Anirban saw a few tennis balls lying around. He wondered, “How many tennis balls can fit in an empty classroom?”. The classroom is of dimensions 30 ft × 30 ft × 15 ft. Estimate how many tennis balls of radius 3 cm fit in this classroom. Before calculating, make a guess.
Step-by-step solution
To find: Estimated number of balls that fit
Idea: Room volume ÷ ball volume is too many, because spheres cannot fill space without gaps. Counting balls in rows and layers (each ball needs a 6 cm cube of space) gives a realistic estimate.
- Guess first: a room is millions of cm3 and a ball is about 100 cm3, so perhaps a few million.½ mark
- Volume of one ball = 43π × 33 ≈ 113 cm3. Volume of room = 30 × 30 × 15 = 13 500 ft3 ≈ 13 500 × 28 000 = 37,80,00,000 cm3.1 mark
- 37,80,00,000113 ≈ 33,45,132. This is only an upper bound: balls leave gaps, so fewer than this can fit.1 mark
- First configuration (each ball directly on top of another): along the length, 900 cm6 cm = 150 balls (30 ft ≈ 900 cm); the same along the width; along the height 4506 = 75.1 mark
- Number ≈ 150 × 150 × 75 = 16,87,500 balls (about 17 lakh). Other ways of packing fit more, as they leave smaller gaps.½ mark
Check: Each ball in the first arrangement takes up a 6 cm cube (216 cm3), and 900 × 900 × 450 ÷ 216 = 16,87,500 ✓. That is about half the upper bound, as expected when balls fill about 52% of the space ✓.
Answer to write in the exam
Volume of ball = 43π × 33 ≈ 113 cm3
Volume of room = 30 × 30 × 15 = 13 500 ft3 ≈ 37,80,00,000 cm3
Upper bound = 37,80,00,000113 ≈ 33,45,132
Ball on ball: 30 ft ≈ 900 cm ⇒ 9006 = 150 per row; height 450 cm ⇒ 75 layers
∴ Number ≈ 150 × 150 × 75 = 16,87,500
Common mistakes that cost marks
- Giving the volume quotient 33,45,132 as the answer. Spheres always leave gaps, so that is only an upper limit.
- Dividing the room length by the radius (3 cm) instead of the diameter (6 cm).
- Mixing feet and centimetres without converting.
How this can come in the exam
Balls of diameter 10 cm are stacked directly on top of each other in a box 1 m × 1 m × 0.5 m. How many fit?
- 500
- 1000
- 5000
- 10 000
Show answer
(A) 500
100 ÷ 10 = 10, 10, and 50 ÷ 10 = 5: 10 × 10 × 5 = 500.
Assertion (A): Volume of a room ÷ volume of a ball gives the exact number of balls that fit in the room.
Reason (R): Spheres packed together always leave gaps between them.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(D) A is false but R is true.
A is false: the quotient is only an upper bound. R is true, and it is the reason A fails.
Try one yourself
Estimate how many footballs of diameter 22 cm fit in a room 6.6 m × 4.4 m × 3.3 m if stacked ball on ball.
Show answer
660 ÷ 22 = 30, 440 ÷ 22 = 20, 330 ÷ 22 = 15: 30 × 20 × 15 = 9000 balls.
More questions like this
- Make an estimate based on the second and third configurations.
- Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.
- A cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.
- Solve the following:
- You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?