Enforcement Officer / Accounts Officer (EO/AO) 2017 recruitment test. Transcribed from the Commission's own booklet (series A, GRS-A-TNS); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

What is the maximal number of spherical balls of radius 1 cm each that can be placed inside a cubical box of height 10 cm?

(a)25
(b)125
(c)250
(d)1000

Answer

Why

Correct — B, (b) 125. The step that decides this item is the one everybody skips: the space a ball occupies along an edge is its diameter, not its radius. Each ball has radius 1 cm, so it is 2 cm across, and along a 10 cm edge of the box exactly 10 ÷ 2 = 5 balls fit end to end with nothing left over.

That count holds in all three directions, because the box is a cube of height 10 cm and its other two edges are the same. Arrange the balls in straight rows touching one another: five along the length, five along the breadth and five layers stacked to the top. The number is therefore 5 × 5 × 5 = 125, which is option (b).

Two things make the fit exact and worth noticing. First, 10 is a whole multiple of 2, so no strip of unused space is left along any edge; had the box been 11 cm high the answer would still have been five layers, because a sixth would need another 2 cm and only 1 cm would remain. That is why counting of this kind is always a whole-number division with the remainder thrown away. Second, the balls touch each other and touch the walls, and the empty space between them — the gaps at the corners of every group of eight — is real but irrelevant to the count. A packing question asks how many objects fit, and the answer comes from how many fit along each dimension, one dimension at a time. It never comes from dividing one volume by another.

Why the others are wrong

(a)25 — Twenty-five is 5 × 5, the number of balls in a single layer covering the floor of the box, and it is what a candidate produces by solving the problem in two dimensions and forgetting that the box has a height. The stem gives that height explicitly — a cubical box of height 10 cm — so the third dimension is in plain sight, and the word cubical settles that the other two edges are 10 cm as well. The error is worth naming because it recurs across the whole topic of packing and stacking: a candidate visualises the arrangement by looking down on it, counts what can be seen, and never lifts the picture into a stack. The guard against it is to write the three counts down separately, one per direction, before multiplying anything, so that a missing factor is visible on the page rather than only in the head.
(c)250 — Two hundred and fifty has no packing arrangement behind it, but it sits close to the figure that the wrong method produces, and that is why it is dangerous. A candidate who divides volumes computes the box as 10 × 10 × 10 = 1,000 cm³ and each ball as four-thirds π times the cube of the radius, which is about 4·19 cm³, and arrives at roughly 238 balls — a number that would make an option in this region look reasonable. The method is invalid because spheres cannot fill space: however cleverly equal spheres are arranged, the densest possible packing occupies only about 74 per cent of the space available, and rows of touching balls in a rectangular grid occupy appreciably less than that. Dividing volumes therefore always overstates the count, and on a container this small it overstates it badly.
(d)1000 — A thousand is 10 × 10 × 10, and it is the answer to the question of how many 1 cm cubes fit in the box. It comes from using the radius where the diameter belongs — treating each ball as though it occupied a cell 1 cm across, because 1 cm is the number printed in the stem. Every ball actually needs a cell 2 cm across, so the true count is smaller by a factor of 2 × 2 × 2, that is by eight, and 1,000 ÷ 8 = 125. The relation between the two figures is itself the lesson: doubling the space each object needs along every edge divides the number that fit by eight. Radius and diameter are the commonest confusion in mensuration generally, and packing problems punish it most heavily, because the error is cubed.

Concept

Packing problems ask how many whole objects fit inside a container, and they are answered dimension by dimension rather than by volume. For identical spheres arranged in straight rows, the space each one occupies along any direction is its diameter, so the count along an edge is the length of that edge divided by the diameter, with any fraction discarded — a partly fitting ball does not count. Multiplying the three counts gives the total. The reason the volume method fails is that spheres leave gaps: place eight of them at the corners of a cube of side equal to their diameter and there is an empty pocket at the centre. In a simple cubic arrangement of the kind this question describes, the balls occupy about 52 per cent of the box; the densest packing known and proved for equal spheres reaches about 74 per cent, and no arrangement can do better. Dividing the container's volume by the object's volume therefore always gives a count that cannot be achieved. The same reasoning applies with a different constant to cylinders and to boxes; the one case in which volume division is exact is the packing of rectangular blocks whose edges divide the container's edges, because those leave no gaps at all.

Mensuration in an EPFO paper is rarely about a formula and almost always about a modelling decision. Here the formula for the volume of a sphere is not needed at all, and the candidate who reaches for it first has already gone wrong. What the item tests is whether a candidate can see that the constraint is geometric — a ball needs a certain width, the box offers a certain width, and the count follows — and whether the difference between radius and diameter survives under time pressure. Both are practical habits. The Commission places two mensuration items in this last block, this one and the spherical container at question 115, and they test opposite instincts: this one punishes a candidate who reaches for the volume formula, and that one rewards a candidate who realises the formula's constant cancels. Reading what the question actually constrains, before computing anything, is the skill common to both.

Key facts

A ball of radius 1 cm occupies 2 cm of length along any direction, because the space it needs is its diameter.
Along a 10 cm edge, 10 ÷ 2 = 5 balls fit exactly, with no unused strip left over.
The box is cubical with height 10 cm, so the same count of five holds in all three directions.
The total is 5 × 5 × 5 = 125, arranged as five layers of twenty-five balls each.
Counts of this kind are whole-number divisions: a fraction of a ball does not count, so any remainder is discarded.
Dividing the container's volume by a sphere's volume is invalid, because spheres cannot fill space and always leave gaps.
The densest possible packing of equal spheres occupies about 74 per cent of the space; simple rows of touching spheres occupy about 52 per cent.
Using the radius where the diameter belongs multiplies the count by eight, which turns 125 into 1,000.

Count dimension by dimension; never divide one volume by another

A ball of radius 1 cm occupies 2 cm of length→The space it takes up along any direction is its diameter, not its radius. This is the step everybody skips, and it decides the item.

10 ÷ 2 = 5 along each edge→The box is cubical with height 10 cm, so the same count holds in all three directions. Ten is a whole multiple of two, so no strip of unused space is left; had the box been 11 cm high the answer would still be five layers, because a sixth would need another 2 cm and only 1 cm would remain.

5 × 5 × 5 = 125→Five along the length, five along the breadth, and five layers stacked to the top. The empty pockets between the balls are real but irrelevant: a packing question asks how many objects fit, and the answer comes from how many fit along each dimension.

(d) 1000 — radius used where diameter belongs→10 × 10 × 10 is the count of 1 cm cubes, reached by treating each ball as occupying a cell 1 cm across because 1 cm is the number printed in the stem. Each ball needs a cell 2 cm across, so the true count is smaller by a factor of eight. The error is cubed, which is why packing problems punish this confusion most heavily.

(c) 250 — the volume method→The box is 1,000 cm³ and each ball about 4·19 cm³, giving roughly 238 — a figure that makes an option in this region look reasonable. But spheres cannot fill space: the densest possible packing of equal spheres occupies about 74 per cent, and simple rows of touching balls about 52 per cent, so dividing volumes always overstates the count.

(a) 25 — the third dimension forgotten→5 × 5 is one layer covering the floor of the box. The stem gives the height explicitly and the word cubical settles the other two edges. Write the three counts down separately, one per direction, so that a missing factor is visible on the page rather than only in the head.

Study next

Packing and stacking problems generally — how many boxes, tins or bricks fit in a containerVolume and surface area of the sphere, the cube, the cuboid and the cylinderScaling: what happens to length, area and volume when a solid's linear dimensions are multiplied by a factorThe relation between a sphere and the cube or cylinder that just contains itProblems in which a solid is melted and recast, where volume is conserved but surface area is not

Common traps

Using the radius instead of the diameter as the space a ball needs, which multiplies the count by eight.
Counting only one layer and forgetting that the container has a height as well as a floor.
Dividing the volume of the box by the volume of a ball, which always gives a count that cannot actually be achieved.
Keeping a fractional ball in the count instead of discarding the remainder at each dimension.
Reaching for the sphere's volume formula on an item where no volume is needed at all.

Mensuration items in EPFO papers divide into two families and this belongs to the counting one. That family asks how many small objects fit into a large container, how many small cubes a large cube can be cut into, or how many tiles cover a floor, and it is answered by dividing lengths and discarding remainders. The other family asks what happens to areas and volumes when dimensions change, and is answered by the square and cube laws of scaling. Both are set in nearly every paper, and both reward the candidate who identifies which quantity is actually constrained — width in the first family, ratio in the second — before writing a formula. Where a paper offers an option that matches the volume-division result, as this one arguably does, the safest response is to remember that the method is wrong in principle rather than merely inaccurate.

Related PYQs

EPFO_EOAO_2023_Q55

An ice cube with 10 cm side is divided into eight smaller cubes, each with same side. Which one of the following statements is correct in this context ?

(a) Total volume will increase and total surface area will decrease.
(b) Total volume will decrease and total surface area will increase.
(c) Total volume will remain the same and total surface area will increase.
(d) Total volume will increase and total surface area will remain the same.

Answer(c) Total volume will remain the same and total surface area will increase.

The companion counting item from a later paper — a ten-centimetre ice cube divided into eight smaller cubes, and what happens to the total volume and the total surface area.

EPFO_APFC_2016_Q113

If the radius of a circle is reduced by 50%, its area will be reduced by

(a) 30%
(b) 50%
(c) 60%
(d) 75%

Answer(d) 75%

The scaling law in two dimensions: by how much the area of a circle falls when its radius is reduced by half.

Practice

practice — not a real PYQ
How many solid cubes of edge 2 cm can be packed completely inside a rectangular box measuring 8 cm by 6 cm by 4 cm?
- (a)12
- (b)24
- (c)48
- (d)96
Answer(b) 24
practice — not a real PYQ
Spherical balls of radius 2 cm each are placed in straight rows inside a cubical box of edge 12 cm. What is the number of balls that can be placed in this way?
- (a)9
- (b)18
- (c)27
- (d)36
Answer(c) 27

Why