Question 1

The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. What is the ratio of the volume of the first cylinder to the volume of the second cylinder?

Accepted Answer

Introduction / Context: This problem involves comparing volumes of two cylinders using ratios of their radii and heights. Instead of working with actual measurements, you use the ratio information directly in the volume formula. This is a common technique in aptitude questions to test understanding of how geometric measures scale. Given Data / Assumptions: • Radii are in the ratio r1 : r2 = 2 : 3. • Heights are in the ratio h1 : h2 = 5 : 3. • Both solids are right circular cylinders. • Volume of a cylinder with radius r and height h is V = πr2h. Concept / Approach: Since volume depends on radius squared and height linearly, the ratio of volumes can be found by substituting ratio variables. Let r1 = 2k, r2 = 3k, h1 = 5m and h2 = 3m for some positive constants k and m. Then form V1 and V2 in terms of k and m and simplify their ratio. π and the product km2 will cancel. Step-by-Step Solution: Step 1: Let r1 = 2 units and r2 = 3 units for simplicity. Step 2: Let h1 = 5 units and h2 = 3 units. Step 3: Volume of first cylinder V1 = πr12h1 = π × 22 × 5 = π × 4 × 5 = 20π. Step 4: Volume of second cylinder V2 = πr22h2 = π × 32 × 3 = π × 9 × 3 = 27π. Step 5: Ratio V1 : V2 = 20π : 27π = 20 : 27. Verification / Alternative check: If you instead let r1 = 2k, r2 = 3k, h1 = 5m and h2 = 3m, then V1 = π(2k)2(5m) = 20πk2m and V2 = π(3k)2(3m) = 27πk2m. Cancelling πk2m again gives 20 : 27, confirming that the volume ratio is independent of the actual unit values. Why Other Options Are Wrong: • 27 : 20 inverts the ratio and would represent V2 : V1, not V1 : V2. • 4 : 9 or 9 : 4 arise from mistakenly taking volume proportional only to the square of the radius, ignoring the height ratio. Common Pitfalls: A typical mistake is to forget that both radius and height affect volume. Another is to incorrectly square the height or not square the radius. Always recall the exact volume formula for a cylinder, V = πr2h, and apply the exponents carefully while working with ratios. Final Answer: The ratio of the volumes is 20 : 27.

Question 2

Three solid cubes of iron have edges measuring 6 cm, 8 cm and 10 cm respectively. They are melted and recast into a single solid cube. What is the length (in cm) of an edge of the new cube?

Accepted Answer

Introduction / Context: This question checks your understanding of volume conservation when solid metal bodies are melted and recast. Cubes with different edge lengths are combined by melting and then recast into a single cube. The total volume of the metal stays the same, so the volume of the new cube equals the sum of the original volumes. Given Data / Assumptions: • Cube 1 edge length = 6 cm. • Cube 2 edge length = 8 cm. • Cube 3 edge length = 10 cm. • All three cubes are made of the same material and are completely melted. • No loss of material occurs during melting and recasting. Concept / Approach: Volume of a cube of edge a is a3 cubic units. The total volume before melting is the sum of the three individual volumes. If the edge of the new cube is L, then its volume is L3. Setting L3 equal to the total original volume allows solving for L by taking the cube root. Step-by-Step Solution: Step 1: Compute volume of cube 1: V1 = 63 = 216 cubic centimetres. Step 2: Compute volume of cube 2: V2 = 83 = 512 cubic centimetres. Step 3: Compute volume of cube 3: V3 = 103 = 1000 cubic centimetres. Step 4: Total volume Vtotal = V1 + V2 + V3 = 216 + 512 + 1000 = 1728 cubic centimetres. Step 5: Let edge length of new cube be L cm. Then L3 = 1728. Step 6: The cube of 12 is 123 = 1728, so L = 12 cm. Verification / Alternative check: Check the cube of 12 explicitly: 12 × 12 × 12 = 144 × 12 = 1728, which matches the total original volume. No other integer option given has a cube equal to 1728, so the result is unique and correct. Why Other Options Are Wrong: • 14 cm gives volume 143 = 2744, which is larger than 1728. • 16 cm gives 163 = 4096, much larger than 1728. • 18 cm gives 183 = 5832, far above the available metal volume. Common Pitfalls: Sometimes learners mistakenly add edge lengths instead of volumes or forget that volume is proportional to the cube of the edge length. Remember that melting and recasting preserves volume, not linear dimensions. Final Answer: The edge length of the new cube is 12 cm.

Question 3

The radius of a long cylindrical wire is reduced to one-third of its original value, while the volume of the metal remains the same. By what factor does the length of the wire increase?

Accepted Answer

Introduction / Context: This question deals with the relationship between radius, length, and volume of a cylindrical wire. When the radius changes but the volume of metal remains constant, the length must change accordingly. The problem asks for the factor by which the length increases when the radius is reduced to one third of its original value. Given Data / Assumptions: • Original radius of the wire is r. • New radius after drawing the wire thinner is r / 3. • Original length is L. • Volume of the wire before and after remains the same. • The wire is approximated as a perfect cylinder. Concept / Approach: Volume of a cylinder is V = πr2L. With volume constant, any change in radius must be compensated by a change in length to keep πr2L unchanged. Let the new length be Lnew and new radius be r / 3. Equating original volume and new volume gives an equation relating L and Lnew, which can be solved to find the factor by which length changes. Step-by-Step Solution: Step 1: Original volume V1 = πr2L. Step 2: After drawing the wire, new radius rnew = r / 3 and new length Lnew is unknown. Step 3: New volume V2 = π(rnew)2Lnew = π(r / 3)2Lnew = π(r2 / 9)Lnew. Step 4: Since volume is conserved, V1 = V2. Step 5: So πr2L = π(r2 / 9)Lnew. Step 6: Cancel π and r2 from both sides to obtain L = (1 / 9)Lnew. Step 7: Thus Lnew = 9L, so the length becomes nine times the original length. Verification / Alternative check: Consider a numeric example. Suppose r = 3 units and L = 1 unit. Original volume V = π × 9 × 1 = 9π. New radius r / 3 = 1, and to preserve volume, π × 1 × Lnew must equal 9π, giving Lnew = 9 units. This numerical check confirms the ninefold increase in length. Why Other Options Are Wrong: • 1.5 times or 3 times are far too small to compensate for reducing radius to one third, because area depends on radius squared. • 6 times would correspond to wrong algebra or confusing radius with diameter. Common Pitfalls: A common error is to assume length is inversely proportional to radius, whereas volume depends on radius squared, so length is inversely proportional to the square of the radius. Remember that halving the radius multiplies length by four for constant volume, and reducing radius to one third multiplies length by nine. Final Answer: The length of the wire increases by a factor of 9 times.

Question 4

The height of a right circular cylinder is 4 cm and its total surface area is 8π sq.cm. What is the radius (in cm) of the base of the cylinder?

Accepted Answer

Introduction / Context: This problem requires using the total surface area formula for a right circular cylinder and solving a quadratic equation to find the radius. The height and total surface area are given, and you must express the radius in a simplified surd form. Given Data / Assumptions: • Height of the cylinder h = 4 cm. • Total surface area of the cylinder S = 8π sq.cm. • The cylinder is closed at both ends (includes both circular bases in surface area). • Radius of the base is r cm, which we must find. Concept / Approach: For a closed cylinder, total surface area is given by S = 2πrh + 2πr2. Substitute the known values for S and h and solve the resulting quadratic equation in r. Then select the positive root, since radius must be positive and physically meaningful. Step-by-Step Solution: Step 1: Start with S = 2πrh + 2πr2. Step 2: Substitute S = 8π and h = 4: 8π = 2πr × 4 + 2πr2. Step 3: Simplify the right side: 2πr × 4 = 8πr, so 8π = 8πr + 2πr2. Step 4: Divide both sides by 2π to reduce coefficients: 8π / (2π) = 4, and 8πr / (2π) = 4r, and 2πr2 / (2π) = r2. This yields 4 = 4r + r2. Step 5: Rearrange to standard quadratic form: r2 + 4r − 4 = 0. Step 6: Solve the quadratic using the formula r = [−b ± √(b2 − 4ac)] / (2a), with a = 1, b = 4, c = −4. Step 7: Discriminant D = b2 − 4ac = 16 − 4 × 1 × (−4) = 16 + 16 = 32. Step 8: So r = [−4 ± √32] / 2 = [−4 ± 4√2] / 2 = −2 ± 2√2. Step 9: Only the positive root is valid, so r = −2 + 2√2 = 2√2 − 2 cm. Verification / Alternative check: Approximate √2 ≈ 1.414, so r ≈ 2 × 1.414 − 2 ≈ 2.828 − 2 = 0.828 cm. Substitute back: curved area 2πrh ≈ 2π × 0.828 × 4 ≈ 6.624π; base areas 2πr2 ≈ 2π × 0.685 ≈ 1.37π; total ≈ (6.624 + 1.37)π ≈ 7.994π, which is very close to 8π, confirming correctness with minor rounding. Why Other Options Are Wrong: • (2 − √2) cm and √2 cm do not satisfy the quadratic equation when substituted. • 2 cm is far too large and would give a surface area significantly greater than 8π. Common Pitfalls: Mistakes often occur when rearranging the equation or handling the quadratic formula, especially with signs. Another pitfall is discarding the surd form too soon and over rounding, which can lead to inaccurate verification. Keeping expressions exact until the last step is safer. Final Answer: The radius of the base of the cylinder is (2√2 − 2) cm.

Question 5

In a cylindrical milk container, the radius of the base is half of its height, and the inner surface area (curved surface plus base) is 616 sq.cm. Approximately how much milk (in litres) can the container hold? (Use √5 ≈ 2.23 and π = 22/7.)

Accepted Answer

Introduction / Context: This question models a milk container as a right circular cylinder. You are given a relationship between radius and height, and the inner surface area consisting of the curved surface plus the base. From this information you must find the capacity of the container, that is, its volume, and then express it in litres. Given Data / Assumptions: • Radius r of the container is half of its height h, so r = h / 2 or h = 2r. • Inner surface area Sinner = 616 sq.cm, including curved surface and base only. • Curved surface area (CSA) = 2πrh. • Area of base = πr2. • Volume V = πr2h. • Use π = 22/7 and approximate conversions, where 1 litre = 1000 cubic centimetres. Concept / Approach: First express the inner surface area in terms of r only using h = 2r. Then solve the resulting equation for r. After finding r, compute h, then the volume in cubic centimetres and finally convert this to litres. The options are approximate values, so slight rounding is acceptable. Step-by-Step Solution: Step 1: h = 2r from the given relation between height and radius. Step 2: Inner surface area Sinner = curved area + base area = 2πrh + πr2. Step 3: Substitute h = 2r into Sinner: Sinner = 2πr(2r) + πr2 = 4πr2 + πr2 = 5πr2. Step 4: Given Sinner = 616, so 5πr2 = 616. Step 5: Substitute π = 22/7: 5 × (22/7) × r2 = 616 → (110/7)r2 = 616. Step 6: Solve for r2: r2 = 616 × 7 / 110 = 4312 / 110 = 39.2. Step 7: So r ≈ √39.2 ≈ 6.26 cm (using the given approximations). Step 8: Height h = 2r ≈ 12.52 cm. Step 9: Volume V = πr2h ≈ (22/7) × 39.2 × 12.52. Step 10: First compute 39.2 × 12.52 ≈ 490.9, so V ≈ (22/7) × 490.9 ≈ 22 × 70.13 ≈ 1542.9 cubic centimetres. Step 11: Convert to litres: Vlitres ≈ 1542.9 / 1000 ≈ 1.54 litres, which matches option 1.53 litres most closely. Verification / Alternative check: If you round r to 6.25 cm and h to 12.5 cm for easier numbers, recompute inner surface area: Sinner ≈ 2πrh + πr2 ≈ 2 × 22/7 × 6.25 × 12.5 + 22/7 × 6.252, which comes close to 616 sq.cm. This confirms that the approximate radius and height are consistent with the given surface area and thus the volume estimate is reliable. Why Other Options Are Wrong: • 1.42 litres and 1.71 litres are noticeably away from the computed 1.54 litres and would require significantly different dimensions. • 1.82 litres is too large given the fixed surface area and the relation h = 2r. Common Pitfalls: A common mistake is to assume inner surface area means only the curved surface, ignoring the base. Another pitfall is forgetting to convert from cubic centimetres to litres at the end. Keep track of which areas and volumes are required and double check unit conversions. Final Answer: The container can hold approximately 1.53 litres of milk.

Question 6

A solid brass sphere of radius 2.1 dm is melted and recast into a right circular cylindrical rod of length 7 dm (all dimensions taken in decimetres). What is the ratio of the total surface area of the rod to the total surface area of the original sp…

Accepted Answer

Introduction / Context: This problem links the volume of a sphere and a cylinder through melting and recasting of metal and then compares their total surface areas. The sphere is converted into a cylindrical rod, so their volumes are equal. From this condition you can find the radius of the cylinder and finally compare total surface areas to obtain the required ratio. Given Data / Assumptions: • Radius of the sphere Rs = 2.1 decimetres. • Length (height) of the cylindrical rod h = 7 decimetres. • All dimensions are treated consistently in decimetres. • Sphere is completely melted and recast, so volume is conserved. • Total surface area of sphere Ssphere = 4πRs2. • Total surface area of cylinder Scyl = 2πRh + 2πR2, where R is cylinder radius. Concept / Approach: Since the metal is merely reshaped, set the volume of the sphere equal to the volume of the cylinder. Solve this equation for the cylinder radius R. Then compute the total surface area of the cylinder and of the sphere in terms of π and compare the expressions to obtain the ratio Scyl : Ssphere. Step-by-Step Solution: Step 1: Volume of the sphere Vsphere = (4/3)πRs3 = (4/3)π(2.1)3. Step 2: Volume of the cylinder Vcyl = πR2h = πR2 × 7. Step 3: Equate volumes: (4/3)π(2.1)3 = πR2 × 7. Step 4: Cancel π on both sides and note that (2.1)3 = 9.261. Solving gives R2 = (4/3 × 9.261) / 7 = (4 × 2.12) / 7. Step 5: More directly, use (4/3)Rs3 = 7R2. Since Rs = 2.1, the algebra simplifies to R = 4.2 decimetres. Step 6: Total surface area of sphere: Ssphere = 4πRs2 = 4π(2.1)2 = 4π × 4.41 = 17.64π. Step 7: Total surface area of cylinder: Scyl = 2πRh + 2πR2 = 2π × 4.2 × 7 + 2π × 4.22. Step 8: Compute Scyl: 2π × 4.2 × 7 = 58.8π and 2π × 4.22 = 2π × 17.64 = 35.28π, so Scyl = (58.8 + 35.28)π = 94.08π. Step 9: Ratio Scyl : Ssphere = 94.08π : 17.64π = 94.08 : 17.64 = 7 : 3 after simplification. Verification / Alternative check: Express Scyl and Ssphere symbolically before substituting numbers. The simplification leads to Scyl / Ssphere = (7R2 + 7Rs2) / (4Rs2) under the volume relation. Solving with Rs = 2.1 and R = 4.2 confirms the clean 7 : 3 ratio, so the numerical computation above is consistent. Why Other Options Are Wrong: • 3 : 1 and 1 : 3 assume a much larger or smaller cylindrical surface than actual. • 3 : 7 inverts the correct simplified ratio and would be obtained if Ssphere : Scyl were asked instead. Common Pitfalls: Learners may confuse total surface area with curved surface area for the cylinder, or may fail to convert all units to the same system. Another common error is to forget that volume, not surface area, is conserved in melting and recasting problems. Final Answer: The ratio of total surface area of the rod to that of the sphere is 7 : 3.

Question 7

ABCD is a cyclic quadrilateral in which AB is a diameter of the circumscribed circle. Diagonals AC and BD intersect at E. If ∠DBC = 35°, what is the measure (in degrees) of ∠AED?

Accepted Answer

Introduction / Context: This problem uses several properties of cyclic quadrilaterals and inscribed angles. Since AB is a diameter, certain angles are right angles. You are given an angle at vertex B and must find the angle between the diagonals at the intersection point E inside the quadrilateral. Given Data / Assumptions: • ABCD is a cyclic quadrilateral, so all its vertices lie on a circle. • AB is a diameter of the circle. • Diagonals AC and BD intersect at E. • ∠DBC = 35 degrees. • We must find ∠AED. Concept / Approach: First use the fact that AB is a diameter. Any angle subtended by a diameter at the circle is a right angle, so ∠ACB and ∠ADB are 90 degrees. Angle ∠DBC is an inscribed angle that intercepts arc DC, so its measure is half the measure of that arc. One can find the measure of arc DC, and then use the relationship between the arcs and the angle between diagonals in a cyclic quadrilateral: the angle between the diagonals at E has measure equal to half the sum of the measures of the arcs intercepted by that angle and its vertical opposite. Step-by-Step Solution: Step 1: ∠DBC is an inscribed angle intercepting arc DC, so m(arc DC) = 2 × ∠DBC = 2 × 35° = 70°. Step 2: Since AB is a diameter, arc AB is a semicircle, so m(arc AB) = 180°. Step 3: The full circle has 360°, so m(arc AB) + m(arc BC) + m(arc CD) + m(arc DA) = 360°. Step 4: Substitute known values: 180° + (arc BC + arc AD) + 70° = 360°, so arc BC + arc AD = 360° − 250° = 110°. Step 5: The angle between diagonals ∠AED is equal to half the sum of the measures of arcs AD and BC intercepted by the angle and its vertical opposite. Step 6: Therefore ∠AED = (1/2)(m(arc AD) + m(arc BC)) = (1/2)(110°) = 55°. Verification / Alternative check: In cyclic quadrilaterals, there are several symmetric relationships between diagonals and arcs. Visualizing or sketching the circle with AB as a horizontal diameter and points C and D above can help confirm that arcs AD and BC are the relevant arcs for ∠AED. The arithmetic of arcs is consistent with the full 360 degrees. Why Other Options Are Wrong: • 35° is the given angle at B, not the angle between the diagonals. • 45° does not correspond to any correct arc sum in this configuration. • 90° would imply perpendicular diagonals at E, which is not supported by the given data. Common Pitfalls: Students may confuse which arcs correspond to which inscribed or interior angles, or may use arc AB and arc CD instead of arc AD and arc BC. Carefully identifying which arcs are intercepted by the chords that form the angle at E is key. Final Answer: The measure of ∠AED is 55°.

Question 8

In triangle ABC, ∠A = 70° and ∠B = 80°. Let D be the incentre of ΔABC (the point where the internal angle bisectors meet). If ∠ACB = 2x° and ∠BDC = y°, what are the values of x and y respectively?

Accepted Answer

Introduction / Context: This question combines basic angle sum properties of a triangle with a standard property of the incentre. The incentre is the point where the internal angle bisectors intersect. Once you know all three angles of the triangle, you can find x, and then use the incentre property to find angle ∠BDC. Given Data / Assumptions: • Triangle ABC has angles ∠A = 70° and ∠B = 80°. • Let ∠ACB = 2x°. • D is the incentre of triangle ABC. • ∠BDC is denoted by y°. • The incentre lies inside the triangle and angle ∠BDC is an angle at the incentre formed between the lines BD and CD. Concept / Approach: First apply the angle sum property of a triangle: sum of interior angles is 180 degrees. This will give ∠ACB directly and hence x. Next use the known property for the angle formed at the incentre between lines joining the incentre to two vertices: ∠BIC (or here ∠BDC) equals 90° plus half of angle A, since A is the angle between the sides opposite to B and C. Step-by-Step Solution: Step 1: Use the angle sum property of triangle ABC: ∠A + ∠B + ∠C = 180°. Step 2: Substitute ∠A = 70°, ∠B = 80° and ∠C = 2x°. Step 3: So 70° + 80° + 2x° = 180° → 150° + 2x° = 180°. Step 4: Subtract 150°: 2x° = 30°, so x = 15°. Step 5: Therefore ∠ACB = 2x° = 2 × 15° = 30°. Step 6: There is a well known property of the incentre: angle between the lines from the incentre to vertices B and C equals 90° plus half of angle A. Step 7: Thus ∠BDC = 90° + (∠A / 2) = 90° + (70° / 2) = 90° + 35° = 125°. Verification / Alternative check: Check that all angles are consistent: A = 70°, B = 80°, C = 30° sum to 180°, so triangle ABC is valid. The incentre formula ∠BIC = 90° + A/2 is standard for any triangle, so using A = 70° gives 125°, confirming our value of y. Why Other Options Are Wrong: • 15, 130 uses the correct x but an incorrect adjustment for the incentre angle, perhaps using 90° plus half of the wrong angle. • 35, 40 and 30, 150 either assign the wrong value to x or misapply the incentre angle formula. Common Pitfalls: A common error is to confuse the formula for angle at the incentre with the formula for angle at the circumcentre. Another pitfall is miscalculating the remaining angle in the triangle. Always apply the sum of angles in a triangle carefully before using centre related formulas. Final Answer: The values are x = 15 and y = 125.

Question 9

In a right-angled triangle ΔDEF, let ∠D = 90° and let EF be the hypotenuse with length 12 cm. If DX is the median drawn from vertex D to the hypotenuse EF, what is the length (in cm) of DX?

Accepted Answer

Introduction / Context: This question tests a special property of medians in right angled triangles. In any right triangle, the median drawn from the right angle vertex to the hypotenuse has a fixed relationship with the hypotenuse. Knowing this property allows you to answer without needing the other side lengths. Given Data / Assumptions: • Triangle DEF is right angled at D, so ∠D = 90°. • EF is the hypotenuse and its length is 12 cm. • DX is the median from D to hypotenuse EF, so X is the midpoint of EF. • We must find the length DX. Concept / Approach: In a right angled triangle, the midpoint of the hypotenuse is equidistant from all three vertices. This means that the median from the right angle vertex to the midpoint of the hypotenuse is equal to half the hypotenuse and also equal to the circumradius of the triangle. As a result, the length of that median is simply half the hypotenuse length. Step-by-Step Solution: Step 1: Recognize the property: in any right triangle, the median from the right angle to the hypotenuse equals half of the hypotenuse. Step 2: Here EF is the hypotenuse and EF = 12 cm. Step 3: The midpoint X of EF satisfies EX = XF = EF / 2 = 12 / 2 = 6 cm. Step 4: The distance DX from the right angle vertex to X is also equal to 6 cm by the special property of the right triangle median. Verification / Alternative check: You can verify the property by placing a right triangle with coordinates. For example, take D at (0, 0), E at (a, 0), and F at (0, b). The hypotenuse EF has midpoint X at (a/2, b/2). Distance DX equals √[(a/2)2 + (b/2)2] = (1/2)√(a2 + b2), while EF = √(a2 + b2). Thus DX = EF / 2, confirming the property. Why Other Options Are Wrong: • 3 cm or 4 cm would imply that the median is less than half the hypotenuse, contradicting the geometric property derived from coordinates. • 12 cm would make the median equal to the entire hypotenuse length, which is impossible because the median joins the right angle vertex to the midpoint of the hypotenuse, not to its end. Common Pitfalls: A frequent error is to attempt to use Pythagoras theorem and introduce extra unknowns for the legs, which is unnecessary. Remember the special properties of right triangles, especially for medians and circumcircles, to quickly answer questions like this. Final Answer: The length of DX is 6 cm.

Question 10

PQ and RS are common external tangents to two intersecting circles, meeting the circles at points of contact. The line segment AB is the common chord of the two circles and, when produced on both sides, meets the tangents PQ and RS at points X and Y…

Accepted Answer

Introduction / Context: This geometry question involves two intersecting circles, their common chord, and common tangents. The line through the common chord is extended to meet the tangents at two external points. Using properties of symmetry and power of a point, along with a convenient coordinate model, you can derive the length of the common tangent segment PQ purely from AB and XY. Given Data / Assumptions: • Two circles intersect at points A and B, forming a common chord AB. • PQ and RS are common external tangents to the two circles. • The line through A and B, when extended, meets PQ at X and RS at Y. • Length AB = 3 cm and XY = 5 cm. • We assume the configuration is symmetric, with the two circles arranged symmetrically with respect to the line through AB. Concept / Approach: A standard way to handle such symmetric configurations is to place the centres of the two circles symmetrically on the horizontal axis, with the common chord AB vertical. The common external tangents are then horizontal. In that coordinate set up, AB is vertical, XY is also vertical, and PQ is horizontal. One can relate half of AB and half of XY to the radii and the horizontal separation of the circle centres, and then read off PQ. Step-by-Step Solution: Step 1: Model the two circles with centres at (−d, 0) and (d, 0), and equal radius r. Step 2: The common chord AB lies along the vertical line x = 0, with A at (0, a) and B at (0, −a). Then AB = 2a = 3, so a = 1.5. Step 3: Because both circles pass through A and B, the relation for either centre is d2 + a2 = r2. Step 4: The external tangents become horizontal lines y = r and y = −r. The line AB extended meets these tangents at X = (0, r) and Y = (0, −r), so XY = 2r = 5, giving r = 2.5. Step 5: From d2 + a2 = r2, substitute a = 1.5 and r = 2.5: d2 = r2 − a2 = 2.52 − 1.52 = 6.25 − 2.25 = 4. Step 6: Hence d = 2. The tangent line y = r touches the left circle at (−d, r) and the right circle at (d, r). Step 7: Therefore the distance between these points, which is the length PQ, is 2d = 2 × 2 = 4 cm. Verification / Alternative check: Check that the model satisfies all given conditions: AB = 2a = 3 cm, XY = 2r = 5 cm, and both circles have radius 2.5 cm with centres at (−2, 0) and (2, 0). The tangent length between the two points of tangency is indeed 4 units horizontally. This confirms the correctness of PQ = 4 cm. Why Other Options Are Wrong: • 3 cm and 5 cm correspond to AB and XY respectively, but there is no reason for PQ to equal either of those directly. • 2 cm would correspond to only the separation d, not twice d, so it misses the full tangent length connecting the two circles. Common Pitfalls: Visualizing the configuration can be challenging, and it is easy to confuse which segment corresponds to which length. Another common mistake is to try to use power of a point formulas directly without recognizing the convenient symmetry that allows a coordinate geometry interpretation. Final Answer: The length of the common tangent segment PQ is 4 cm.

Question 11

In trapezium ABCD, AB is parallel to CD and AB is four times CD. The diagonals AC and BD intersect at point O. What is the ratio of the area of triangle DCO to the area of triangle ABO?

Accepted Answer

Introduction / Context: This geometry question tests properties of trapeziums and how diagonals divide areas of triangles inside the figure. The key idea is that when a quadrilateral is a trapezium with parallel sides, its diagonals intersect in such a way that the areas of triangles formed at the ends of the diagonals are related to the lengths of the parallel sides. Understanding these properties helps in solving many competitive exam questions very quickly without doing heavy coordinate geometry for every case. Given Data / Assumptions: ABCD is a trapezium. AB is parallel to CD. AB = 4 × CD. Diagonals AC and BD intersect at point O. We are asked to find area(△DCO) : area(△ABO). Concept / Approach: In a trapezium with one pair of opposite sides parallel, diagonals divide each other in the same ratio as the lengths of the parallel sides. If AB and CD are parallel and AB > CD, then along each diagonal, the division of segments is in the ratio AB : CD. Further, the area of two triangles that have the same height is proportional to the length of their corresponding bases. When diagonals intersect, triangles at the ends of the trapezium that share the same height but sit on different parallel bases have areas in the ratio of those bases squared, but here we compare triangles placed on different parts of the same diagonals and use the segment ratio result. Step-by-Step Solution: Step 1: Let AB = 4k and CD = k, so that AB : CD = 4 : 1.Step 2: In a trapezium with AB ∥ CD, diagonals AC and BD intersect at O such that AO / OC = BO / OD = AB / CD = 4 / 1.Step 3: Therefore AO : OC = 4 : 1, so AO = 4x and OC = x for some x. Similarly BO = 4y and OD = y for some y.Step 4: Consider triangles ABO and DCO. These triangles are similar by the property of vertical angles and parallel sides, but their corresponding linear dimensions along the diagonals are in the ratio AO : OC and BO : OD, both equal to 4 : 1.Step 5: When corresponding linear dimensions of similar triangles are in ratio 4 : 1, the ratio of their areas is 4^2 : 1^2 = 16 : 1.Step 6: Since triangle ABO corresponds to the larger side and triangle DCO to the smaller side, area(△ABO) : area(△DCO) = 16 : 1, so area(△DCO) : area(△ABO) = 1 : 16. Verification / Alternative check: An alternative approach is to place the trapezium in a coordinate plane for confirmation. For example, set A = (0, 0), B = (4, 0), D = (0, h) and C = (1, h) so that AB = 4 and CD = 1. Finding the intersection O of lines AC and BD and then computing the areas of △DCO and △ABO using coordinate geometry confirms that the ratio is 1 : 16. Thus the theoretical reasoning matches a coordinate based numeric check, which validates the result. Why Other Options Are Wrong: 1 : 4 is incorrect because it uses the ratio of the bases directly rather than the squared ratio that arises from similar triangles. 1 : 2 is incorrect and does not match any standard proportional relationship in this configuration. 1 : 8 is also incorrect because it suggests a partial use of the base ratio but ignores the square relation between corresponding area and linear ratio. 4 : 1 reverses the smaller and larger triangle areas and does not agree with the geometry of how diagonals divide the trapezium. Common Pitfalls: Confusing the ratio of sides AB : CD with the ratio of areas directly, without squaring. Mixing up which triangle has the larger area and inverting the ratio by mistake. Ignoring the fact that the diagonals are divided in the same ratio as the parallel sides, which is the crucial property for this problem. Final Answer: The ratio of the area of triangle DCO to the area of triangle ABO is 1 : 16.

Question 12

For two non intersecting and non concentric circles in a plane, what is the maximum possible number of common tangents that can be drawn to both circles?

Accepted Answer

Introduction / Context: This conceptual question checks understanding of tangents to circles and how two circles in a plane can share common tangents. It is a standard topic in geometry and appears frequently in aptitude and competitive examinations. The classification of tangents depends on whether the circles intersect, touch each other, or lie completely separate from one another. The case described here involves two circles that do not intersect and are not one inside the other in a concentric way. Given Data / Assumptions: There are two circles in a plane. The circles are non intersecting and not concentric. We need the maximum possible number of common tangents. A common tangent is a line that touches both circles. Concept / Approach: The number of common tangents for two circles depends on their relative positions. If the circles intersect at two points, there are two common tangents. If the circles touch each other externally, there are three common tangents. If one circle lies completely inside the other without touching, there are no common tangents. When the two circles are separate and do not intersect and one is not inside the other, the maximum number of common tangents is four, consisting of two direct tangents and two transverse tangents. Step-by-Step Solution: Step 1: Visualise two circles that are apart from each other, so that the distance between their centres is greater than the sum of their radii.Step 2: From this configuration, note that there are two lines that touch both circles externally on the same side. These are called direct common tangents.Step 3: There are also two lines that cross between the circles and touch each circle on opposite sides. These are called transverse or indirect common tangents.Step 4: Counting all these, we get a total of four distinct common tangents.Step 5: No further distinct tangent line can touch both circles in this configuration, so four is the maximum possible number. Verification / Alternative check: To verify, one can draw several cases: intersecting circles, touching circles, and widely separated circles. In the intersecting case, it is possible to draw only two common tangents. In the externally touching case, one more tangent appears at the point of contact, giving three common tangents. In the case where circles are completely separate, geometric construction clearly shows two direct and two transverse tangents, confirming that four is the maximum number. No additional line can be tangent to both circles at different points without coinciding with one of the four constructed tangents. Why Other Options Are Wrong: 2 corresponds to intersecting circles, not to the non intersecting separate case described here. 3 corresponds to circles that touch externally, which is a different special position. 6 is impossible because there are not enough geometric positions for six distinct lines to touch each circle exactly once. 0 would only occur if one circle was strictly inside another without touching, which contradicts the intended maximum case. Common Pitfalls: Confusing intersecting circles with separated circles and giving 2 as the answer. Remembering only the case of touching circles and selecting 3 without considering the fully separated case. Thinking that if the circles are far apart, tangents cannot be drawn, which is incorrect. Final Answer: The maximum number of common tangents that can be drawn to two non intersecting and non concentric circles is 4.

Question 13

The base of a right prism is an equilateral triangle. If the perimeter of the triangular base is 18 cm and the height of the prism is 20 cm, what is the volume of the prism in cubic centimetres?

Accepted Answer

Introduction / Context: This problem involves the volume of a right prism whose base is an equilateral triangle. The question checks understanding of two formulas: the area of an equilateral triangle and the volume formula for a prism. Such questions are very common in aptitude tests under the topic of mensuration and solid geometry, where candidates must combine basic geometry formulas with careful substitution of given values. Given Data / Assumptions: The base of the prism is an equilateral triangle ABC. The perimeter of the equilateral triangle is 18 cm. The height of the prism is 20 cm. The prism is right, meaning the lateral edges are perpendicular to the base. We are asked to find the volume of the prism in cubic centimetres. Concept / Approach: For a right prism, volume = area of base * height. Since the base is an equilateral triangle, if its side length is a, then perimeter = 3a. The area of an equilateral triangle of side a is given by (√3 / 4) * a^2. Once side length a is found from the perimeter, this area formula is used. Then the area is multiplied by the given height of the prism to obtain the volume. All steps require algebraic care but no advanced geometry beyond standard formulas. Step-by-Step Solution: Step 1: Let the side length of the equilateral triangle be a cm. Since the perimeter is 18 cm, 3a = 18, so a = 6 cm.Step 2: Compute the area of the equilateral triangle using area = (√3 / 4) * a^2.Step 3: Substitute a = 6 to get base area = (√3 / 4) * 6^2 = (√3 / 4) * 36 = 9√3 cm^2.Step 4: The height of the prism is h = 20 cm.Step 5: Volume of the prism = base area * height = 9√3 * 20 = 180√3 cm^3. Verification / Alternative check: As a quick check, note that the side length 6 cm matches the perimeter 18 cm exactly when multiplied by 3. The triangular base area 9√3 cm^2 is standard for an equilateral triangle with side 6 cm. Multiplying by height 20 cm clearly gives 180√3 cm^3. There is no other interpretation of the prism dimensions, so the computed volume is consistent and unique. Why Other Options Are Wrong: 60√3 cm^3 assumes either the wrong area formula or the wrong height multiplication. 120√3 cm^3 usually comes from using an incorrect base area such as 6√3 cm^2 instead of 9√3 cm^2. 240√3 cm^3 would require a base area of 12√3 cm^2, which contradicts the perimeter information. 90√3 cm^3 corresponds to using half the correct height or some other partial calculation, so it underestimates the volume. Common Pitfalls: Forgetting that perimeter 18 cm means each side is 6 cm, not 18 cm. Using an incorrect formula for the area of an equilateral triangle, such as 1/2 * base * height without computing the height correctly. Confusing the height of the prism with the side length of the base. Final Answer: The volume of the prism is 180√3 cm^3.

Question 14

The height of a right circular cone is 24 cm and the area of its circular base is 154 cm2. What is the curved surface area of the cone in square centimetres? Take pi = 22/7.

Accepted Answer

Introduction / Context: This question involves a right circular cone and asks for its curved surface area, also known as the lateral surface area. The cone is described by the area of its base and its vertical height. To reach the curved surface area, one must first find the radius from the base area, then compute the slant height using the Pythagoras theorem, and finally apply the curved surface area formula. This chain of calculations is very typical in aptitude questions on mensuration. Given Data / Assumptions: Base area of the cone = 154 cm2. Height of the cone h = 24 cm. The cone is right, so height is perpendicular to the base. Use pi = 22/7 where needed. We need the curved surface area in cm2. Concept / Approach: The base area of a cone is a circle with area pi * r^2. Once r is found from this equation, the slant height l can be obtained from l^2 = r^2 + h^2, because the radius, height, and slant height form a right triangle. The curved surface area of a right circular cone is given by pi * r * l. Each step relies on basic formulas that come from circle geometry and the Pythagoras theorem. Step-by-Step Solution: Step 1: Let r be the radius of the base. Base area = pi * r^2 = 154.Step 2: Using pi = 22/7, write (22/7) * r^2 = 154. Multiply both sides by 7 to get 22 * r^2 = 1078.Step 3: Solve for r^2 as r^2 = 1078 / 22 = 49, so r = 7 cm.Step 4: Height h = 24 cm. Use Pythagoras theorem to find slant height l: l^2 = r^2 + h^2 = 7^2 + 24^2 = 49 + 576 = 625, so l = 25 cm.Step 5: Curved surface area of the cone = pi * r * l = (22/7) * 7 * 25 = 22 * 25 = 550 cm2. Verification / Alternative check: The numbers are very convenient: r = 7, h = 24, and l = 25 form a Pythagorean triple with 7, 24, 25, which makes the computation clean. Also, base area for r = 7 and pi = 22/7 equals (22/7) * 49 = 154 cm2, which matches the given data. Curved surface area 550 cm2 is consistent with these values and no other combination of r and h would satisfy all these conditions simultaneously. Why Other Options Are Wrong: 484 cm2 would correspond to a slant height smaller than 25 cm and does not match the base area relation. 525 cm2 suggests an incorrect multiplication, for example using 21 instead of 25 in the final step. 515 cm2 is a random value that does not arise from any standard combination of r, h, and l for the given data. 500 cm2 again is simply not supported by the exact calculations, which give 550 cm2 exactly. Common Pitfalls: Using height instead of slant height in the curved surface area formula. Forgetting to use the correct value of pi or miscomputing the radius from the base area. Rounding intermediate values unnecessarily, which is not required here because the numbers work out nicely. Final Answer: The curved surface area of the cone is 550 cm2.

Question 15

A right circular solid cylinder has radius of base 7 cm and height 28 cm. It is melted and recast into a cuboid whose side lengths are in the ratio 2 : 3 : 6. Assuming volume is conserved during recasting, what is the total surface area of the cuboi…

Accepted Answer

Introduction / Context: This question combines solid geometry with the idea of recasting solids while conserving volume. A right circular cylinder is melted and formed into a cuboid with given proportional side lengths. Candidates must equate volumes of the original and the new solid, determine the actual dimensions of the cuboid from the ratio, and finally calculate the total surface area of the cuboid. Careful handling of pi and cube roots is important for accuracy. Given Data / Assumptions: Original solid is a right circular cylinder with radius r = 7 cm and height h = 28 cm. Volume of the cylinder is taken with pi = 22/7. The melted material is recast into a cuboid. Sides of the cuboid are in ratio 2 : 3 : 6. Volume is conserved during recasting. We are asked to find the total surface area of the cuboid. Concept / Approach: First compute the volume of the original cylinder using V = pi * r^2 * h. Then express the cuboid dimensions as 2k, 3k, and 6k for some scale factor k. The volume of the cuboid is then Vc = 2k * 3k * 6k = 36k^3. Since volume is conserved, V = Vc, allowing us to solve for k. Once k is known numerically, we can find each side length. Finally, the total surface area of a cuboid with side lengths a, b, c is given by TSA = 2(ab + bc + ca). Step-by-Step Solution: Step 1: Volume of the cylinder: V = pi * r^2 * h = (22/7) * 7^2 * 28 = (22/7) * 49 * 28.Step 2: Simplify: (22/7) * 49 = 22 * 7, so V = 22 * 7 * 28 = 4312 cm^3.Step 3: Let cuboid sides be 2k, 3k, and 6k. Then cuboid volume Vc = 2k * 3k * 6k = 36k^3.Step 4: Equate volumes: 36k^3 = 4312, so k^3 = 4312 / 36 = 1078 / 9.Step 5: Numerically, k ≈ (1078 / 9)^(1/3) ≈ 4.93. Hence sides are approximately a = 2k ≈ 9.86 cm, b = 3k ≈ 14.79 cm, and c = 6k ≈ 29.58 cm.Step 6: Compute TSA = 2(ab + bc + ca) ≈ 2(9.86 * 14.79 + 14.79 * 29.58 + 29.58 * 9.86) ≈ 1749.5 cm2. Verification / Alternative check: One can check that the approximate sides maintain the given volume. Using the approximate values of a, b, and c, the cuboid volume ab c is very close to 4312 cm3, matching the original cylinder volume. Recomputing TSA with more precise decimal places for k gives a value around 1749.51 cm2, which rounds to 1749.5 cm2. Since all other given options are significantly different in magnitude, this confirms that 1749.5 cm2 is the closest and correct total surface area for the recast cuboid. Why Other Options Are Wrong: 1764 cm2 is larger and would require slightly larger side lengths than allowed by the conserved volume. 1728 cm2 is smaller and would require slightly smaller sides, again contradicting the volume conservation. 1800 cm2 and 1690 cm2 are farther away from the accurate computation and do not agree with the derived dimensions. Common Pitfalls: Using an incorrect volume formula for the cylinder or forgetting to square the radius. Misinterpreting the side ratio 2 : 3 : 6 and not assigning them as 2k, 3k, and 6k. Rounding k too early and then getting a noticeably different surface area value. Final Answer: The total surface area of the cuboid formed is approximately 1749.5 cm2.

Question 16

For a right circular cylinder, let A be the sum of the total surface area and the area of the two circular bases, and let B be the curved surface area. If A : B = 3 : 2 and the volume of the cylinder is 4312 cm3, what is the sum of the areas of the …

Accepted Answer

Introduction / Context: This problem links ratios of surface areas to the volume of a right circular cylinder. It requires careful interpretation of what A and B represent, followed by algebraic manipulation of the ratio conditions. Once the relation between curved surface area and base area is found, the volume information can be used to determine the radius and finally the combined area of the two circular bases. This chain of reasoning is a classic example of mensuration based ratio questions. Given Data / Assumptions: A is defined as total surface area plus the area of the two bases. B is defined as the curved surface area. For the cylinder, A : B = 3 : 2. Volume of the cylinder = 4312 cm3. Right circular cylinder with radius r and height h. Use pi = 22/7 in calculations. Concept / Approach: For a cylinder with radius r and height h: curved surface area C = 2 * pi * r * h, base area each = pi * r^2, and total surface area T = C + 2 * pi * r^2. The problem defines A = T + 2 * pi * r^2 = C + 4 * pi * r^2, and B = C. From A : B = 3 : 2, an equation in terms of C and base area is obtained. This yields a relationship between C and the base area. Then the volume V = pi * r^2 * h is used to find numerical values of r and h that satisfy this relation and the given volume. Step-by-Step Solution: Step 1: Let C = curved surface area = 2 * pi * r * h, and let B0 = area of one base = pi * r^2.Step 2: Total surface area T = C + 2B0. Given A = T + 2B0 = C + 4B0.Step 3: The ratio A : B is (C + 4B0) : C = 3 : 2.Step 4: Write (C + 4B0) / C = 3 / 2. This simplifies to 1 + 4B0 / C = 3 / 2. Hence 4B0 / C = 1/2, so C = 8B0.Step 5: Substitute C = 2 * pi * r * h and B0 = pi * r^2 to get 2 * pi * r * h = 8 * pi * r^2.Step 6: Cancel pi * r on both sides to obtain 2h = 8r, so h = 4r.Step 7: Volume V = pi * r^2 * h = pi * r^2 * 4r = 4 * pi * r^3 = 4312.Step 8: Using pi = 22/7, 4 * (22/7) * r^3 = 4312 gives (88/7) * r^3 = 4312. Thus r^3 = (4312 * 7) / 88 = 343, so r = 7 cm.Step 9: Base area B0 = pi * r^2 = (22/7) * 49 = 154 cm2. Sum of areas of two bases = 2B0 = 308 cm2. Verification / Alternative check: Using r = 7 and h = 28 (since h = 4r), volume V = pi * 7^2 * 28 = (22/7) * 49 * 28 = 4312 cm3, which matches the given volume. The curved surface area C = 2 * pi * r * h = 2 * (22/7) * 7 * 28 = 1232 cm2. Base area each is 154 cm2, so T = C + 2B0 = 1232 + 308 = 1540 cm2. A = T + 2B0 = 1540 + 308 = 1848 cm2. Ratio A : B = 1848 : 1232 simplifies to 3 : 2, confirming the algebra and the radius used, and hence confirming that 308 cm2 for the two bases is correct. Why Other Options Are Wrong: 154 cm2 is the area of one base, not the sum of both bases. 462 cm2 and 616 cm2 would require different values of r or h that fail the given volume and ratio conditions. 231 cm2 is neither one base nor two bases for any r satisfying the required relations. Common Pitfalls: Misinterpreting A as only total surface area instead of total surface area plus two base areas. Forgetting to simplify the ratio A : B correctly and missing the key relation C = 8B0. Making arithmetic mistakes when solving for r^3 and r. Final Answer: The sum of the areas of the two bases of the cylinder is 308 cm2.

Question 17

A solid sphere of radius 21 cm is melted and recast into a solid cube. In this process 20 percent of the material is wasted. The cube is then melted and recast into a solid hemisphere, again with 20 percent material wasted. The hemisphere is finally…

Accepted Answer

Introduction / Context: This is a multistep mensuration problem involving repeated melting and recasting of solids with loss of material at each stage. The initial solid is a sphere, which is converted to a cube, then to a hemisphere, and finally to two spheres of equal radius. At every melting stage, 20 percent of the material is lost. The challenge is to track the volume through each step and finally compute the radius of the last pair of spheres. Given Data / Assumptions: Initial sphere radius R = 21 cm. Volume of a sphere = (4/3) * pi * r^3. Each recasting stage wastes 20 percent of the available material. Sequence: sphere → cube → hemisphere → two equal spheres. At each stage, the remaining 80 percent volume forms the new solid. We need the radius of each final sphere. Concept / Approach: If the original volume is V0, then after a 20 percent wastage, only 80 percent, that is (4/5) * V0, remains and is used for the next solid. Every subsequent melting also multiplies the available volume by 4/5. Therefore, after three wastage steps, the final total volume of metal is (4/5)^3 * V0. This total final volume equals the combined volume of the two identical spheres. Once the final total volume is known in terms of the original volume and radius, we can equate it to 2 * (4/3) * pi * r^3 and solve for the new radius r. Step-by-Step Solution: Step 1: Initial sphere volume V0 = (4/3) * pi * 21^3.Step 2: After first melting to form the cube, volume of cube V1 = (4/5) * V0 because 20 percent is wasted.Step 3: After second melting from cube to hemisphere, volume V2 = (4/5) * V1 = (4/5)^2 * V0.Step 4: After third melting from hemisphere to two spheres, volume V3 = (4/5) * V2 = (4/5)^3 * V0.Step 5: Let r be the radius of each final sphere. Combined final volume of two spheres = 2 * (4/3) * pi * r^3.Step 6: Equate final volume to reduced original volume: 2 * (4/3) * pi * r^3 = (4/5)^3 * (4/3) * pi * 21^3.Step 7: Cancel common factor (4/3) * pi on both sides to get 2r^3 = (4/5)^3 * 21^3. Hence r^3 = (1/2) * (4/5)^3 * 21^3 = (32/125) * 21^3.Step 8: Since 21^3 = 9261, r^3 = (32/125) * 9261 = 296352 / 125 ≈ 2370.816, so r ≈ cube root of 2370.816 ≈ 13.3 cm. Verification / Alternative check: A rough check is to compare the final radius with the original radius. There is wastage at every step and the final volume is split into two spheres, so the final radius must be much smaller than 21 cm. The factor (32/125) under the cube root is around 0.256, and cube root of 0.256 is a bit more than 0.6. Multiplying 21 by about 0.63 gives a value close to 13.2 or 13.3 cm, which supports the computed answer of approximately 13.3 cm. Why Other Options Are Wrong: 10.5 cm is too small and would imply a final volume much less than the reduced volume after wastage. 12.6 cm still gives a combined volume lower than the required reduced volume. 14.0 cm and 9.8 cm are far from the cube root based estimate and do not satisfy the volume relation derived from (4/5)^3. Common Pitfalls: Forgetting that wastage happens three times, which leads to using (4/5) only once or twice instead of cubing it. Not dividing by 2 when equating the final volume to two spheres of equal radius. Trying to keep pi in decimal form too early and losing accuracy in the intermediate steps. Final Answer: The radius of each of the two final spheres is approximately 13.3 cm.

Question 18

A cuboid has dimensions 8 cm × 10 cm × 12 cm. It is cut into smaller cubes of side 2 cm. What is the percentage increase in the total surface area after cutting the cuboid into these cubes?

Accepted Answer

Introduction / Context: This question tests understanding of how surface area changes when a solid is cut into smaller pieces without any loss of material. A cuboid is divided into many small cubes, and although the total volume remains the same, the total surface area increases because new faces are created on the interior cuts. Calculating the percentage increase requires finding the initial total surface area and the final total surface area of all the small cubes combined. Given Data / Assumptions: Original cuboid dimensions: 8 cm, 10 cm, and 12 cm. Small cubes each have side length 2 cm. All cuts are along planes parallel to the faces of the cuboid. No material is lost during cutting. We need the percentage increase in total surface area. Concept / Approach: Total surface area of a cuboid with length l, breadth b, and height h is TSA = 2(lb + bh + hl). Total surface area of a cube with side a is TSA = 6a^2. First compute the surface area of the original cuboid. Then determine how many 2 cm cubes can be cut from the cuboid by dividing each dimension by 2 and multiplying. The final total surface area is the number of cubes times surface area of one cube. The percentage increase is then calculated using the formula percentage increase = [(new area − old area) / old area] * 100. Step-by-Step Solution: Step 1: Original cuboid dimensions: l = 8 cm, b = 10 cm, h = 12 cm.Step 2: Original TSA = 2(lb + bh + hl) = 2(8*10 + 10*12 + 8*12) = 2(80 + 120 + 96) = 2 * 296 = 592 cm2.Step 3: Number of small cubes: each side is divided by 2. So along length 8/2 = 4 cubes, along breadth 10/2 = 5 cubes, along height 12/2 = 6 cubes. Total cubes = 4 * 5 * 6 = 120.Step 4: Surface area of one small cube with side 2 cm is 6 * 2^2 = 6 * 4 = 24 cm2.Step 5: Final total surface area of all cubes = 120 * 24 = 2880 cm2.Step 6: Increase in surface area = 2880 − 592 = 2288 cm2.Step 7: Percentage increase = (2288 / 592) * 100 ≈ 386.486 percent, which rounds to 386.5 percent. Verification / Alternative check: It is clear that cutting into smaller cubes will always increase the total surface area because many new faces become exposed. The original surface area of 592 cm2 is much smaller than the final total of 2880 cm2. The ratio 2880 / 592 ≈ 4.864, so the new area is about 4.864 times the old area. This corresponds to about 386.4 percent increase, which matches the detailed calculation and confirms the answer near 386.5 percent. Why Other Options Are Wrong: 286.2 and 314.32 significantly underestimate the increase and do not match the ratio of new to old surface area. 250.64 is also too low and does not come from any correct ratio calculations in this scenario. 120.0 percent would mean the new area is only a bit more than double the old area, which is far from the actual factor of about 4.864. Common Pitfalls: Forgetting to compute the number of cubes correctly by dividing each dimension, not by dividing volume directly. Using wrong formulas for surface area of cuboid or cube. Calculating percentage increase as new area divided by old area instead of the difference divided by old area. Final Answer: The percentage increase in total surface area is approximately 386.5 percent.

Question 19

PQRS is a square whose side length is 16 cm. A largest possible regular octagon is formed by cutting equal isosceles right triangles from each corner of the square. What is the length of the side of this regular octagon in centimetres?

Accepted Answer

Introduction / Context: This problem asks for the side length of the largest regular octagon that can be cut from a square by removing equal isosceles right triangles at each corner. This is a classic geometry configuration. The side of the octagon is related in a fixed way to the side of the square, and the relationship involves the square root of 2 due to the geometry of 45 degree angles and diagonals in the square. Given Data / Assumptions: PQRS is a square with side length 16 cm. Equal isosceles right triangles are cut off at each of the four corners. The remaining central shape is a regular octagon. The octagon is the largest that can be inscribed in this way. We need the side length of this regular octagon. Concept / Approach: Let the side of the square be a and let the length cut off along each side be x. Then each corner triangle has legs of length x, and the cut removes x from each adjacent side. The middle portion between two consecutive cuts forms one side of the octagon. The remaining length on each original side is a − 2x. The sloping edges are formed along the diagonals of the corner triangles, which are of length x√2, but that diagonal becomes part of the octagon side in a symmetrical configuration. For the largest regular octagon, the octagon side s is equal to the remaining straight part between the cuts, and there is a known relation s = a(√2 − 1). Step-by-Step Solution: Step 1: Let the side of the square be a. Here a = 16 cm.Step 2: Consider one corner where a right isosceles triangle is cut. If the perpendicular legs along the square sides are x and x, then the hypotenuse, which becomes one side of the octagon, has length x√2.Step 3: After cutting, the straight portion of the original side between adjacent cut points has length a − 2x.Step 4: For a regular octagon inscribed in this way, the side length s of the octagon satisfies s = a(√2 − 1). This standard result can be derived by equating the geometry of the diagonals and the equal side lengths but is often memorised for such questions.Step 5: Substitute a = 16 to obtain s = 16(√2 − 1) = 16√2 − 16 cm. Verification / Alternative check: To double check, note that if a = 1, the formula gives s = √2 − 1, which matches the derived relation in many geometry texts for a unit square. Here, with a scale factor 16, all linear dimensions scale by 16, so the octagon side becomes 16(√2 − 1). Numerically, √2 is about 1.414, so s ≈ 16(1.414 − 1) = 16 * 0.414 ≈ 6.624 cm, which is a reasonable side length less than 16 cm, consistent with the octagon fitting inside the square. Why Other Options Are Wrong: 8√2 − 4 cm and 8(√2 − 1) cm correspond to a square of side 8 cm, not 16 cm. 4(2√2 − 1) cm is another incorrect scaling that does not fit the standard formula for this construction. 16 − 8√2 cm is negative when computed numerically, which is impossible for a side length. Common Pitfalls: Assuming that the octagon side equals the remaining straight length a − 2x without relating x to a using right triangle geometry. Mixing up the factor √2 − 1 with 2 − √2 or other similar looking expressions. Trying to compute everything from scratch without using the known result s = a(√2 − 1), which can lead to algebra errors. Final Answer: The side length of the largest regular octagon that can be cut from the square is 16√2 − 16 cm.

Question 20

A right prism has a regular hexagonal base with side length 6 cm. If the total surface area of the prism is 216√3 cm2, what is the height of the prism in centimetres?

Accepted Answer

Introduction / Context: This question focuses on the surface area of a right prism with a regular hexagonal base. The base side length and the total surface area are known, and the task is to find the height of the prism. It combines area formula for a regular hexagon with the general surface area formula for a prism, which is an essential combination in solid geometry and aptitude exams. Given Data / Assumptions: The base is a regular hexagon with side length a = 6 cm. Total surface area (TSA) of the prism = 216√3 cm2. The prism is right, so lateral faces are rectangles with height equal to the prism height. We need the height h of the prism. Concept / Approach: For a right prism, total surface area T is given by T = 2 * (area of base) + (perimeter of base) * height. For a regular hexagon of side a, the perimeter P = 6a, and the area A_base = (3√3 / 2) * a^2. After computing base area and perimeter using a = 6 cm, substitute into the TSA expression and solve the resulting linear equation for h. This direct substitution approach avoids unnecessary complications. Step-by-Step Solution: Step 1: For a regular hexagon of side a, area A_base = (3√3 / 2) * a^2 and perimeter P = 6a.Step 2: Substitute a = 6 cm. Then P = 6 * 6 = 36 cm.Step 3: Compute base area: A_base = (3√3 / 2) * 6^2 = (3√3 / 2) * 36 = 54√3 cm2.Step 4: Total surface area T = 2 * A_base + P * h = 2 * 54√3 + 36h = 108√3 + 36h.Step 5: Given T = 216√3, set 108√3 + 36h = 216√3.Step 6: Subtract 108√3 from both sides to get 36h = 108√3, so h = (108√3) / 36 = 3√3 cm. Verification / Alternative check: Substitute h = 3√3 back into the TSA formula. T = 108√3 + 36 * 3√3 = 108√3 + 108√3 = 216√3 cm2, which matches the given total surface area. Thus the height value is consistent and unique for the given base dimensions and total surface area. Why Other Options Are Wrong: 6√3 cm and 6 cm correspond to different TSA values and do not satisfy T = 216√3 cm2 when substituted. 3 cm is too small; it would give much less surface area than required. 9 cm produces a TSA larger than 216√3 cm2, so it is not correct either. Common Pitfalls: Using a wrong formula for area of a regular hexagon, such as treating it as a simple rectangle or triangle. Forgetting the factor of 2 multiplying the base area in the total surface area formula. Making algebra mistakes when solving the linear equation for h. Final Answer: The height of the prism is 3√3 cm.

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