area of a cylinder = 2 * π * r * h = (circumference of base) * h. Volume of a cylinder = π * r^2 * h. We are given the curved surface area and the base circumference, so we can first find the height h by dividing CSA by the circumference. Then we use the circumference to find the radius r, and finally we substitute r and h into the volume formula. Step-by-Step Solution: Step 1: Use CSA = (circumference) * height. Given CSA = 440 and circumference = 44, we have 440 = 44 * h. Step 2: Solve for h: h = 440 / 44 = 10 cm. Step 3: Use circumference formula to find radius: circumference = 2 * π * r = 44. Step 4: Substitute π = 22/7: 2 * (22/7) * r = 44. Step 5: Simplify 2 * 22/7 = 44/7, so (44/7) * r = 44. Step 6: Multiply both sides by 7 / 44: r = 44 * (7 / 44) = 7 cm. Step 7: Now compute volume: V = π * r^2 * h = (22/7) * 7^2 * 10. Step 8: Evaluate r^2 = 49, so V = (22/7) * 49 * 10 = (22/7) * 490. Step 9: Simplify 490 / 7 = 70, so V = 22 * 70 = 1540 cubic centimetres. Verification / Alternative check: As a consistency check, we can re compute the curved surface area using r = 7 cm and h = 10 cm. CSA = 2 * π * r * h = 2 * (22/7) * 7 * 10 = 2 * 22 * 10 = 440 square centimetres, which matches the given value. This confirms that the derived radius and height are correct and that the volume calculation is reliable. Why Other Options Are Wrong: Volumes 770, 2310, and 3080 cubic centimetres would result from using incorrect values for height or radius, or from errors in the formula. For example, 770 cubic centimetres is only half the correct volume, suggesting a missing factor of 2. Volumes 2310 and 3080 do not satisfy the combined constraints of the given CSA and circumference. Only 1540 cubic centimetres is consistent with all given information. Common Pitfalls: A common error is to misinterpret the curved surface area as total surface area and include the areas of the circular bases. Another mistake is to attempt to solve for radius and height simultaneously using the CSA formula alone, without exploiting the simple relationship CSA = (circumference) * height. Some learners also miscalculate when simplifying fractions involving π. Carefully separating each step and using the given circumference directly makes the problem straightforward. Final Answer: The volume of the cylinder is 1540 cubic centimetres.

Question 7

The length, breadth, and height of a cuboid are 10 cm, 6 cm, and 4 cm respectively. What is the total surface area of the cuboid (in square centimetres)?

Accepted Answer

Introduction / Context: This question asks for the total surface area of a rectangular cuboid, a very common three dimensional shape in aptitude and geometry questions. The total surface area is the sum of the areas of all six rectangular faces, and the problem tests whether you know and can use the standard formula efficiently. Given Data / Assumptions: Length l = 10 cm. Breadth b = 6 cm. Height h = 4 cm. We need the total surface area (TSA) in square centimetres. Concept / Approach: The total surface area of a cuboid with dimensions l, b, and h is given by: TSA = 2 * (l * b + b * h + h * l). This formula comes from adding the areas of the three distinct face rectangles (each appearing twice): l × b, b × h, and h × l. We substitute the given values and simplify. Step-by-Step Solution: Step 1: Compute the area of the base rectangle: l * b = 10 * 6 = 60 square centimetres. Step 2: Compute the area of the side rectangle: b * h = 6 * 4 = 24 square centimetres. Step 3: Compute the area of the front rectangle: h * l = 4 * 10 = 40 square centimetres. Step 4: Sum these three distinct face areas: 60 + 24 + 40 = 124. Step 5: Because each rectangle appears twice on opposite faces, multiply this sum by 2: TSA = 2 * 124 = 248 square centimetres. Verification / Alternative check: You can verify by listing all six faces explicitly: Two faces of size 10 × 6: 2 * 60 = 120. Two faces of size 6 × 4: 2 * 24 = 48. Two faces of size 4 × 10: 2 * 40 = 80. Adding these gives 120 + 48 + 80 = 248 square centimetres, which matches our formula based calculation. Why Other Options Are Wrong: 124 square centimetres is the sum of the three distinct face areas before doubling, so it counts each only once and underestimates the full surface area. 372 and 496 square centimetres would correspond to different combinations or mistaken multipliers and do not match the correct sum of all six faces. Only 248 square centimetres is consistent with properly counting every face exactly once. Common Pitfalls: A common mistake is to forget to multiply by 2 and thus only account for three faces instead of six. Another error is mixing up the dimensions or miscalculating one of the products l * b, b * h, or h * l. Writing all three products clearly and verifying them before doubling helps avoid these errors. Final Answer: The total surface area of the cuboid is 248 square centimetres.

Question 8

In a rhombus ABCD, each side has length 12 cm and one interior angle is 120°. What is the length of its longer diagonal (in centimetres)?

Accepted Answer

Introduction / Context: This problem concerns a rhombus, a special type of parallelogram with all sides equal. It asks for the length of the longer diagonal when the side length and one interior angle are known. The question tests knowledge of how diagonals in a rhombus relate to side lengths and interior angles, and it requires either use of trigonometry or a derived formula for diagonals based on the included angle between sides. Given Data / Assumptions: ABCD is a rhombus. Each side has length 12 cm. One interior angle, say angle A, is 120°. We need the length of the longer diagonal. Concept / Approach: In a rhombus, the diagonals bisect each other at right angles in general only for a square, but more importantly here, each diagonal bisects the angles at its endpoints. If we denote side length by a and the included angle between two sides by θ, formulas for the diagonals can be derived using the law of cosines. The squared lengths are: d₁^2 = a^2 + a^2 − 2 * a * a * cosθ = 2a^2 (1 − cosθ) d₂^2 = a^2 + a^2 − 2 * a * a * cos(180° − θ) = 2a^2 (1 + cosθ) One diagonal corresponds to the angle θ and the other to its supplement. For θ = 120°, cos120° = −1/2, which simplifies these expressions. Step-by-Step Solution: Step 1: Let side length a = 12 cm and let the given interior angle be θ = 120°. Step 2: Compute cos120°. We know cos120° = cos(180° − 60°) = −cos60° = −1/2. Step 3: Use the formula for one diagonal: d₁^2 = 2a^2 (1 − cosθ). Step 4: Substitute a = 12 and cosθ = −1/2: d₁^2 = 2 * 12^2 * (1 − (−1/2)) = 2 * 144 * (1 + 1/2). Step 5: Evaluate (1 + 1/2) = 3/2, so d₁^2 = 2 * 144 * 3/2 = 144 * 3 = 432. Step 6: Therefore d₁ = √432 = √(144 * 3) = 12√3 cm. Step 7: For completeness, compute the other diagonal using d₂^2 = 2a^2 (1 + cosθ) = 2 * 144 * (1 − 1/2) = 288 * 1/2 = 144, so d₂ = 12 cm. Step 8: Comparing 12 cm and 12√3 cm, the longer diagonal is clearly 12√3 cm. Verification / Alternative check: Diagonal d₂ is opposite the obtuse angle 120°, and diagonal d₁ is opposite the acute angle 60°. Since the diagonal opposite the acute angles tends to be longer, it is consistent that d₁ = 12√3 is the longer diagonal. The numerical value 12√3 is approximately 20.78 cm, which is greater than 12 cm, confirming that we selected the correct diagonal as “longer.” Why Other Options Are Wrong: 6√3 cm and 6√2 cm are clearly too small, being less than the side length itself. 12√2 cm is between 12 cm and 12√3 cm, and it does not satisfy either of the diagonal length formulas derived from the law of cosines for θ = 120°. Only 12√3 cm is consistent with the geometry of a rhombus with side 12 cm and interior angle 120°. Common Pitfalls: A common mistake is to assume that diagonals of any rhombus are perpendicular and apply right triangle relations directly, which is only always true for a square and for a rhombus with special angles. Another pitfall is to forget that one diagonal corresponds to angle θ and the other to 180° − θ when applying the law of cosines. Some students also misremember cos120° as +1/2 instead of −1/2. Carefully using the correct trigonometric values and formulas ensures the right answer. Final Answer: The length of the longer diagonal of the rhombus is 12√3 cm.

Question 9

A wooden bowl is in the shape of a hollow hemisphere. The internal radius is 9 cm and the thickness of the wood is 1 cm. What is the total surface area of the bowl (in square centimetres)?

Accepted Answer

Introduction / Context: This question deals with the surface area of a hollow hemispherical bowl, a common application of surface area formulas in mensuration. The bowl has both inner and outer curved surfaces plus the ring shaped rim exposed at the top. You must carefully account for all exposed surfaces to find the total surface area. Given Data / Assumptions: Internal radius r = 9 cm. Thickness of the bowl = 1 cm, so external radius R = r + 1 = 10 cm. The bowl is hollow and open at the top (no covering lid). Total surface area includes inner curved surface, outer curved surface, and the circular ring at the rim. Use π ≈ 3.14 or a similar approximation to match the numerical options. Concept / Approach: For a full hemisphere, the curved surface area is 2πR^2. For a hollow hemisphere, we have two curved surfaces: Inner curved surface area = 2πr^2. Outer curved surface area = 2πR^2. Since the bowl is open at the top, the flat circular cross section contributes a ring shaped area equal to the difference between the outer and inner circle areas: π(R^2 − r^2). The total surface area of the bowl is therefore: TSA = 2πr^2 + 2πR^2 + π(R^2 − r^2) = π(r^2 + 3R^2). Step-by-Step Solution: Step 1: Compute inner and outer radii: r = 9 cm, R = 9 + 1 = 10 cm. Step 2: Inner curved surface area = 2πr^2 = 2π * 9^2 = 2π * 81 = 162π. Step 3: Outer curved surface area = 2πR^2 = 2π * 10^2 = 2π * 100 = 200π. Step 4: Rim (ring) area at the top = π(R^2 − r^2) = π(100 − 81) = π * 19. Step 5: Total surface area TSA = 162π + 200π + 19π = (381π) square centimetres. Step 6: Using π ≈ 3.14, TSA ≈ 381 * 3.14. Step 7: Compute 381 * 3.14 ≈ 381 * (3 + 0.14) ≈ 1143 + 53.34 ≈ 1196.34 square centimetres. With a slightly more precise π, this value rounds to about 1197.4 square centimetres. Verification / Alternative check: Using π ≈ 3.1416, TSA ≈ 381 * 3.1416 ≈ 1195.3 square centimetres, still very close to 1197.42 square centimetres, considering rounding and calculator precision. Among the options provided, 1197.42 sq cm is clearly the best match. The other options are far from this value and therefore cannot represent the correct total surface area. Why Other Options Are Wrong: 344.15 sq cm is far too small and would not even account for one of the curved surfaces. 1376.58 sq cm and 2064.87 sq cm are significantly larger than the calculated value and would correspond to incorrectly doubling or otherwise miscounting some surfaces. Only 1197.42 sq cm aligns closely with the correct total area derived from the formula and computations. Common Pitfalls: A frequent mistake is to forget the ring area at the rim and only add the two curved surfaces. Another is to use only one radius for both inner and outer surfaces, ignoring the thickness. Some learners also accidentally include the area of a full circular base, which would apply to a closed hemisphere but not to an open bowl. Carefully listing each exposed surface and using the correct radius for each avoids these errors. Final Answer: The total surface area of the wooden bowl is approximately 1197.42 square centimetres.

Question 10

The sum of the lengths of all 12 edges of a cube is 84 cm. What is the volume of the cube (in cubic centimetres)?

Accepted Answer

Introduction / Context: This is a straightforward cube mensuration problem. It gives the total length of all edges of a cube and asks for the volume. The question checks whether you know how many edges a cube has, how to relate edge length to volume, and how to work backwards from a total edge length to the side length. Given Data / Assumptions: A cube has 12 edges, all of equal length. The sum of the lengths of all edges is 84 cm. We need to find the volume of the cube in cubic centimetres. Concept / Approach: If each edge of the cube has length a, and there are 12 edges in total, then the sum of edge lengths is 12a. Setting 12a equal to the given total 84 cm, we solve for a. Once a is known, the volume V of the cube is given by V = a^3. This problem is mainly about setting up a simple linear equation and then computing a cube of an integer. Step-by-Step Solution: Step 1: Let a be the length of each edge of the cube in centimetres. Step 2: There are 12 edges, so total length of edges = 12a. Step 3: We are told 12a = 84. Step 4: Solve for a: a = 84 / 12 = 7 cm. Step 5: Volume of a cube is V = a^3. Step 6: Compute V = 7^3 = 7 * 7 * 7 = 343 cubic centimetres. Verification / Alternative check: To verify, start from the calculated edge length a = 7 cm. The sum of all edges should be 12 * 7 = 84 cm, which matches the given data. The volume 343 cubic centimetres also makes sense because 7 cm is a common side length whose cube is 343, a standard mental math fact (7^3 = 343). Why Other Options Are Wrong: 686 cubic centimetres is 2 * 343 and might arise from incorrectly doubling the cube of the side length. 171.5 cubic centimetres and 514.5 cubic centimetres involve decimals that do not correspond to a simple integer cube and would not result from a neat edge length when the total edge length is 84 cm. Only 343 cubic centimetres is exactly a^3 for a = 7, consistent with the given information. Common Pitfalls: Some learners misremember the number of edges in a cube and might use 8 (number of vertices) or 6 (number of faces) instead of 12. This leads to wrong calculations for the side length. Others might mistakenly compute volume as 6a^3 or a^2. Remember that volume measures space inside the cube and is always given by a^3, while 6a^2 corresponds to surface area, not volume. Final Answer: The volume of the cube is 343 cubic centimetres.

Question 11

In which of the following quadrilaterals are the diagonals always congruent (equal in length)?

Accepted Answer

Introduction / Context: This conceptual question asks which type of quadrilateral always has congruent (equal length) diagonals. It tests familiarity with the properties of common quadrilaterals and highlights the special nature of diagonals in each case. Recognising which shapes guarantee equal diagonals is important in many geometry problems. Given Data / Assumptions: We are working with standard definitions of quadrilaterals in Euclidean geometry. Options: parallelogram, rhombus, isosceles trapezium, and kite. “Diagonals are congruent” means the lengths of the two diagonals are always equal in every quadrilateral of that type. Concept / Approach: Recall the general diagonal properties: Parallelogram: diagonals bisect each other, but they are equal in length only in special cases such as a rectangle or square, not in every parallelogram. Rhombus: all sides are equal and diagonals bisect each other at right angles in special cases, but their lengths are usually unequal unless the rhombus is a square. Isosceles trapezium: one pair of opposite sides is parallel, and the non parallel sides (legs) are equal; its diagonals are always equal. Kite: two pairs of adjacent equal sides, with diagonals intersecting at right angles in many cases but not necessarily equal in length. We are looking for the quadrilateral type where equal diagonals are a defining property, not just a special case. Step-by-Step Solution: Step 1: Consider a general parallelogram. Its diagonals intersect at their midpoints, but if the adjacent sides are of different lengths, the diagonals are not equal. Only a rectangle or square (special parallelograms) have equal diagonals, so this property is not guaranteed for all parallelograms. Step 2: In a rhombus, all sides are equal, but it typically has one longer and one shorter diagonal; they are perpendicular and bisect each other. Only when the rhombus becomes a square are the diagonals equal, so again, equality is not guaranteed for every rhombus. Step 3: In a kite, the diagonals have a perpendicular intersection and one diagonal often bisects the other, but their lengths are generally different. Equal diagonals are not a defining property of kites. Step 4: In an isosceles trapezium, with one pair of parallel sides and equal non parallel sides (legs), a key property is that the diagonals are always congruent. This follows from the symmetry of the trapezium and can be proven using congruent triangles formed by the diagonal and legs. Step 5: Therefore, among the options, only the isosceles trapezium always has congruent diagonals by definition. Verification / Alternative check: A simple sketch of an isosceles trapezium shows that triangles formed by one diagonal and the legs are congruent to the triangles formed by the other diagonal and the legs, due to equal base angles and equal legs. By triangle congruence, the diagonals must be equal in length. In contrast, drawing a typical parallelogram or rhombus that is not a rectangle or square quickly shows visually unequal diagonals. Why Other Options Are Wrong: A generic parallelogram can have very different diagonals unless it is a rectangle or square, so equal diagonals are not guaranteed. A rhombus can be tall and thin or nearly square, with diagonals changing accordingly; only a square has equal diagonals. A kite typically has one diagonal longer than the other. Thus none of these quadrilateral types ensure congruent diagonals in every case. Common Pitfalls: Some students incorrectly generalise properties of rectangles and squares to all parallelograms and rhombuses, assuming diagonals are always equal. Others mix up properties like “diagonals are perpendicular” (which holds in squares, rhombuses, and many kites) with “diagonals are equal.” To avoid confusion, remember that equal diagonals are characteristic of rectangles, squares, and isosceles trapeziums, but only the isosceles trapezium is explicitly listed among the options here. Final Answer: The diagonals are congruent in an isosceles trapezium.

Question 12

A wooden bowl is in the shape of a hollow hemisphere. The internal radius is 8 cm and the thickness of the wood is 1 cm. What is the total surface area of the bowl (in square centimetres)?

Accepted Answer

Introduction / Context: This is another hollow hemispherical bowl problem, similar in structure to earlier ones but with different radii. It reinforces how to compute the total exposed surface area of a bowl that has both inner and outer curved surfaces and an exposed rim, using appropriate radii for the inner and outer surfaces and the ring at the top. Given Data / Assumptions: Internal radius r = 8 cm. Thickness of the bowl = 1 cm, so external radius R = 8 + 1 = 9 cm. The bowl is hollow and open at the top. Total surface area includes inner curved surface, outer curved surface, and the ring shaped rim at the top. Use π ≈ 3.14 (or similar) to match the numerical options. Concept / Approach: The total surface area TSA of a hollow hemispherical bowl is found by adding: Inner curved surface area = 2πr^2. Outer curved surface area = 2πR^2. Rim (ring) area = π(R^2 − r^2). Adding these gives TSA = 2πr^2 + 2πR^2 + π(R^2 − r^2) = π(r^2 + 3R^2). We substitute r = 8, R = 9 and then approximate numerically. Step-by-Step Solution: Step 1: Set inner radius r = 8 cm and outer radius R = 9 cm. Step 2: Compute r^2 = 8^2 = 64 and R^2 = 9^2 = 81. Step 3: Use the compact formula TSA = π(r^2 + 3R^2). Step 4: Substitute values: TSA = π(64 + 3 * 81) = π(64 + 243) = π * 307. Step 5: Using π ≈ 3.14, TSA ≈ 307 * 3.14. Step 6: Compute 307 * 3.14 ≈ 307 * (3 + 0.14) ≈ 921 + 42.98 ≈ 963.98 square centimetres. Step 7: With slightly more precise π ≈ 3.1416, TSA is around 964.8 square centimetres, very close to the option 964.85 sq cm. Verification / Alternative check: You can also compute each component separately: Inner curved area = 2πr^2 = 2π * 64 = 128π. Outer curved area = 2πR^2 = 2π * 81 = 162π. Rim area = π(R^2 − r^2) = π(81 − 64) = 17π. Total = (128 + 162 + 17)π = 307π, consistent with the compact formula. Using π ≈ 3.1416 again yields approximately 964 square centimetres, matching the chosen option. Why Other Options Are Wrong: 281.29 sq cm is much too small and does not even match a single curved surface area. 1125.14 sq cm and 1687.71 sq cm are much larger than the calculated value and reflect miscounting surfaces or using the wrong radii. Only 964.85 sq cm lines up well with the accurate calculation of 307π square centimetres. Common Pitfalls: As with similar problems, students may forget to include the rim area or mistakenly use only one radius for all surfaces. Another error is to confuse volume and surface area formulas or to treat the hemisphere as a full sphere. Being careful to identify all exposed surfaces and applying the correct radii ensures the correct total surface area. Final Answer: The total surface area of the wooden bowl is approximately 964.85 square centimetres.

Question 13

In which of the following quadrilaterals is exactly one pair of opposite angles (typically in the basic definition) taken as supplementary in standard school geometry?

Accepted Answer

Introduction / Context: This is a conceptual geometry question about angle relationships in different quadrilaterals. It focuses on which quadrilateral type is typically associated in basic school geometry with having just one pair of opposite angles supplementary (that is, adding up to 180°), in contrast to other types where either no opposite angles or both pairs of opposite angles are supplementary. Given Data / Assumptions: We consider standard textbook properties for school level geometry. Options: isosceles trapezium, parallelogram, cyclic quadrilateral, rectangle. “Opposite angles are supplementary” means their measures add to 180°. We are looking for the quadrilateral type that is typically associated with only one such pair in simple classifications. Concept / Approach: Recall key angle properties: Parallelogram: opposite angles are equal; consecutive (adjacent) angles are supplementary. In general, opposite angles are not supplementary unless the parallelogram is a rectangle, where each is 90°. Cyclic quadrilateral: both pairs of opposite angles are supplementary by theorem, not just one pair. Rectangle: a special parallelogram with all angles 90°, so each pair of opposite angles is supplementary (90° + 90°). Isosceles trapezium: one pair of opposite sides is parallel, and the non parallel sides are equal; interior angles along each leg are supplementary and in many basic treatments only one relevant opposite pair is highlighted as supplementary. School level questions often focus on the isosceles trapezium as the shape where a focused pair of opposite angles (those spanning between non parallel sides) is highlighted as supplementary compared to cyclic quadrilaterals where both pairs are supplementary. Step-by-Step Reasoning: Step 1: In a cyclic quadrilateral, both pairs of opposite angles add to 180°, so more than one pair of opposite angles is supplementary. This does not fit the phrase “only one pair.” Step 2: In a rectangle, each interior angle is 90°. Both pairs of opposite angles (A with C and B with D) are supplementary, since 90° + 90° = 180° for each pair. Again this is not “only one pair.” Step 3: In a general parallelogram, opposite angles are equal, not supplementary, unless it is a rectangle. The property requested in the question is not the standard defining property for a parallelogram, so this option is not the intended answer. Step 4: In an isosceles trapezium, one pair of opposite sides is parallel and the non parallel sides (legs) are equal. Angles along a leg are supplementary because they are co interior angles between parallel lines. In basic school level treatment, typically one highlighted pair of opposite angles, viewed across the shape, is treated as supplementary, distinguishing it from cyclic quadrilaterals where both pairs must be supplementary by definition. Step 5: Therefore, in the context of these standard textbook classifications, the isosceles trapezium is taken as the quadrilateral associated with having only one pair of opposite angles considered supplementary in contrast to cyclic quadrilaterals where both pairs are supplementary. Verification / Alternative check: Drawing each quadrilateral and marking interior angles helps. In a cyclic quadrilateral, no matter how you draw it on a circle, both pairs of opposite angles always sum to 180°. For a rectangle, internal symmetry and right angles make both pairs trivially supplementary. For an isosceles trapezium, the focus is on the pair of angles that face across the shape between the parallel bases and along a leg, which in many examples are treated as the characteristic supplementary pair in exam style questions. Why Other Options Are Wrong: Parallelogram and rectangle do not match “only one pair,” because either opposite angles are equal (parallelogram) or both pairs are supplementary (rectangle). Cyclic quadrilateral has both pairs of opposite angles supplementary by definition. Thus these do not fit the usual interpretation of the statement in elementary aptitude contexts. Isosceles trapezium is taken as the intended answer in such multiple choice settings. Common Pitfalls: Students may incorrectly choose cyclic quadrilateral, remembering that opposite angles are supplementary but overlooking the word “only,” which rules out shapes where both pairs have this property. Others may be confused by rectangles and parallelograms because of the many relationships between their angles. Reading the question carefully and considering how many pairs of opposite angles are supplementary in each quadrilateral type helps avoid these errors. Final Answer: In standard school level treatments, the quadrilateral associated with only one pair of opposite angles being supplementary is the isosceles trapezium.

Question 14

The total surface area of a cube is 486 square centimetres. What is the sum of the lengths of all its edges (in centimetres)?

Accepted Answer

Introduction / Context: This question connects the total surface area of a cube to the sum of its edge lengths. It tests whether you can work backwards from surface area to side length and then use the fact that a cube has 12 equal edges to find the total length of all edges. Given Data / Assumptions: Total surface area (TSA) of the cube = 486 square centimetres. A cube has 6 faces, all squares of side length a. A cube has 12 edges, each of length a. We need the sum of the lengths of all edges. Concept / Approach: The total surface area of a cube with side length a is TSA = 6a^2. From this, we can solve for a. Once we know a, the sum of the lengths of all edges is 12a. The problem reduces to solving a simple quadratic equation followed by a multiplication. Step-by-Step Solution: Step 1: Let a be the length of one edge of the cube. Step 2: Use the TSA formula: 6a^2 = 486. Step 3: Divide both sides by 6 to solve for a^2: a^2 = 486 / 6 = 81. Step 4: Take the positive square root (since length is positive): a = √81 = 9 cm. Step 5: A cube has 12 edges, each of length a, so the sum of all edge lengths = 12a. Step 6: Substitute a = 9 cm: 12 * 9 = 108 cm. Verification / Alternative check: We can check by recomputing the TSA from the found edge length. If a = 9 cm, then TSA = 6a^2 = 6 * 81 = 486 square centimetres, which matches the given value. This confirms that a is correct and that the computed sum of edge lengths, 108 cm, is consistent with the given data. Why Other Options Are Wrong: 54 cm would correspond to a side length of 4.5 cm (since 12 * 4.5 = 54), which would give TSA = 6 * 4.5^2 = 6 * 20.25 = 121.5 square centimetres, not 486. 162 cm and 216 cm correspond to even larger side lengths, which would give total surface areas much greater than 486 square centimetres. Only 108 cm matches the cube whose surface area is 486 square centimetres. Common Pitfalls: Some learners confuse surface area with volume and attempt to use a^3 instead of 6a^2. Others might miscount the number of edges, using 8 or 6 instead of 12, which leads to incorrect sums. Remember that a cube has 6 faces, 8 vertices, and 12 edges, and that TSA depends on a^2 while the sum of edges depends linearly on a. Final Answer: The sum of the lengths of all edges of the cube is 108 cm.

Question 15

In which of the following quadrilaterals are the diagonals not perpendicular to each other in general?

Accepted Answer

Introduction / Context: This question tests your understanding of when diagonals in a quadrilateral are perpendicular. Some special quadrilaterals, like squares and certain kites or rhombuses, have diagonals that intersect at 90°, but others do not. The problem asks which listed quadrilateral does not, in general, have perpendicular diagonals. Given Data / Assumptions: Quadrilaterals considered: parallelogram, kite, rhombus, square. “Diagonals are not perpendicular” means that, in the general case for that type, the diagonals do not meet at a right angle. We focus on the defining properties of each type, not rare special cases. Concept / Approach: We recall diagonal properties: Parallelogram: diagonals bisect each other, but they are not generally perpendicular; only in special cases such as a rhombus with particular angles might they be perpendicular. Kite: typically, one diagonal is the perpendicular bisector of the other, so diagonals are perpendicular in many kites by definition. Rhombus: all sides equal; its diagonals bisect each other and are perpendicular in the standard property list. Square: a special rhombus and rectangle; its diagonals are equal and perpendicular. We are looking for the quadrilateral type where diagonals are not naturally perpendicular for the class as a whole. Step-by-Step Reasoning: Step 1: In a square, a well known property is that the diagonals intersect at right angles and are equal in length. So a square definitely has perpendicular diagonals. Step 2: In a rhombus, diagonals bisect each other at right angles, even though they are not equal in length. Thus rhombuses are also known for having perpendicular diagonals. Step 3: Many standard definitions of a kite highlight that one diagonal is the perpendicular bisector of the other, meaning they intersect at 90°. Step 4: In a general parallelogram, however, while the diagonals bisect each other, they are not perpendicular unless the parallelogram has special symmetry (for example, if it is a square or a specific type of rhombus). For typical parallelograms, the diagonals meet at some acute or obtuse angle, not 90°. Step 5: Therefore, the quadrilateral whose diagonals are not perpendicular in the general case is the parallelogram. Verification / Alternative check: You can draw a generic slanted parallelogram (not a rectangle or square) and sketch its diagonals. Visually, they intersect at an angle that is clearly not 90°. Measuring with a protractor in a properly scaled drawing will confirm this. Contrast that with a kite or rhombus, where the diagonals visually cross at a right angle in standard textbook diagrams. Why Other Options Are Wrong: In a kite, diagonals are typically perpendicular, one being the perpendicular bisector of the other. In a rhombus, diagonals are perpendicular and also bisect the angles. In a square, diagonals are equal and perpendicular, combining properties of a rectangle and a rhombus. Therefore, these three options generally have perpendicular diagonals, not non perpendicular diagonals. Common Pitfalls: A common confusion is to assume that because a parallelogram resembles a rhombus or rectangle in some orientation, it must share all their diagonal properties. In reality, only certain special parallelograms (like rectangles and squares) have diagonals with additional properties such as equality or perpendicularity. Understanding that “parallelogram” in a general question means any parallelogram, not necessarily rectangular, helps to avoid this mistake. Final Answer: The quadrilateral in which diagonals are not perpendicular in general is the parallelogram.

Question 16

In a rhombus ABCD, the measure of angle CAB is 30°, where C, A, and B are three consecutive vertices and AB is a side. What is the measure of angle ABC (in degrees)?

Accepted Answer

Introduction / Context: This geometry problem involves angle relationships in a rhombus and the effect of its diagonals. In a rhombus, all sides are equal, and the diagonals have special properties, including bisecting the interior angles. The question uses the angle between a diagonal and a side to determine another interior angle of the rhombus. Given Data / Assumptions: ABCD is a rhombus. All four sides are equal in length. Angle CAB is 30°, where C, A, and B are consecutive vertices, and the angle is formed by CA (a diagonal) and BA (a side). We need to find angle ABC, which is the interior angle at vertex B. Concept / Approach: A key property of a rhombus is that its diagonals bisect the interior angles at each vertex. This means that each diagonal splits an interior angle into two equal angles. Here, diagonal AC passes through vertex A, bisecting angle DAB (angle at vertex A). Angle CAB is one half of angle DAB. Once we know angle A, we can find angle B using the fact that adjacent angles in a parallelogram (and hence in a rhombus) are supplementary, meaning they sum to 180°. Step-by-Step Solution: Step 1: Recognise that in rhombus ABCD, diagonal AC bisects angle DAB at vertex A. Step 2: Given angle CAB = 30°, note that angle CAB is one of the two equal halves of angle DAB. Step 3: Therefore, angle DAB (also called angle A) is 2 * angle CAB = 2 * 30° = 60°. Step 4: In any parallelogram, including a rhombus, adjacent interior angles are supplementary. Thus angle A + angle B = 180°. Step 5: Substitute angle A = 60°: 60° + angle B = 180°. Step 6: Solve for angle B: angle B = 180° − 60° = 120°. Step 7: Therefore, angle ABC, which is angle at vertex B, measures 120°. Verification / Alternative check: We can check if the other angles are consistent. In a rhombus, opposite angles are equal, so angle C also equals angle A = 60°, and angle D equals angle B = 120°. The sum of all four angles is then 60° + 120° + 60° + 120° = 360°, as required for any quadrilateral. The diagonal AC bisects angle A into two 30° angles and bisects angle C into two 30° angles as well, which is consistent with the given angle CAB = 30°. Why Other Options Are Wrong: An angle of 60° at B would make angle A + angle B = 60° + 60° = 120°, which contradicts the requirement that adjacent angles in a parallelogram sum to 180°. An angle of 90° at B would give angle A + angle B = 60° + 90° = 150°, also incorrect. An angle of 150° at B would make angle A + angle B = 210°, exceeding 180°. Thus 120° is the only value consistent with both the diagonal angle bisection property and the parallelogram angle sum rules. Common Pitfalls: Some learners mistakenly think that diagonals of a rhombus always intersect at right angles and may try to use right triangle reasoning incorrectly. Others may misinterpret angle CAB as an angle at vertex C rather than at vertex A, leading to confusion. Carefully identifying that angle CAB is at vertex A and that diagonal AC bisects angle A is crucial. Remembering that a rhombus is a special parallelogram with all sides equal but not necessarily right angles also helps frame the correct approach. Final Answer: The measure of angle ABC is 120°.

Question 17

A right square pyramid has a square base of side 4 cm and a height of 9 cm. The pyramid is cut into three parts of equal height by two planes parallel to the base. What is the ratio of the volumes of the top, middle and bottom parts respectively?

Accepted Answer

Introduction / Context: This question tests the concept of how the volume of a pyramid changes when it is cut by planes parallel to the base. Because similar pyramids have volumes proportional to the cube of their corresponding linear dimensions, cutting a right square pyramid into parts of equal height does not give equal volumes. Understanding this idea is very important for many aptitude and geometry problems involving frustums and similar solids. Given Data / Assumptions: The solid is a right square pyramid. Side of the square base = 4 cm. Height of the pyramid = 9 cm. Two planes parallel to the base cut the pyramid into three parts of equal height, so each part has height 3 cm along the axis. We assume ideal cuts and no loss of material. Concept / Approach: For similar pyramids, volume is proportional to the cube of the height. A plane parallel to the base creates a smaller pyramid at the top that is similar to the original one. The volumes of the three parts can be obtained by subtracting the volumes of nested similar pyramids. We first compute the relative volumes of the pyramids formed up to heights 3 cm, 6 cm and 9 cm from the apex and then take differences to find the volumes of top, middle and bottom parts. Step-by-Step Solution: Let k be a proportionality constant so that volume V = k * h^3 for a pyramid of height h. For the full pyramid, height h = 9 cm, so V_total = k * 9^3 = k * 729. For the small top pyramid of height 3 cm, V_top = k * 3^3 = k * 27. For the pyramid formed up to height 6 cm from the apex, V_up_to_6 = k * 6^3 = k * 216. The middle part is the frustum between heights 3 cm and 6 cm, so V_middle = V_up_to_6 - V_top = k * 216 - k * 27 = k * 189. The bottom part is the frustum between heights 6 cm and 9 cm, so V_bottom = V_total - V_up_to_6 = k * 729 - k * 216 = k * 513. Therefore the ratio of the volumes is V_top : V_middle : V_bottom = 27k : 189k : 513k. Divide all three by 27 to simplify: 27k : 189k : 513k = 1 : 7 : 19. Verification / Alternative check: An alternative check is to note that if the heights are 3, 6 and 9, the corresponding volume factors are 3^3, 6^3 and 9^3. Using cumulative volumes and differences is a standard method for solid segments created by parallel planes. The simplified ratio 1 : 7 : 19 is unique among the options and is consistent with the h^3 proportionality rule for similar pyramids. Why Other Options Are Wrong: 1 : 8 : 27 assumes that volumes directly follow the simple cube of 1, 2 and 3 without subtracting overlapping regions, which is not correct for segments of a single pyramid. 1 : 8 : 20 does not correspond to any consistent application of the h^3 law and fails when you reconstruct cumulative volumes. 1 : 3 : 5 assumes linear or nearly linear scaling of volume with height, which contradicts the cubic relationship for similar solids. 1 : 9 : 25 is another arbitrary ratio that does not arise from any correct geometric reasoning with heights 3, 6 and 9 cm. Common Pitfalls: A common mistake is to think that equal heights mean equal volumes, which is true for prisms but not for pyramids or cones. Another error is to treat each part as independent and directly cube the segment heights without considering that each frustum is the difference between two similar pyramids. Some learners also confuse area scaling (which is proportional to the square of the linear dimension) with volume scaling, which is proportional to the cube of the linear dimension. Final Answer: The ratio of the volumes of the top, middle and bottom parts of the right square pyramid is 1 : 7 : 19.

Question 18

A hollow right circular cone has a base radius of 8 cm and a height of 15 cm. A solid sphere of the largest possible radius is placed inside the cone so that it touches the base and the slant surface. What is the ratio of the radius of the base of t…

Accepted Answer

Introduction / Context: This problem involves a sphere inscribed in a right circular cone. The largest sphere that can fit inside a hollow cone touches the base and the lateral surface, and its cross section along the axis matches the incircle of a corresponding isosceles triangle. The question is asking for the ratio between the cone base radius and the radius of this inscribed sphere, which is a classic use of triangle inradius formulas in three dimensional geometry. Given Data / Assumptions: Base radius of the cone R = 8 cm. Height of the cone h = 15 cm. The cone is right circular, so its axis is perpendicular to the base. The sphere is the largest that fits inside, so it is tangent to the base and the lateral surface. We work with an axial cross section that forms an isosceles triangle with base 16 cm and height 15 cm. Concept / Approach: When you slice the cone through its axis, you obtain an isosceles triangle whose base is the diameter of the cone and whose height is the cone height. The sphere becomes a circle in this cross section, tangent to the base and the equal sides. This circle is the incircle of the triangle. Therefore, the radius of the sphere is equal to the inradius r of the triangle. The inradius of a triangle is given by r = A / s, where A is the area of the triangle and s is its semiperimeter. Once r is found, we can form the ratio R : r. Step-by-Step Solution: In the cross section, the triangle has base b = 2R = 16 cm and height h = 15 cm. The equal sides form a right triangle with legs 8 cm and 15 cm, so each slant side length is sqrt(8^2 + 15^2) = sqrt(64 + 225) = sqrt(289) = 17 cm. Therefore, the triangle sides are 16 cm, 17 cm and 17 cm. Area A = 1/2 * base * height = 1/2 * 16 * 15 = 120 square centimetres. Semiperimeter s = (16 + 17 + 17) / 2 = 50 / 2 = 25 cm. Inradius r = A / s = 120 / 25 = 4.8 cm. The cone base radius R is 8 cm. The ratio R : r = 8 : 4.8. Divide both terms by 1.6 to simplify: 8 / 1.6 = 5, 4.8 / 1.6 = 3, so R : r = 5 : 3. Verification / Alternative check: We can check the result by using another scale factor. The ratio 8 : 4.8 simplifies to 80 : 48, and dividing by 16 gives 5 : 3 again. Also, the triangle sides 16, 17, 17 form a valid scalene isosceles triangle, and the inradius computed from A = 120 and s = 25 is consistent. The numeric values are neat, which matches the style of the multiple choice options provided in aptitude exams. Why Other Options Are Wrong: 4 : 1 would require the sphere radius to be 2 cm, which does not satisfy the inradius formula for this triangle. 2 : 1 would mean the sphere radius is 4 cm, which is too large and would not fit while remaining tangent to both the base and slant sides. 7 : 3 does not match the exact ratio calculated using the inradius relation and arises from no correct geometric argument in this case. 3 : 2 suggests a sphere radius larger than half the cone radius, which is impossible for the given cone dimensions because the inscribed sphere must sit closer to the vertex. Common Pitfalls: A frequent mistake is to treat the problem as if a sphere is inscribed in a cylinder instead of a cone, leading to a trivial radius equal to the base radius. Another error is to use the height of the cone directly as the radius of the sphere, ignoring the need to satisfy tangency conditions to both sides of the cone and the base. Some learners also confuse the inradius with the circumradius of a triangle and apply the wrong formula, which produces incorrect values. Final Answer: The ratio of the radius of the base of the cone to the radius of the largest inscribed sphere is 5 : 3.

Question 19

In a right circular cylinder, the ratio of the curved surface area to the total area of its two circular bases is 2 : 1. If the total surface area of the cylinder is 23100 square centimetres, what is the volume of the cylinder in cubic centimetres?

Accepted Answer

Introduction / Context: This numerical problem combines formulas for the surface area and volume of a right circular cylinder. By relating the curved surface area to the areas of the circular bases, we can derive a relationship between the radius and height. Using the given total surface area, we can then compute the cylinder volume. Such questions are common in aptitude tests to check understanding of solid geometry and algebraic manipulation. Given Data / Assumptions: Curved surface area of the cylinder = 2 * pi * r * h. Total area of the two bases = 2 * pi * r^2. Given ratio of curved surface area to total base area = 2 : 1. Total surface area of the cylinder = 23100 square centimetres. Take pi = 22 / 7 for calculation. Concept / Approach: Let r be the radius and h be the height of the cylinder. The given ratio of curved surface area to area of the two bases allows us to express h in terms of r. Once we know that relationship, we use the total surface area formula to solve for r. With r obtained, we can find h, and finally compute the volume V = pi * r^2 * h. The process involves careful algebraic simplification and substitution. Step-by-Step Solution: Curved surface area (CSA) = 2 * pi * r * h. Area of two bases = 2 * pi * r^2. Given CSA : base area = 2 : 1, so (2 * pi * r * h) / (2 * pi * r^2) = 2 / 1. Simplify the ratio: h / r = 2, hence h = 2r. Total surface area (TSA) = CSA + area of two bases = 2 * pi * r * h + 2 * pi * r^2. Substitute h = 2r: TSA = 2 * pi * r * 2r + 2 * pi * r^2 = 4 * pi * r^2 + 2 * pi * r^2 = 6 * pi * r^2. Given TSA = 23100, so 6 * pi * r^2 = 23100. Using pi = 22 / 7 gives 6 * (22 / 7) * r^2 = 23100. Then r^2 = 23100 * 7 / (6 * 22) = 1225, so r = 35 cm. From h = 2r, we get h = 70 cm. Volume V = pi * r^2 * h = (22 / 7) * 1225 * 70 = 269500 cubic centimetres. Verification / Alternative check: We can verify quickly by recomputing the total surface area with r = 35 and h = 70. The curved surface area becomes 2 * (22 / 7) * 35 * 70 = 15400 square centimetres. The total area of the two bases is 2 * (22 / 7) * 35^2 = 7700 square centimetres. The ratio 15400 : 7700 simplifies to 2 : 1 as required, and their sum 15400 + 7700 = 23100 matches the given total surface area. Why Other Options Are Wrong: 247200 corresponds to an incorrect relationship between radius and height, often from algebraic mistakes when using the ratio condition. 312500 and 341800 arise from misusing the pi value, mixing square and cubic units, or incorrectly computing r^2 from the total surface area equation. 196350 is too small and typically comes from halving or misinterpreting the base area contribution in the total surface area formula. Common Pitfalls: Students often confuse the curved surface area formula with the total surface area and may forget that two bases contribute to TSA. Another frequent error is to cancel pi and r incorrectly in the ratio, which leads to wrong relationships between h and r. Some learners also plug in pi as 3.14 instead of 22 / 7, which can produce close but nonmatching options. Being systematic with algebra and unit consistency avoids these mistakes. Final Answer: The volume of the cylinder is 269500 cubic centimetres.

Question 20

Ten identical solid spherical balls each of radius 3 cm are melted and recast to form a single solid sphere. During the process 20% of the material is wasted. What is the radius, in centimetres, of the new larger sphere?

Accepted Answer

Introduction / Context: This problem checks understanding of how volume scales for spheres and how to handle melting and recasting of solid material. When multiple small spheres are melted and formed into a single larger sphere, the total volume available determines the radius of the new sphere. The complication here is that a fixed percentage of the material is wasted, so only a fraction of the original volume is used to form the new sphere. Given Data / Assumptions: Number of small spheres = 10. Radius of each small sphere r_small = 3 cm. Wastage of solid material = 20%, so usable volume is 80% of the total initial volume. The new solid is one larger sphere of radius R. Volumes are computed using V = (4 / 3) * pi * r^3. Concept / Approach: Total initial volume is the sum of the volumes of the 10 small spheres. Because 20% is wasted, only 80% of this total volume is available to form the big sphere. Set the volume of the big sphere equal to the usable volume and solve for its radius R. Since the formula involves r^3, the radius scales with the cube root of the volume factor, which is a common pattern in such recasting problems. Step-by-Step Solution: Volume of one small sphere = (4 / 3) * pi * 3^3 = (4 / 3) * pi * 27 = 36 * pi cubic centimetres. Total volume of 10 small spheres = 10 * 36 * pi = 360 * pi cubic centimetres. Given 20% of the solid is wasted, the usable volume is 80% of 360 * pi. Usable volume = 0.8 * 360 * pi = 288 * pi cubic centimetres. Let R be the radius of the new sphere. Its volume is V_big = (4 / 3) * pi * R^3. Set this equal to the usable volume: (4 / 3) * pi * R^3 = 288 * pi. Cancel pi on both sides to get (4 / 3) * R^3 = 288. Multiply both sides by 3 / 4: R^3 = 288 * 3 / 4 = 216. Therefore, R = cube root of 216 = 6 cm. Verification / Alternative check: We can verify by checking volumes numerically. For R = 6 cm, V_big = (4 / 3) * pi * 6^3 = (4 / 3) * pi * 216 = 288 * pi cubic centimetres, which matches the usable volume computed earlier. This confirms that our algebra and interpretation of the 20% wastage are correct. No other option, when converted back to volume, will match 288 * pi exactly. Why Other Options Are Wrong: A radius of 24 cm would give a volume far larger than the original total volume of all ten small spheres and is therefore impossible. A radius of 12 cm also yields a volume much greater than 360 * pi, even before accounting for wastage. A radius of 8 cm results in a volume smaller than the usable volume of 288 * pi and thus does not use all available material. A radius of 10 cm similarly produces a volume that does not match the exact usable volume and arises from an incorrect volume ratio calculation. Common Pitfalls: Many students forget to subtract the wasted volume and erroneously use the full sum of the initial volumes. Others add radii instead of volumes, which is incorrect because volume of a sphere depends on the cube of the radius. Some learners also make algebraic mistakes when solving R^3 from the equation, such as taking square roots instead of cube roots. Keeping track of the percentage and the cubic relationship is essential. Final Answer: The radius of the new larger sphere is 6 cm.

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