Make an estimate based on the second and third configurations.

Step-by-step solution

Idea: The tighter the balls nestle into the gaps, the larger the fraction of the room they fill. Number of balls ≈ (fraction filled) × (room volume ÷ volume of one ball), so scale up the first estimate by the ratio of the fractions filled.

1. First configuration: each ball sits in its own 6 cm cube, so the balls fill 113216 ≈ 52% of the space (exactly π6). That gave 16,87,500 balls.½ mark
2. Second configuration: one ball rests in the middle of each group of eight, so layers interlock. Such packing fills about 68% of the space. Estimate ≈ 16,87,500 × 6852.4 ≈ 21,90,000, about 22 lakh balls.1 mark
3. Check by counting: balls 6.93 cm apart in each square layer (130 × 130 = 16 900), with an in-between layer (129 × 129 = 16 641); layers 3.46 cm apart give 129 layers: 65 × 16 900 + 64 × 16 641 ≈ 21,64,000 ✓.½ mark
4. Third configuration: each layer sits in the hollows of a tightly packed layer below (like oranges on a cart). This is the closest packing, about 74%. Estimate ≈ 16,87,500 × 7452.4 ≈ 23,80,000, about 24 lakh balls. (Counting layers 4.24 cm apart, alternately 150 × 150 and 149 × 149 balls, gives about 23.5 lakh.)1 mark

Check: Every estimate stays below the upper bound of about 33 lakh (which would need 100% filling) ✓, and C > B > A as the gaps shrink ✓.

Answer to write in the exam

Common mistakes that cost marks

• Expecting a tighter packing to fit more than the volume quotient. No packing of balls can fill 100% of the space.
• Counting the in-between layers as full layers of 150 × 150; they sit in the gaps and hold slightly fewer.

How this can come in the exam

Try one yourself

Marbles of volume 4 cm³ are poured into a jar of 1 litre and settle so that they fill about 64% of the jar. Estimate the number of marbles.

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