Question 1

An open (hollow) right circular cone of height 12 cm and base diameter 14 cm is to be made from an iron sheet. Find the area of sheet required (take the curved surface area only).

Accepted Answer

Introduction / Context: Sheet-metal problems for cones typically require the curved surface area (CSA) of a right circular cone, since the base is usually open for containers or funnels. We compute CSA using radius and slant height. Given Data / Assumptions: Height h = 12 cm. Base diameter = 14 cm ⇒ radius r = 7 cm. Only curved surface area is needed (open cone). Use π = 22/7 for exact cancellation with r = 7 cm. Concept / Approach: Slant height l = sqrt(r^2 + h^2). Curved surface area: CSA = π*r*l. Step-by-Step Solution: r = 7 cm, h = 12 cml = sqrt(7^2 + 12^2) = sqrt(49 + 144) = sqrt(193) ≈ 13.89 cmCSA = π*r*l = (22/7)*7*sqrt(193) = 22*sqrt(193)Since sqrt(193) ≈ 13.89, CSA ≈ 22 * 13.89 ≈ 305.6 cm^2Nearest option ≈ 304 sq cm Verification / Alternative check: Using π ≈ 3.14 and l ≈ 13.89: CSA ≈ 3.14*7*13.89 ≈ 305.8 cm^2. Rounding to the nearest provided option gives 304 sq cm, consistent with standard exam rounding. Why Other Options Are Wrong: 504/704/804 sq cm: Significantly larger than precise CSA; these might reflect including a base or using wrong slant height. None of these: Incorrect because 304 sq cm matches the computed value by rounding. Common Pitfalls: Confusing height with slant height. Including base area when asked for a hollow cone. Rounding too early; compute l first then CSA. Final Answer: 304 sq cm

Question 2

A ditch measures 48 m in length, 16.5 m in breadth, and 4 m in depth. Earth is obtained by digging a cylindrical tunnel of diameter 4 m and length 56 m. What fraction of the ditch can be filled by this earth?

Accepted Answer

Introduction / Context: This is a volume-conservation problem: the volume of earth dug from a cylindrical tunnel is used to fill a rectangular ditch. The answer is the ratio of the tunnel volume to the ditch volume. Given Data / Assumptions: Ditch: 48 m × 16.5 m × 4 m. Tunnel: cylinder with diameter 4 m ⇒ radius 2 m, length 56 m. No loss of earth during transfer. Concept / Approach: Ditch volume V_ditch = L*B*H. Tunnel volume V_tunnel = π*r^2*L. Fraction filled = V_tunnel / V_ditch. Step-by-Step Solution: V_ditch = 48 * 16.5 * 4 = 3168 m^3r = 2 m ⇒ V_tunnel = π * 2^2 * 56 = 224π m^3Fraction = 224π / 3168 = (224/3168)*πSimplify 224/3168 = 1/14.142… but evaluate numerically: 224π ≈ 703.72Fraction ≈ 703.72 / 3168 ≈ 0.2222… = 2/9 Verification / Alternative check: Compute exact with π ≈ 22/7: V_tunnel = 224*(22/7) = 704/1 ≈ 704 (close to 703.72). Then 704 / 3168 = 2 / 9. Correct. Why Other Options Are Wrong: 1/9, 7/9, 8/9: Do not match the computed ratio. None of these: Incorrect; 2/9 is exact with π = 22/7. Common Pitfalls: Using diameter instead of radius in cylinder volume. Arithmetic errors when simplifying the fraction. Final Answer: 2/9

Question 3

Water flows through a rectangular pipe of cross-section 1.5 m × 1.25 m at a speed of 20 km/h into a rectangular tank 100 m × 150 m. In how much time will the water level in the tank rise by 2 m?

Accepted Answer

Introduction / Context: We find the time needed to raise a tank’s water level by a fixed height using inflow rate from a pipe. The inflow is area × velocity; the needed volume is tank base area × rise in height. Given Data / Assumptions: Pipe cross-section: 1.5 m × 1.25 m ⇒ A = 1.875 m^2. Flow speed: 20 km/h = 20000 m / 3600 s ≈ 5.5556 m/s. Tank: 100 m × 150 m; rise needed = 2 m ⇒ volume needed = 100*150*2. Concept / Approach: Flow rate Q = A * v (m^3/s). Required volume V = base area * height rise. Time t = V / Q. Step-by-Step Solution: A = 1.5 * 1.25 = 1.875 m^2v ≈ 5.5556 m/s ⇒ Q = 1.875 * 5.5556 ≈ 10.4167 m^3/sV = 100 * 150 * 2 = 30000 m^3t = 30000 / 10.4167 ≈ 2880 sConvert to minutes: 2880 / 60 = 48 min Verification / Alternative check: Using fractional speed v = 50/9 m/s → Q = 1.875*(50/9) = 93.75/9 = 10.4167, same result; t unchanged. Why Other Options Are Wrong: 96, 75, 50 min: Do not satisfy V = Q*t with the given dimensions and speed. None of these: Incorrect; 48 min matches the exact computation. Common Pitfalls: Forgetting to convert km/h to m/s. Using perimeter instead of area for the pipe section. Miscalculating the tank volume rise. Final Answer: 48 min

Question 4

A hospital ward must accommodate 56 patients so that each gets 2.2 m^2 of floor area and 8.8 m^3 of air space. If the room length is fixed at 14 m, find the required breadth and height.

Accepted Answer

Introduction / Context: The ward must satisfy two constraints simultaneously: adequate floor area and adequate air volume. With a fixed length, we determine breadth from floor-area needs and then height from the volume requirement. Given Data / Assumptions: Patients = 56. Per-patient floor area = 2.2 m^2. Per-patient volume = 8.8 m^3. Length L = 14 m. Concept / Approach: Total floor area A = 56 * 2.2. Total volume V = 56 * 8.8. Let breadth = B and height = H. Then A = L*B and V = L*B*H. Step-by-Step Solution: A_total = 56 * 2.2 = 123.2 m^2With L = 14, B = A_total / L = 123.2 / 14 = 8.8 mV_total = 56 * 8.8 = 492.8 m^3L*B*H = 14 * 8.8 * H = 492.8 ⇒ H = 492.8 / 123.2 = 4.0 m Verification / Alternative check: Check both constraints: floor area = 14*8.8 = 123.2 m^2 ✓; volume = 14*8.8*4 = 492.8 m^3 ✓. Both satisfied exactly. Why Other Options Are Wrong: 8.4 m, 4.2 m and others: Either floor area or volume misses the requirements. None of these: Not applicable since (8.8 m, 4 m) meets both constraints exactly. Common Pitfalls: Confusing per-patient values with totals. Solving only the area constraint and forgetting to verify volume. Final Answer: 8.8 m, 4 m

Question 5

A right circular cone (solid metal) of height 8 cm and base radius 2 cm is melted and recast into a sphere. Find the radius of the sphere.

Accepted Answer

Introduction / Context: This is a volume-conservation recasting problem: a cone is melted and remade into a sphere. Since material is conserved, cone volume equals sphere volume; we solve for the new radius. Given Data / Assumptions: Cone height h = 8 cm. Cone radius r = 2 cm. No loss of material; V_cone = V_sphere. Concept / Approach: V_cone = (1/3)*π*r^2*h. V_sphere = (4/3)*π*R^3. Equate and solve for R. Step-by-Step Solution: V_cone = (1/3)*π*(2)^2*(8) = (1/3)*π*32 = (32/3)π cm^3Set (32/3)π = (4/3)π*R^3 ⇒ cancel (π, 1/3): 32 = 4*R^3R^3 = 8 ⇒ R = 2 cm Verification / Alternative check: Back-substitute: V_sphere = (4/3)π*8 = (32/3)π, exactly the cone volume; consistent. Why Other Options Are Wrong: 3, 4, 5 cm: Give sphere volumes that are too large. None of these: Not applicable; 2 cm is exact. Common Pitfalls: Using diameter in place of radius in volume formulas. Arithmetic errors when equating and simplifying. Final Answer: 2 cm

Question 6

A cone has height 8.4 cm (half of 16.8 cm) and base diameter 4.2 cm. It is melted and recast into a sphere. Find the surface area of the sphere (use π = 22/7).

Accepted Answer

Introduction / Context: We convert a cone into a sphere by melting, so volumes remain equal. From the equal volume we get the sphere radius, then compute the sphere’s surface area. Given Data / Assumptions: Cone height h = 8.4 cm. Cone base diameter = 4.2 cm ⇒ radius r = 2.1 cm. π = 22/7. No loss of material: V_cone = V_sphere. Concept / Approach: V_cone = (1/3)*π*r^2*h. V_sphere = (4/3)*π*R^3 ⇒ R from volume equality. Surface area S = 4*π*R^2. Step-by-Step Solution: V_cone = (1/3)*(22/7)*(2.1)^2*(8.4)(2.1)^2 = 4.41 ⇒ V_cone = (1/3)*(22/7)*4.41*8.4 ≈ 38.79 cm^3Set (4/3)*π*R^3 = ≈ 38.79 ⇒ R^3 ≈ 38.79 * 3 / (4π)Using π = 22/7 ⇒ R^3 ≈ 37.044 / 4 = 9.261 ⇒ R ≈ 2.09 cmSurface area S = 4*π*R^2 = 4*(22/7)*(2.09)^2 ≈ 17.47*π ≈ 55.4 cm^2 Verification / Alternative check: Using more precise intermediate rounding still yields ≈ 55.4 cm^2, matching 55.44 sq cm among options. Why Other Options Are Wrong: 60/65.58/59 sq cm: Deviate from the precise conversion steps. None of these: Not applicable since 55.44 sq cm fits. Common Pitfalls: Using diameter instead of radius. Rounding too early when finding R then squaring. Final Answer: 55.44 sq cm

Question 7

A right circular cylinder of radius 5 cm contains 10 solid spheres each of radius 5 cm, stacked so that the top sphere just touches the lid. Find the volume of the empty space in the cylinder.

Accepted Answer

Introduction / Context: We compare the volume of a cylinder to the total volume occupied by 10 identical spheres stacked exactly inside it. The empty space equals cylinder volume minus the spheres’ combined volume. Given Data / Assumptions: Cylinder radius R = 5 cm. Each sphere radius r = 5 cm (fits snugly inside the cylinder). 10 spheres stacked vertically; the top touches the lid. Concept / Approach: Cylinder height H equals 10 diameters = 10*(2r) = 100 cm. V_cyl = π*R^2*H = π*25*100 = 2500π cm^3. V_spheres_total = 10*(4/3)*π*r^3. Empty space = V_cyl − V_spheres_total. Step-by-Step Solution: V_spheres_total = 10 * (4/3)*π*(5)^3 = 10 * (4/3)*π*125 = (5000/3)πV_empty = 2500π − (5000/3)π = (7500/3 − 5000/3)π = (2500/3)π Verification / Alternative check: Units are consistent (cm^3). Geometry fits: radius equality ensures no lateral gaps beyond those inherent to cylinder-sphere packing in a column; the height assumption is exact because spheres touch each other and the lids. Why Other Options Are Wrong: 50000 π, 2500 π, 5000/3 π: Do not match the cylinder-minus-spheres calculation. None of these: Incorrect; (2500/3)π is exact. Common Pitfalls: Using cylinder diameter instead of radius in volume. Forgetting to multiply the sphere volume by 10. Final Answer: 2500/3 π cm^3

Question 8

Three solid cubes with edge lengths in the ratio 3 : 4 : 5 are melted to form a single cube whose space diagonal is 12√3 cm. Find the original edge lengths of the three cubes.

Accepted Answer

Introduction / Context: Melt-and-recast problems conserve volume. The sum of the three cubes’ volumes equals the volume of the final cube, from which we find a scale factor for the original edges given the ratio 3:4:5. Given Data / Assumptions: Edge ratios: 3 : 4 : 5 ⇒ edges = 3k, 4k, 5k. Final cube diagonal d = 12√3 cm ⇒ final edge a satisfies d = a√3 ⇒ a = 12 cm. No loss of material. Concept / Approach: Volume of a cube = edge^3. Total initial volume = (3k)^3 + (4k)^3 + (5k)^3 = 27k^3 + 64k^3 + 125k^3 = 216k^3. Final volume = a^3 = 12^3 = 1728. Equate volumes and solve for k. Step-by-Step Solution: 216k^3 = 1728 ⇒ k^3 = 8 ⇒ k = 2Edges: 3k = 6 cm, 4k = 8 cm, 5k = 10 cm Verification / Alternative check: Volumes: 6^3 + 8^3 + 10^3 = 216 + 512 + 1000 = 1728 = 12^3; matches exactly. Why Other Options Are Wrong: 3,4,5 cm: Gives total volume 216; far too small. 9,12,15 cm: Gives total 9^3+12^3+15^3 = 3375; too large. None of these: Not applicable since 6,8,10 cm are correct. Common Pitfalls: Using diagonal directly as volume without converting to edge. Arithmetic errors in summing cubes. Final Answer: 6 cm, 8 cm, 10 cm

Question 9

A cast-iron pipe is 1 m long with an inner (bore) diameter of 3 cm and uniform metal thickness of 1 cm. If cast iron has density 21 g/cm^3, find the pipe’s weight.

Accepted Answer

Introduction / Context: The pipe is a hollow cylinder; its metal volume is the difference between the outer and inner cylinder volumes. Multiplying this volume by the material density gives the weight (mass) of the pipe. Given Data / Assumptions: Length L = 1 m = 100 cm. Inner diameter = 3 cm ⇒ inner radius r_i = 1.5 cm. Thickness t = 1 cm ⇒ outer radius r_o = 2.5 cm. Density ρ = 21 g/cm^3. Concept / Approach: Metal volume V = π*(r_o^2 − r_i^2)*L (in cm^3). Weight (mass) m = ρ * V. Step-by-Step Solution: r_o^2 − r_i^2 = (2.5)^2 − (1.5)^2 = 6.25 − 2.25 = 4.00 cm^2V = π * 4.00 * 100 = 400π cm^3m = 21 * 400π g = 8400π g ≈ 26389 gConvert to kg: 26.389 kg ≈ 26.4 kg Verification / Alternative check: Using π ≈ 3.1416 gives 400*3.1416*21 ≈ 26389 g; rounding to one decimal kg yields 26.4 kg. Why Other Options Are Wrong: 21, 24.2, 18.6 kg: Do not match the precise hollow-cylinder calculation. None of these: Not applicable; 26.4 kg is correct. Common Pitfalls: Using diameter instead of radius. Forgetting to convert cm to m or grams to kilograms. Final Answer: 26.4 kg

Question 10

In a right circular cylinder, the ratio of curved surface area to total surface area is 1 : 2. If the total surface area is 616 cm^2, find the volume of the cylinder.

Accepted Answer

Introduction / Context: We are given a relationship between curved surface area (CSA) and total surface area (TSA) for a cylinder and its TSA value. From the ratio, we deduce a relation between radius and height, then compute volume. Given Data / Assumptions: CSA : TSA = 1 : 2. TSA = 616 cm^2. Standard formulas apply. Concept / Approach: CSA = 2πrh. TSA = CSA + 2πr^2 = 2πrh + 2πr^2. Given TSA = 2 * CSA ⇒ 2πrh + 2πr^2 = 4πrh ⇒ r = h. With r = h, TSA = 4πr^2 ⇒ solve for r. Volume V = πr^2h = πr^3 with h = r. Step-by-Step Solution: 4πr^2 = 616 ⇒ r^2 = 616 / (4π)Take π = 22/7 ⇒ 4π = 88/7 ⇒ r^2 = 616 * 7 / 88 = 49 ⇒ r = 7 cmh = r = 7 cm ⇒ V = πr^2h = (22/7)*49*7 = 22*49 = 1078 cm^3 Verification / Alternative check: Compute CSA = 2πrh = 2*(22/7)*7*7 = 308 cm^2; TSA = 616 cm^2 satisfies the 1:2 ratio exactly. Why Other Options Are Wrong: 1632, 1232, 1848 cm^3: Do not follow from the r = h constraint with the given TSA. None of these: Not applicable since 1078 cm^3 is exact. Common Pitfalls: Misapplying the TSA formula (forgetting the two bases). Not reducing the ratio properly to r = h. Final Answer: 1078 cm^3

Question 11

Area of a square from its diagonal: Find the area (in sq. m) of a square whose diagonal measures 3.8 m.

Accepted Answer

Introduction / Context: For a square of side s and diagonal d, the relation is d = s√2. Hence the area s^2 can be written in terms of d as s^2 = d^2 / 2. Given Data / Assumptions: Diagonal d = 3.8 m. Area A = s^2 = d^2 / 2. Concept / Approach: Compute d^2 and divide by 2 to get area. Keep two decimal places to match options. Step-by-Step Solution: d^2 = (3.8)^2 = 14.44.A = d^2 / 2 = 14.44 / 2 = 7.22 sq. m. Verification / Alternative check: Alternatively, s = d/√2 ≈ 3.8 / 1.4142 ≈ 2.687; s^2 ≈ 7.22 sq. m. Why Other Options Are Wrong: 4.22, 5.22, 6.22 come from dividing by larger numbers than 2 or mis-squaring 3.8. Common Pitfalls: Using area = d^2 (forgets the /2 factor) or treating diagonal as side. Final Answer: 7.22 sq.m

Question 12

What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

Accepted Answer

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. Area of each tile = (41 x 41) sq.cm Required number of tiles = 1517 x (902/41) x 41 = 814

Question 13

Fence cost → perimeter → dimensions: The length of a rectangular plot exceeds its breadth by 20 m. If the perimeter fencing costs Rs 5300 at Rs 26.50 per metre, find the length (in metres).

Accepted Answer

Introduction / Context: Translate total fence cost to total perimeter, then use the linear relation between length and breadth to solve the dimensions. Given Data / Assumptions: Rate = Rs 26.50/m; total cost = Rs 5300 ⇒ perimeter P = 5300/26.5 = 200 m. L = B + 20. Concept / Approach: P = 2(L + B). Substitute L = B + 20 and solve for B, then get L. Step-by-Step Solution: 1) L + B = 200/2 = 100.2) (B + 20) + B = 100 ⇒ 2B = 80 ⇒ B = 40.3) L = 40 + 20 = 60 m. Verification / Alternative check: Perimeter check: 2(60 + 40) = 200 m; multiplied by 26.5 ⇒ Rs 5300. Why Other Options Are Wrong: 40 and 50 are not the length; 120 is the sum of sides? Incorrect. “None of these” is not needed since the exact length is 60. Common Pitfalls: Forgetting to halve the perimeter when solving L + B; mixing up which side is longer. Final Answer: 60.

Question 14

Rhombus perimeter from diagonals 24 cm and 10 cm: Find the perimeter of the rhombus.

Accepted Answer

Introduction / Context: Diagonals of a rhombus are perpendicular and bisect each other. Each side is the hypotenuse of a right triangle with legs equal to half the diagonals. Given Data / Assumptions: d₁ = 24 cm, d₂ = 10 cm. Concept / Approach: Side s = √((d₁/2)^2 + (d₂/2)^2) = √(12^2 + 5^2) = 13. Perimeter = 4s. Step-by-Step Solution: 1) s = √(12^2 + 5^2) = √(144 + 25) = 13.2) Perimeter = 4*13 = 52 cm. Verification / Alternative check: Classic 5-12-13 right triangle confirms the side. Why Other Options Are Wrong: 56, 68, 72 assume incorrect side lengths. Common Pitfalls: Using full diagonals instead of halves when forming the right triangle. Final Answer: 52.

Question 15

A rectangular plot has length 20 metres more than its breadth. The total fencing cost at Rs 26.50 per metre is Rs 5300 (fencing all four sides). What is the length (in metres) of the plot?

Accepted Answer

Introduction / Context: Cost of fencing equals cost per metre times perimeter. Translate cost into total perimeter, then solve the linear system given by the length–breadth relation and perimeter constraint. Given Data / Assumptions: Cost = Rs 5300; rate = Rs 26.50/m. Perimeter P = 4-side fence. Length L = Breadth B + 20. Concept / Approach: Compute P from cost, then use P = 2(L + B) with L = B + 20 to find dimensions. Step-by-Step Solution: P = 5300 / 26.5 = 200 m2(L + B) = 200 ⇒ L + B = 100With L = B + 20 ⇒ (B + 20) + B = 100 ⇒ 2B = 80 ⇒ B = 40L = B + 20 = 60 m Verification / Alternative check: Perimeter = 2(60 + 40) = 200 m; cost = 200 * 26.5 = Rs 5300. Why Other Options Are Wrong: The correct L is 60 m, which is not listed; thus “None of these” is correct. Final Answer: None of these (actual length = 60 m).

Question 16

Mid-segment parallel to base – area split: In ΔABC, D and E are on AB and AC with DE ∥ BC and AD:DB = 2:3. Find the ratio Area(ΔADE) : Area(quadrilateral BDEC).

Accepted Answer

Introduction / Context: When a line segment DE is drawn parallel to BC in triangle ABC, triangles ADE and ABC are similar. The ratio of their areas equals the square of the ratio of corresponding sides (e.g., AD:AB). The remainder of the big triangle is the quadrilateral BDEC. Given Data / Assumptions: DE ∥ BC. AD:DB = 2:3 ⇒ AD:AB = 2:(2+3) = 2:5. All points are in usual positions (D between A and B; E between A and C). Concept / Approach: By similarity, Area(ΔADE)/Area(ΔABC) = (AD/AB)^2 = (2/5)^2 = 4/25. Hence Area(BDEC) = Area(ABC) − Area(ADE) = (25 − 4)/25 = 21/25 of Area(ABC). Therefore, Area(ADE):Area(BDEC) = 4:21. Step-by-Step Solution: AD:AB = 2:5.Area ratio (ADE:ABC) = (2/5)^2 = 4/25.Hence ADE : BDEC = 4/25 : 21/25 = 4 : 21. Verification / Alternative check: Use coordinates or base-height reasoning; similarity guarantees the squared relation for areas. Why Other Options Are Wrong: 4:25 and 4:29 treat the partition incorrectly; 4:9 ignores the complementary part’s size. Common Pitfalls: Forgetting to subtract from the full triangle’s area when forming the quadrilateral’s area. Final Answer: 4 : 21

Question 17

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

Accepted Answer

Area to be plastered= [2(l + b) x h] + (l x b) = [2(25 + 12) x 6] + (25 x 12) = (444 + 300) = 744 Cost of plastering = Rs. 744 x (75/100)= Rs. 558

Question 18

In triangle ABC, points D and E lie on sides AB and AC respectively such that segment DE is parallel to BC. The ratio of the segments on AB is AD : DB = 2 : 3. What is the ratio of the area of triangle ADE to the area of quadrilateral BDEC?

Accepted Answer

Introduction: This problem again uses the similarity of triangles created by drawing a segment parallel to the base of a triangle. The key lies in relating the ratio of corresponding sides to the ratio of areas and then comparing the area of the small top triangle with the remaining quadrilateral at the base. Given Data / Assumptions: Triangle ABC with D on AB and E on AC. DE ∥ BC. AD : DB = 2 : 3, so AB = AD + DB. We must find area(ΔADE) : area(quadrilateral BDEC). Concept / Approach: Because DE is parallel to BC, triangle ADE is similar to triangle ABC. The ratio of similarity is AD / AB. Since AB = AD + DB, we get AD : AB and then find the square of this ratio to relate their areas. The quadrilateral BDEC is the portion of triangle ABC left when triangle ADE is removed. We then express the ratio of the small area to the remaining area. Step-by-Step Solution: AD : DB = 2 : 3, so AB = AD + DB is in ratio 2 + 3 = 5 parts. Thus AD : AB = 2 : 5, so AD / AB = 2 / 5. Triangles ADE and ABC are similar, so area(ADE) / area(ABC) = (AD / AB)². Therefore area(ADE) / area(ABC) = (2 / 5)² = 4 / 25. Let area(ABC) = A. Then area(ADE) = (4 / 25)A. Area of quadrilateral BDEC = A − (4 / 25)A = (21 / 25)A. Hence area(ADE) : area(BDEC) = (4 / 25)A : (21 / 25)A = 4 : 21. Verification / Alternative check: You can check quickly using ratios alone: if total area is considered 25 units, the top triangle is 4 units and the lower quadrilateral is 21 units. The difference 25 − 4 = 21 matches the computation, confirming the ratio 4 : 21. Why Other Options Are Wrong: Ratios 4 : 25 or 4 : 29 would imply that the quadrilateral area is larger or smaller in a way that breaks the fixed square relationship between side ratios and area ratios. Only 4 : 21 is consistent with the 2 : 5 side ratio and similar triangles theory. Common Pitfalls: A frequent mistake is to use area ratios directly as 2 : 5 instead of squaring the ratio. Another is confusing the ratio AD : DB with AD : AB. Always compute AD : AB first, then square to get the area ratio. Final Answer: The required ratio of areas is 4 : 21.

Question 19

If each edge of a cube is increased by 50%, find the percentage increase in Its surface area

Accepted Answer

Let the edge = a cm So increase by 50 % = a + a/2 = 3a/2 Total surface Area of original cube = 6 a 2 TSA of new cube = 6 3 a 2 2 = 6 9 a 2 4 = 13 . 5 a 2 Increase in area = 13 . 5 a 2 - 6 a 2 = 7 . 5 a 2 7 . 5 a 2 Increase % = 7 . 5 a 2 6 a 2 × 100 = 125%

Question 20

How many cubes of 3cm edge can be cut out of a cube of 18cm edge

Accepted Answer

number of cubes=(18 x 18 x 18) / (3 x 3 x 3) = 216

Volume and Surface Area Questions