Question 1

ABCD is a quadrilateral in which AC and BD are its diagonals. Which of the following inequalities is always true for any quadrilateral?

Accepted Answer

Introduction / Context: This question explores the relationship between the sides and diagonals of a quadrilateral. It asks which inequality always holds, regardless of the specific shape of the quadrilateral. This is an extension of the triangle inequality to quadrilaterals and tests your understanding of how distances behave in polygons. Such conceptual questions are often used in aptitude and geometry exams to probe reasoning rather than computation. Given Data / Assumptions: - ABCD is any quadrilateral, not necessarily convex or cyclic, but usually considered simple and non self intersecting. - AC and BD are its diagonals, joining opposite vertices. - AB, BC, CD and DA are the four side lengths of the quadrilateral. - We want to know how the sum of the side lengths compares with the sum of the diagonal lengths AC and BD. - No special angle or side length conditions are given, so the statement must hold for all quadrilaterals. Concept / Approach: The triangle inequality states that in any triangle, the sum of the lengths of any two sides is greater than the length of the third side. We can apply this idea to triangles formed inside the quadrilateral by its diagonals. For instance, diagonal AC splits the quadrilateral into triangles ABC and ADC. Applying triangle inequalities in these triangles yields inequalities involving AB, BC, AC and also AD, CD, AC. Adding these inequalities allows comparison of the perimeter with 2AC, and similar reasoning with diagonal BD shows that the total perimeter is greater than the sum of diagonals. Step-by-Step Solution: Step 1: Consider diagonal AC. It divides quadrilateral ABCD into triangles ABC and ADC.Step 2: In triangle ABC, triangle inequality gives AB + BC > AC.Step 3: In triangle ADC, triangle inequality gives AD + DC > AC.Step 4: Add the two inequalities: (AB + BC) + (AD + DC) > AC + AC = 2AC.Step 5: Thus AB + BC + CD + DA > 2AC.Step 6: Similar reasoning using diagonal BD and triangles ABD and CBD yields AB + BC + CD + DA > 2BD.Step 7: From the two results, it follows that AB + BC + CD + DA is greater than both 2AC and 2BD, and therefore greater than AC + BD as well. So AB + BC + CD + DA > AC + BD is always true. Verification / Alternative check: Consider a simple example like a square of side 1 unit. Its perimeter is 4 units. The diagonal length is square root of 2 units, so AC + BD = 2 * square root of 2 which is approximately 2.828. Clearly 4 is greater than 2.828. You can also test with a rectangle of sides 2 and 1: perimeter is 6, diagonal is square root of 5, so AC + BD = 2 * square root of 5 which is about 4.472. Again 6 is greater than 4.472. These examples support the general inequality. Why Other Options Are Wrong: The inequality AB + BC + CD + DA < AC + BD contradicts even simple examples such as a square. Equality AB + BC + CD + DA = AC + BD or AB + BC + CD + DA = 2(AC + BD) would require special rigid conditions that do not hold for arbitrary quadrilaterals. The stronger inequality AB + BC + CD + DA > 2(AC + BD) is also not guaranteed; for some shapes the sum of diagonals can be relatively large compared to the sides, so this statement need not always be true. Common Pitfalls: A frequent mistake is to assume that diagonals might be longer when connecting distant vertices, leading to the belief that the sum of diagonals could exceed the perimeter. Another error is to misapply triangle inequality or to think it only holds for identical triangles. Remember that each diagonal creates two triangles, and applying the triangle inequality in each gives solid bounds relating sides and diagonals. Final Answer: The inequality that always holds is AB + BC + CD + DA > AC + BD.

Question 2

O is the centre of a circle of radius 5 cm. From a point P at a distance of 13 cm from O, two tangents PQ and PR are drawn to the circle. What is the area of the quadrilateral PQOR in square centimetres?

Accepted Answer

Introduction / Context: This problem deals with tangents drawn from an external point to a circle and asks for the area of a quadrilateral formed by the point, the two tangent points and the centre of the circle. It tests your understanding of properties of tangents, right triangles involving radii and tangents, and basic area calculation using triangles. Such questions are standard in geometry sections of competitive exams. Given Data / Assumptions: - Circle has centre O and radius 5 cm. - A point P lies outside the circle at a distance OP = 13 cm. - Tangents PQ and PR are drawn from P to the circle, touching it at Q and R respectively. - Quadrilateral PQOR is formed by joining P, Q, O and R in sequence. - We need the area of quadrilateral PQOR in square centimetres. Concept / Approach: A radius drawn to the point of tangency is perpendicular to the tangent. Therefore, OQ is perpendicular to PQ, and OR is perpendicular to PR. This means triangles OPQ and OPR are right triangles. Also, PQ equals PR since tangents from an external point to a circle are equal in length. The quadrilateral PQOR can be decomposed into two congruent right triangles OPQ and OPR. Finding the area of one triangle and doubling it gives the area of the quadrilateral. Step-by-Step Solution: Step 1: In triangle OPQ, OP is the hypotenuse, OQ is the radius and PQ is the tangent.Step 2: Given OP = 13 cm and OQ = 5 cm. Angle at Q is 90 degrees because the radius is perpendicular to the tangent.Step 3: By Pythagoras theorem, PQ^2 = OP^2 - OQ^2.Step 4: Compute PQ^2 = 13^2 - 5^2 = 169 - 25 = 144, so PQ = 12 cm.Step 5: Area of triangle OPQ = (1 / 2) * OQ * PQ = (1 / 2) * 5 * 12 = 30 sq. cm.Step 6: Triangle OPR is congruent to triangle OPQ, so its area is also 30 sq. cm.Step 7: Area of quadrilateral PQOR = area of triangle OPQ + area of triangle OPR = 30 + 30 = 60 sq. cm. Verification / Alternative check: You can visualise PQOR as a kite with equal sides OP and OP on each side, and radii OQ and OR. Because OP is fixed and both tangents have equal length, the shape is symmetric about line OP. Computation of each right triangle area is straightforward, and the sum gives the quadrilateral area. There is no other part to add or subtract, so 60 sq. cm is consistent and unique. Why Other Options Are Wrong: 30 sq. cm would be the area of only one of the triangles, not the entire quadrilateral. 32.5 sq. cm and 65 sq. cm arise from misusing wrong values or mixing radius and tangent lengths. 90 sq. cm would require each triangle to have area 45 sq. cm, which is not compatible with the given side lengths. Only 60 sq. cm correctly accounts for both congruent right triangles formed by the tangents and radii. Common Pitfalls: Some students forget that the radius to the point of tangency is perpendicular to the tangent, so they do not treat OPQ as a right triangle. Others mistakenly use OP as a leg and PQ as the hypotenuse. Another common error is to compute the area of only one triangle and overlook the need to double it for the quadrilateral. Always carefully identify right angles and count all constituent regions in composite figures. Final Answer: The area of the quadrilateral PQOR is 60 sq. cm.

Question 3

Forty men take a dip in a rectangular swimming pool 30 m long and 25 m broad. If each man displaces on average 5 cubic metres of water, by how many centimetres will the water level in the pool rise?

Accepted Answer

Introduction / Context: This is a practical volume and level rise question related to a swimming pool. When people enter the pool, they displace water, and the level of water rises. The problem assesses understanding of volume conservation: the total volume displaced must equal the increase in volume of water in the pool. It combines basic mensuration of a cuboid with unit conversion from metres to centimetres. Given Data / Assumptions: - The swimming pool is rectangular with length 30 m and breadth 25 m. - Forty men enter the pool and each displaces 5 cubic metres of water on average. - Total displaced volume is the sum of the volumes displaced by all men. - The water spreads uniformly over the entire base area of the pool, raising the water level by the same amount everywhere. - We need to find the rise in water level in centimetres. Concept / Approach: The volume increase in a pool is given by rise in water level multiplied by the base area of the pool. The total volume of water displaced by the men becomes this volume increase. Therefore, rise in height h can be computed as total displaced volume V divided by the base area A. After finding h in metres, we convert it to centimetres by multiplying by 100. Step-by-Step Solution: Step 1: Base area of the pool A = length * breadth = 30 * 25 = 750 square metres.Step 2: Each man displaces 5 cubic metres of water.Step 3: Total number of men = 40. Total volume displaced V = 40 * 5 = 200 cubic metres.Step 4: Let the rise in water level be h metres.Step 5: Volume increase in the pool = base area * rise in height = A * h = 750 * h.Step 6: This volume must equal the displaced volume, so 750 * h = 200.Step 7: Therefore, h = 200 / 750 = 2 / 7.5 = 0.2666 recurring, which is approximately 0.2667 metres or 26.67 centimetres. Verification / Alternative check: You can double check the conversion: 0.2667 m multiplied by 100 equals 26.67 cm. Also, if the pool had area 100 square metres instead, the same volume would raise the level by 2 metres. Since our pool has a much larger area of 750 square metres, the rise must be much smaller than 2 metres, so a value around 0.27 metres is quite reasonable. Why Other Options Are Wrong: 25.0 cm and 27.33 cm are close but not equal to the precise computed value and usually result from rounding mistakes or approximating the fraction 200 / 750 incorrectly. 28.5 cm and 30.0 cm overestimate the rise and would require a smaller base area or a larger displaced volume. Only 26.67 cm matches the exact calculation based on given data. Common Pitfalls: A common error is to use the pool perimeter instead of its area when relating volume to level rise. Another mistake is to forget to convert the final height from metres to centimetres, leaving the answer as 0.2667 instead of 26.67 cm. Some learners also miscalculate the total displaced volume or mis-handle fractions when dividing 200 by 750. Careful use of units and arithmetic avoids these issues. Final Answer: The water level in the pool rises by approximately 26.67 cm.

Question 4

The length, breadth and height of a cuboid are in the ratio 19 : 11 : 13. If the length is 30 cm more than the height, what is the volume of the cuboid in cubic centimetres?

Accepted Answer

Introduction / Context: This problem combines ratio of dimensions with an additional linear condition to determine actual dimensions of a cuboid. Once the real length, breadth and height are found, the volume can be computed by simple multiplication. Such questions are common in aptitude tests to check whether you can move from proportional relationships to specific values using extra conditions. Given Data / Assumptions: - Length : breadth : height of a cuboid are in the ratio 19 : 11 : 13. - Let the common multiplying factor be x centimetres. - So length = 19x, breadth = 11x and height = 13x. - It is given that length is 30 cm more than height. - We need to find the volume of the cuboid in cubic centimetres. Concept / Approach: When dimensions are in a given ratio, you can represent them as multiples of a common variable. An additional equation, such as a fixed difference between two dimensions, allows you to solve for this variable. Once the exact numerical values of length, breadth and height are known, the volume is given by V = length * breadth * height. Careful substitution and multiplication lead to the final volume. Step-by-Step Solution: Step 1: Represent the dimensions using the ratio and a common factor x.Step 2: Length L = 19x, breadth B = 11x, height H = 13x.Step 3: Given that length is 30 cm more than height, so L = H + 30.Step 4: Substitute: 19x = 13x + 30.Step 5: Simplify: 19x - 13x = 30 which gives 6x = 30.Step 6: Therefore, x = 30 / 6 = 5.Step 7: Compute actual dimensions: L = 19 * 5 = 95 cm, B = 11 * 5 = 55 cm, H = 13 * 5 = 65 cm.Step 8: Volume V = L * B * H = 95 * 55 * 65 cubic centimetres.Step 9: First 95 * 55 = 5225, then 5225 * 65 = 339,625 cubic centimetres. Verification / Alternative check: Check that the difference between length and height equals 30 cm. With L = 95 cm and H = 65 cm, L - H = 95 - 65 = 30 cm, which matches the given condition. Also, ratios of 95 : 55 : 65 reduce by dividing each term by 5 to 19 : 11 : 13, confirming that the solution respects the original ratio. Thus the computed volume 339,625 cubic centimetres is consistent with all given information. Why Other Options Are Wrong: Volumes such as 81,510 or 89,665 cubic centimetres correspond to incorrect values of x or arithmetic errors when multiplying the dimensions. 195,300 and 410,000 cubic centimetres are far from the calculated value and usually result from misreading the ratio or forgetting the height condition. None of these options satisfy both the ratio and the 30 cm difference requirement simultaneously. Common Pitfalls: Students sometimes misinterpret the ratio as length : height : breadth or rearrange the order, leading to wrong dimension assignments. Another common mistake is to subtract the wrong way when applying the condition L = H + 30, or to miscalculate products like 95 * 55. Always keep the ratio order consistent, handle the linear equation carefully and perform multiplications step by step to avoid errors. Final Answer: The volume of the cuboid is 339,625 cubic centimetres.

Question 5

In triangle ABC, AD is a median and G is the centroid of the triangle lying on AD. What is the ratio AG : AD?

Accepted Answer

Introduction / Context: This is a pure concept question in triangle geometry about the centroid and medians. The centroid is one of the most important centres of a triangle, and it has a specific ratio property along medians. The question asks for the ratio in which the centroid divides a median drawn from a vertex to the midpoint of the opposite side. Understanding this property is essential for many geometric proofs and coordinate geometry applications. Given Data / Assumptions: - ABC is any triangle. - AD is a median, so D is the midpoint of side BC and AD connects vertex A to D. - G is the centroid of triangle ABC and lies on the median AD. - By definition, the centroid is the intersection point of the three medians of the triangle. - We are asked to find the ratio AG : AD, that is, how much of the median lies between A and G compared to the whole median from A to D. Concept / Approach: A well known property of the centroid is that it divides each median in the ratio 2 : 1, counting from the vertex to the midpoint of the opposite side. That is, if we label the median from A to the midpoint D, then AG : GD = 2 : 1. Since AD is the entire median and is composed of AG plus GD, we can express AD as three equal parts, with AG being two of those parts. This immediately gives the ratio AG : AD. Step-by-Step Solution: Step 1: Recall the definition of a median: AD is a line segment from vertex A to the midpoint D of side BC.Step 2: Recall the definition of the centroid G: the common intersection point of the three medians of a triangle.Step 3: A key property of the centroid is that it divides each median in the ratio 2 : 1, with the longer part closer to the vertex.Step 4: On median AD, this means AG : GD = 2 : 1.Step 5: Let each unit in this ratio represent one part. Then AD = AG + GD = 2 parts + 1 part = 3 parts.Step 6: Therefore AG is 2 parts out of the total 3 parts of AD.Step 7: Hence AG : AD = 2 : 3. Verification / Alternative check: You can visualise this by placing triangle ABC in coordinate geometry. Suppose A, B and C have coordinates and D is the midpoint of BC. The centroid G is the average of the coordinates of A, B and C. If you compute vector AG and compare it to AD, you will find that AG is always two thirds of AD, regardless of the triangle shape. This coordinate demonstration confirms the ratio AG : AD = 2 : 3. Why Other Options Are Wrong: The ratio 3 : 2 reverses the correct ordering, implying that the centroid is closer to the midpoint than to the vertex, which is not true. Ratios 3 : 1 and 1 : 2 would suggest that the centroid either lies much closer to the vertex than two thirds of the way or exactly at the midpoint, both of which contradict the established centroid property. The ratio 1 : 3 also has no basis in standard geometry for centroids. Common Pitfalls: Learners often confuse properties of the centroid with those of other centres like the incenter or orthocentre. Some mistakenly think the centroid lies at the midpoint of each median, giving a 1 : 2 ratio rather than 2 : 1. It is important to remember that the centroid is located two thirds of the distance from each vertex to the midpoint of the opposite side, not halfway along the median. Final Answer: The required ratio is AG : AD = 2 : 3.

Question 6

A point P lies inside a triangle and is equidistant from all three vertices of the triangle. Which special point of the triangle does P represent?

Accepted Answer

Introduction / Context: This question is about identifying special points associated with a triangle: centroid, incenter, orthocentre, circumcentre and excentre. The key information is that the point inside the triangle is equidistant from all three vertices. The problem tests whether you can correctly match this geometric property with the name of the corresponding centre, which is a crucial concept in Euclidean geometry and many competitive exams. Given Data / Assumptions: - P is a point located inside a given triangle. - Distances from P to the vertices A, B and C are equal, that is PA = PB = PC. - No information is provided about distances from P to the sides or about right angles or medians. - We must determine which triangle centre P represents among centroid, incenter, orthocentre, circumcentre and excentre. Concept / Approach: Each special centre of a triangle has a characteristic definition: - The centroid is the intersection of medians and is the centre of mass of the triangle. - The incenter is equidistant from all three sides and is the centre of the inscribed circle. - The orthocentre is the intersection of altitudes and is not defined by equal distances to vertices or sides. - The circumcentre is the intersection of perpendicular bisectors of the sides and is equidistant from all three vertices. - Excentres are centres of external circles tangent to one side and extensions of the other two sides and are not equidistant from all vertices. The only centre that is equidistant from all vertices is the circumcentre. Step-by-Step Solution: Step 1: Recall that the circumcentre is defined as the point that is equidistant from all three vertices of the triangle and is the centre of the circumcircle.Step 2: The problem states that P is equidistant from all three vertices, so PA = PB = PC.Step 3: This exactly matches the definition of the circumcentre.Step 4: The centroid, incenter and orthocentre do not have this property of equal vertex distances.Step 5: The incenter is equidistant from sides, not vertices.Step 6: The orthocentre is defined by altitudes and their intersection, not by any distance equality.Step 7: Therefore, the correct identification of P is the circumcentre. Verification / Alternative check: Visualise a triangle and its circumcircle. The circumcentre is the centre of that circle and lies at the intersection of perpendicular bisectors of the sides. Since all vertices lie on the circumcircle, the distance from the circumcentre to each vertex is the same, equal to the circumradius. No other standard centre maintains equal distances to all three vertices. This confirms that any point equidistant from the vertices must be the circumcentre. Why Other Options Are Wrong: The centroid is the average of the vertex positions and lies two thirds of the way along each median, but its distances to the vertices are generally different. The incenter is equidistant from all three sides, not from the vertices. The orthocentre can lie inside, outside or on the triangle, depending on the type of triangle, and has no symmetry in distances to vertices. Excentres lie outside the triangle and are defined relative to sides and their extensions, not equal vertex distances. Therefore none of these match the condition given. Common Pitfalls: Students often confuse circumcentre and incenter because both relate to circles associated with the triangle. Remember the guiding keyword: circumcentre means circle passing through the vertices, so equal distances to vertices; incenter means circle touching the sides, so equal distances to sides. Mistaking one for the other is a frequent error in multiple choice questions like this. Final Answer: Hence, the point P is the Circumcentre of the triangle.

Question 7

Two equal circles each have radius 3 cm and the distance between their centres is 10 cm. What is the length of one transverse common tangent between the circles?

Accepted Answer

Introduction / Context: This aptitude geometry question tests your understanding of common tangents to two circles, especially the formula for the length of a transverse common tangent. Problems like this are common in competitive exams because they combine circle geometry, Pythagorean thinking and remembering the difference between direct and transverse tangents. Knowing which formula applies in which configuration helps you quickly solve many similar tangent based questions. Given Data / Assumptions: • Two equal circles, each with radius r = 3 cm. • Distance between their centres d = 10 cm. • We are asked for the length of one transverse common tangent. • Circles are assumed to be in the same plane and non intersecting because d is greater than 2r. Concept / Approach: The key concept is that there are two types of common tangents to two circles: direct common tangents and transverse common tangents. For two circles with radii r1 and r2 and distance between centres d, the length of a direct common tangent is sqrt(d^2 - (r1 - r2)^2), while the length of a transverse common tangent is sqrt(d^2 - (r1 + r2)^2). Because the circles here are equal, r1 = r2 = r, which simplifies the expression. The required length can be found directly using the Pythagoras theorem in the right triangle formed by joining the centres and the tangent segment. Step-by-Step Solution: 1. Let r1 = r2 = r = 3 cm and distance between centres d = 10 cm. 2. For a transverse common tangent, use the formula: length L = sqrt(d^2 - (r1 + r2)^2). 3. Compute r1 + r2 = 3 + 3 = 6 cm. 4. Compute d^2 = 10^2 = 100. 5. Compute (r1 + r2)^2 = 6^2 = 36. 6. Substitute into the formula: L = sqrt(100 - 36) = sqrt(64). 7. Therefore, L = 8 cm. Verification / Alternative check: One quick validation is to compare the value of the tangent length with the distance between centres and the radii. The sum of the radii is 6 cm and the distance between centres is 10 cm, so the tangent segment must be shorter than 10 cm but longer than 4 cm to form a valid right triangle. The computed length 8 cm clearly lies between 4 cm and 10 cm, which is consistent with a Pythagorean triple 6, 8, 10. This gives a strong confidence that the calculation is correct. Why Other Options Are Wrong: • 7 cm: This is slightly less than the correct value and does not satisfy the relation d^2 = L^2 + (r1 + r2)^2. • 9 cm: This is too large; if L = 9, then 9^2 + 6^2 would exceed 10^2, which is impossible. • 10 cm: The tangent length cannot equal the distance between centres; that would force the radii sum to be zero. • 6 cm: This would mean d^2 = 6^2 + 6^2 = 72, which is not equal to 100. Common Pitfalls: Many students confuse direct and transverse common tangents and mistakenly use sqrt(d^2 - (r1 - r2)^2) even when the question explicitly says transverse common tangent. Another frequent error is forgetting that the radii add for the transverse case, leading to an incorrect smaller value. Some learners also neglect units, but it is important to keep everything in centimetres and state the final answer with units. Final Answer: The length of one transverse common tangent between the two circles is 8 cm.

Question 8

A sphere has the same curved surface area as a right circular cone with vertical height 40 cm and base radius 30 cm. What is the radius of the sphere in centimetres?

Accepted Answer

Introduction / Context: This question checks your understanding of curved surface areas of common 3D solids, specifically a right circular cone and a sphere. In many aptitude and engineering entrance exams, you are asked to relate areas or volumes of different solids that are melted or equated. Recognising the correct formulas and equating them is the key to answering quickly and correctly. Given Data / Assumptions: • A right circular cone has vertical height h = 40 cm. • Radius of the cone's base r = 30 cm. • A sphere has the same curved surface area as this cone. • We are asked to find the radius R of the sphere. Concept / Approach: The curved surface area of a right circular cone is given by CSA_cone = π * r * l, where r is the base radius and l is the slant height. The curved surface area of a sphere is given by CSA_sphere = 4 * π * R^2, where R is the radius of the sphere. Since the question states that these curved surface areas are equal, we equate the two formulas and solve for R. First, we must compute the slant height l of the cone using the Pythagoras theorem in the right triangle formed by radius, height and slant height. Step-by-Step Solution: 1. Height of the cone h = 40 cm and radius r = 30 cm. 2. Find the slant height l of the cone: l = sqrt(h^2 + r^2). 3. Compute h^2 = 40^2 = 1600 and r^2 = 30^2 = 900. 4. Then l = sqrt(1600 + 900) = sqrt(2500) = 50 cm. 5. Curved surface area of cone: CSA_cone = π * r * l = π * 30 * 50 = 1500π. 6. Curved surface area of sphere: CSA_sphere = 4 * π * R^2. 7. Set CSA_cone = CSA_sphere: 4 * π * R^2 = 1500π. 8. Cancel π on both sides to get 4 * R^2 = 1500. 9. So R^2 = 1500 / 4 = 375. 10. Write 375 as 25 * 15, so R = sqrt(25 * 15) = 5 * sqrt(15). Verification / Alternative check: You can verify by substituting R = 5 * sqrt(15) back into the sphere's curved surface area formula. Then CSA_sphere = 4 * π * (5 * sqrt(15))^2 = 4 * π * 25 * 15 = 1500π, which exactly matches the cone's curved surface area. This confirms that the derived radius is consistent. Also, note that the radius is slightly less than 5 * 4 = 20 cm, which is reasonable compared to the cone dimensions. Why Other Options Are Wrong: • 5√5 cm: This would give R^2 = 25 * 5 = 125, and CSA_sphere = 4 * π * 125 = 500π, which is too small. • 5√3 cm: Here R^2 = 75, so CSA_sphere = 300π, still much less than 1500π. • 5√10 cm: This gives R^2 = 250, so CSA_sphere = 1000π, which is still less than 1500π. • 10 cm: Then CSA_sphere = 4 * π * 100 = 400π, again far from 1500π. Common Pitfalls: A frequent mistake is to equate the volumes instead of the curved surface areas, which completely changes the equation. Another common error is to forget to compute or correctly use the slant height of the cone, leading to wrong values of area. Some students also forget to cancel π from both sides and end up carrying unnecessary factors through the calculation. Staying organised with formulas and algebraic steps avoids these errors. Final Answer: The radius of the sphere is 5√15 cm.

Question 9

Three solid spheres of radii 3 cm, 4 cm and 5 cm are melted and recast into a single solid sphere. What percentage decrease occurs in the total surface area?

Accepted Answer

Introduction / Context: This question combines volume conservation with surface area comparison, which is a classic theme in aptitude and engineering exams. When multiple solids are melted and recast into a new solid, volume is conserved but surface area generally changes. You are asked to compute how much the total surface area decreases when three smaller spheres are melted into a single larger sphere. Given Data / Assumptions: • Radii of three original spheres: 3 cm, 4 cm and 5 cm. • They are melted and recast into one solid sphere. • Volume is conserved during melting and recasting. • We are asked to find the percentage decrease in total surface area. Concept / Approach: The volume V of a sphere of radius r is given by V = (4 / 3) * π * r^3. When multiple spheres are melted and recast into a single sphere, the total initial volume equals the final volume. So we first compute the sum of volumes of the three small spheres and equate it to the volume of the new sphere to find its radius R. Then we compute the initial total surface area, which is the sum of 4 * π * r^2 for each sphere, and compare it with the surface area of the new sphere 4 * π * R^2. Finally, we use the percentage decrease formula. Step-by-Step Solution: 1. Volume of first sphere: V1 = (4 / 3) * π * 3^3 = (4 / 3) * π * 27. 2. Volume of second sphere: V2 = (4 / 3) * π * 4^3 = (4 / 3) * π * 64. 3. Volume of third sphere: V3 = (4 / 3) * π * 5^3 = (4 / 3) * π * 125. 4. Sum the volumes: V_total = (4 / 3) * π * (27 + 64 + 125) = (4 / 3) * π * 216. 5. Simplify: (4 / 3) * 216 = 288, so V_total = 288π. 6. Let the radius of the new sphere be R. Then (4 / 3) * π * R^3 = 288π. 7. Cancel π and multiply both sides by 3 / 4: R^3 = 288 * (3 / 4) = 216. 8. Thus R = cube root of 216 = 6 cm. 9. Total initial surface area: S_initial = 4 * π * (3^2 + 4^2 + 5^2) = 4 * π * (9 + 16 + 25) = 4 * π * 50 = 200π. 10. Final surface area: S_final = 4 * π * R^2 = 4 * π * 6^2 = 4 * π * 36 = 144π. 11. Decrease in surface area: ΔS = S_initial - S_final = 200π - 144π = 56π. 12. Percentage decrease = (ΔS / S_initial) * 100 = (56π / 200π) * 100 = 28 %. Verification / Alternative check: As a quick reasonableness check, notice that the final radius 6 cm is larger than any of the original radii 3 cm, 4 cm and 5 cm. Because surface area grows with the square of radius and volume grows with the cube, combining smaller spheres into one larger sphere almost always decreases the total surface area. The drop from 200π to 144π is significant but not extreme, making a decrease around 25 to 30 percent reasonable. Our calculation of 28 percent fits nicely in this expected range. Why Other Options Are Wrong: • 12 %: This would give a very small change, inconsistent with the large difference between 200π and 144π. • 14 %: Still too small given that 56π is more than a quarter of 200π. • 16 %: This corresponds to a much smaller surface area drop than actually computed. • 32 %: This would require an even larger drop than 56π, which does not match the exact values. Common Pitfalls: A common mistake is to add radii directly instead of adding volumes, which violates volume conservation. Another error is to compute only one of the surface areas and forget to compare initial and final values. Also, some students forget to include π consistently or cancel it improperly. Being systematic about using r^3 for volume and r^2 for surface area helps avoid these traps. Final Answer: The percentage decrease in total surface area is 28 %.

Question 10

In triangle ABC, a line from vertex A meets side BC at point D. If BC = 9 cm and DC = 3 cm, what is the ratio of the areas of triangles ABD and ADC respectively?

Accepted Answer

Introduction / Context: This geometry problem tests your understanding of how the areas of triangles that share the same altitude are related to the lengths of their bases. It appears frequently in aptitude exams because it can be solved quickly using proportional reasoning without any complicated formulas, provided you remember the basic area relation for triangles. Given Data / Assumptions: • Triangle ABC has a side BC of length 9 cm. • Point D lies on BC such that DC = 3 cm. • A line from vertex A meets BC at D, forming triangles ABD and ADC. • We need the ratio of areas of triangles ABD and ADC. Concept / Approach: The area of a triangle is given by (1 / 2) * base * height. If two triangles share the same altitude (height) from a common vertex to a common base line, then their areas are directly proportional to their bases. In this problem, both triangle ABD and triangle ADC have the same altitude from vertex A to line BC. Therefore, the ratio of their areas equals the ratio of their bases on BC, which are BD and DC. So we first find BD by subtracting DC from BC, then compare BD and DC directly. Step-by-Step Solution: 1. Given BC = 9 cm and DC = 3 cm. 2. Since D lies on BC, we have BD + DC = BC. 3. Compute BD = BC - DC = 9 - 3 = 6 cm. 4. Area of triangle ABD is (1 / 2) * BD * height from A to BC. 5. Area of triangle ADC is (1 / 2) * DC * the same height from A to BC. 6. Let the common height from A to BC be h. Then area(ABD) = (1 / 2) * 6 * h and area(ADC) = (1 / 2) * 3 * h. 7. The ratio of the areas is area(ABD) : area(ADC) = (6 * h) : (3 * h). 8. Cancel h and common factor 3 to get 6 : 3 = 2 : 1. Verification / Alternative check: You can verify with simple numbers. Assume the height from A to BC is, for example, 4 cm. Then area(ABD) = (1 / 2) * 6 * 4 = 12 square centimetres and area(ADC) = (1 / 2) * 3 * 4 = 6 square centimetres. The ratio 12 : 6 simplifies to 2 : 1, which matches our algebraic result. This confirms that the reasoning using proportional bases is correct and does not depend on the actual height value. Why Other Options Are Wrong: • 1 : 1: This would mean BD and DC are equal, but BD = 6 cm and DC = 3 cm, so they are not. • 3 : 1: This would require BD to be three times DC, that is 9 cm versus 3 cm, which is not the case. • 4 : 1: This would correspond to BD = 12 cm and DC = 3 cm or a similar proportion, which does not match BC = 9 cm. • 5 : 1: Again, this does not reflect the actual ratio of 6 cm to 3 cm. Common Pitfalls: Some students mistakenly assume that the ratio of areas is related to the lengths of AD or to some angle measures, which is not needed. Others might incorrectly think that additional data is required. Remember that when two triangles share the same altitude to a common base line, the base lengths alone determine the area ratio. Drawing a rough figure often helps to visualise this clearly. Final Answer: The ratio of the areas of triangles ABD and ADC is 2 : 1.

Question 11

In right angled triangle PQR with angle R = 90 degrees, RS is drawn perpendicular to hypotenuse PQ. If PR = 3 cm and RQ = 4 cm, what is the length of RS in centimetres?

Accepted Answer

Introduction / Context: This question involves a standard right triangle configuration and the altitude drawn from the right angle to the hypotenuse. Such problems frequently appear in aptitude tests to check if you remember special formulas and can apply Pythagoras theorem quickly. The relation between the legs, hypotenuse and the altitude from the right angle is very useful in many geometric situations. Given Data / Assumptions: • Triangle PQR is right angled at R, so ∠R = 90 degrees. • PR = 3 cm and RQ = 4 cm are the legs of the right triangle. • RS is drawn perpendicular to the hypotenuse PQ, so RS is the altitude from the right angle. • We need to find the length of RS. Concept / Approach: For a right triangle with legs a and b and hypotenuse c, the altitude h from the right angle to the hypotenuse satisfies the relation h = (a * b) / c. This can be derived from similarity of triangles or from area formulas. First we find the hypotenuse PQ using the Pythagoras theorem: c^2 = a^2 + b^2. Then we use the altitude formula to compute RS directly. This approach is efficient and avoids more complicated constructions. Step-by-Step Solution: 1. Identify the legs: PR = 3 cm and RQ = 4 cm. 2. Compute the hypotenuse PQ using Pythagoras theorem: PQ^2 = PR^2 + RQ^2. 3. So PQ^2 = 3^2 + 4^2 = 9 + 16 = 25. 4. Therefore, PQ = sqrt(25) = 5 cm. 5. Let RS = h, the altitude from R to PQ. 6. Use the formula h = (a * b) / c, where a = PR, b = RQ and c = PQ. 7. Substitute the values: h = (3 * 4) / 5. 8. Compute h = 12 / 5 cm. 9. So RS = 12 / 5 cm, which is 2.4 cm in decimal form. Verification / Alternative check: Another way to verify is to compare areas. The area of triangle PQR is (1 / 2) * PR * RQ = (1 / 2) * 3 * 4 = 6 square centimetres. The same area can also be expressed as (1 / 2) * PQ * RS = (1 / 2) * 5 * RS. Equating these gives 6 = (1 / 2) * 5 * RS, which leads to RS = 12 / 5 cm. This matches the earlier result and confirms the correctness of our method. Why Other Options Are Wrong: • 36/5 cm: This is much larger than the hypotenuse and cannot be the altitude of such a small triangle. • 5 cm: This equals the hypotenuse length, which is impossible for a perpendicular segment inside the triangle. • 2.5 cm: Close to 2.4 cm but not exact; it comes from rough rounding rather than correct calculation. • 3 cm: This matches one leg but has no geometric justification as the altitude to the hypotenuse. Common Pitfalls: Some students mistakenly treat the altitude as equal to one of the legs or use the wrong formula h^2 = a * b without checking the context. Another common error is to forget to find the hypotenuse first and plug incorrect values into the altitude formula. Always remember to apply Pythagoras theorem carefully and keep track of which side is which. Final Answer: The length of RS is 12/5 cm.

Question 12

If four distinct lines are drawn in a plane, which of the following cannot be the number of distinct points of intersection of these lines?

Accepted Answer

Introduction / Context: This problem is about combinatorial geometry and tests your understanding of how many intersection points can be formed by a given number of lines in a plane. Such questions are common in aptitude tests to check logical thinking, imagination and basic combinatorics, rather than heavy computation. Given Data / Assumptions: • We have exactly four distinct lines in a plane. • Lines may intersect, be parallel or even concurrent (many passing through one point). • We are asked which given number cannot represent the count of distinct intersection points. Concept / Approach: For n distinct lines in a plane, if no two are parallel and no three are concurrent, the maximum possible number of intersection points is nC2, because each pair of lines intersects in exactly one point. For n = 4, the maximum is 4C2 = 6. By adjusting whether some lines are parallel or concurrent, we can obtain fewer intersection points. Any suggested number greater than 6 would be impossible. Additionally, we should reason whether smaller numbers like 0, 4 or 5 can be constructed by choosing a suitable arrangement of the lines. Step-by-Step Solution: 1. For four lines, compute the maximum intersections if all pairs meet: 4C2 = (4 * 3) / 2 = 6. 2. Therefore, there can never be more than 6 distinct intersection points. 3. Check if 0 is possible: take all four lines parallel to each other. Then no two lines meet, so 0 intersections is possible. 4. Check if 4 is possible: take three concurrent lines (all passing through a single point) and a fourth line that intersects each of these three at different points. We then have 3 + 1 = 4 distinct intersection points. 5. Check if 5 is possible: take three lines in general position (forming 3 intersections), and a fourth line parallel to one of the three but intersecting the other two. This gives 3 existing intersections plus 2 new ones, for a total of 5. 6. Finally, consider 7: this exceeds the maximum possible 6, so 7 intersections is impossible with only four lines. Verification / Alternative check: You can visualise different configurations or quickly sketch them. For maximum intersections, draw four lines so that no two are parallel and no three pass through the same point; you will count exactly six intersection points. This confirms that 6 is the absolute maximum. Any number less than or equal to 6 may or may not be possible, so we must construct examples or reason as above, but any number greater than 6 is automatically impossible. Why Other Options Are Wrong: • 0: Possible when all four lines are mutually parallel. • 5: Possible with three lines in general position and a fourth line parallel to one and intersecting the other two. • 4: Possible with three concurrent lines plus a fourth line intersecting them at distinct points. • 6: This is actually the maximum possible and is achievable when all pairs of lines intersect uniquely. Common Pitfalls: A common error is to think that if four lines are in general position they always produce 4 intersections, which is incorrect. Another mistake is to ignore the possibility of all lines being parallel or many lines being concurrent. Remember to use the formula nC2 for the maximum intersection points and then think creatively about arrangements that reduce the count. Final Answer: The number of intersection points that cannot occur is 7.

Question 13

Three solid spheres with radii r1, r2 and r3 are melted and recast into a single solid sphere. What is the radius of the new sphere expressed in terms of r1, r2 and r3?

Accepted Answer

Introduction / Context: This is a symbolic version of the volume conservation idea used in mensuration. Instead of specific numerical radii, the radii r1, r2 and r3 are kept as variables and you must express the radius of the resulting sphere in terms of these variables. Such algebraic geometry questions appear in higher level aptitude and engineering entrance exams. Given Data / Assumptions: • Three solid spheres have radii r1, r2 and r3. • They are completely melted and recast into one solid sphere. • Volume is conserved during the melting and recasting process. • We must express the radius R of the new sphere in terms of r1, r2 and r3. Concept / Approach: The volume V of a sphere with radius r is V = (4 / 3) * π * r^3. When several spheres are melted together to form one new sphere, the total volume remains constant. Therefore, the sum of the volumes of the original three spheres will equal the volume of the new sphere of radius R. We write this equation using the general formula and then solve for R in terms of r1, r2 and r3. This will naturally lead to a cube root expression. Step-by-Step Solution: 1. Volume of the first sphere: V1 = (4 / 3) * π * r1^3. 2. Volume of the second sphere: V2 = (4 / 3) * π * r2^3. 3. Volume of the third sphere: V3 = (4 / 3) * π * r3^3. 4. Total initial volume: V_total = V1 + V2 + V3 = (4 / 3) * π * (r1^3 + r2^3 + r3^3). 5. Let the radius of the new sphere be R. Then its volume is V_new = (4 / 3) * π * R^3. 6. By volume conservation, V_total = V_new. 7. So (4 / 3) * π * (r1^3 + r2^3 + r3^3) = (4 / 3) * π * R^3. 8. Cancel the common factor (4 / 3) * π from both sides to get r1^3 + r2^3 + r3^3 = R^3. 9. Take cube root on both sides: R = (r1^3 + r2^3 + r3^3)^(1/3). Verification / Alternative check: To verify the formula, you can test it with specific values. For example, if r1 = r2 = r3 = r, then the three equal spheres are melted into one sphere. The new volume is three times the original volume of one sphere, so R^3 = 3 * r^3 and R = (3^(1/3)) * r. Plugging r1 = r2 = r3 = r into the formula gives R = (r^3 + r^3 + r^3)^(1/3) = (3 * r^3)^(1/3) = (3^(1/3)) * r, which matches the direct reasoning. Why Other Options Are Wrong: • r1 + r2 + r3: This simply adds the radii linearly and ignores the cubic relationship of volume. • (r1^2 + r2^2 + r3^2)^(1/2): This is a root mean square style expression and fits area scaling, not volume scaling. • (r1^4 + r2^4 + r3^4)^(1/4): This relates to a fourth power and has no connection to the volume formula. • r1 * r2 * r3: Multiplying radii directly does not correspond to the sum of volumes and has no geometric meaning here. Common Pitfalls: A common mistake is to assume that the new radius is just the sum or average of the old radii. Another error is to forget that volume scales with the cube of the radius, not with r or r^2. Always start from the exact volume formula, sum the volumes correctly and then isolate R^3 before taking the cube root. Final Answer: The radius of the new sphere is (r1^3 + r2^3 + r3^3)^(1/3).

Question 14

The extension in a spring is directly proportional to the weight attached to it. If a weight of 5 kg produces an extension of 0.4 cm, what weight will produce an extension of 5 cm?

Accepted Answer

Introduction / Context: This is a direct proportion problem based on Hooke's law, which states that, within the elastic limit, the extension in a spring is proportional to the load applied. Such questions are common in quantitative aptitude because they test your ability to use proportional reasoning with very little calculation. Given Data / Assumptions: • Extension in the spring is directly proportional to the weight attached. • A weight of 5 kg produces an extension of 0.4 cm. • We want the weight that will produce an extension of 5 cm. Concept / Approach: If one quantity is directly proportional to another, we can write W1 / x1 = W2 / x2, where W1 and W2 are the weights and x1 and x2 are the corresponding extensions. Equivalently, W = k * x for some constant k. First, we find the constant of proportionality using the given data, and then apply it to the required extension. It is important to keep units consistent, but here both extensions are in centimetres, so we can work directly. Step-by-Step Solution: 1. Let W be the weight in kilograms and x be the extension in centimetres. 2. Given the direct proportionality, W = k * x for some constant k. 3. From the given data, when W1 = 5 kg, x1 = 0.4 cm, so 5 = k * 0.4. 4. Solve for k: k = 5 / 0.4. 5. Compute 5 / 0.4 = 5 / (4 / 10) = (5 * 10) / 4 = 50 / 4 = 12.5. 6. Therefore, k = 12.5 kg per centimetre. 7. For an extension x2 = 5 cm, the required weight W2 = k * x2. 8. Substitute: W2 = 12.5 * 5 = 62.5 kg. Verification / Alternative check: You can verify by using the ratio form of direct proportion. Since W1 / x1 = W2 / x2, we have 5 / 0.4 = W2 / 5. Therefore, W2 = 5 * (5 / 0.4) = 25 / 0.4 = 62.5 kg. This matches the value we obtained using the constant k method. Both methods show consistent results, confirming the correctness of the calculation. Why Other Options Are Wrong: • 6.25 kg: This would correspond to a much smaller extension than 5 cm when using the same proportionality constant. • 4 kg: This is less than the original 5 kg, yet the extension desired is much larger than 0.4 cm, so it cannot be correct. • 40 kg: This gives W2 / x2 = 40 / 5 = 8, which does not match 5 / 0.4 = 12.5. • 12.5 kg: This corresponds to an extension of 1 cm, not 5 cm, under the same proportionality constant. Common Pitfalls: A frequent mistake is to treat the relationship as inverse proportion by accident, or to mis-handle the division by a decimal such as 0.4. Another pitfall is forgetting to keep both extensions in the same unit, but here that is already consistent. Writing the proportion clearly before substituting numbers helps avoid errors. Final Answer: The required weight is 62.5 kg.

Question 15

If PA and PB are tangents to a circle with centre O such that angle APB = 50 degrees, what is the measure of angle OAB?

Accepted Answer

Introduction / Context: This geometry question focuses on properties of tangents to a circle and the relationship between angles at an external point, the centre and points of tangency. Understanding these angle relations helps in many circle related questions in competitive exams. Given Data / Assumptions: • A circle has centre O. • From an external point P, two tangents PA and PB are drawn to the circle. • Angle APB, the angle between the tangents, is 50 degrees. • We are asked to find angle OAB, an angle at point A inside triangle OAB. Concept / Approach: Two main concepts are used here. First, the radii OA and OB are perpendicular to the tangents PA and PB at the points of contact, so OA ⟂ PA and OB ⟂ PB. Second, in quadrilateral AOPB, the sum of interior angles is 360 degrees. This gives a direct link between angle APB and angle AOB, the central angle between the radii to the tangency points. Finally, triangle AOB is isosceles with OA = OB, so its base angles at A and B are equal and can be found from its angle sum. Step-by-Step Solution: 1. Consider quadrilateral AOPB formed by joining O to A and B. 2. At point A, OA is a radius and PA is a tangent, so angle OAP = 90 degrees. 3. Similarly, at point B, OB is a radius and PB is a tangent, so angle OBP = 90 degrees. 4. Let angle AOB at the centre be x degrees. 5. Sum of interior angles of quadrilateral AOPB is 360 degrees, so 90 + 90 + 50 + x = 360. 6. Simplify: 230 + x = 360, so x = 130 degrees. 7. Triangle AOB is isosceles with OA = OB, so its base angles at A and B are equal. 8. Sum of angles in triangle AOB is 180 degrees, so angle OAB + angle OBA + angle AOB = 180. 9. Let angle OAB = angle OBA = y degrees. Then y + y + 130 = 180, so 2y = 50, giving y = 25 degrees. 10. Therefore, angle OAB = 25 degrees. Verification / Alternative check: You can also think of this as a known relation: the angle between two tangents from an external point is supplementary to the central angle subtended by the points of tangency. That is, angle APB + angle AOB = 180, giving angle AOB = 130 degrees. Then the isosceles triangle reasoning leads again to angle OAB = (180 - 130) / 2 = 25 degrees. Both methods are consistent. Why Other Options Are Wrong: • 30°: This would make the angles at A and B sum to 60 degrees, leaving only 120 degrees for angle AOB, contradicting the 130 degree result. • 40°: Then triangle AOB would have angles 40, 40 and 100 degrees, and angle APB would not be 50 degrees. • 50°: This would leave no angle for the centre consistent with the tangent quadrilateral sum. • 35°: Again, this does not satisfy the relationship between central and base angles in triangle AOB. Common Pitfalls: A frequent error is to assume angle OAB equals half of angle APB directly, which is not a general rule. Others forget that radii are perpendicular to tangents and miss the 90 degree angles at A and B. Drawing a clear diagram and marking all right angles usually prevents such mistakes. Final Answer: The measure of angle OAB is 25°.

Question 16

A trapezium has two parallel sides of lengths 18 m and 24 m, and its height is 12 m. What is the area of the trapezium in square metres?

Accepted Answer

Introduction / Context: This is a straightforward mensuration question on the area of a trapezium. Trapeziums appear regularly in aptitude tests, and knowing the basic area formula saves time and avoids confusion with rectangles or triangles. Given Data / Assumptions: • The two parallel sides of the trapezium are 18 m and 24 m. • The perpendicular distance between these parallel sides (height) is 12 m. • We are asked to find the area in square metres. Concept / Approach: The area A of a trapezium with parallel sides a and b and height h is given by A = (1 / 2) * (a + b) * h. This formula can be remembered as the average of the parallel sides times the height. Here, the parallel sides are 18 m and 24 m, and the height is 12 m, so we simply substitute these values into the formula. Step-by-Step Solution: 1. Let a = 18 m and b = 24 m be the lengths of the parallel sides. 2. Let h = 12 m be the height of the trapezium. 3. Use the trapezium area formula: A = (1 / 2) * (a + b) * h. 4. Compute a + b = 18 + 24 = 42 m. 5. Multiply by the height: (a + b) * h = 42 * 12 = 504 square metres. 6. Now multiply by 1 / 2: A = (1 / 2) * 504 = 252 square metres. Verification / Alternative check: You can think of the area as the average of the two bases times the height. The average of 18 m and 24 m is (18 + 24) / 2 = 42 / 2 = 21 m. Multiplying this by the height 12 m gives 21 * 12 = 252 square metres, which matches our earlier calculation. This alternative perspective confirms the result. Why Other Options Are Wrong: • 126 sq.m: This corresponds to taking half of the correct area and arises if you mistakenly divide twice by 2. • 504 sq.m: This is the value before multiplying by 1 / 2, so it ignores the full trapezium formula. • 1024 sq.m: This is unrelated to the given numbers and likely a distractor. • 378 sq.m: This does not follow from any reasonable combination of the sides and height. Common Pitfalls: Many students confuse the trapezium formula with triangle area, or forget the 1 / 2 factor entirely. Others mistakenly use only one base instead of the sum. Always recall that a trapezium can be thought of as having an effective base equal to the average of the two parallel sides. Final Answer: The area of the trapezium is 252 sq.m.

Question 17

PQRS is a cyclic quadrilateral in a circle and PQ is a diameter of the circle. If angle RPQ = 38 degrees, what is the measure of angle PSR?

Accepted Answer

Introduction / Context: This question tests properties of cyclic quadrilaterals and angles subtended by diameters in a circle. Recognising that a diameter subtends a right angle is crucial, and so is the property that opposite angles of a cyclic quadrilateral sum to 180 degrees. Given Data / Assumptions: • PQRS is a cyclic quadrilateral, so all four vertices lie on a circle. • PQ is a diameter of the circle. • Angle RPQ, at P between PR and PQ, is 38 degrees. • We are asked to find angle PSR. Concept / Approach: First, use the fact that an angle subtended by a diameter at any point on the circle is a right angle. So angle PRQ is 90 degrees because it subtends chord PQ which is a diameter. Then in triangle PQR, we can find angle PQR using the angle sum property. Once we know angle PQR, we use the property of a cyclic quadrilateral: opposite angles of a cyclic quadrilateral sum to 180 degrees. Here, angle PSR is opposite angle PQR. Step-by-Step Solution: 1. Since PQ is a diameter, angle PRQ is a right angle, so angle PRQ = 90 degrees. 2. In triangle PQR, we know angle RPQ = 38 degrees and angle PRQ = 90 degrees. 3. Use the triangle angle sum: angle RPQ + angle PRQ + angle PQR = 180 degrees. 4. Substitute: 38 + 90 + angle PQR = 180. 5. Combine known angles: 128 + angle PQR = 180. 6. So angle PQR = 180 - 128 = 52 degrees. 7. In cyclic quadrilateral PQRS, opposite angles are supplementary, so angle PQR + angle PSR = 180 degrees. 8. Substitute angle PQR = 52 degrees: 52 + angle PSR = 180. 9. Thus angle PSR = 180 - 52 = 128 degrees. Verification / Alternative check: You can sketch a circle, mark PQ as a diameter and place R and S on the circle. By roughly measuring angles in your diagram, you will see that angle PSR is obtuse and larger than 120 degrees, which makes 128 degrees a reasonable value. The calculations use standard circle theorems and triangle angle sums, so the result is well supported. Why Other Options Are Wrong: • 52°: This equals angle PQR itself and would make the opposite angles sum to only 104 degrees instead of 180 degrees. • 77°: When added to 52 degrees, the sum is 129 degrees, not 180 degrees, so it violates the cyclic quadrilateral property. • 142°: This would make the sum with 52 degrees equal to 194 degrees, which is impossible for opposite angles. • 90°: This would imply angle PQR is 90 degrees, contradicting the triangle angle sum we already used. Common Pitfalls: Some students forget that a diameter subtends a right angle, or they apply the 180 degree sum property to adjacent instead of opposite angles in the cyclic quadrilateral. Another pitfall is to mix up which angle is at P versus at Q or R. Carefully labelling the diagram and writing down known values helps to avoid these errors. Final Answer: The measure of angle PSR is 128°.

Question 18

The smaller diagonal of a rhombus is equal to the length of each of its sides. If each side is 6 cm long, what is the area of an equilateral triangle whose side is equal to the larger diagonal of the rhombus?

Accepted Answer

Introduction / Context: This problem links properties of a rhombus with those of an equilateral triangle. Rhombus diagonals and equilateral triangle area formulas are standard mensuration topics. Here, you must first use the rhombus geometry to find the larger diagonal, then treat that diagonal as the side of an equilateral triangle and compute its area. Given Data / Assumptions: • Each side of the rhombus is 6 cm. • The smaller diagonal of the rhombus is equal to the side length, so smaller diagonal d1 = 6 cm. • The larger diagonal d2 is unknown. • An equilateral triangle has its side equal to d2, and we are asked to find its area. Concept / Approach: In a rhombus, all sides are equal and the diagonals intersect at right angles and bisect each other. If the diagonals are d1 and d2, then each half diagonal has lengths d1 / 2 and d2 / 2. A side of the rhombus forms the hypotenuse of a right triangle whose legs are these half diagonals. Thus, we can use Pythagoras theorem to find d2. Once we know d2, we use the formula for the area of an equilateral triangle with side a: area = (sqrt(3) / 4) * a^2. Step-by-Step Solution: 1. Let side of the rhombus be a = 6 cm. 2. Smaller diagonal is d1 = 6 cm, so half of it is d1 / 2 = 3 cm. 3. Let d2 be the larger diagonal; half of it is d2 / 2. 4. In one of the right triangles formed by the diagonals, the legs are d1 / 2 and d2 / 2, and the hypotenuse is the side a. 5. Apply Pythagoras theorem: a^2 = (d1 / 2)^2 + (d2 / 2)^2. 6. Substitute a = 6 and d1 / 2 = 3: 6^2 = 3^2 + (d2 / 2)^2. 7. Compute: 36 = 9 + (d2^2 / 4). 8. Rearranging: d2^2 / 4 = 36 - 9 = 27. 9. Multiply by 4: d2^2 = 108. 10. Hence d2 = sqrt(108) = sqrt(36 * 3) = 6 * sqrt(3). 11. Now, an equilateral triangle has side a_equilateral = d2 = 6 * sqrt(3) cm. 12. Area of an equilateral triangle with side a is (sqrt(3) / 4) * a^2. 13. So area = (sqrt(3) / 4) * (6 * sqrt(3))^2 = (sqrt(3) / 4) * (36 * 3) = (sqrt(3) / 4) * 108. 14. Simplify: 108 / 4 = 27, so area = 27 * sqrt(3) square centimetres. Verification / Alternative check: You can check the reasonableness by noting that d2 is longer than the side of the rhombus, which is consistent with a longer diagonal. The equilateral triangle with side 6 * sqrt(3) is significantly larger than a triangle with side 6, and its area should be larger than 6^2. Since 27 * sqrt(3) is approximately 27 * 1.732, which is around 46.8, it fits a realistic area for such a side length. Why Other Options Are Wrong: • 18√3 sq.cm: This corresponds to a smaller effective side length and does not match the calculated d2. • 32√3 sq.cm: This does not follow from the computed d2^2 = 108 and would require a different side. • 36√3 sq.cm: This would correspond to an equilateral triangle with even larger side than 6 * sqrt(3). • 24√3 sq.cm: Again, this value does not come from the formula with a^2 = 108. Common Pitfalls: Common mistakes include assuming the diagonals of a rhombus are equal (which is only true for a square), or forgetting that the diagonals are perpendicular and bisect each other. Some students incorrectly use the entire diagonals instead of half diagonals when applying Pythagoras theorem. Being careful with halves and the right triangle structure is essential. Final Answer: The area of the equilateral triangle is 27√3 sq.cm.

Question 19

A solid cone of height 24 cm and base radius 8 cm is melted and recast into a solid cylinder of radius 6 cm and height 6 cm. What percentage of the original material is wasted in the process?

Accepted Answer

Introduction / Context: This is a classic volume comparison problem where one solid is melted and recast into another solid of a different shape. The task is to use volume formulas for a cone and a cylinder, compare the initial and final volumes and interpret the difference as wasted material. Questions like this test both formula recall and percentage calculations. Given Data / Assumptions: • Original solid: a right circular cone with height h1 = 24 cm and base radius r1 = 8 cm. • New solid: a cylinder with radius r2 = 6 cm and height h2 = 6 cm. • The cone is melted and recast into the cylinder. • Any material not fitting into the cylinder is considered wasted. Concept / Approach: The volume of a cone is V_cone = (1 / 3) * π * r^2 * h. The volume of a cylinder is V_cyl = π * r^2 * h. First compute the volume of the cone and of the cylinder using the given dimensions. The difference V_cone - V_cyl is the volume of material that is not used in the cylinder and is therefore wasted. Finally, compute the percentage waste as (wasted volume / original volume) * 100. Step-by-Step Solution: 1. Volume of the cone: V_cone = (1 / 3) * π * r1^2 * h1. 2. Substitute r1 = 8 cm and h1 = 24 cm: V_cone = (1 / 3) * π * 8^2 * 24. 3. Compute 8^2 = 64 and 64 * 24 = 1536, so V_cone = (1 / 3) * π * 1536. 4. Simplify: 1536 / 3 = 512, so V_cone = 512π cubic centimetres. 5. Volume of the cylinder: V_cyl = π * r2^2 * h2. 6. Substitute r2 = 6 cm and h2 = 6 cm: V_cyl = π * 6^2 * 6 = π * 36 * 6. 7. Compute 36 * 6 = 216, so V_cyl = 216π cubic centimetres. 8. Wasted volume: V_waste = V_cone - V_cyl = 512π - 216π = 296π. 9. Percentage of material wasted = (V_waste / V_cone) * 100. 10. Substitute: Percentage wasted = (296π / 512π) * 100 = (296 / 512) * 100. 11. Simplify 296 / 512 by dividing numerator and denominator by 8: 296 / 8 = 37, 512 / 8 = 64, so the fraction is 37 / 64. 12. Compute 37 / 64 ≈ 0.578125. Multiply by 100 to get approximately 57.8125 percent. 13. Rounded to one decimal place, this is about 57.8 %. Verification / Alternative check: We can also compute the percentage of material used instead of wasted. The used fraction is V_cyl / V_cone = 216π / 512π = 27 / 64, which is about 0.421875 or 42.1875 percent. Subtracting from 100 percent gives the wasted fraction: 100 - 42.1875 = 57.8125 percent. This matches the earlier calculation and confirms that the percentage waste is about 57.8 percent. Why Other Options Are Wrong: • 48.5 %: This would correspond to a smaller difference between the cone and cylinder volumes than actually computed. • 37.5 %: This is close to the fraction V_cyl / V_cone, not the wasted portion. • 64 %: This overestimates the waste and would imply the cylinder uses only 36 % of the volume, contrary to the exact ratio 27 / 64. • 52 %: This is an arbitrary mid value and does not match the precise ratio 37 / 64. Common Pitfalls: Many students mistakenly equate volumes and forget that part of the melted material does not fit into the new solid. Another common error is to use incorrect dimensions in the formulas or to confuse the cone and cylinder formulas. Additionally, rounding too early can cause small discrepancies; it is best to simplify fractions first and then convert to a percentage. Final Answer: The percentage of material wasted is approximately 57.8 %.

Question 20

In triangle ABC, angle ABC = 90 degrees and BP is drawn perpendicular to hypotenuse AC. If angle BAP = 50 degrees, what is the measure of angle PBC?

Accepted Answer

Introduction / Context: This geometry question explores angle relationships in a right triangle when the altitude is drawn from the right angle to the hypotenuse. It uses triangle similarity and basic angle sum properties. Such configurations appear often in Olympiad and aptitude problems because one right triangle can be split into two smaller similar triangles. Given Data / Assumptions: • Triangle ABC is right angled at B, so angle ABC = 90 degrees. • AC is the hypotenuse. • BP is drawn perpendicular to AC, so P is the foot of the altitude from B to AC. • Angle BAP = 50 degrees. • We are asked to find angle PBC. Concept / Approach: In a right triangle, the altitude from the right angle to the hypotenuse splits the original triangle into two smaller right triangles, both similar to the original triangle and to each other. Angle BAP is an angle at A between AB and AP, where AP lies along the hypotenuse AC. Since AP is part of AC, the angle between AB and AP equals the angle at A of the original triangle. Therefore angle BAP is the same as angle A. Once we know angle A, we can find angle C, and by considering triangle PBC we can deduce angle PBC. Step-by-Step Solution: 1. In triangle ABC, angle ABC = 90 degrees. 2. Angle BAP is given as 50 degrees. Since AP lies along AC, angle BAP equals the angle at A of triangle ABC, so angle A = 50 degrees. 3. Using the angle sum of triangle ABC: angle A + angle B + angle C = 180 degrees. 4. Substitute angle A = 50 degrees and angle B = 90 degrees: 50 + 90 + angle C = 180. 5. Simplify: 140 + angle C = 180, so angle C = 40 degrees. 6. Now consider triangle PBC. At point P, BP is perpendicular to AC and C lies on AC, so angle BPC = 90 degrees. 7. In triangle PBC, we already know angle C = angle BCA = 40 degrees, because CP is a segment of AC. 8. Using angle sum in triangle PBC: angle PBC + angle BPC + angle PCB = 180 degrees. 9. Substitute angle BPC = 90 degrees and angle PCB = 40 degrees: angle PBC + 90 + 40 = 180. 10. So angle PBC + 130 = 180, giving angle PBC = 50 degrees. Verification / Alternative check: You can also use triangle similarity. Triangles ABC, ABP and PBC are all similar. Angle A in the original triangle equals 50 degrees and corresponds to angle PBC in the smaller triangle at B. This similarity based approach leads directly to angle PBC = 50 degrees without redoing the angle sum in triangle PBC, reinforcing the same result. Why Other Options Are Wrong: • 30°: Too small to satisfy the angle sum in triangle PBC with its 90 degree and 40 degree angles. • 45°: Would give a total of 45 + 90 + 40 = 175 degrees, which is less than 180 degrees. • 60°: Would make the sum 60 + 90 + 40 = 190 degrees, exceeding 180 degrees. • 40°: This equals angle C, not angle PBC at B. Common Pitfalls: Students often misinterpret angle BAP, thinking it involves some special construction instead of recognising it as the main angle at A. Another mistake is to forget that triangle PBC is right angled at P, not at B, and to use the wrong angle in the sum. Carefully identifying which angles belong to which triangles helps to keep the reasoning consistent. Final Answer: The measure of angle PBC is 50°.

Volume and Surface Area Questions