In a race of 1 km, A can beat B by 40 m and B can beat C by 50 m. With how much distance can A beat C in a 0·5 km race ?
- (a)42 m
- (b)43 m
- (c)44 m
- (d)45 m
Answer
Why
Correct — C, (c) 44 m.
THE STEP THAT DECIDES THIS QUESTION is that margins in a chained race COMPOSE MULTIPLICATIVELY, not additively. Forty plus fifty is ninety, and ninety is wrong. Here is why.
FIRST LINK. Over 1 km, A beats B by 40 m: when A has run 1000 metres, B has run 960. So B's speed is 960/1000 of A's, that is 0·96 of it.
SECOND LINK. Over 1 km, B beats C by 50 m: when B has run 1000 metres, C has run 950. So C's speed is 950/1000 of B's, that is 0·95 of it.
CHAIN THEM. C's speed compared with A's is the PRODUCT of the two fractions, because C is 0·95 of B and B is 0·96 of A:
0·96 × 0·95 = 0·912
So when A runs 1000 metres, C runs 912, and A beats C by 88 metres over a kilometre — not by 90.
SCALE TO THE RACE ACTUALLY ASKED ABOUT. The margin between two runners of fixed speeds is proportional to the length of the race, because the ratio of their speeds does not change. Over half the distance the margin is half as large:
88 ÷ 2 = 44 metres over 0·5 km
and that is option (c).
WHERE THE MISSING TWO METRES GO. The margins would add exactly if C's loss were measured against the full course; instead it is measured against the shorter distance B has already covered. When A has finished, B is at 960, and C is 5 per cent behind B's 960 rather than 5 per cent behind 1000 — that is 48 metres behind B, not 50. So the combined margin is 40 + 48 = 88.
THIS IS THE SUCCESSIVE-PERCENTAGE RULE IN DISGUISE. Express the margins as percentages: A beats B by 4 per cent and B beats C by 5 per cent. The combined shortfall is
4 + 5 − (4 × 5)/100 = 9 − 0·2 = 8·8 per cent
the same formula used for two successive discounts. Over 1 km that is 88 metres; over 0·5 km it is 44. The interaction term −0·2 per cent is small, and it is always NEGATIVE, so the true margin is always a little LESS than the sum of the two.
THAT DIRECTION IS THE WHOLE OPTION LIST. Half of the naive sum is 45 metres, and the correction over half a kilometre is 0·2 per cent of 500, which is exactly 1 metre. The four options are 45, 44, 43 and 42 — the uncorrected figure, the figure with the correction applied once, and then the same correction applied twice and three times over. Knowing that the correction exists, that it is downward, and that it is small, identifies 44 without any multiplication at all.
The distance in the stem is printed as 0·5 km with a raised middle dot, and the option units are plain metres.
Why the others are wrong
- (a)42 m — Forty-two metres is three units below the naive figure of 45, which means the correction has been applied three times over. There is no route to it by any consistent chaining of the two given margins: it would require A to beat C by 8·4 per cent over the course, whereas the product of the two ratios gives 8·8 per cent exactly. It is worth seeing how the option list is built, because the construction recurs. Half the additive answer is 45 metres; the true interaction term costs exactly 1 metre over half a kilometre; and the four options are 45, 44, 43 and 42, that is the uncorrected value followed by the same correction applied once, twice and three times. A candidate who has grasped that a downward adjustment is needed but not how large it is will find several plausible-looking homes for his uncertainty. The remedy is to compute the product 0·96 × 0·95 rather than to estimate the adjustment, since the multiplication takes no longer than the guess.
- (b)43 m — Forty-three metres is two units below the naive figure, so it corresponds to applying the correction twice — a combined margin of 8·6 per cent instead of 8·8. Like the option below it, it cannot be produced by any correct handling of the two ratios. Its presence on the list is instructive about how these items are set: the setter has arranged the options so that a candidate who knows the additive answer is wrong, but does not know by how much, gains nothing from that knowledge. Three of the four options lie below 45, and only one of them is right. The lesson is that a qualitative insight — margins do not simply add — has to be completed by the quantitative one. Multiply the two fractions, 0·96 × 0·95 = 0·912, read the shortfall as 0·088 of the course, and apply it to the 500 metres actually asked about. The whole computation is one multiplication and one halving.
- (d)45 m — Forty-five metres is the additive answer and the trap this question was built for. It comes from adding the two margins to get 40 + 50 = 90 metres over a kilometre and then halving for the half-kilometre race. The addition is wrong because the two margins are measured against DIFFERENT baselines. C's fifty-metre deficit is a deficit behind B over a full kilometre; but when A finishes, B has run only 960 metres, and C is therefore 5 per cent of 960 behind him, which is 48 metres and not 50. Adding 40 and 48 gives the true 88. The same point in percentage terms is the familiar successive-discount rule: two reductions of 4 per cent and 5 per cent combine to 8·8 per cent, never to 9, because the second reduction applies to what is left after the first. That interaction term is always negative, so the additive answer is always an overstatement — which means this option can be rejected on direction alone, before its size is even considered.
Concept
CHAINED COMPARISONS COMPOSE BY MULTIPLICATION, and this is one of the most transferable ideas in the whole quantitative syllabus. Whenever a quantity is expressed as a fraction of a second quantity, which is in turn a fraction of a third, the relation between the first and the third is the PRODUCT of the two fractions. It never is their sum, and it is only approximately their sum when both fractions are close to one.
In this problem the fractions are speeds. B runs at 0·96 of A's speed and C at 0·95 of B's, so C runs at 0·96 × 0·95 = 0·912 of A's. The shortfall of 0·088 is the margin, expressed as a fraction of the course.
THE SAME STRUCTURE WEARING OTHER CLOTHES:
SUCCESSIVE DISCOUNTS. Twenty per cent off, then ten per cent off, leaves 0·8 × 0·9 = 0·72 of the marked price, an effective discount of 28 per cent rather than 30. SUCCESSIVE PERCENTAGE CHANGES. A rise of 10 per cent followed by a fall of 10 per cent leaves 1·1 × 0·9 = 0·99, a net LOSS of 1 per cent — the classic result that a rise and an equal fall do not cancel. POPULATION OR PRICE GROWTH over two periods, which is compounding. EFFICIENCY CHAINS, where one machine works at a fraction of another's rate and that one at a fraction of a third's.
The general formula for two successive percentage changes of a and b per cent is a + b + ab/100, with the sign of each term following the direction of its change. For two reductions both terms are negative and the product term is positive relative to them, which is why two discounts always combine to less than their sum. For a reduction and a rise the product term works the other way. Rather than memorising sign rules, it is safer to multiply the multipliers: 0·96 × 0·95 leaves nothing to get wrong.
THE SECOND IDEA IS THAT A MARGIN SCALES WITH THE COURSE. Two runners with fixed speeds keep a constant RATIO between them, so the gap between them at the finish is proportional to how far they have run. If A beats C by 88 metres over a kilometre, he beats him by 44 over half a kilometre, by 8·8 over a hundred metres, and by 176 over two kilometres. That proportionality is what allows a result obtained for a convenient distance to be transferred to the distance the question asks about, which is usually the quickest way to organise the working: solve the chain over the distance in which the data are given, then rescale once at the end.
This item and the one printed immediately before it are the paper's pair on races, and they are pitched at different skills. The earlier one tests careful translation of a head start and a time margin; this one tests whether the candidate knows that comparisons chain multiplicatively. The second is the more conceptual of the two and the more widely useful, because the same idea governs successive discounts, compound growth and every problem in which one quantity is described relative to another that is itself described relatively.
The option list is the most carefully built on this stretch of the paper, and it is worth studying as a specimen. The four values are 45, 44, 43 and 42 metres, one metre apart. Forty-five is the answer obtained by adding the margins; the true correction is exactly one metre over the half-kilometre; and the remaining two options are that same correction applied twice and three times. So the list is constructed to defeat both the candidate who does not know a correction is needed and the candidate who knows one is needed but guesses its size. There is no way through except to compute the product.
Two habits follow. First, when a stem gives a relationship in stages, write each stage as a MULTIPLIER — 0·96, 0·95 — rather than as a difference, and multiply. Second, if a question asks about a distance different from the one in which the data are given, solve in the data's units and rescale once at the end; converting at every step invites arithmetic slips.
The paper prints the shorter distance as 0·5 km with a raised middle dot rather than a full stop, and gives the margins and the options in plain metres.
Key facts
- A beats B by 40 m in 1 km, so B runs 960 metres while A runs 1000 and B's speed is 0·96 of A's.
- B beats C by 50 m in 1 km, so C's speed is 950/1000 = 0·95 of B's.
- Chaining the two gives C at 0·96 × 0·95 = 0·912 of A's speed, so A beats C by 88 metres over a kilometre and not by 90.
- A margin between runners of fixed speeds is proportional to the length of the race, so over 0·5 km the margin is half of 88, that is 44 metres.
- In percentage terms the margins are 4 and 5 per cent and they combine by the successive-percentage rule to 4 + 5 − (4 × 5)/100 = 8·8 per cent.
- The interaction term is always negative for two shortfalls, so the true combined margin is always a little LESS than the sum of the two margins.
- The missing two metres arise because C's five per cent deficit is measured against the 960 metres B has covered, not against the full 1000 — 5 per cent of 960 is 48, and 40 + 48 = 88.
- The four options are the naive 45 metres and the same one-metre correction applied once, twice and three times, so guessing the size of the correction cannot succeed.
- The same multiplicative rule governs successive discounts, compound growth, and the classic result that a 10 per cent rise followed by a 10 per cent fall leaves a net loss of 1 per cent.
Study next
Common traps
- Adding the two margins to get 90 metres. They are measured against different baselines and compose by multiplication instead.
- Knowing that a correction is needed but guessing its size. Three of the four options lie below the additive figure and only one is right.
- Forgetting to rescale to the half-kilometre race after solving the chain over a kilometre, or rescaling twice.
- Measuring C's deficit against the full course. When A finishes, B has run 960 metres and C is five per cent behind that, which is 48 metres.
- Treating a percentage of a percentage as a simple sum, which is the same error in the setting of discounts, growth rates and efficiency chains.
Chained-comparison problems are a favourite of quantitative setters because a single stem can test proportional reasoning, multiplicative composition and unit scaling all at once, and because the additive answer is so easy to construct as an option. On EPFO papers they appear as races, as successive discounts on a marked price, as compound growth over two periods, and occasionally as work-and-efficiency chains.
Expect the naive additive figure to be on the option list every time. Expect the other options to be clustered near it rather than scattered, so that a candidate who knows only the direction of the correction gains nothing. And expect the distance, price or period asked about to differ from the one in which the data are supplied, which adds a scaling step at the end.
The preparation that pays is to internalise the multiplier habit. Convert every relative statement into a multiplier as soon as it is read — beaten by 4 per cent becomes 0·96, a discount of 20 per cent becomes 0·8, a rise of 10 per cent becomes 1·1 — and then multiply the multipliers. That single discipline handles the entire family without any formula, keeps the direction of every change correct, and reduces most of these questions to one multiplication and one subtraction from unity.
Related PYQs
EPFO_APFC_2016_Q103A and B run a 1 km race. A gives B a start of 50 m and still beats him by 15 seconds. If A runs at 8 km/h, what is the speed of B ?
- (a) 4·4 km/h
- (b) 5·4 km/h
- (c) 6·4 km/h
- (d) 7·4 km/h
Answer(d) 7·4 km/h
The race item printed immediately before this one, where a 50 metre head start and a 15 second margin have to be converted into a speed.
EPFO_APFC_2016_Q113If 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%
A percentage-change item on this paper — halving a circle's radius cuts its area by three-quarters — where the effect again compounds rather than adding.
EPFO_APFC_2016_Q93A man buys apples at a certain price per dozen and sells them at 8 times that price per hundred. What percentage does he gain or lose ?
- (a) 4% profit
- (b) 6% profit
- (c) 4% loss
- (d) 6% loss
Answer(c) 4% loss
The apples item on this paper, where a profit-and-loss comparison must be made between counts expressed in different units before any percentage is taken.
Practice
- practice — not a real PYQ
In a race of 100 metres, A beats B by 10 metres and B beats C by 10 metres. By how much does A beat C over the same 100 metres ?
- (a)18 m
- (b)19 m
- (c)20 m
- (d)21 m
Answer(b) 19 m — B runs at 0·9 of A's speed and C at 0·9 of B's, so C runs at 0·81 of A's. When A finishes the 100 metres C has covered 81, and the margin is 19 metres rather than the 20 that adding the two margins would suggest.
- practice — not a real PYQ
Two successive discounts of 20 per cent and 10 per cent are allowed on the marked price of an article. What single discount is equivalent to the two of them together ?
- (a)26 per cent
- (b)28 per cent
- (c)30 per cent
- (d)32 per cent
Answer(b) 28 per cent — the two reductions leave 0·8 × 0·9 = 0·72 of the marked price, so 28 per cent has been taken off in all. The sum of 30 per cent overstates it, because the second discount applies only to what is left after the first.