EPFO APFC 2023 General Studies · Question 72 of 120

Correct — C, (c) 3. The volume of a hemisphere of radius r is two-thirds of pi times r cubed, so when the radius is multiplied by 1·5 the volume is multiplied by 1·5 cubed, which is 3·375. The larger container therefore holds exactly 3·375 times as much water as one of the smaller ones, and the constant two-thirds of pi never has to be evaluated because it cancels in the ratio. Now fill the large container, emptying as many small ones as possible. Three small containers can be poured in whole, contributing 3 units of the 3·375 the large one takes. The remaining 0·375 of a unit has to come from a fourth small container, which gives up rather more than a third of its water and keeps the rest. When the large container is full, therefore, three of the seven have been emptied out, one has been partly drained, and the other three are untouched. The question asks how many remain fully filled, and the answer is those three untouched containers. The fourth still holds water, but it is no longer full, so it is not counted. Two things generalise from this. First, the shape barely matters: any two similar solids have volumes in the ratio of the cube of their linear scale, so the same 3·375 would appear for spheres, cones, cubes or any other family of similar containers, and the word hemispherical is there to make the problem concrete rather than to make it harder. Second, when a ratio like 3·375 is not a whole number, the fractional part always implicates one extra container, and the wording of the question then decides whether that container is counted or not.

• (a)5 — This is the answer produced by scaling the volume by 1·5 instead of by 1·5 cubed — that is, by treating a container of one and a half times the radius as holding one and a half times as much water. On that arithmetic the large container takes one whole small container and half of a second, so two containers are touched and five remain full. The error is the commonest one in mensuration and it is worth naming: linear scale, area and volume scale by the first, second and third powers respectively, so a container that is half as large again in radius holds not one and a half times but three and three-eighths times the water. The same mistake explains why doubling the radius of a tank multiplies its capacity by eight rather than by two, and it is the single fact most often tested in problems about similar solids.
• (b)4 — Four is the number of small containers that still hold some water at the end, and it is the answer to a question that was not asked. Three containers are emptied completely and a fourth gives up 0·375 of its contents; that fourth container is still on the table with water in it, so a candidate who counts vessels that are not empty arrives at four. The stem asks how many remain fully filled, and a container that has given away more than a third of its water does not qualify. The same figure also arises from a second and unrelated error, that of scaling the volume by 1·5 squared rather than 1·5 cubed, which gives a ratio of 2·25, empties two containers and part of a third, and again leaves four. When two different mistakes converge on one option, that option is the one to be most careful about.
• (d)2 — Two is what the count returns if the volume ratio is taken as 4·5, which is what a candidate gets by computing 1·5 cubed as 1·5 multiplied by 3 rather than as 1·5 multiplied by itself three times. On that figure four small containers are emptied whole and a fifth is half drained, leaving two full. The lesson is arithmetical rather than geometrical: cubing a number is not multiplying it by three, and 1·5 cubed is 3·375, comfortably less than four. It is worth developing a rough check for such values before doing the arithmetic. Since 1·5 lies between 1 and 2, its cube must lie between 1 and 8, and since 1·5 is nearer to 1 than to 2, the cube should be well below the midpoint of that range. A figure of 4·5 fails that sanity check and would have been caught before it reached the count.

Two solids are similar when one is a uniform enlargement of the other, and for any two similar solids the ratio of corresponding lengths, the ratio of corresponding areas and the ratio of volumes are related as k, k squared and k cubed, where k is the linear scale factor. That single relation is the whole content of this question. The volume of a hemisphere of radius r is two-thirds of pi times r cubed, half the volume of the sphere of the same radius, whose surface area is four pi r squared and whose volume is four-thirds pi r cubed; a hemisphere's curved surface is two pi r squared, with a flat circular face of pi r squared on top of that if it is closed. But none of these formulae has to be recalled to answer the question, because both containers are hemispheres and every constant cancels when their capacities are compared. What is left is the cube of 1·5, which is 3·375. The second idea in the question is the treatment of a non-integral ratio when the objects being counted are indivisible. Pouring 3·375 containers' worth of water means emptying three containers and taking part of a fourth, so the count of containers involved is the ceiling of the ratio while the count of containers wholly emptied is its floor, and the number left full is the total minus the ceiling. Keeping those three quantities distinct is what makes the difference between three and four here.

The quantitative section of this paper favours problems that look like they need a formula and in fact need only a ratio, and mensuration of similar solids is the classic instance. A candidate who reaches for the volume of a hemisphere, writes out two-thirds pi R cubed and two-thirds pi times 1·5 R cubed and then divides has done more work than necessary but will get there; a candidate who recognises at once that only the cube of the scale factor matters will finish the problem in a few seconds and have time for the harder items. The second skill the item tests is reading the exact question. The stem is carefully worded — it asks how many containers remain fully filled, not how many still hold water and not how many were emptied — and the option set rewards each of those readings with a different number, so the arithmetic can be perfectly correct and the answer still wrong. That pattern recurs throughout this paper's quantitative block, where the difference between the quantity computed and the quantity asked for is often the whole difficulty. Read the final sentence of the stem again after the arithmetic is done, and check which of the numbers now sitting in front of you it is asking for. Note also that this booklet prints decimals with a raised middle dot, so 1·5 in the stem is one and a half.

• For similar solids, corresponding lengths scale as k, areas as k squared and volumes as k cubed; multiplying the radius of a container by 1·5 therefore multiplies its capacity by 1·5 cubed, which is 3·375.
• The volume of a hemisphere of radius r is two-thirds of pi times r cubed, half that of the sphere of the same radius, but the constant cancels when two hemispheres are compared, so the formula is not needed to answer this question.
• Pouring water from seven full containers into a vessel holding 3·375 of them empties three completely and partly drains a fourth, so four containers are involved and three of the original seven remain untouched and full.
• The number of vessels wholly emptied is the whole-number part of the ratio, the number of vessels involved is that figure plus one whenever the ratio is not a whole number, and the number left full is the total minus the number involved.
• The cube of 1·5 is 3·375 and not 4·5; cubing multiplies a number by itself three times, and since 1·5 lies between 1 and 2 its cube must lie between 1 and 8.

• Scaling volume by the linear factor rather than by its cube, which turns 3·375 into 1·5 and changes the answer completely
• Computing the cube of a decimal as the number multiplied by three; 1·5 cubed is 3·375, not 4·5
• Counting the containers that still hold water instead of those that remain fully filled; the partly drained fourth container falls between the two counts
• Evaluating two-thirds pi R cubed in full when both vessels are hemispheres and every constant cancels in the ratio
• Reading the raised middle dot in this booklet's numbers as anything other than a decimal point, so that 1·5 is one and a half

The quantitative block of this paper sets one or two mensuration items in each group of five, and they are almost always about the relation between a linear dimension and a capacity rather than about recalling an unusual formula. Typical forms are the recasting of a sphere into smaller spheres or into a wire, the number of small containers that fill a large one, the effect on capacity of changing a radius or a height by a stated percentage, and the comparison of two similar vessels. The formulae needed are the standard ones for the sphere, hemisphere, cylinder and cone, and in most of these questions they cancel. The examination-hall technique is to write the ratio first and evaluate only what survives, and then to reread the final clause of the stem, because these items are frequently decided by whether the candidate reports the number emptied, the number involved or the number left full.

No directly related past PYQ was found.

• practice — not a real PYQ

  The radius of a spherical vessel is increased by 20 per cent. By what percentage does its capacity increase?

  • (a)20 per cent
  • (b)44 per cent
  • (c)60 per cent
  • (d)72·8 per cent

  Answer(d) 72·8 per cent — capacity scales as the cube of the linear factor, and 1·2 cubed is 1·728, an increase of 72·8 per cent. The figure of 44 per cent is what the surface area would increase by, since 1·2 squared is 1·44, and 20 per cent is the increase in the radius itself; each of the three powers answers a different question about the same enlargement.

• practice — not a real PYQ

  A hemispherical bowl of radius 2r is filled with water poured from identical full hemispherical cups of radius r. How many such cups are emptied completely before the bowl is full?

  • (a)2
  • (b)4
  • (c)6
  • (d)8

  Answer(d) 8 — the volumes of similar solids are in the ratio of the cube of their linear scale, so a bowl of twice the radius holds two cubed, that is eight, times as much as one cup. The ratio is a whole number here, so exactly eight cups are emptied and none is left partly drained, which is what makes this version simpler than the item it is modelled on.