CBSE Class 9 Maths Ganita Manjari Part 2 Ch 14 Math of Space Surface Area and Volume Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. A cube of side 12 cm and a cuboid measuring 27 cm by 8 cm by 8 cm are made from the same material.
(i) Show that the two solids have the same volume.
(ii) Find the total surface area of each.
(iii) State which of the two would need less material to wrap, and explain why this matters when choosing the shape of a carton.
Answer:
(i) Cube: 12³ = 1728 cm³. Cuboid: 27 × 8 × 8 = 1728 cm³. Equal.
(ii) Cube: 6 × 144 = 864 cm². Cuboid: 2(216 + 64 + 216) = 992 cm².
(iii) The cube needs 128 cm² less (about 13%). Cartons of the same capacity can use very different amounts of card; the more cube-like box is cheaper to make and wrap.
In simple words: Same volume, but the cube has less surface.

Teacher's Note:
1. Equal volume ≠ equal surface.
2. Link to biology: thin shapes lose heat faster.

Question 2. Find every cuboid whose edges are a whole number of centimetres and whose total surface area is exactly 52 cm².
(i) Find one such cuboid.
(ii) Find all of them.
(iii) Explain how you know that you have found them all.
Answer:
Need lw + wh + hl = 26 with l ≤ w ≤ h.
(i) 2 × 3 × 4: 6 + 12 + 8 = 26 ✓.
(ii) Exactly two: 1 × 2 × 8 and 2 × 3 × 4.
(iii) Each product is at least l², so 3l² ≤ 26, giving l ≤ 2.
l = 1: w + wh + h = 26 → (w + 1)(h + 1) = 27 → only 3 × 9, so w = 2, h = 8.
l = 2: 2w + wh + 2h = 26 → (w + 2)(h + 2) = 30 with both factors ≥ 4 → only 5 × 6, so w = 3, h = 4.
All cases are checked, so the list is complete.
In simple words: Limit the smallest edge, then factorise to find every case.

Teacher's Note:
1. The factorising trick is the key idea.
2. Ordering l ≤ w ≤ h avoids repeats.

Question 3. A cube of side 6 cm is painted on all six faces and then cut into 216 cubes of side 1 cm. Find how many of the small cubes have exactly three faces painted, exactly two, exactly one, and no face painted. Verify that the four counts add up to 216, and explain where in the cube each of the four kinds is found.
Answer:
Three faces: 8 (the corners).
Two faces: 12 edges × (6 − 2) = 48 (along edges, not corners).
One face: 6 faces × (6 − 2)² = 96 (inside each face).
None: (6 − 2)³ = 64 (the hidden core).
Check: 8 + 48 + 96 + 64 = 216 ✓.
In simple words: Corners, edges, faces and core give the four kinds.

Teacher's Note:
1. Give the general formulas for side n.
2. Link each count to a part of the cube.

Question 4. The radius of the base of a cylinder is increased by 10%, and at the same time its height is decreased by x%. Given that the volume of the cylinder is unchanged, find the value of x, correct to two decimal places.
Answer:
π(1.1r)² × h(1 − x/100) = πr²h.
1.21(1 − x/100) = 1 → 1 − x/100 = 1/1.21 = 100/121.
x/100 = 21/121 → x = 2100/121 ≈ 17.36.
In simple words: A 21% rise from the radius needs about a 17.36% cut in height.

Teacher's Note:
1. Not 10% and not 21%.
2. Cancel πr²h.

Question 5. A sphere of radius r fits exactly inside the smallest cylinder that can contain it.
(i) Write down the radius and the height of that cylinder.
(ii) Prove that the volume of the sphere is \( \frac{2}{3} \) of the volume of the cylinder.
(iii) Prove that the surface area of the sphere is equal to the curved surface area of the cylinder.
This pair of results is the discovery Archimedes is said to have wanted carved on his tomb.
Answer:
(i) Radius r, height 2r.
(ii) V(sphere) = (4/3)πr³; V(cylinder) = πr² × 2r = 2πr³. Ratio = (4/3) ÷ 2 = 2/3.
(iii) S(sphere) = 4πr²; CSA(cylinder) = 2πr × 2r = 4πr². Equal.
Both hold for every r, which is why Archimedes valued them so much.
In simple words: The sphere fills two thirds of its cylinder and has the same area as its curved side.

Teacher's Note:
1. Draw the figure.
2. Tell the tombstone story.

Question 6. The surface area of a sphere of radius 5 cm is five times the curved surface area of a cone of base radius 4 cm. Find the height of the cone and its volume in terms of π.
Answer:
Sphere: 4π × 25 = 100π cm².
100π = 5 × π × 4 × l → l = 5 cm.
h = √(25 − 16) = 3 cm.
V = ⅓π × 16 × 3 = 16π cm³.
In simple words: Find the slant height from the area condition, then the height by Pythagoras.

Teacher's Note:
1. 3–4–5 cone.
2. Keep π as a symbol.

Question 7. Suppose the earth is a perfect sphere. A string is wrapped tightly around the equator. A second string, one metre longer, is placed around the earth so that it forms a larger circle standing the same distance above the ground all the way round.
(i) Find how high above the ground the second string stands.
(ii) Repeat the calculation for a volleyball instead of the earth.
(iii) Explain why the two answers are the same, and why this is surprising.
Answer:
(i) 2π(R + d) = 2πR + 1 → 2πd = 1 → d = 1/(2π) ≈ 0.159 m ≈ 15.9 cm.
(ii) Same working with radius ρ: d = 1/(2π) ≈ 15.9 cm.
(iii) R cancels in the first line: circumference grows by 2π for every unit of radius, whatever the size. It feels surprising because we compare 1 m with the huge earth, but size does not matter here.
In simple words: One extra metre of string always lifts it by about 16 cm, whatever the ball.

Teacher's Note:
1. A great discussion question.
2. Linear relation: only the rate matters.

Question 8. Take the earth to be a sphere of radius 6370 km, Jupiter a sphere of radius 69 900 km, and the sun a sphere of radius 695 700 km. Compute approximately (i) the ratio of the volume of Jupiter to the volume of the earth; (ii) the ratio of the volume of the sun to the volume of the earth. State clearly the property of the volume formula that makes these ratios easy to find. (A calculator may be used for this question.)
Answer:
Volume ∝ r³, so the ratio of volumes = (ratio of radii)³; the constant (4/3)π cancels.
(i) 69 900 ÷ 6370 ≈ 10.97; 10.97³ ≈ 1320.
(ii) 695 700 ÷ 6370 ≈ 109.2; 109.2³ ≈ 1.3 million.
In simple words: Divide the radii once, then cube.

Teacher's Note:
1. The sun is only 109 times wider but 1.3 million times bigger.
2. Calculator allowed.

Question 9. A cylinder has base radius r and height h. A pyramid on a square base is fitted inside it, so that the four corners of the square base lie on the circular base of the cylinder and the apex of the pyramid lies on the top face of the cylinder. Find the ratio of the volume of the pyramid to the volume of the cylinder.
Answer:
The square's diagonal is a diameter 2r, so s² + s² = 4r², giving base area s² = 2r².
Pyramid height = h, so V = ⅓ × 2r² × h = \( \frac{2}{3} \)r²h.
Cylinder: πr²h. Ratio = \( \frac{2}{3\pi} \) ≈ 0.212 (about 21%).
In simple words: The square covers 2/π of the circle, and the pyramid keeps one third of that prism.

Teacher's Note:
1. Only the base area is needed, not the side.
2. Two losses multiply.

Question 10. Find, in each case, the change in the volume when one dimension is altered by a single unit.
(i) The length of a cuboid with dimensions l, w, h is increased by 1 unit.
(ii) The radius of a cylinder with radius r and height h is increased by 1 unit.
(iii) The radius of a sphere of radius r is decreased by 1 unit.
Answer:
(i) (l + 1)wh − lwh = wh (one extra slab, independent of l).
(ii) π(r + 1)²h − πr²h = πh(2r + 1) (depends on r).
(iii) (4/3)π[r³ − (r − 1)³] = (4/3)π(3r² − 3r + 1) decrease, close to 4πr² for large r.
In simple words: Adding length adds a slab; adding radius adds a shell that grows with r.

Teacher's Note:
1. Expand (r − 1)³ carefully.
2. Link (iii) to surface area × thickness.

Question 11. A cube has volume V and total surface area S.
(i) Express S in terms of V.
(ii) Express V in terms of S.
(iii) Use your first formula to find the surface area of a cube of volume 729 cm³, and your second to find the volume of a cube of surface area 150 cm².
Answer:
(i) a = V^1/3, so S = 6V^2/3.
(ii) a = √(S/6), so V = (S/6)^3/2.
(iii) V = 729: a = 9, S = 6 × 81 = 486 cm². S = 150: a = 5, V = 125 cm³.
In simple words: Go through the side length a in both directions.

Teacher's Note:
1. S is not proportional to V.
2. Doubling V multiplies S by about 1.59.

Question 12. Ice cream fills a cuboidal container measuring 10 cm by 15 cm by 20 cm. It is served in scoops, each a hemisphere of radius 2.5 cm. Estimate how many whole scoops the container will give. State your assumptions, and say why the real number served would be smaller.
Answer:
Container: 3000 cm³. Scoop: (2/3) × \( \frac{22}{7} \) × 15.625 ≈ 32.74 cm³.
3000 ÷ 32.74 ≈ 91.6, so about 91 whole scoops.
Assumptions: container full with no air, every scoop an exact hemisphere, all ice cream used.
Real number is smaller: some sticks to the walls and corners, the last bits do not make full scoops, and some melts.
In simple words: Volume division gives an upper bound, not an exact count.

Teacher's Note:
1. State assumptions in a list.
2. Estimates are upper bounds here.

Question 13. A classroom measures 8 m by 6 m by 3.5 m. Tennis balls of radius 3.5 cm are to be packed into it.
(i) Estimate the number of balls by dividing the volume of the room by the volume of one ball.
(ii) Estimate it again by counting how many balls fit along each edge of the room.
(iii) Explain which of your two answers is an upper bound and which is the better estimate, and why they differ.
Answer:
Room = 168 000 000 cm³; ball ≈ 179.67 cm³.
(i) 168 000 000 ÷ 179.67 ≈ 935 000.
(ii) Diameter 7 cm: 114 × 85 × 50 = 484 500.
(iii) (i) is an upper bound: it assumes balls fill every gap like liquid. (ii) is the better estimate for neat rows. The balls fill only about 52% (π/6) of the space, so the answers differ by nearly half.
In simple words: Spheres always leave gaps, so dividing volumes overestimates.

Teacher's Note:
1. Floor each edge count.
2. Staggered packing would reach about 74%.

Question 14. A toy is made by mounting a cone of base radius 7 cm and height 24 cm on a hemisphere of the same radius, as shown.
Find the total surface area of the toy and its volume. Explain carefully which surfaces are counted and which are not.
Answer:
l = √(576 + 49) = 25 cm.
Counted: cone's curved surface and hemisphere's curved surface. Not counted: the two flat circles pressed together at the join.
TSA = πrl + 2πr² = 550 + 308 = 858 cm².
Volume = ⅓πr²h + (2/3)πr³ = 1232 + 718.67 = 1950.67 cm³.
In simple words: Surfaces at the join vanish, but volumes simply add.

Teacher's Note:
1. Common error: adding 2πr² extra.
2. Draw and shade the visible surfaces.

Question 15. A solid metallic sphere of radius 10.5 cm is melted and recast into small cones, each of base radius 3.5 cm and height 3 cm. Find how many cones are obtained, and state the principle that makes the calculation possible.
Answer:
Principle: conservation of volume (melting changes shape, not amount).
Sphere: (4/3) × \( \frac{22}{7} \) × 1157.625 = 4851 cm³.
Cone: ⅓ × \( \frac{22}{7} \) × 12.25 × 3 = 38.5 cm³.
Number = 4851 ÷ 38.5 = 126 (exact).
In simple words: Total volume is shared out equally among the cones.

Teacher's Note:
1. Surface area is not conserved.
2. Check the division is exact.

Question 16. A carton in the shape of a cuboid with a square base is to hold 1000 cm³.
(i) Find the total surface area of each of these three cartons: 10 cm by 10 cm by 10 cm; 20 cm by 20 cm by 2.5 cm; and 5 cm by 5 cm by 40 cm.
(ii) Verify that all three hold 1000 cm³.
(iii) State which uses the least material, and explain what this suggests about the most economical shape for a box of given capacity.
Answer:
(i) 10 × 10 × 10: 600 cm²; 20 × 20 × 2.5: 1000 cm²; 5 × 5 × 40: 850 cm².
(ii) 1000, 1000 and 1000 cm³ ✓.
(iii) The cube uses least. The closer a box is to a cube, the less card it needs for a given capacity; very flat or very long boxes waste material.
In simple words: For a fixed capacity, the cube-shaped box is the most economical.

Teacher's Note:
1. Compare with the sphere result.
2. Real packaging has other constraints too.

Question 17. A school is choosing between two designs for a rectangular swimming pool. Pool A: 15 m long, 6 m wide and 2.5 m deep. Pool B: 30 m long, 4.5 m wide and 1.8 m deep.
(i) Find the volume of water each pool holds.
(ii) Find the area to be tiled in each pool, counting the floor and the four walls.
(iii) Discuss which pool you would recommend, naming the quantities you have compared.
Answer:
(i) A: 15 × 6 × 2.5 = 225 m³. B: 30 × 4.5 × 1.8 = 243 m³.
(ii) A: floor 90 + walls 105 = 195 m². B: floor 135 + walls 124.2 = 259.2 m².
(iii) B holds 8% more water but needs 33% more tiling, so A is cheaper per m³ of water (about 1.15 m³ per m² tiled against 0.94).
But B is 30 m long (good for lane swimming) and shallower (safer for beginners). Recommend A on cost; B if training and safety matter more. Any well-argued choice is acceptable.
In simple words: Compare volume and tiled area, then weigh cost against use.

Teacher's Note:
1. No top surface is tiled.
2. Reward reasoning, not a single 'right' pool.

Question 18. Estimate whether a person could drink enough water in a lifetime to fill a classroom measuring 8 m by 6 m by 3.5 m. Assume a person drinks 2.5 litres a day and lives for 75 years. State your assumptions, give your answer as a fraction of the room, and say which of your assumptions the answer is most sensitive to.
Answer:
Lifetime: 2.5 × 365 × 75 = 68 437.5 litres ≈ 68.4 m³.
Room: 8 × 6 × 3.5 = 168 m³.
Fraction ≈ 68.4 ÷ 168 ≈ 0.41 (about two fifths). So no; filling it would need about 6.1 litres a day.
Assumptions: steady 2.5 L every day from birth, 75 years, leap years ignored, only drinking water counted.
Most sensitive to the daily intake, which can vary from about 1.5 to 4 litres (roughly a quarter to two thirds of the room).
In simple words: A lifetime of drinking fills only about two fifths of a classroom.

Teacher's Note:
1. Both inputs act linearly.
2. The less certain input matters more.

Question 19. The water on the earth's surface has an estimated volume of 1.38 billion km³. Suppose the earth is a perfect sphere of radius 6371 km and that all this water forms a uniform layer completely covering its surface.
(i) Write an expression for the thickness of the layer.
(ii) Estimate its value. (A calculator may be used.)
(iii) Comment on how thin this layer is compared with the radius of the earth.
Answer:
(i) For a thin layer, V ≈ 4πR² × t, so t ≈ \( \frac{V}{4\pi R^2} \).
(ii) 4π(6371)² ≈ 5.10 × 10⁸ km²; t ≈ 1.38 × 10⁹ ÷ 5.10 × 10⁸ ≈ 2.7 km.
(iii) 2.7 ÷ 6371 ≈ 0.0004 (0.04%). On a globe of radius 1 m the oceans would be under half a millimetre thick.
In simple words: Thickness = volume ÷ area; the oceans are a very thin film.

Teacher's Note:
1. Thin shell: area × thickness.
2. Relate to a real globe.

Question 20. Imagine a cone of base radius r and slant height l cut along a slant edge and unrolled flat.
(i) Explain why the unrolled surface is a sector of a circle of radius l.
(ii) By equating the arc length of that sector with the circumference of the base of the cone, show that the angle of the sector satisfies \( \frac{\theta}{360^\circ} \) = \( \frac{r}{l} \), and hence that the curved surface area is πrl.
(iii) Write down the total surface area, and apply your formulas to a cone of base radius 10 cm and slant height 26 cm, finding also its height and volume in terms of π.
Answer:
(i) Every point of the rim is l from the apex, and the surface flattens without stretching. So the flat shape is bounded by an arc of radius l and two straight edges: a sector.
(ii) Arc = 2πr and also \( \frac{\theta}{360^\circ} \) × 2πl, so \( \frac{\theta}{360^\circ} \) = \( \frac{r}{l} \).
Area = \( \frac{\theta}{360^\circ} \) × πl² = \( \frac{r}{l} \) × πl² = πrl.
(iii) TSA = πr(l + r). For r = 10, l = 26: CSA = 260π cm², TSA = 360π cm².
h = √(676 − 100) = 24 cm; V = ⅓π × 100 × 24 = 800π cm³. (Sector angle ≈ 138.5°.)
In simple words: The unrolled cone is a sector whose share of the full circle is r/l.

Teacher's Note:
1. Make a paper cone and unroll it.
2. 5–12–13 scaled by 2.

Case Study Based Questions

Question 21. The Village Water Tank: A closed cylindrical water tank has base radius 1.4 m and height 2.5 m. Its curved outer surface is to be painted.
(a) Find the volume of the tank in cubic metres.
(b) Find its capacity in litres.
(c) Find the area of its curved surface.
(d) Find the cost of painting that surface at ₹25 per square metre.
Answer:
(a) \( \frac{22}{7} \) × 1.96 × 2.5 = 15.4 m³.
(b) 15 400 litres.
(c) 2 × \( \frac{22}{7} \) × 1.4 × 2.5 = 22 m².
(d) 22 × 25 = ₹550.
In simple words: Only the curved surface is painted.

Teacher's Note:
1. 1 m³ = 1000 L.
2. Ends are not painted.

Question 22. The Ice Cream Cone: An ice cream cone is a right circular cone of base radius 3.5 cm and depth 12 cm. It is filled level with the rim, and a hemisphere of ice cream of the same radius is placed on top.
(a) Find the volume of ice cream inside the cone.
(b) Find the volume of the hemisphere on top.
(c) Find the total volume of ice cream.
(d) State which of the two parts holds more, and by roughly how much.
Answer:
(a) ⅓ × \( \frac{22}{7} \) × 12.25 × 12 = 154 cm³.
(b) (2/3) × \( \frac{22}{7} \) × 42.875 ≈ 89.83 cm³.
(c) ≈ 243.83 cm³.
(d) The cone, by about 64 cm³ (roughly 1.7 times as much).
In simple words: The deep cone holds more than the scoop on top.

Teacher's Note:
1. Depth outweighs the narrowing.
2. 3.5³ = 42.875.

Question 23. Bricks for a Wall: A wall is to be built 5 m long, 3 m high and 0.2 m thick, using bricks measuring 20 cm by 10 cm by 8 cm.
(a) Find the volume of the wall in cubic centimetres.
(b) Find the volume of one brick.
(c) Estimate the number of bricks needed, ignoring the mortar.
(d) Explain whether the true number of bricks is more or fewer than your estimate, and why.
Answer:
(a) 500 × 300 × 20 = 3 000 000 cm³.
(b) 20 × 10 × 8 = 1600 cm³.
(c) 1875 bricks.
(d) Fewer: mortar in the joints takes up part of the wall's volume.
In simple words: Mortar takes space, so fewer bricks are needed.

Teacher's Note:
1. Convert metres to cm first.
2. Cut bricks affect buying, not the count in the wall.

Question 24. Painting the Dome: A building is topped by a hemispherical dome of radius 7 m.
(a) Find the curved surface area of the dome.
(b) Find its total surface area, counting the circular floor beneath it.
(c) Find the volume of air enclosed by the dome.
(d) Find the cost of painting only the curved surface at ₹40 per square metre.
Answer:
(a) 2πr² = 308 m².
(b) 3πr² = 462 m².
(c) (2/3)πr³ ≈ 718.67 m³.
(d) 308 × 40 = ₹12 320.
In simple words: Curved surface 2πr²; add πr² for the floor.

Teacher's Note:
1. πr² = 154 m² here.
2. Paint only the curved part.

Question 25. Packing the Balls: Balls of radius 2.5 cm are to be packed into a cubical box of edge 20 cm, sitting in neat rows and layers.
(a) Find how many balls fit along one edge of the box.
(b) Find the total number of balls the box holds.
(c) Find the total volume of those balls.
(d) Find what percentage of the box is filled by the balls, and say where the rest of the space is.
Answer:
(a) 20 ÷ 5 = 4.
(b) 4³ = 64.
(c) 64 × 65.48 ≈ 4190.5 cm³.
(d) 4190.5 ÷ 8000 ≈ 52.4%. The rest is air in the gaps around and between the balls.
In simple words: Balls in rows fill only about 52% of the box.

Teacher's Note:
1. Use the diameter, not the radius.
2. π/6 ≈ 0.524 for any size.

Question 26. The Cylindrical Pillars: A hall is supported by 12 cylindrical pillars, each of base radius 0.35 m and height 4 m. The curved surface of every pillar is to be polished.
(a) Find the curved surface area of one pillar.
(b) Find the total curved surface area of all 12.
(c) Find the cost of polishing them at ₹60 per square metre.
(d) Find the total volume of concrete in all 12 pillars.
Answer:
(a) 2 × \( \frac{22}{7} \) × 0.35 × 4 = 8.8 m².
(b) 12 × 8.8 = 105.6 m².
(c) 105.6 × 60 = ₹6336.
(d) 12 × 1.54 = 18.48 m³.
In simple words: Find one pillar, then multiply by 12.

Teacher's Note:
1. 0.35² = 0.1225.
2. Area for polishing, volume for concrete.

Question 27. Two Containers, One Capacity: A manufacturer must choose between two container shapes of the same capacity: a cube of edge 10 cm, and a cuboid measuring 25 cm by 8 cm by 5 cm.
(a) Find the volume of each container.
(b) Find the total surface area of each.
(c) State which needs less material to make.
(d) Give one reason a manufacturer might still choose the other shape.
Answer:
(a) Both 1000 cm³.
(b) Cube 600 cm²; cuboid 730 cm².
(c) The cube (130 cm² less).
(d) For example: a larger face for printing and shelf display, or a better fit for the product (like biscuits).
In simple words: The cube saves material, but other factors can decide the shape.

Teacher's Note:
1. Accept any sensible reason in (d).
2. Cost is not only material.

Question 28. Archimedes and his Tombstone: A sphere of radius r sits exactly inside the smallest cylinder that can contain it, as shown.
(a) Write down the surface area of the sphere in terms of r.
(b) Write down the curved surface area of the cylinder in terms of r.
(c) State what Archimedes noticed on comparing your two answers.
(d) Find the ratio of the volume of the sphere to the volume of the cylinder.
Answer:
(a) 4πr².
(b) 2πr × 2r = 4πr².
(c) They are exactly equal.
(d) (4/3)πr³ : 2πr³ = 2 : 3.
In simple words: Equal areas and a 2 : 3 volume ratio: Archimedes' favourite result.

Teacher's Note:
1. Height of cylinder is 2r.
2. Tell the story briefly.

Question 29. The Heap of Grain: Grain poured onto a threshing floor forms a heap in the shape of a right circular cone of base diameter 24 m and height 5 m.
(a) Find the radius of the base of the heap.
(b) Find the volume of grain in the heap.
(c) Find the slant height of the heap.
(d) Find the area of canvas needed to cover the heap completely.
Answer:
(a) 12 m.
(b) ⅓ × \( \frac{22}{7} \) × 144 × 5 = \( \frac{5280}{7} \) ≈ 754.29 m³.
(c) √(25 + 144) = 13 m.
(d) πrl = \( \frac{22}{7} \) × 12 × 13 ≈ 490.29 m².
In simple words: Cover only the curved surface; the base sits on the floor.

Teacher's Note:
1. 5–12–13 triple.
2. Halve the diameter.

Question 30. Two Estimates, Two Answers: A cylindrical jar of base radius 9 cm and height 20 cm is to be filled with small spherical sweets. Two students estimate how many sweets the jar will hold, making different assumptions. Student 1 assumes each sweet has radius 1.5 cm and that the sweets occupy 60% of the jar. Student 2 assumes each sweet has radius 1.2 cm and that the sweets occupy 65% of the jar.
(a) Find the volume of the jar.
(b) Find Student 1's estimate of the number of sweets.
(c) Find Student 2's estimate.
(d) The two estimates differ by more than a factor of two. State which assumption is chiefly responsible, and what this shows about guesstimate problems.
Answer:
(a) \( \frac{22}{7} \) × 81 × 20 ≈ 5091.43 cm³.
(b) Sweet ≈ 14.14 cm³; 0.6 × 5091.43 ÷ 14.14 ≈ 216.
(c) Sweet ≈ 7.24 cm³; 0.65 × 5091.43 ÷ 7.24 ≈ 457.
(d) The radius: (1.5/1.2)³ ≈ 1.95, while the packing fractions differ only by about 8%. A cubed quantity magnifies errors, so state assumptions clearly and give a range.
In simple words: A small change in radius makes a big change in the count.

Teacher's Note:
1. Neither student is wrong.
2. Guesstimates need stated assumptions.