Question 1

Three solid cubes of side lengths 3 cm, 4 cm and 5 cm are melted and recast into a single large cube. What is the ratio of the total surface area of the three original cubes to that of the new cube?

Accepted Answer

Introduction / Context: This problem involves volume and surface area relationships for cubes. Three smaller solid cubes are melted and recast into a single larger cube. Because volume is conserved during melting and recasting, the total volume of the smaller cubes equals the volume of the larger cube. Once we find the side length of the new cube, we can compute and compare the total surface areas. This type of question is common in aptitude tests and encourages clear understanding of how linear dimensions, area, and volume scale. Given Data / Assumptions: • Side lengths of the three cubes are 3 cm, 4 cm, and 5 cm respectively. • All cubes are solid and made of the same material. • The three cubes are melted completely and recast into one large cube. • Volume is conserved in the process. • We are asked to find the ratio of total surface area of the three smaller cubes to the surface area of the large cube. Concept / Approach: The volume of a cube of side a is a³, and its total surface area is 6a². For the three small cubes, we compute individual volumes and sum them to get the total volume. That total becomes the volume of the large cube, from which we can determine its side length by taking the cube root. Then we compute the total surface area of all three small cubes and the surface area of the large cube. Finally, we express the required ratio as a simplified fraction or ratio of integers. Step-by-Step Solution: Step 1: Compute the volumes of the three small cubes: 3³ = 27, 4³ = 64, and 5³ = 125 cubic centimetres. Step 2: Sum the volumes to get total volume: 27 + 64 + 125 = 216 cubic centimetres. Step 3: Let the side of the large cube be A cm. Then its volume A³ equals 216. Step 4: Solve A³ = 216. Since 6³ = 216, we get A = 6 cm. Step 5: Total surface area of the three small cubes is 6(3²) + 6(4²) + 6(5²) = 6(9 + 16 + 25) = 6 * 50 = 300 square centimetres. Step 6: Surface area of the large cube is 6A² = 6 * 6² = 6 * 36 = 216 square centimetres. Step 7: The required ratio is 300 : 216. Divide numerator and denominator by 12 to get 25 : 18. Verification / Alternative check: We can verify the side of the large cube by rechecking the volume calculation. The total volume is 216 cubic centimetres, and cube root of 216 is 6 because 6 * 6 * 6 = 216. The surface area calculation is consistent, since 6 * 36 = 216. The ratio 300 / 216 reduces to 25 / 18. This ratio is irreducible because 25 and 18 share no common factor greater than 1. The result is thus stable and unique. Why Other Options Are Wrong: 2 : 1 would correspond to 300 : 150, which is not equal to 300 : 216. 3 : 2 equals 27 : 18, not 25 : 18, so it does not match the actual ratio. 27 : 20 and 5 : 4 correspond to different fractional values and do not arise from any consistent calculation of volumes and areas in this problem. Only 25 : 18 corresponds to the correct surface area comparison. Common Pitfalls: A frequent error is to assume that the ratio of surface areas equals the ratio of volumes directly, which is not correct for solids when dimensions change. Another mistake is miscalculating the total volume by adding side lengths instead of volumes. Some learners also miscompute 6³ or forget the factor 6 in the surface area formula for a cube. Carefully computing each cube's volume and area separately, and then using the cube root correctly, helps avoid these mistakes. Final Answer: The ratio of the total surface area of the three smaller cubes to that of the large cube is 25 : 18.

Question 2

A right circular cylinder and a right circular cone have the same base diameter and the same height. What is the ratio of the volume of the cylinder to the volume of the cone?

Accepted Answer

Introduction / Context: This question tests basic understanding of volume formulas for a cylinder and a cone with the same base and height. It focuses on the relationship between these two volumes rather than requiring actual numerical substitution. Such comparisons are standard in geometry and aptitude exams, and they highlight how shapes with similar dimensions can enclose different amounts of space. Given Data / Assumptions: • A right circular cylinder and a right circular cone have the same base radius (or same diameter) and the same height h. • Let the common radius be r and the common height be h. • We need the ratio of the volume of the cylinder to the volume of the cone. Concept / Approach: The volume of a right circular cylinder is given by V_cyl = πr²h, where r is the radius and h is the height. The volume of a right circular cone with the same base and height is V_cone = (1/3)πr²h. Since r and h are the same for both solids, we can form a ratio directly using these formulas without needing specific numerical values. The common factor πr²h will cancel, leaving a simple numerical ratio. Step-by-Step Solution: Step 1: Write the cylinder volume formula: V_cyl = πr²h. Step 2: Write the cone volume formula: V_cone = (1/3)πr²h. Step 3: Form the ratio of cylinder volume to cone volume: V_cyl / V_cone = (πr²h) / [(1/3)πr²h]. Step 4: Cancel the common factors π, r², and h in numerator and denominator. Step 5: The ratio simplifies to 1 / (1/3) = 3. Step 6: Therefore, the ratio of volumes is 3 : 1. Verification / Alternative check: To verify, we can assign simple values, for example r = 1 and h = 1. Then the cylinder volume becomes π * 1² * 1 = π. The cone volume becomes (1/3) * π * 1² * 1 = π/3. The ratio π : π/3 simplifies to 1 : 1/3, which is 3 : 1. Any other common choice of r and h would scale both volumes by the same factor, leaving the ratio unchanged, so the result 3 : 1 is robust and independent of actual dimensions. Why Other Options Are Wrong: 2 : 1, 4 : 1, and 5 : 1 imply different numerical relationships that are not supported by the volume formulas and would require the cone to have a different factor than 1/3 in its formula. 1 : 3 reverses the correct ratio and suggests that the cone volume is greater than the cylinder volume, which is not possible when both share the same base and height because the cone occupies only a third of the corresponding cylinder's volume. Common Pitfalls: Some learners misremember the cone volume formula and use 1/2 instead of 1/3, leading to incorrect ratios. Others forget that the base and height are the same and attempt unnecessary calculations with arbitrary values. A few students also confuse ratio order and write the cone to cylinder ratio instead of cylinder to cone. Remembering the correct formula and the requested order ensures the correct ratio is reported. Final Answer: The ratio of the volume of the cylinder to the volume of the cone is 3 : 1.

Question 3

A square pyramid has a base that is a square of side 6 cm and a slant height of 5 cm. What is its slant (lateral) surface area (in square centimetres)?

Accepted Answer

Introduction / Context: This question asks for the slant or lateral surface area of a square pyramid. The base is a square and each of the four lateral faces is a congruent isosceles triangle with given base and slant height. Lateral surface area is important in applications involving roof design, packaging, or any structure where the sides are triangular faces attached to a polygonal base. Given Data / Assumptions: • The base is a square with side length 6 cm. • The slant height of the pyramid (height of each triangular face measured along its altitude) is 5 cm. • We are asked for the slant surface area, meaning the total area of the four triangular faces, not including the base. Concept / Approach: Each lateral face is an isosceles triangle whose base equals the side of the square base and whose height equals the slant height of the pyramid. The area of one triangle is (1/2) * base * height. Since there are four such identical triangles, the total lateral area is 4 times the area of one triangle. There is no need to find the vertical height or total surface area, because the question specifically asks for slant surface area only. Step-by-Step Solution: Step 1: Base side length a = 6 cm. Step 2: Slant height l = 5 cm. Step 3: Area of one triangular face = (1/2) * base * height = (1/2) * 6 * 5. Step 4: Compute this area: (1/2) * 6 * 5 = 3 * 5 = 15 square centimetres. Step 5: There are four such triangular faces, so total slant area = 4 * 15 = 60 square centimetres. Verification / Alternative check: We can also use the general formula for the lateral surface area of a square pyramid: Lateral area = 2 * a * l, where a is the base side and l is the slant height. Substituting a = 6 and l = 5, we get 2 * 6 * 5 = 60 square centimetres. This matches the earlier calculation based on four triangular faces, so the answer is consistent and correct. Why Other Options Are Wrong: 30 square centimetres corresponds to only two faces or half of the lateral surface area, so it is too small. 40 and 45 square centimetres do not correspond to any correct combination of triangular areas given the specified dimensions. 50 square centimetres is also not a multiple of the single triangle area of 15 square centimetres and therefore cannot represent the total lateral area of four identical faces. Common Pitfalls: A common mistake is to confuse the slant surface area with the total surface area and either add or subtract the base area improperly. Another error is to misinterpret the slant height as a vertical height or to use the wrong formula. Some learners also forget to multiply by 4 after computing the area of one triangular face. Using the simple formula 2 * a * l or carefully counting the triangles avoids these issues. Final Answer: The slant surface area of the square pyramid is 60 sq.cm.

Question 4

The radii of two cylinders are in the ratio 3 : 2 and their heights are in the ratio 3 : 7. What is the ratio of their volumes?

Accepted Answer

Introduction / Context: This question checks the understanding of how the volume of a cylinder depends on its radius and height. We are given the ratios of the radii and heights of two cylinders and asked to find the ratio of their volumes. It illustrates how to combine given ratios and apply them to formula based relationships, an important skill in quantitative reasoning. Given Data / Assumptions: • Let the radii of the two cylinders be r₁ and r₂ with r₁ : r₂ = 3 : 2. • Let their heights be h₁ and h₂ with h₁ : h₂ = 3 : 7. • Volumes of the cylinders are V₁ and V₂ respectively. • We need the ratio V₁ : V₂. Concept / Approach: The volume of a right circular cylinder is V = πr²h. For two cylinders, we form the ratio V₁ / V₂ = (πr₁²h₁) / (πr₂²h₂). The π cancels, leaving a ratio involving r₁² : r₂² and h₁ : h₂. We use the given simple ratios to express r₁ and r₂ and h₁ and h₂ in proportional terms, and then compute the resulting ratio of volumes. Finally, we simplify the fraction to lowest terms. Step-by-Step Solution: Step 1: Use the ratio r₁ : r₂ = 3 : 2. We can take r₁ = 3k and r₂ = 2k for some positive k. Step 2: Use the ratio h₁ : h₂ = 3 : 7. We can take h₁ = 3m and h₂ = 7m for some positive m. Step 3: Volume of cylinder 1 is V₁ = π r₁² h₁ = π (3k)² (3m) = π * 9k² * 3m = 27πk²m. Step 4: Volume of cylinder 2 is V₂ = π r₂² h₂ = π (2k)² (7m) = π * 4k² * 7m = 28πk²m. Step 5: Form the ratio V₁ : V₂ = 27πk²m : 28πk²m. Step 6: Cancel π, k², and m from numerator and denominator to get 27 : 28. Verification / Alternative check: We can also substitute specific values consistent with the ratios, such as r₁ = 3, r₂ = 2, h₁ = 3, and h₂ = 7. Then V₁ = π * 3² * 3 = π * 9 * 3 = 27π and V₂ = π * 2² * 7 = π * 4 * 7 = 28π. The ratio 27π : 28π simplifies directly to 27 : 28. This confirms the earlier symbolic calculation and shows that the exact proportional constants k and m do not affect the ratio. Why Other Options Are Wrong: 4 : 7 and 7 : 4 come from combining radii and heights incorrectly, for example by ignoring the square on the radius or by adding exponents wrongly. 28 : 27 is the reverse of the correct ratio, implying the second cylinder is slightly smaller, which is not supported by the calculated volumes. 9 : 7 reflects only the square of the radius ratio (3² : 2²) but ignores the height ratio, so it is incomplete and incorrect. Common Pitfalls: A typical error is to forget that the radius is squared in the volume formula, leading to using r instead of r² in the ratio. Another pitfall is mixing up the order of the ratio or failing to incorporate the height ratio. Some learners attempt to add or multiply ratios incorrectly without grounding them in the formula. The correct approach is to substitute proportional values and compute the resulting volumes before forming the ratio. Final Answer: The ratio of the volumes of the two cylinders is 27 : 28.

Question 5

A cylindrical vessel of radius 6 cm is partially filled with water. If a solid sphere of radius 5 cm is completely immersed in the vessel, by how much will the water level rise? (Take π = 22/7.)

Accepted Answer

Introduction / Context: This question concerns volume displacement. When a solid object such as a sphere is fully immersed in a liquid, it displaces a volume of liquid equal to its own volume. In a cylindrical vessel, this displaced volume causes the water level to rise. By equating the volume of the sphere to the volume increase of water in the cylinder, we can compute the height by which the water level rises. This principle is widely used in physics and engineering, especially in fluid mechanics and measurement methods based on displacement. Given Data / Assumptions: • The radius of the cylindrical vessel, R, is 6 cm. • The radius of the solid sphere, r, is 5 cm. • The sphere is completely immersed in the water. • π is to be taken as 22/7. • The water level rise is h cm, which we must find. • The sphere is fully below the water surface and does not touch the bottom in a way that changes the effective volume relationship. Concept / Approach: The volume of a sphere is V_sphere = (4/3)πr³. When this sphere is immersed in the cylindrical vessel, it displaces an equal volume of water. The cylinder volume associated with a height increase h is V_cylinder = πR²h. Setting these equal gives (4/3)πr³ = πR²h. After cancelling π, we can solve for h. Finally, substitute the numeric values and simplify to obtain the rise in water level. The question options are given in centimetres rounded to two decimal places. Step-by-Step Solution: Step 1: Write the displaced volume equality: (4/3)πr³ = πR²h. Step 2: Cancel π from both sides to get (4/3)r³ = R²h. Step 3: Substitute r = 5 cm and R = 6 cm: (4/3) * 5³ = 6² * h. Step 4: Compute 5³ = 125 and 6² = 36, so (4/3) * 125 = 36h. Step 5: Evaluate (4/3) * 125 = 500/3. Step 6: So we have 500/3 = 36h, giving h = (500/3) / 36 = 500 / 108. Step 7: Simplify the fraction: divide numerator and denominator by 4 to get h = 125 / 27. Step 8: Evaluate 125 / 27 ≈ 4.63 cm when rounded to two decimal places. Verification / Alternative check: We can check the reasonableness of 4.63 cm by approximating. The sphere volume is roughly (4/3) * 3.14 * 5³ ≈ 4.19 * 125 ≈ 523 cubic centimetres. The cylinder cross section area is πR² ≈ 3.14 * 36 ≈ 113 cubic centimetres. Dividing 523 by 113 gives approximately 4.6 to 4.7 cm, which agrees closely with 4.63 cm. This consistency between detailed fraction based computation and rough estimation confirms the correctness of the result. Why Other Options Are Wrong: 4.50 cm is slightly smaller than the correct value and would underestimate the displaced volume when multiplied by the cylinder's cross sectional area. 5.56 cm, 6.67 cm, and 6.94 cm are all larger and would imply displaced volumes greater than that of the sphere. None of these match the exact fraction 125 / 27 derived from equating the volumes. Common Pitfalls: Students sometimes forget to cancel π from both sides and carry it incorrectly through the equation. Another frequent mistake is to use the diameter instead of the radius in the sphere volume or cylinder area formulas. Errors in cube and square calculations, such as miscomputing 5³ or 6², also lead to wrong answers. Finally, some may round too early and get inaccurate final values. Keeping exact fractions and only rounding at the final step helps maintain accuracy. Final Answer: The water level in the cylindrical vessel rises by approximately 4.63 cm.

Question 6

Three consecutive angles of a cyclic quadrilateral are in the ratio 1 : 4 : 5. What is the measure of the fourth angle (in degrees)?

Accepted Answer

Introduction / Context: This problem is about angle properties of a cyclic quadrilateral, which is a quadrilateral whose vertices all lie on a circle. In such a quadrilateral, opposite angles are supplementary, meaning they add up to 180 degrees. Using the given ratio of three consecutive angles, we can find the actual angles and then determine the fourth angle. This kind of question checks knowledge of both ratio manipulation and special properties of cyclic quadrilaterals. Given Data / Assumptions: • The quadrilateral is cyclic, so opposite interior angles sum to 180 degrees. • Three consecutive angles are in the ratio 1 : 4 : 5. • Let the three consecutive angles be A, B, and C, with A : B : C = 1 : 4 : 5. • The fourth angle is D, which we must find in degrees. Concept / Approach: Assign a variable x to represent the common ratio unit. Then the three consecutive angles can be written as x, 4x, and 5x. In a quadrilateral, the sum of all four interior angles is 360 degrees. In a cyclic quadrilateral, each pair of opposite angles also sums to 180 degrees. If we take A, B, C, D in order, then A and C are opposite angles. We can write equations using both the 360 degree sum and the 180 degree opposite angle property to solve for x and then find D. Step-by-Step Solution: Step 1: Let A = x, B = 4x, and C = 5x according to the given ratio 1 : 4 : 5. Step 2: Since the quadrilateral is cyclic, opposite angles sum to 180 degrees. Angles A and C are opposite, so A + C = 180 degrees. Step 3: Substitute A = x and C = 5x to get x + 5x = 180. Step 4: Simplify to 6x = 180, so x = 180 / 6 = 30 degrees. Step 5: Now A = x = 30 degrees, B = 4x = 120 degrees, and C = 5x = 150 degrees. Step 6: The sum of all interior angles is 360 degrees, so A + B + C + D = 360. Step 7: Substitute values: 30 + 120 + 150 + D = 360, so 300 + D = 360. Step 8: Solve for D: D = 360 - 300 = 60 degrees. Verification / Alternative check: We can verify the opposite angle property using the found values. A is 30 degrees and C is 150 degrees; their sum is 30 + 150 = 180 degrees, satisfying the cyclic quadrilateral condition. B is 120 degrees and D is 60 degrees, and their sum is also 120 + 60 = 180 degrees, confirming that they form the other pair of opposite angles. The total of all angles is 30 + 120 + 150 + 60 = 360 degrees, which matches the general quadrilateral angle sum. This consistent set of checks confirms that D = 60 degrees is correct. Why Other Options Are Wrong: 120 degrees would make B + D = 120 + 120 = 240 degrees, which violates the 180 degree opposite angle condition for a cyclic quadrilateral. 30 degrees would force two angles to be equal and break the given 1 : 4 : 5 ratio structure when the full quadrilateral sum is considered. 80 and 100 degrees also fail either the total angle sum of 360 degrees or the requirement that opposite angles in a cyclic quadrilateral must be supplementary. They do not fit the consistent set of values that we derived. Common Pitfalls: A common mistake is to use only the 360 degree sum and ignore the special cyclic property, which can lead to incorrect systems of equations. Others may misidentify which angles are opposite or may assign the ratio incorrectly to nonconsecutive angles. Miscalculating the ratio sum or arithmetic errors when solving for x can also occur. Carefully naming the angles in order and explicitly using both the 360 degree and 180 degree rules helps avoid these traps. Final Answer: The measure of the fourth angle of the cyclic quadrilateral is 60°.

Question 7

In triangle ABC, angle ABC is 15 degrees. Point D lies on side BC such that AD = BD. What is the measure of angle ADC (in degrees)?

Accepted Answer

Introduction / Context: This is a geometry problem involving special angle constructions in a triangle. Triangle ABC has angle ABC equal to 15 degrees, and a point D is chosen on side BC such that AD equals BD. We are asked to find angle ADC. This is a nontrivial configuration and requires careful geometric reasoning or construction, often using properties of isosceles triangles and angle chasing. Such questions appear in higher level aptitude and Olympiad style geometry sections. Given Data / Assumptions: • Triangle ABC is any triangle with angle ABC = 15 degrees. • Point D lies on side BC of triangle ABC. • Length AD equals length BD (AD = BD). • We must determine the measure of angle ADC in degrees. • The configuration implicitly determines D uniquely so that AD = BD is satisfied. Concept / Approach: Because AD = BD, triangle ABD is isosceles with AB as the base. The given angle at B in triangle ABC is small (15 degrees), which suggests that known angle splitting patterns like 15, 30, 45, and 60 degrees may arise. A common approach is to place the triangle on coordinates or use classical angle chasing. If we place B at the origin, take BC along the x axis, and choose A at an appropriate angle, we can solve using analytic geometry, enforcing AD = BD to determine D, and then compute angle ADC. It turns out that this configuration leads to angle ADC equal to 30 degrees. Step-by-Step Solution: Step 1: Conceptually, let us place B at (0, 0) and C along the x axis. Let BC = 1 unit for convenience, so C is at (1, 0). Step 2: Angle ABC is 15 degrees, so side BA makes an angle of 15 degrees with BC. Take AB = 1 unit for convenience, so A has coordinates (cos 15°, sin 15°). Step 3: Let D be at (t, 0) along BC, with 0 < t < 1. Then BD = t and AD is the distance between A and D. Step 4: Impose the condition AD = BD. Squaring both distances gives an equation in t. Solving this yields t ≈ 0.5176, which lies between 0 and 1, so D is indeed on BC. Step 5: With the coordinates of A, D, and C known, compute the angle at D formed by segments DA and DC using the vector or dot product method. The resulting angle ADC is found to be exactly 30 degrees. Verification / Alternative check: Instead of coordinates, one can use synthetic geometry. Since AD = BD, triangle ABD is isosceles with angles at A and B equal. If angle B in triangle ABC is 15 degrees, then angle ABD equals angle BAD = 15 degrees. With appropriate constructions and focusing on triangle ADC, angle chasing using known sums of angles in triangles can show that angle ADC must be 30 degrees. The coordinate method and synthetic approach both agree on 30 degrees, so the result is confirmed. Why Other Options Are Wrong: 15 degrees would make angle ADC equal to angle ABC, which contradicts the internal isosceles structure created by AD = BD. 45 and 60 degrees do not fit the precise angle relations that come from AD = BD and ABC = 15 degrees when checked by detailed angle chasing or coordinate geometry. 75 degrees is too large given the geometry and does not arise from any consistent configuration respecting all conditions. Only 30 degrees is compatible with both the distance equality and the angle structure. Common Pitfalls: A major pitfall is assuming D is some simple point like the midpoint of BC without verifying the AD = BD condition. Another error is applying angle bisector theorems where they do not apply, or assuming AD bisects some angle. Mismanaging the trigonometric or coordinate computations can also lead to incorrect numerical values. Drawing an accurate diagram, using either coordinate geometry or systematic angle chasing, and carefully enforcing AD = BD are key to solving this problem correctly. Final Answer: The measure of angle ADC is 30 degrees.

Question 8

Two circles with diameters 15 cm and 13 cm intersect so that their common chord has length 12 cm. What is the distance between the centres of the circles (in cm)?

Accepted Answer

Introduction / Context: This question involves intersecting circles and a common chord. The length of the common chord and the radii of the two circles are given, and we are asked to find the distance between the centres. This is a standard geometry configuration: the line joining the centres is perpendicular to the common chord and bisects it. Using right triangle relationships, we can relate the radii, the half chord length, and the distance from each centre to the chord to obtain the distance between centres. Given Data / Assumptions: • Diameter of the first circle is 15 cm, so its radius r₁ = 7.5 cm. • Diameter of the second circle is 13 cm, so its radius r₂ = 6.5 cm. • The length of the common chord is 12 cm. • The chord is common to both circles and lies between their centres. • The line joining the centres is perpendicular to the common chord and bisects it. • We need the distance d between the two centres. Concept / Approach: If two circles intersect, their common chord is perpendicular to the line joining the centres and is bisected by that line. Let the midpoint of the chord be M. Then the distance from each centre to M, say d₁ and d₂, can be found from right triangles using Pythagoras theorem: d₁² = r₁² - (chord/2)² and d₂² = r₂² - (chord/2)². The distance between the centres is then d = d₁ + d₂ because the centres, the midpoint of the chord, and the chord itself form a straight configuration with M lying between the centres. Step-by-Step Solution: Step 1: Compute the radii: r₁ = 15 / 2 = 7.5 cm, r₂ = 13 / 2 = 6.5 cm. Step 2: The chord length is 12 cm, so half the chord length is 12 / 2 = 6 cm. Step 3: For the first circle, let d₁ be the distance from its centre to the chord midpoint. Use Pythagoras: d₁² = r₁² - 6². Step 4: Compute r₁² = 7.5² = 56.25, so d₁² = 56.25 - 36 = 20.25, giving d₁ = 4.5 cm. Step 5: For the second circle, let d₂ be the distance from its centre to the chord midpoint. Then d₂² = r₂² - 6². Step 6: Compute r₂² = 6.5² = 42.25, so d₂² = 42.25 - 36 = 6.25, giving d₂ = 2.5 cm. Step 7: The centres lie on opposite sides of the chord midpoint, so the distance between the centres is d = d₁ + d₂ = 4.5 + 2.5 = 7 cm. Verification / Alternative check: We can verify by visualising the configuration. The line of centres is a straight line that passes through the chord midpoint M, with distances d₁ and d₂ from the centres. Because the chord is common and perpendicular to this line, and its half length is 6 cm, the right triangles formed in each circle are consistent. Squaring the sums or recomputing the distances ensures there is no arithmetic mistake. Since both right triangles satisfy Pythagoras theorem with the same half chord, adding the distances 4.5 cm and 2.5 cm naturally gives the full separation between centres as 7 cm, which is a simple and plausible result. Why Other Options Are Wrong: 3.5 cm is too small and would not allow a chord of length 12 cm to exist with the given radii. 7√2 cm and 14 cm are significantly larger distances, which would change the geometry so that either the circles do not intersect or the chord length changes. 5 cm does not match the sum of the distances d₁ and d₂ computed from the radii and chord length. Only 7 cm works with both radii and the 12 cm chord consistently. Common Pitfalls: Learners may forget that the chord is bisected by the line joining the centres and may incorrectly use the full chord length instead of half. Another common mistake is to subtract distances in the wrong order and attempt d = |d₁ - d₂| instead of d₁ + d₂. Errors in squaring the radii or half chord length can also lead to incorrect values for d₁ and d₂. Keeping track of the geometry and the positions of the centres relative to the chord is essential. Final Answer: The distance between the centres of the two circles is 7 centimetres.

Question 9

A solid sphere of diameter 7 cm is cut into two equal hemispheres. By how many square centimetres does the total surface area increase?

Accepted Answer

Introduction / Context: This question compares the total surface area of a solid sphere with that of two hemispheres obtained by cutting the sphere into two equal halves. Cutting a sphere exposes additional flat circular surfaces, increasing the total surface area. Understanding how surface areas change when solids are cut or recombined is important in geometry and practical applications such as material coating or heat transfer calculations. Given Data / Assumptions: • The original solid is a sphere with diameter 7 cm, so its radius r = 3.5 cm. • The sphere is cut into two equal hemispheres. • Surface area of a sphere is 4πr². • Surface area of a hemisphere includes both the curved surface and the flat circular base. • We need the increase in total surface area after cutting, in square centimetres. Concept / Approach: First compute the total surface area of the original sphere using 4πr². Then consider each hemisphere. A hemisphere has curved surface area equal to half of the full sphere surface, that is 2πr², plus the area of the base circle, which is πr². So the surface area of one hemisphere is 3πr². For two hemispheres, multiply by 2 to get 6πr². The increase in total surface area is the difference between 6πr² and 4πr², which simplifies to 2πr². Substituting r = 3.5 cm and a standard approximation for π gives the numerical answer. Step-by-Step Solution: Step 1: Radius of the sphere r = diameter / 2 = 7 / 2 = 3.5 cm. Step 2: Total surface area of the original sphere is S_sphere = 4πr². Step 3: Compute r² = 3.5² = 12.25, so S_sphere = 4π * 12.25 = 49π square centimetres. Step 4: Surface area of one hemisphere is S_hemi = curved area + base area = 2πr² + πr² = 3πr². Step 5: For two hemispheres, the total surface area is S_two = 2 * 3πr² = 6πr² = 6π * 12.25 = 73.5π square centimetres. Step 6: The increase in surface area is ΔS = S_two - S_sphere = 73.5π - 49π = 24.5π. Step 7: Take π = 22/7. Then ΔS = 24.5 * 22 / 7. Since 24.5 = 49 / 2, we get ΔS = (49 / 2) * 22 / 7 = (49 * 22) / (14) = (49 / 14) * 22 = 3.5 * 22 = 77 square centimetres. Verification / Alternative check: We can check the computations directly. Using r = 3.5 cm and π = 22/7, S_sphere = 4πr² = 4 * 22/7 * 12.25. Since 12.25 = 49 / 4, this becomes 4 * 22/7 * 49 / 4 = 22 * 7 = 154 square centimetres. For two hemispheres, S_two = 6πr² = 6 * 22/7 * 12.25, which simplifies to 6 * 22/7 * 49/4 = 3 * 22 * 7 / 7 = 3 * 22 * 7 / 4 when computed carefully, giving 231 square centimetres. The difference 231 - 154 equals 77 square centimetres, confirming the earlier result. Why Other Options Are Wrong: 154 represents the total surface area of the original sphere, not the increase in area. 38.5 and 87 are smaller values that do not equal the exact computed increase 24.5π when converted using π = 22/7. 96 does not match any clear combination of sphere and hemisphere surface areas. Only 77 square centimetres is the correct difference between the two total surface areas. Common Pitfalls: A common mistake is to forget to add the base circles when computing the surface area of hemispheres, leading to an underestimate. Another error is to subtract in the wrong order or to miscalculate r². Some learners also confuse diameter and radius when applying formulas. Keeping track of which surfaces are exposed before and after cutting, and using the formulas consistently, helps avoid these errors. Final Answer: The total surface area increases by 77 square centimetres when the sphere is cut into two hemispheres.

Question 10

In triangle PQR, angle QPR is 45 degrees. The internal bisectors of angles PQR and PRQ intersect at point O. What is the measure of angle QOR (in degrees)?

Accepted Answer

Introduction / Context: This question is about angles formed at the incenter of a triangle. In triangle PQR, the internal angle bisectors from vertices Q and R intersect at O, which is the incenter. We are given that angle QPR (the angle at P) is 45 degrees and asked to find angle QOR, the angle at the incenter between the bisectors from Q and R. This uses a standard property of incenters that relates the angle between two internal bisectors to the opposite vertex angle. Given Data / Assumptions: • Triangle PQR is any triangle with angle QPR = 45 degrees. • The internal bisector of angle PQR and the internal bisector of angle PRQ intersect at O, the incenter of triangle PQR. • We must determine the measure of angle QOR in degrees. • All angles mentioned are internal angles in the triangle or at the incenter. Concept / Approach: In any triangle, the incenter is the intersection point of the three internal angle bisectors. A key property is that the angle at the incenter formed by the bisectors of two angles is equal to 90 degrees plus half of the remaining angle. Specifically, the angle between the bisectors of angles at Q and R is 90° + (angle P)/2. In this problem, angle P is given as 45 degrees, so we can directly apply this formula to find angle QOR. Step-by-Step Solution: Step 1: Note that the bisectors from Q and R meet at O, making O the incenter of triangle PQR. Step 2: Recall the standard result: angle QOR = 90° + (angle QPR)/2, where angle QPR is the remaining vertex angle at P. Step 3: Given angle QPR = 45°, substitute this value into the formula. Step 4: Compute half of angle QPR: (angle QPR)/2 = 45° / 2 = 22.5°. Step 5: Add 90° to this half angle: 90° + 22.5° = 112.5°. Step 6: Therefore, angle QOR is 112.5 degrees. Verification / Alternative check: We can verify this property by recalling the general relationships at the incenter. The three angles at the incenter formed by pairs of adjacent angle bisectors sum to 360 degrees. These angles are 90° + (A/2), 90° + (B/2), and 90° + (C/2) for a triangle with angles A, B, and C at the vertices. Their sum is 270° + (A + B + C)/2 = 270° + 180° / 2 = 270° + 90° = 360°, which is consistent. Applying this specifically with A = angle P, the angle between the bisectors at Q and R must be 90° + A/2. Substituting A = 45° gives 112.5° as found above. Why Other Options Are Wrong: 107.5°, 117.5°, and 122.5° differ from 112.5° by 5° or 10°, and none of them equal 90° + 45° / 2. 135° would correspond to 90° + 45°, which would be the wrong application of the formula, using the full angle instead of half the angle at P. Therefore, these values do not match the incenter angle property for this triangle. Common Pitfalls: A common error is to use 90° + angle P instead of 90° + angle P/2, leading to 135° in this case. Others may confuse which vertex angle is used in the formula, sometimes taking half of Q or R instead of P. Another issue is forgetting that the bisectors intersect at the incenter, not some arbitrary point. Remembering the specific formula for angles at the incenter and carefully identifying the opposite vertex angle avoids these errors. Final Answer: The measure of angle QOR is 112.5 degrees.

Question 11

ABCD is a quadrilateral with AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm, and angle ABC is a right angle. What is the area of triangle ADC (in square centimetres)?

Accepted Answer

Introduction / Context: This question involves a quadrilateral with one right angle and known side lengths. We are asked to find the area of one of the triangles formed by its vertices, namely triangle ADC. By placing the figure on a coordinate plane in a convenient way and using distance relations, we can determine the coordinates of point D and then compute the area of triangle ADC using base and height. This approach highlights how coordinate geometry can simplify area calculations for composite figures. Given Data / Assumptions: • Quadrilateral ABCD has side lengths AB = 9 cm, BC = 40 cm, CD = 28 cm, and DA = 15 cm. • Angle ABC is a right angle (90 degrees). • We are required to find the area of triangle ADC. • No other angles are explicitly given, but the distances and right angle condition determine the geometry uniquely enough for this area. Concept / Approach: Because angle ABC is 90 degrees, we can choose a coordinate system with B at the origin, BC on the x axis, and BA on the y axis. Then we can locate points A and C easily. Point D must satisfy two distance conditions: its distance from C is 28 cm and from A is 15 cm. Solving these conditions shows that D lies on segment BC. Once we know the coordinates of A, D, and C, triangle ADC has a horizontal base DC and a vertical height equal to the y coordinate of A, making area computation straightforward using the formula (1/2) * base * height. Step-by-Step Solution: Step 1: Place point B at (0, 0). Let BC lie along the x axis with C at (40, 0) so that BC = 40 cm. Step 2: Since angle ABC is 90 degrees and AB = 9 cm, let A be at (0, 9) on the y axis. Step 3: Point D must satisfy CD = 28 cm and DA = 15 cm. Let D = (x, 0) on the x axis. Then distance CD = |40 - x| must equal 28, and distance AD must be 15. Step 4: From CD = 28, we get |40 - x| = 28, so x = 12 or x = 68. Point D must lie between B and C, so x = 12 is valid and D is at (12, 0). Step 5: Verify AD: distance from A(0, 9) to D(12, 0) is √[(12 - 0)² + (0 - 9)²] = √(144 + 81) = √225 = 15 cm, which matches the given DA. Step 6: Now triangle ADC has vertices A(0, 9), D(12, 0), and C(40, 0). The base DC lies along the x axis with length DC = 40 - 12 = 28 cm. Step 7: The height of triangle ADC from A to base DC is the vertical distance from A to the x axis, which is 9 cm. Step 8: Area of triangle ADC = (1/2) * base * height = (1/2) * 28 * 9 = 14 * 9 = 126 square centimetres. Verification / Alternative check: We can verify the area using the coordinate formula for area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃). Applying that formula to A(0, 9), D(12, 0), and C(40, 0) gives area = (1/2) * |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|. Substituting the coordinates and simplifying yields (1/2) * |0(0 - 0) + 12(0 - 9) + 40(9 - 0)| = (1/2) * |0 - 108 + 360| = (1/2) * 252 = 126. This matches the earlier base height method and confirms the answer. Why Other Options Are Wrong: 124, 122, 120, and 130 square centimetres are all close but do not match the exact computed area of 126 square centimetres. They might arise from arithmetic mistakes in computing distances or in applying the area formula, but they are not consistent with either the base height approach or the coordinate geometry formula. Common Pitfalls: Typical errors include misplacing the points in the coordinate plane, miscomputing the length of DC, or incorrectly calculating the area using wrong base or height. Some learners might also attempt to split the quadrilateral in a different way and end up with more complicated calculations. Using the right angle at B to set up simple coordinates and then applying basic triangle area formulas is the cleanest method and avoids unnecessary complexity. Final Answer: The area of triangle ADC is 126 SQ.CM.

Question 12

In a circle of radius 8 cm, AB and AC are two equal chords, each measuring 12 cm. If B and C are the other endpoints of these chords on the circle, what is the length of the chord BC that joins points B and C?

Accepted Answer

Introduction / Context: In this geometry question, we work with chords in a circle and use basic circle properties to find the length of an unknown chord. The circle has radius 8 cm, and two equal chords AB and AC measure 12 cm each. Points B and C lie on the circle, and we are asked to determine the length of the chord BC. This tests understanding of the relation between chord length and the central angle that subtends the chord, as well as some simple trigonometry inside a circle. Given Data / Assumptions: - Radius of the circle r = 8 cm. - Chords AB and AC are equal: AB = AC = 12 cm. - B and C lie on the circle, and BC is the chord joining them. - The centre of the circle is denoted by O. - Standard chord length formula is used: chord length = 2 * r * sin(central angle / 2). Concept / Approach: To solve this, we use the fact that equal chords subtend equal central angles at the centre of a circle. Let the central angle subtended by chord AB be θ. Then chord AB has length 2 * r * sin(θ / 2). From AB = 12 cm and r = 8 cm, we can compute sin(θ / 2). Once we know θ, we can find the central angle between B and C, which is the angle subtending chord BC. Because AB and AC are symmetric around OA, if OB and OC are drawn, the angle BOC becomes 2θ. Then we again use the chord formula for BC with angle BOC to get its length. Step-by-Step Solution: Step 1: Let O be the centre of the circle. Draw OA, OB, and OC so that OA, OB, and OC are radii of length 8 cm. Step 2: Let the central angle AOB be θ. Since AB is a chord of length 12 cm, use AB = 2 * r * sin(θ / 2). Step 3: Substitute AB = 12 and r = 8: 12 = 2 * 8 * sin(θ / 2) which simplifies to 12 = 16 * sin(θ / 2). Step 4: Therefore sin(θ / 2) = 12 / 16 = 3 / 4. Step 5: Compute cos(θ / 2) using the identity sin^2(x) + cos^2(x) = 1. So cos(θ / 2) = √(1 - (3 / 4)^2) = √(1 - 9 / 16) = √(7 / 16) = √7 / 4. Step 6: Now find sin(θ) using the double angle formula sin(θ) = 2 * sin(θ / 2) * cos(θ / 2) = 2 * (3 / 4) * (√7 / 4) = (3 * √7) / 8. Step 7: Points B and C are symmetric with respect to OA, so the central angle BOC is 2θ. Step 8: Chord BC subtends angle BOC = 2θ at the centre. The chord length formula uses half of this angle, so BC = 2 * r * sin(BOC / 2) = 2 * r * sin(θ). Step 9: Substitute r = 8 and sin(θ) = (3 * √7) / 8 to get BC = 2 * 8 * (3 * √7 / 8) = 16 * (3 * √7 / 8) = 6√7 cm. Verification / Alternative check: An alternative method is to construct triangle OAB and use right triangle geometry by dropping a perpendicular from O to chord AB, then repeating a similar idea for BC. However, the trigonometric method used above is standard and efficient. Numeric substitution gives a positive value for BC, and the units are consistent in centimetres, so the result is reasonable. Why Other Options Are Wrong: Option 2√6 cm is too small and would correspond to a much smaller central angle than the one created by points B and C. Option 3√6 cm also does not match the value of BC found using the correct chord and angle relations. Option 3√7 cm is close in form but is only half of the correct value 6√7 cm. Common Pitfalls: Many learners forget that the chord formula uses half the central angle, leading to mistakes when relating chord length to the angle. Others may assume BC is equal to AB simply because AB and AC are equal, which is not true. It is also easy to confuse angle θ with 2θ when moving from chords AB and AC to chord BC. Careful use of the double angle identity for sine is important here. Final Answer: The length of chord BC in the given circle is 6√7 cm.

Question 13

Consider a triangle in plane geometry. 1) The perimeter of a triangle is greater than the sum of the lengths of its three medians. 2) In any triangle ABC, if D is any point on side BC, then AB + BC + CA is greater than 2AD. Which of the above statem…

Accepted Answer

Introduction / Context: This is a conceptual geometry question that asks you to judge the validity of two statements about triangles. Both statements involve inequalities between side lengths and related segments such as medians and line segments drawn from a vertex to a point on the opposite side. Such questions are common in aptitude and competitive exams because they test understanding of triangle inequality and properties of medians without requiring numerical calculation. Given Data / Assumptions: - We consider a general triangle with sides of positive length. - Statement 1: The perimeter of a triangle is greater than the sum of its three medians. - Statement 2: For triangle ABC and any point D on side BC, AB + BC + CA is greater than 2AD. - No special type of triangle is assumed unless implied by standard geometry (triangle is non degenerate). Concept / Approach: For Statement 1, we use a known inequality involving medians: in any triangle, the sum of the medians is less than the perimeter. This follows from the fact that each median is shorter than the average of the two sides it connects in an indirect way and from geometric constructions that compare medians with sides. For Statement 2, we apply the triangle inequality carefully to two smaller triangles formed when we choose a point D on side BC. By writing the triangle inequality in triangles ABD and ACD and then adding the inequalities, we derive AB + BC + CA greater than 2AD. Step-by-Step Solution: Step 1: Let the triangle have sides a, b, c and medians ma, mb, mc drawn to those sides. Step 2: It is a known geometric result that ma + mb + mc is less than a + b + c. Geometrically, one way to see this is by constructing the triangle formed by the three medians and comparing its perimeter to the original triangle, or by more advanced inequality proofs. The key is that each median is always less than half the sum of the two sides that form the endpoints of the median. Step 3: Since a + b + c is the perimeter of the triangle, and ma + mb + mc is less than this perimeter, Statement 1, which claims that the perimeter is greater than the sum of medians, is correct. Step 4: For Statement 2, consider triangle ABC and let D be any point on side BC. Step 5: In triangle ABD, the triangle inequality gives AB + BD greater than AD. Step 6: In triangle ACD, the triangle inequality gives AC + CD greater than AD. Step 7: Add these two inequalities: (AB + BD) + (AC + CD) greater than AD + AD which simplifies to AB + AC + (BD + CD) greater than 2AD. Step 8: Since D lies on BC, BD + CD equals BC. So we have AB + AC + BC greater than 2AD. Step 9: Therefore Statement 2 is also correct. Verification / Alternative check: You can test Statement 2 with a simple numerical example. Consider a right triangle with AB = 3, AC = 4, BC = 5. Choose D as the midpoint of BC so that BD = CD = 2.5. Compute AD using the distance formula or Pythagoras in coordinates. You will find that AB + BC + CA is comfortably larger than 2AD, confirming the inequality numerically. Since the proof using triangle inequalities was general, the statement holds for any choice of D on BC. Why Other Options Are Wrong: Option 1 only would mean Statement 2 is false, but we have shown that the triangle inequality directly proves Statement 2. Option 2 only would mean Statement 1 is false, which contradicts the known result that the sum of medians is always less than the perimeter of the triangle. Option Neither 1 nor 2 is incorrect because both statements are valid geometric facts. Common Pitfalls: Students sometimes misremember properties of medians and may think that the sum of medians can exceed the perimeter, which is not true. Another common mistake is to apply the triangle inequality incorrectly, for example writing AB + AC greater than BC + AD incorrectly or mixing up which segments belong to which triangle. It is important to write triangle inequalities separately for each triangle, then add them carefully. Also, some learners mistakenly think that D must be the midpoint of BC; however, Statement 2 holds for any point D on BC, not just the midpoint. Final Answer: Both statements are correct, so the right choice is Both 1 and 2.

Question 14

In a triangle, each of the three sides is reduced to half of its original length. By what percentage does the area of the triangle decrease as a result of halving all the sides?

Accepted Answer

Introduction / Context: This aptitude question checks understanding of how the area of a triangle changes when all side lengths are scaled by a common factor. Instead of working with specific numbers, we look at the relationship between linear dimensions and area. Questions like this are common in geometry and mensuration, especially in exams where students must quickly see how scaling affects area or volume without recomputing from scratch. Given Data / Assumptions: - We have an original triangle with some base and height or with some sides. - Each side of the triangle is halved, so linear dimensions are scaled by a factor of 1/2. - We are asked to find the percentage decrease in area, not the new area itself. - The shape remains a similar triangle after scaling, only the size changes. Concept / Approach: The key concept is similarity and scaling. When all the sides of a triangle are multiplied by a factor k, the area is multiplied by k^2. Here, the factor is 1/2. Therefore the new area becomes (1/2)^2 times the original area. Once we know the ratio of new area to old area, we can easily compute the percentage decrease by comparing the old area and the new area and expressing the reduction as a percentage of the original area. Step-by-Step Solution: Step 1: Let the original area of the triangle be A square units. Step 2: Each side is halved. That means every linear dimension of the triangle is multiplied by 1/2. Step 3: For similar figures, area scales with the square of the linear scale factor. So new area = A * (1/2)^2. Step 4: Compute (1/2)^2 = 1/4. Step 5: Therefore the new area is A / 4, which is one fourth of the original area. Step 6: The amount of decrease in area is original area minus new area which is A - A / 4 = 3A / 4. Step 7: To find the percentage decrease, divide the decrease by the original area and multiply by 100 percent. Step 8: Percentage decrease = (3A / 4) / A * 100% = (3 / 4) * 100% = 75%. Verification / Alternative check: Take a simple example for verification. Suppose a right triangle has base 4 units and height 6 units. Its area is (1/2) * 4 * 6 = 12 square units. Now halve both base and height: new base = 2 units and new height = 3 units, so new area = (1/2) * 2 * 3 = 3 square units. Compare: original area 12, new area 3. The decrease is 9 square units. Compute percentage decrease: 9 / 12 * 100% = 75%. This matches the general derivation. Why Other Options Are Wrong: Option 50% would be correct if area depended linearly on side length, but area depends on the square of the scale factor, not the first power. Option 25% confuses the remaining area with the decrease. The new area is 25% of the original, not the decrease. Option No change is incorrect because the triangle clearly becomes smaller when sides are halved, so the area must reduce. Common Pitfalls: Many learners treat area as if it scales in direct proportion to side length rather than with the square of the side length. Another mistake is mixing up the remaining percentage and the percentage decrease. Here the new area is 25% of the original, so the percentage decrease is 75%, not 25%. Always carefully distinguish between what remains and what is lost. Remember that scaling laws for area and volume are powerful shortcuts for many exam problems. Final Answer: When each side of the triangle is halved, the area decreases by 75%.

Question 15

In a circle of radius 10 cm, two parallel chords have lengths 12 cm and 16 cm respectively. What is the distance between these two chords inside the circle (considering both possible positions of the chords)?

Accepted Answer

Introduction / Context: This problem involves two parallel chords in a circle and asks for the distance between them. When chords are parallel, their distances from the centre of the circle determine how far apart they are. Because chords can lie on the same side of the centre or on opposite sides of the centre, two possible distances often arise. This question tests understanding of the relationship between chord length, radius, and perpendicular distance from the centre of the circle to the chord. Given Data / Assumptions: - Radius of the circle r = 10 cm. - Length of one chord is 12 cm. - Length of the other chord is 16 cm. - Chords are parallel and lie inside the same circle. - Perpendicular distance from the centre to a chord is used to find how far chords are from the centre and from each other. Concept / Approach: For a circle of radius r, if a chord of length L is at a perpendicular distance d from the centre, we can use the right triangle formed by the radius, half of the chord, and the distance d. The relation is r^2 = d^2 + (L / 2)^2. From this we get d = √(r^2 - (L / 2)^2). Once we find the distances of both chords from the centre, the distance between the chords is either the difference of these distances (if chords are on the same side of the centre) or the sum of these distances (if chords lie on opposite sides of the centre). Step-by-Step Solution: Step 1: For the chord of length 12 cm, half the chord is L1 / 2 = 6 cm. Step 2: Use r^2 = d1^2 + (L1 / 2)^2 with r = 10. So 10^2 = d1^2 + 6^2 which gives 100 = d1^2 + 36. Step 3: Therefore d1^2 = 100 - 36 = 64, so d1 = √64 = 8 cm. Step 4: For the chord of length 16 cm, half the chord is L2 / 2 = 8 cm. Step 5: Use r^2 = d2^2 + (L2 / 2)^2 again. So 10^2 = d2^2 + 8^2 which gives 100 = d2^2 + 64. Step 6: Therefore d2^2 = 100 - 64 = 36, so d2 = √36 = 6 cm. Step 7: If both chords lie on the same side of the centre, the distance between them is the difference of their distances from the centre: |d1 - d2| = |8 - 6| = 2 cm. Step 8: If the chords lie on opposite sides of the centre, the distance between them is the sum of their distances from the centre: d1 + d2 = 8 + 6 = 14 cm. Step 9: Hence there are two possible distances between the two parallel chords: 2 cm or 14 cm. Verification / Alternative check: You can sketch a circle with centre O, place one chord closer to the centre and the other further out, and visually check that the computed distances 8 cm and 6 cm from the centre are reasonable for chords of lengths 12 cm and 16 cm. Since both distances are less than the radius 10 cm, both chords indeed lie inside the circle. Considering them on the same side or on opposite sides of the centre naturally gives the two possible distances, confirming our reasoning. Why Other Options Are Wrong: Option 1 cm or 7 cm does not match the correct distances derived from the radius and chord relations. Option 3 cm or 21 cm and Option 4 cm or 28 cm both give values that are inconsistent with the maximum possible separation within a circle of radius 10 cm. In particular, a distance of 21 cm or 28 cm would exceed the diameter, which is impossible for two chords inside the circle. Common Pitfalls: A common mistake is to forget to divide the chord length by 2 before applying the Pythagoras relation with the radius and the perpendicular distance. Another error is to think that there is only one possible distance between parallel chords, ignoring the possibility that chords can be on opposite sides of the centre. Learners may also incorrectly add or subtract distances without considering the geometry of where each chord lies relative to the centre. Final Answer: The distance between the two parallel chords can be either 2 cm or 14 cm depending on their positions relative to the centre of the circle.

Question 16

The co ordinates of the centroid of triangle ABC are given as (−1, 4). If the co ordinates of vertices A and B are (−3, −1) and (3, 5) respectively, what are the co ordinates of vertex C?

Accepted Answer

Introduction / Context: This coordinate geometry problem involves the centroid of a triangle. The centroid is the point where the three medians of a triangle intersect, and it has a simple formula in terms of the vertex co ordinates. In this question, the centroid and two vertices are known, and the task is to determine the co ordinates of the third vertex. This type of problem is very standard in analytic geometry and tests the ability to use and rearrange the centroid formula. Given Data / Assumptions: - Triangle ABC has centroid G with co ordinates (−1, 4). - Vertex A has co ordinates (−3, −1). - Vertex B has co ordinates (3, 5). - Vertex C has co ordinates (xC, yC), which we must find. - The centroid formula for a triangle with vertices (xA, yA), (xB, yB), (xC, yC) is used. Concept / Approach: The centroid of a triangle is the average of the x co ordinates and the average of the y co ordinates of its vertices. If G is the centroid, then xG = (xA + xB + xC) / 3 and yG = (yA + yB + yC) / 3. In this question, xG and yG are known along with xA, yA, xB, and yB. Therefore, we can set up two linear equations in xC and yC, solve them, and match the result with the provided options. Step-by-Step Solution: Step 1: Write down the centroid formulas. Step 2: For the x co ordinate: xG = (xA + xB + xC) / 3. Step 3: Substitute the known values: xG = −1, xA = −3, xB = 3. So we have −1 = (−3 + 3 + xC) / 3. Step 4: Simplify inside the bracket: −3 + 3 + xC = xC. So −1 = xC / 3. Step 5: Multiply both sides by 3 to get xC = −3. Step 6: For the y co ordinate: yG = (yA + yB + yC) / 3. Step 7: Substitute the known values: yG = 4, yA = −1, yB = 5. So 4 = (−1 + 5 + yC) / 3. Step 8: Simplify −1 + 5 + yC = 4 + yC. So 4 = (4 + yC) / 3. Step 9: Multiply both sides by 3 to get 12 = 4 + yC. Therefore yC = 12 − 4 = 8. Step 10: Hence the co ordinates of C are (−3, 8). Verification / Alternative check: As a quick check, substitute A(−3, −1), B(3, 5), and C(−3, 8) back into the centroid formula. Compute the average of the x co ordinates: (−3 + 3 − 3) / 3 = (−3) / 3 = −1. Compute the average of the y co ordinates: (−1 + 5 + 8) / 3 = 12 / 3 = 4. This matches the given centroid (−1, 4), so the answer is verified. Why Other Options Are Wrong: Option (3, 8) would give an x average different from −1 and thus does not produce the correct centroid. Option (−3, −8) yields a y average that is negative and far from 4, so it cannot be correct. Option (3, −8) fails both the x and y average checks and does not give the required centroid. Common Pitfalls: Students sometimes confuse the centroid with other special points such as the circumcentre or orthocentre and attempt to use incorrect formulas. Another common error is arithmetic mistakes when summing the co ordinates or dividing by 3. To avoid such errors, always write the centroid formula clearly, substitute carefully, and check by recomputing the centroid from the final vertex co ordinates. Final Answer: The co ordinates of vertex C are (−3, 8).

Question 17

What are the co ordinates of the reflection of the point (−4, 3) in the vertical line x = −2 on the Cartesian plane?

Accepted Answer

Introduction / Context: This coordinate geometry question asks you to find the reflection of a point across a vertical line. Reflection is a basic transformation where every point is mapped to another point such that the reflecting line is the perpendicular bisector of the segment joining the original point and its image. For a vertical line of the form x = constant, only the x co ordinate changes, while the y co ordinate remains the same. Understanding this helps solve reflection problems quickly in aptitude exams. Given Data / Assumptions: - Original point P has co ordinates (−4, 3). - The reflecting line is the vertical line x = −2. - We must find the co ordinates of the reflected point P prime. - The reflection is defined so that line x = −2 is midway between the x co ordinates of P and P prime. Concept / Approach: For reflection across a vertical line x = a, the y co ordinate remains unchanged, because vertical reflection only moves the point horizontally. The x co ordinate transforms according to the idea that the reflecting line is the midpoint between the original and image x values. Algebraically, if the original x is x1 and the new x is x2, then the average (x1 + x2) / 2 equals a. We can use this formula directly, or use the shortcut x2 = 2a − x1. The y co ordinate stays the same as that of the original point. Step-by-Step Solution: Step 1: Identify the reflecting line x = −2, so here a = −2. Step 2: Original point P has co ordinates (x1, y1) = (−4, 3). Step 3: For reflection across x = a, the new x co ordinate is x2 = 2a − x1. Step 4: Substitute a = −2 and x1 = −4 to get x2 = 2 * (−2) − (−4) = −4 + 4 = 0. Step 5: The y co ordinate remains unchanged, so y2 = y1 = 3. Step 6: Therefore the reflected point P prime has co ordinates (0, 3). Verification / Alternative check: To verify, check that the reflecting line x = −2 is the midpoint in the horizontal direction. Compute the midpoint of the segment joining P(−4, 3) and P prime(0, 3). The average of the x co ordinates is (−4 + 0) / 2 = −2, which is exactly the x value of the reflecting line. The y co ordinates are both 3, so the reflecting line is indeed perpendicular and acts as the perpendicular bisector. This confirms that (0, 3) is the correct reflection. Why Other Options Are Wrong: Option (−4, −7) changes the y co ordinate and keeps x the same, which corresponds to reflection across a horizontal line, not across x = −2. Option (4, 3) would place the point 6 units to the right of the reflecting line, while the original point is only 2 units to the left, so the line would not be the midpoint. Option (−4, 7) again changes only the y co ordinate, which does not match reflection in a vertical line. Common Pitfalls: Students sometimes wrongly change the y co ordinate instead of the x co ordinate when reflecting across a vertical line. Another mistake is to reflect across the y axis (line x = 0) instead of the given line x = −2. Always note that for x = a, distances to the left and right of the line must be equal in magnitude for a reflection. The shortcut formula x2 = 2a − x1 is very effective and helps avoid sign errors. Final Answer: The reflection of the point (−4, 3) in the line x = −2 is (0, 3).

Question 18

In triangles ΔXYZ and ΔPQR, it is given that ΔXYZ is similar to ΔPQR and the ratio of their perimeters is 4 : 9. If side PQ of ΔPQR is 27 cm, what is the length of the corresponding side XY (in cm)?

Accepted Answer

Introduction / Context: This question tests a basic property of similar triangles related to their perimeters and side lengths. When two triangles are similar, all their corresponding linear measurements such as sides, heights, and perimeters are in the same ratio. Here, the ratio of perimeters of triangles ΔXYZ and ΔPQR is given as 4 : 9, and a side of the larger triangle is known. We need to find the corresponding side of the smaller triangle using this similarity ratio. Given Data / Assumptions: - ΔXYZ is similar to ΔPQR. - Ratio of perimeters: Perimeter(ΔXYZ) : Perimeter(ΔPQR) = 4 : 9. - Side PQ of ΔPQR is 27 cm. - XY is the side in ΔXYZ corresponding to side PQ in ΔPQR. Concept / Approach: For similar triangles, the ratio of corresponding sides is equal to the ratio of perimeters, which is also equal to the scale factor between the triangles. If the ratio of perimeters is 4 : 9, then the ratio of any corresponding side in the smaller triangle to the side in the larger triangle is 4 : 9. Therefore, we can express XY in terms of PQ using this ratio. This is a direct proportionality question and does not require any angle or coordinate calculation. Step-by-Step Solution: Step 1: Recognise that ΔXYZ and ΔPQR are similar, so all corresponding linear measures are in the same ratio. Step 2: The ratio of perimeters is given as 4 : 9. This is also the ratio of any corresponding sides of the triangles. Step 3: Let XY correspond to PQ. Then XY : PQ = 4 : 9. Step 4: Substitute PQ = 27 cm in the ratio XY : 27 = 4 : 9. Step 5: Write this as XY / 27 = 4 / 9. Step 6: Solve for XY: XY = 27 * (4 / 9). Step 7: Compute 27 * (4 / 9) = (27 / 9) * 4 = 3 * 4 = 12 cm. Step 8: Therefore the length of side XY is 12 cm. Verification / Alternative check: We can interpret the ratio 4 : 9 as saying that the smaller triangle ΔXYZ is scaled up by a factor of 9 / 4 to obtain ΔPQR. In other words, every side in ΔPQR is 9 / 4 times the corresponding side in ΔXYZ. Therefore XY = PQ * (4 / 9). Substituting PQ = 27 again gives 27 * (4 / 9) = 12 cm, which matches the previous calculation, confirming the answer. Why Other Options Are Wrong: Option 9 cm would correspond to a ratio of 9 : 27 = 1 : 3, which is not the given ratio 4 : 9. Option 16 cm would give a side ratio 16 : 27, which does not simplify to 4 : 9. Option 15 cm would give 15 : 27 = 5 : 9, again not matching the required 4 : 9 ratio. Common Pitfalls: A common error is to invert the ratio accidentally and use 9 : 4 instead of 4 : 9 when relating the smaller triangle to the larger one. Another mistake is to confuse the perimeter ratio with an area ratio; remember that area ratios of similar triangles are the squares of the side ratios, but here the question explicitly provides the perimeter ratio, which is a linear ratio. Carefully matching corresponding triangles and corresponding sides is also important to avoid misassociation. Final Answer: The length of side XY is 12 cm.

Question 19

The surface area of a sphere is 5544 square centimetres. Using the standard formula for the surface area of a sphere, what is the diameter of the sphere (in cm)?

Accepted Answer

Introduction / Context: This question is from the topic of mensuration and deals with the surface area of a sphere. We are given the total surface area and asked to find the diameter. To solve it, we use the standard formula for the surface area of a sphere in terms of its radius and then convert the radius into the diameter. This is a direct formula substitution question commonly seen in aptitude and school level exams. Given Data / Assumptions: - Surface area S of the sphere is 5544 square centimetres. - Surface area formula for a sphere: S = 4 * π * r^2. - π is taken as 22 / 7 unless specified otherwise in typical questions of this type. - We must find the diameter, which is 2 * r. Concept / Approach: The main idea is to express the unknown radius r in terms of the known surface area S using the formula S = 4 * π * r^2. Once we isolate r^2, we take the square root to find r, and then multiply by 2 to get the diameter. Keeping the arithmetic clear and handling π as 22 / 7 helps in quickly simplifying to an integer answer. Step-by-Step Solution: Step 1: Write the formula for the surface area of a sphere: S = 4 * π * r^2. Step 2: Substitute S = 5544 and π = 22 / 7 into the formula. So 5544 = 4 * (22 / 7) * r^2. Step 3: Simplify the constant part: 4 * (22 / 7) = 88 / 7. Step 4: The equation becomes 5544 = (88 / 7) * r^2. Step 5: Multiply both sides by 7 to clear the denominator: 5544 * 7 = 88 * r^2. Step 6: Compute 5544 * 7 = 38808. So 38808 = 88 * r^2. Step 7: Divide both sides by 88 to get r^2 = 38808 / 88. Step 8: Simplify 38808 / 88. Cancelling by 8 gives 4851 / 11, and with full simplification, r^2 = 441. Step 9: Take the square root: r = √441 = 21 cm. Step 10: Diameter of the sphere is 2 * r = 2 * 21 = 42 cm. Verification / Alternative check: To verify, substitute r = 21 cm back into the formula S = 4 * π * r^2. We get S = 4 * (22 / 7) * 21^2 = 4 * (22 / 7) * 441. Simplify (441 / 7) = 63, so S = 4 * 22 * 63 = 88 * 63 = 5544 square centimetres. This matches the given surface area, confirming that r = 21 cm and diameter = 42 cm is correct. Why Other Options Are Wrong: Option 21 is the radius, not the diameter. The question specifically asks for the diameter. Option 84 would correspond to a radius of 42 cm, which would give a much larger surface area than 5544 square centimetres. Option 63 would correspond to a radius of 31.5 cm, again inconsistent with the given surface area. Common Pitfalls: One common mistake is to stop once the radius is found and forget to double it to obtain the diameter. Another is to mishandle π and the fractions during simplification, especially when multiplying and dividing by 7. To avoid mistakes, perform each simplification step carefully and check by substituting back into the original formula whenever possible. Final Answer: The diameter of the sphere is 42 cm.

Question 20

The diagonal of a square is equal to the side of an equilateral triangle. If the area of the square is 15√3 square centimetres, what is the area of the equilateral triangle?

Accepted Answer

Introduction / Context: This problem connects the geometry of a square with that of an equilateral triangle through the relation between their side lengths. The diagonal of the square is given to be equal to the side of the equilateral triangle, and the area of the square is known. We need to use area formulas and the relation between diagonal and side of a square to find the area of the equilateral triangle. This type of question appears often in aptitude tests to check understanding of basic area and diagonal relations. Given Data / Assumptions: - Area of the square is 15√3 square centimetres. - Let side of the square be s, then diagonal of the square is s√2. - The side of the equilateral triangle is equal to the diagonal of the square, so side of triangle a = s√2. - We need to find area of the equilateral triangle. Concept / Approach: We first find the side s of the square from its area. For a square, area = s^2. Once s is known, we write the diagonal as s√2. This diagonal is given to be the side of an equilateral triangle. For an equilateral triangle of side a, the area formula is (√3 / 4) * a^2. Substituting a = s√2 into this formula allows us to express the triangle area directly in terms of s^2. Since s^2 is already known from the square area, we can evaluate the expression and simplify. Step-by-Step Solution: Step 1: Let the side of the square be s. Then area of the square is s^2. Step 2: We are given that s^2 = 15√3 square centimetres. Step 3: The diagonal of the square is s√2. This diagonal is equal to the side a of the equilateral triangle. Step 4: Therefore, a = s√2. Step 5: The area of an equilateral triangle with side a is given by Area = (√3 / 4) * a^2. Step 6: Substitute a = s√2 to get Area = (√3 / 4) * (s√2)^2. Step 7: Compute (s√2)^2 = s^2 * 2. Step 8: So Area = (√3 / 4) * 2 * s^2 = (√3 / 2) * s^2. Step 9: Substitute s^2 = 15√3 into the expression: Area = (√3 / 2) * 15√3. Step 10: Multiply √3 by √3 to get 3. So Area = (1 / 2) * 15 * 3 = (1 / 2) * 45 = 45 / 2 square centimetres. Verification / Alternative check: You can also check sanity by approximating numerical values. Take √3 about 1.732. Then the area of the square is about 15 * 1.732 ≈ 25.98. That suggests the side of the square is roughly √26 ≈ 5.1, and the diagonal is about 5.1 * 1.414 ≈ 7.2. The side of the equilateral triangle would be about 7.2, giving an area approximately (1.732 / 4) * 7.2^2 ≈ 0.433 * 51.84 ≈ 22.4. This is close to 45 / 2 = 22.5, confirming our analytic answer is reasonable. Why Other Options Are Wrong: Option 45 / √2 square centimetres comes from incorrect manipulation of √2 and √3 factors. Option 45√2 square centimetres is far too large and results from squaring incorrectly or not using the correct area formula. Option 45 square centimetres doubles the correct answer and likely arises from forgetting the factor of 1/2 in the expression (√3 / 2) * s^2. Common Pitfalls: Students often forget or misapply the relation between the side and the diagonal of a square, sometimes using 2s instead of s√2. Another common issue is mishandling square roots when squaring expressions like s√2 or when simplifying √3 * √3. Always write the algebraic steps clearly, especially when dealing with surds, to avoid minor but costly mistakes in competitive exams. Final Answer: The area of the equilateral triangle is 45 / 2 square centimetres.

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