Enforcement Officer / Accounts Officer (EO/AO) 2020 recruitment test, advertisement 421. Transcribed from the Commission's own booklet (series A, NMJX-U-GTA); answer as per the official UPSC final answer key. This question is in Part B. Answer last reviewed 24 August 2026.

If W1 and W2 are weights of two solid iron balls of radii 1/2 metre and 1/3 metre respectively, then which one of the following is equal to W1 : W2 ?

(a)8 : 27
(b)27 : 8
(c)4 : 16
(d)16 : 4

Answer

Why

Correct — B, (b) 27 : 8. The insight that decides this question is that volume scales with the cube of a length. Both balls are solid and both are iron, so they have the same density, and weight is density times volume: the density is a common factor and cancels out of the ratio. The volume of a sphere is four-thirds pi r cubed, so four-thirds and pi cancel too, and what is left is W1 : W2 = r1 cubed : r2 cubed. Substituting the radii, that is one-half cubed to one-third cubed, or 1/8 : 1/27. A ratio is unchanged when both sides are multiplied by the same number, so multiply by 216, the product of 8 and 27, and the ratio becomes 27 : 8. Equivalently, the radii are in the ratio 1/2 : 1/3, which is 3 : 2, and cubing 3 : 2 gives 27 : 8 directly — the quicker route, and the one to practise, because it avoids fractions altogether. Before any of that, there is a check worth two seconds. A radius of half a metre is larger than a radius of a third of a metre, so the first ball is the heavier and W1 : W2 must be greater than one. Two of the four options, 8 : 27 and 4 : 16, are less than one and are eliminated on sight, which leaves a choice between 27 : 8 and 16 : 4 before any cubing is done. Note finally how the booklet sets this item: the two weights are printed with true subscript numerals, as W with a small 1 and W with a small 2, and the two radii are printed as stacked fractions, one over two and one over three, with a horizontal rule. The transcription writes the subscripts on the baseline and the fractions inline with a solidus, and nothing about the mathematics changes.

Why the others are wrong

(a)8 : 27 — 8 : 27 is the right pair of numbers in the wrong order, and it is what a candidate produces by cubing the denominators of the two radii — 2 and 3 — instead of cubing the radii themselves. It is the exact reciprocal of the answer, which is the signature of a ratio written back to front. Two independent checks catch it. The first is the size check: the ball of radius half a metre is the bigger and therefore the heavier, so the first term of W1 : W2 has to be the larger number, and 8 : 27 puts the larger number second. The second is to name the ratio you are computing before you compute it — the question asks for W1 : W2 in that order, and writing the order down at the start of the working is the cheapest defence against inverting it at the end.
(c)4 : 16 — 4 : 16 reduces to 1 : 4 and so, like the first option, claims the larger ball is the lighter one; the size check disposes of it before any arithmetic. It is worth seeing that it does not correspond to the standard wrong idea either. A candidate who used an area relation instead of a volume one — squaring the radii rather than cubing them — would get one-half squared to one-third squared, that is 1/4 : 1/9, which is 9 : 4, and not this. So 4 : 16 matches neither the volume ratio nor the surface-area ratio nor the ratio of the radii themselves, which is 3 : 2. It is a plain filler, and recognising a filler quickly is worth as much in the hall as computing the answer.
(d)16 : 4 — 16 : 4 reduces to 4 : 1 and at least points the right way, with the larger ball heavier, so it survives the size check and is the only distractor here that needs the actual computation. It fails on magnitude. The radii are in the ratio 3 : 2, and cubing that gives 27 : 8, which is 3.375 to 1, not 4 to 1; for the weights to stand at 4 : 1 the radii would have to be in a ratio whose cube is 4, which 3 : 2 is not. This option is where a candidate lands who knows the answer is bigger than one and picks the roundest number available instead of finishing the cube, and it is a reminder that a partial insight on a proportionality question earns nothing unless the power is right.

Concept

The idea being tested is how a quantity scales when a length changes, and it runs through mensuration, physics and everyday estimation alike. For any family of similar shapes, a length scales as the first power of the scale factor, an area as the square, and a volume as the cube. Double the radius of a sphere and its surface area quadruples while its volume becomes eight times as large. That single rule answers most questions of this form without any formula being written out, and it is worth holding as a rule about shapes rather than as a fact about spheres, because it is equally true of cubes, cones and cylinders. Weight enters through density. Weight is proportional to mass, and mass is density times volume, so for two objects made of the same material the weights are in the ratio of the volumes; the density never has to be known, only to be the same. That is why the stem takes the trouble to say both balls are iron and both are solid — remove either word and the question is unanswerable, because different materials would need their densities and a hollow ball would need its wall thickness. The formula behind it, four-thirds pi r cubed for the volume of a sphere and four pi r squared for its surface area, is worth knowing exactly, but on a ratio question every constant in it cancels and only the power of r survives.

This is the mensuration item of the paper's last block, and it is set so that the arithmetic is trivial once the right relation is chosen. That is the characteristic design of proportionality questions on EPFO EO/AO papers: the marks turn on knowing which power to apply, and the numbers are then chosen to be clean. It follows that the reading to do first is not of the radii but of the words around them — solid, iron, weights, radii — each of which is doing a job. The habit worth building is to state the relation in words before touching the numbers: same material, so weight goes as volume; volume goes as the cube of the radius; therefore the weights go as the cube of the radii. Written that way, the item takes one line. The second habit is the direction check. On any ratio question where one object is plainly bigger than the other, deciding which term must be larger takes a moment and here removes half the option list, leaving a decision that only the correct power can settle. Note in passing that the four options are printed with spaces around the colon, and that two of them are unreduced — 4 : 16 and 16 : 4 rather than 1 : 4 and 4 : 1 — so reducing each one mentally is part of reading the list.

Key facts

For similar shapes, length scales as the first power of the scale factor, area as the square and volume as the cube. This one rule answers most questions about how a change in size changes a quantity.
The volume of a sphere is four-thirds pi r cubed and its surface area is four pi r squared. On a ratio question every constant in these formulae cancels, leaving only the power of r.
For two objects of the same material, weight is proportional to volume, because weight goes as mass and mass is density times volume; the common density cancels and never has to be known.
The working here: the radii are 1/2 and 1/3, so their ratio is 3 : 2, and cubing that ratio gives the ratio of the weights as 27 : 8 — equivalently 1/8 : 1/27 multiplied through by 216.
The direction check on any such item: the larger ball must be the heavier, so W1 : W2 has to exceed one, which eliminates 8 : 27 and 4 : 16 before any computation is done.
Had the question asked for the ratio of the surface areas of the same two balls, the answer would have been 9 : 4 — the square of 3 : 2 — which is the standard companion item and the reason for being sure which power is wanted.

Volume scales with the CUBE of a length, and the density cancels

The direction check, before any arithmetic→A radius of half a metre is larger than a radius of a third of a metre, so the first ball is the heavier and W1 : W2 must be greater than one. Two of the four options, 8 : 27 and 4 : 16, are less than one and are eliminated on sight — which leaves a choice between 27 : 8 and 16 : 4 before any cubing is done.

Why only the radii survive→Both balls are solid and both are iron, so they have the same density, and weight is density times volume — the density is a common factor and cancels out of the ratio. The volume of a sphere is four-thirds pi r^3, so the four-thirds and the pi cancel too. What is left is W1 : W2 = r1^3 : r2^3.

The quick route: cube the ratio of the radii→The radii are in the ratio 1/2 : 1/3, which is 3 : 2. Cubing 3 : 2 gives 27 : 8 directly. This avoids fractions altogether, and it is the route to practise.

The same thing the long way→One-half cubed to one-third cubed is 1/8 : 1/27. A ratio is unchanged when both sides are multiplied by the same number, so multiply through by 216, the product of 8 and 27, and it becomes 27 : 8.

Where the two survivable wrong options come from→8 : 27 is the right pair of numbers in the wrong order — what a candidate produces by cubing the DENOMINATORS of the two radii, 2 and 3, instead of cubing the radii themselves. Being the exact reciprocal of the answer is the signature of a ratio written back to front. 16 : 4 reduces to 4 : 1 and at least points the right way, so it survives the size check and is the only distractor here that needs the actual computation: cubing 3 : 2 gives 3.375 to 1, not 4 to 1.

Why the stem says ‘solid’ and ‘iron’→Remove either word and the question is unanswerable — different materials would need their densities, and a hollow ball would need its wall thickness. And be sure which power is wanted: had the question asked for the ratio of the SURFACE AREAS of the same two balls, the answer would be 9 : 4, the square of 3 : 2. For any family of similar shapes, length scales as the first power of the scale factor, area as the square, and volume as the cube.

Study next

Surface area and volume formulae for the sphere, hemisphere, cylinder, cone and cube, together with the effect of scaling each oneProblems in which a solid is melted and recast into smaller solids, where volume is conserved and the count of new pieces follows from the cube ruleDirect, inverse and joint variation stated in words, the language in which proportionality questions are usually set on these papersDensity, relative density and the way weight follows volume for objects of the same material, which links this topic to the flotation questions in the same blockRatio and proportion technique, including reducing a ratio, comparing two ratios and keeping the order of terms straight

Common traps

Using the square instead of the cube. Squaring the radii answers a question about surface area, not about weight, and on this item it would give 9 : 4, which is not even offered — the power is the whole question.
Inverting the ratio. The stem asks for W1 : W2 in that order, and the reciprocal 8 : 27 is printed as an option precisely to catch a candidate who computes correctly and writes the terms the wrong way round.
Forgetting that the balls are solid and of the same material. Those two words are what allow the density and the shape constants to cancel; without them the ratio of the weights could not be found from the radii alone.
Reading an unreduced ratio as though it were in lowest terms. Two of the options here are printed as 4 : 16 and 16 : 4, and reducing them to 1 : 4 and 4 : 1 is part of judging them.

Mensuration on EPFO EO/AO papers arrives as ratio and scaling rather than as computation: the ratio of volumes or areas of two similar solids, the effect of doubling a dimension, or a solid recast into smaller ones. The numbers are chosen so that the answer is exact and the arithmetic is light, because the discrimination is meant to happen at the choice of relation. Options are commonly built as a matched set — the correct ratio, its reciprocal, and one or two ratios drawn from the wrong power — so a candidate who has the right idea and a moment's carelessness still has somewhere wrong to land. Expect the companion items on density and flotation in the same block, which use the same volume relation from the other direction.

Related PYQs

EPFO_EOAO_2020_Q85Open & attempt →

If ‘a’ varies as ‘b’, then which of the following statements is/are correct ? 1. n^th root of a^2b varies as (2n)^th root of a^4b^2. 2. a/b^2 varies inversely as b. Select the correct answer using the code given below :

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Answer(c) Both 1 and 2

EPFO_EOAO_2020_Q120Open & attempt →

A hollow closed cube of side 10 cm, each weighing 200 gm, is made. This cube is placed in water with a horizontal face. How many cm of its height sinks in water if the density of water is 1 gm per cm^3 ?

(a) 1 cm
(b) 1·5 cm
(c) 2 cm
(d) 2·5 cm

Answer(c) 2 cm

The last item of the paper, where a cube floats in water. It uses the same volume-and-density relation from the other side — there the density is what is unknown and the geometry is given, here the geometry is unknown and the material is shared.

Practice

practice — not a real PYQ
Two solid spheres are made of the same metal and the ratio of their radii is 2 : 5. Which one of the following is the ratio of their weights ?
- (a)2 : 5
- (b)4 : 25
- (c)8 : 125
- (d)125 : 8
Answer(c) 8 : 125
practice — not a real PYQ
If the radius of a solid iron sphere is doubled while the material remains the same, its weight becomes how many times the weight of the original sphere ?
- (a)2 times
- (b)4 times
- (c)6 times
- (d)8 times
Answer(d) 8 times

Why