EPFO APFC 2023 General Studies · Question 102 of 120

Correct — C, (c) 43·75. Mixture questions are solved by tracking one ingredient in absolute units and the total volume separately, never by averaging percentages, and once that discipline is fixed the arithmetic here is three lines. Take the oil. Container X holds oil and water in the ratio 1 : 3, so oil is one part out of four and the oil fraction is 1/4; two litres drawn from it carry 2 × 1/4 = 0·5 litre of oil. Container Y holds them in the ratio 2 : 3, so oil is two parts out of five and the oil fraction is 2/5; five litres drawn from it carry 5 × 2/5 = 2 litres of oil. Container Z already holds one litre of pure oil, which is one litre of oil and no water. Add the three: 0·5 + 2 + 1 = 3·5 litres of oil. Now the total. Nothing evaporates and nothing is poured away, so the final volume is simply 1 + 2 + 5 = 8 litres. The percentage of oil is therefore 3·5 ÷ 8 = 0·4375, that is 43·75 per cent, which is exactly option (c). The same computation written as a weighted average of concentrations gives the identical result and is worth seeing once, because it is the form that scales to four or five containers: X is 25 per cent oil, Y is 40 per cent oil and the litre already in Z is 100 per cent oil, so the mixture is (2 × 25 + 5 × 40 + 1 × 100) ÷ 8 = (50 + 200 + 100) ÷ 8 = 350 ÷ 8 = 43·75. The two litres and the five litres and the one litre are the weights; the trap in the weighted-average form is forgetting that the litre of pure oil is a weight as well as a concentration, and it must appear in the denominator too. There is also a structural check that settles this particular option set without finishing the division. Every quantity of oil entering Z is a whole number of half-litres — a half from X, two from Y, one from the pure oil — so the total oil is a multiple of 0·5 litre in 8 litres, and the answer must therefore be a multiple of 0·5 ÷ 8 = 6·25 per cent. Of the four figures offered, only 43·75 is such a multiple, since 43·75 = 7 × 6·25. That check is specific to the numbers in this stem and not a general rule, but on this item it is faster than the division.

• (a)41·25 — This works out to 3·3 litres of oil in the 8 litres, and no clean mis-step in the problem produces 3·3 litres. That is the point of the option: the four figures form a tight cluster of quarter-percent values around the true one, so the item separates candidates who compute exactly from candidates who estimate and then reach for whichever printed number looks close. Rough reasoning lands here easily. A candidate who notes that Y contributes the bulk of the liquid and is 40 per cent oil, then nudges the figure up a little for the pure oil and down a little for the weak mixture from X, will produce something in the low forties and pick this. The safeguard is the multiple-of-6·25 check: 41·25 divided by 6·25 is 6·6, not a whole number, so no combination of the half-litres of oil described in this stem can give it, whatever the arithmetic used to reach it.
• (b)42·75 — This implies 3·42 litres of oil, which again is not the output of any single identifiable error — it is a near-value placed one step below the answer so that a candidate who has done the work but slipped in the final division has somewhere to land. It is also the option that catches a very specific hesitation: having found 3·5 litres of oil in 8 litres, a candidate who divides in a hurry may recall that 3·5 ÷ 8 is 'about 0·43' and then choose the printed figure that begins 42 or 43 without carrying the division out. Carry it out: 8 × 0·4 = 3·2, leaving 0·3; 8 × 0·03 = 0·24, leaving 0·06; 8 × 0·007 = 0·056. The quotient is 0·4375 exactly, because 3·5/8 = 7/16 and 7/16 is a terminating fraction. 42·75 corresponds to no fraction of the form n/16.
• (d)44·25 — This implies 3·54 litres of oil, half a percentage point above the true figure, and it is the mirror of the option below the answer — placed there so that a candidate who over-counts the oil slightly, or who rounds upward, still finds a printed number waiting. One real error does push the estimate in this direction: treating the ratio 2 : 3 in container Y as though oil were two-thirds rather than two-fifths of the liquid inflates Y's contribution, though carried through properly that mistake gives a figure far higher than any option here, which is itself a useful signal that the ratio has been misread. The reliable defence is the same as for the other two wrong figures. The oil in Z is 0·5 + 2 + 1 = 3·5 litres and the volume is 8 litres; 3·5/8 = 7/16 = 0·4375 exactly, and 44·25 is not 7/16 or any other sixteenth.

A mixture problem is a book-keeping problem. The safe method is to convert every ratio into an absolute quantity of one chosen ingredient, add those quantities, add the volumes separately, and divide once at the end. The single most common failure is confusing a ratio with a fraction: in a mixture with oil and water in the ratio 1 : 3 the oil is one part in four, not one part in three, because the ratio compares the two ingredients with each other while the fraction compares one ingredient with the whole. Convert a : b into a/(a + b) the moment it is read and the error cannot recur. The second idea the question uses is the weighted average of concentrations. When volumes V1, V2, V3 with oil concentrations c1, c2, c3 are combined, the resulting concentration is (V1c1 + V2c2 + V3c3)/(V1 + V2 + V3). Pure oil is simply a concentration of 1, and pure water a concentration of 0, so a component that looks like a special case is nothing of the kind — it enters the sum on the same footing as the others. The result of that formula always lies between the smallest and the largest concentration being mixed, which is a free sanity check: here the extremes are 25 per cent and 100 per cent, so any answer outside that band would be wrong on inspection. The third idea, useful when quantities are removed and replaced rather than added, is the replacement rule: if a fraction f of a mixture is removed and replaced by pure water n times, the surviving quantity of the original ingredient is its initial amount multiplied by (1 − f) raised to the power n. That rule is not needed here, but it is the other half of the topic and papers alternate between them.

The quantitative block of this paper is placed so that each item can be finished in about a minute, and the examiner's lever is not difficulty but precision. Here all four options are of the same shape, all four are within three percentage points of one another, and every one ends in a quarter or a three-quarter, so nothing can be picked by its appearance and estimation is worthless. That design tells the candidate something useful about how to spend time: where the options are widely spaced, an approximation is enough and should be used; where they are clustered like this, the arithmetic must be carried to the last digit, and it is faster to accept that at the start than to estimate, hesitate and then compute anyway. One point of print is worth noticing for its own sake. Of the eighty-five stems in this booklet that carry a question mark, this is the only one printed with a space before it, 'container Z ?'; the other eighty-four run the mark straight on to the word. It is the Commission's typesetting, not a transcription slip, and the same applies to the raised middle dot in the four options — 41·25 rather than 41.25. That dot is this booklet's decimal point throughout, appearing in eight of its questions, and a candidate meeting it for the first time under time pressure should read it as nothing more than a decimal separator.

• A ratio a : b between two ingredients makes the first ingredient a/(a + b) of the whole, not a/b of it. Oil and water in the ratio 1 : 3 means the oil is one quarter of the liquid; in the ratio 2 : 3 it is two fifths. Converting each ratio to a fraction of the total at the moment it is read removes the commonest error in the whole topic.
• In this item: two litres from container X give 2 × 1/4 = 0·5 litre of oil; five litres from container Y give 5 × 2/5 = 2 litres of oil; container Z's own litre is pure oil. Total oil 3·5 litres, total volume 1 + 2 + 5 = 8 litres, so the oil is 3·5/8 = 7/16 = 43·75 per cent.
• The general rule for combining mixtures is the weighted average of concentrations: mixing volumes V1, V2, V3 of oil concentrations c1, c2, c3 gives (V1c1 + V2c2 + V3c3)/(V1 + V2 + V3). Pure oil enters as concentration 1 and pure water as concentration 0, and the litre of pure oil must appear in the denominator as well as the numerator.
• The result of any such mixing must lie between the lowest and the highest concentration combined. Here those are 25 per cent from container X and 100 per cent from the pure oil, so a figure outside that range can be rejected without arithmetic — a check that costs nothing and catches gross slips.
• On this particular item every quantity of oil is a whole number of half-litres in a total of eight litres, so the answer must be a multiple of 0·5/8 = 6·25 per cent. Only 43·75 satisfies that, being 7 × 6·25. The check depends on the numbers in this stem and is not a general law of mixture problems.

• Reading the ratio as a fraction of the whole. Oil and water in the ratio 1 : 3 makes the oil a quarter of the liquid, not a third; the ratio 2 : 3 makes it two fifths, not two thirds. Every ratio in a mixture problem should be converted to a/(a + b) before any multiplication is done.
• Leaving the litre of pure oil out of the total volume. It contributes to both the oil and the mixture, so the denominator is 8 litres, not 7. Counting it in the numerator alone gives 50 per cent, and leaving it out of both gives 31·25 per cent — neither is offered, which is a warning in itself.
• Averaging the concentrations 25 per cent, 40 per cent and 100 per cent without weighting them by the volumes drawn. The unweighted mean is 55 per cent, far outside the band the options occupy; concentrations may only be averaged in proportion to the quantities they arrive in.
• Estimating when the options are clustered. All four figures here lie within three percentage points of one another and all end in a quarter or three-quarter, so an approximate answer cannot be matched to a printed one. The division 3·5 ÷ 8 has to be finished.

Mixture items in EPFO papers keep to a small number of forms. The commonest is this one — two or three known mixtures poured together, with the concentration of the result asked for — and it is sometimes dressed up with a component of pure liquid, which is simply a concentration of nought or one hundred per cent and needs no special treatment. The second form runs backwards: the final concentration is given and the quantity to be added, or the ratio in which two mixtures must be combined, is asked for, which is where alligation earns its keep. The third is the replacement form, in which some of the mixture is drawn off and replaced by water once or repeatedly. All three are answerable with the same two habits — convert every ratio to a fraction of the whole at once, and track one ingredient in absolute units while tracking the total volume separately. Expect option sets as tight as this one, because in a paper where every question carries the same marks the examiner controls difficulty by controlling how much precision the options demand.

No directly related past PYQ was found.

• practice — not a real PYQ

  Container P contains a mixture of milk and water in the ratio 3 : 2, and container Q contains a mixture of milk and water in the ratio 1 : 4. If 6 litres are taken from P and 10 litres from Q and poured into an empty vessel, what is the percentage of milk in the vessel?

  • (a)30
  • (b)32.5
  • (c)35
  • (d)37.5

  Answer(c) 35 — milk is 3/5 of the liquid in P, so 6 litres from P carry 3.6 litres of milk, and milk is 1/5 of the liquid in Q, so 10 litres from Q carry 2 litres. The vessel holds 5.6 litres of milk in 6 + 10 = 16 litres, and 5.6/16 = 0.35, that is 35 per cent. Reading the ratios as thirds and fifths of the whole instead of fifths and fifths is what the other figures are placed for.

• practice — not a real PYQ

  A vessel contains 20 litres of a mixture of oil and water in the ratio 3 : 2. How many litres of water must be added so that the ratio of oil to water in the vessel becomes 3 : 4?

  • (a)4
  • (b)6
  • (c)8
  • (d)10

  Answer(c) 8 — the vessel starts with 3/5 of 20 = 12 litres of oil and 2/5 of 20 = 8 litres of water. Adding water changes nothing about the oil, so the 12 litres of oil must end up as three parts of a 3 : 4 split, making one part 4 litres and the water 16 litres. The water must therefore rise from 8 litres to 16 litres, so 8 litres are added. Holding the unchanged ingredient fixed is the whole technique for this shape of question.