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 the ratio of speeds of ‘A’ and ‘B’ is 5 : 6 and ‘B’ allows ‘A’ a start of 70 metres in a 1·2 km race, who will win the race and by what distance ?

(a)‘A’ wins by 30 m.
(b)‘B’ wins by 200 m.
(c)‘B’ wins by 130 m.
(d)The race finishes in a dead heat.

Answer

Why

Correct — C, (c) ‘B’ wins by 130 m. First convert: 1·2 km is 1200 metres, and the booklet's raised middle dot is an ordinary decimal point. A start of 70 metres means A begins 70 metres up the track, so A has 1200 − 70 = 1130 metres of running to do while B has the full 1200. Now settle the race in the time B takes to finish, because that is the instant the result is decided. Speeds in the ratio 5 : 6 means that over any equal interval of time the distances covered are also in the ratio 5 : 6, so while B covers 1200 metres A covers five-sixths of that, which is 1000 metres. Add A's head start and A stands at 70 + 1000 = 1070 metres from the start line when B breasts the tape. B has won, and A is 1200 − 1070 = 130 metres short of the finish. The same result in one line, and the form worth memorising: over 1200 metres B's extra speed is worth 1200 x (6 − 5) / 6 = 200 metres; the start returns 70 of those to A; 200 − 70 = 130 metres is what is left of B's advantage.

Why the others are wrong

(a)‘A’ wins by 30 m. — A cannot win on these numbers, and the threshold is worth knowing exactly rather than judging by feel. With speeds 5 : 6 over 1200 metres, B's advantage is worth 200 metres; A therefore needs a start of more than 200 metres to win, exactly 200 to tie, and anything less to lose. Seventy metres is barely a third of the requirement. A ratio as close as 5 : 6 makes a 70-metre start sound generous, which is precisely why the ratio has to be converted into metres of this particular race before it is judged — the same 5 : 6 over a 300-metre race would only need a 50-metre start. Nothing in the arithmetic produces a margin in A's favour, so there is no reading of the question on which this option stands.
(b)‘B’ wins by 200 m. — 200 metres is the margin B wins by with no start at all — 1200 x (6 − 5) / 6 — so this option is the answer to the question with its one complication removed. It is what you get by computing B's advantage over the distance and then never spending the head start. The number is worth holding on to for a different reason: on this race 200 metres carries two meanings at once. It is B's winning margin from a level start, and it is also the exact start that would produce a dead heat. Being clear about which of the two you have just calculated is the whole discipline of these items.
(d)The race finishes in a dead heat. — A dead heat is an exact condition, not a near miss, and it needs the start to cancel B's advantage precisely: start = distance x (difference of the ratio terms) / (larger term) = 1200 x 1 / 6 = 200 metres. The paper gives 70. This option is tempting because 5 : 6 is a narrow ratio and 70 metres in a race of over a kilometre feels like it ought to even things out, but a ratio does not work like a fixed handicap — it scales with the distance, so the longer the race the larger the start A needs. If your working does not land on the dead-heat identity exactly, the race is not tied.

Concept

Race problems are ratio problems with one piece of vocabulary attached. 'A gives B a start of x metres' means B begins x metres ahead of the start line and so runs (D − x) metres while A runs the full D. 'A beats B by x metres' means that when A finishes, B is still x metres from the line. The single fact that solves all of them is that in equal time the distances covered stand in the same ratio as the speeds — if speeds are 5 : 6 then in any shared interval the distances are 5 : 6. So the method never changes: pick the moment the faster runner finishes, work out where the other one is at that instant, and compare. From this one identity falls out: the start that produces a dead heat over a distance D with speeds in the ratio m : n (n larger) is D x (n − m) / n, and any start smaller than that loses while any larger one wins. Note that the handicap is proportional to the distance — a ratio is not a fixed number of metres, which is why the race length has to be in the calculation from the start.

This is the second of the paper's quantitative items and it is built to punish reading rather than arithmetic. Three things have to be got right before any sum is done: that 1·2 km is 1200 metres, that the start belongs to A and shortens A's run rather than lengthening B's, and that the ratio has to be applied over the full 1200 metres of B's run and not over A's shortened 1130. Every wrong option on the page corresponds to getting one of those three steps wrong while doing the arithmetic perfectly. EPFO stems mix units deliberately — this one spells 'metres' out in the stem, abbreviates 'km' beside it, and then uses 'm' in the options.

Key facts

Speeds in the ratio 5 : 6 mean distances covered in equal time are also in the ratio 5 : 6.
1·2 km = 1200 metres; a 70-metre start leaves A 1130 metres to run and B the full 1200.
When B finishes, A has run 5/6 of 1200 = 1000 metres and stands at the 1070-metre mark, 130 metres short.
B's advantage over the whole race is 1200 x (6 − 5) / 6 = 200 metres, from which the 70-metre start is subtracted.
The dead-heat start for speeds m : n over distance D is D x (n − m) / n — here 1200 x 1/6 = 200 metres.
A start of more than 200 metres would let A win; exactly 200 ties; less than 200 loses.
'A beats B by x metres' and 'A gives B a start of x metres' are different statements and are set up differently.

Settle the race at the instant the faster runner finishes

Convert, and set the two distances out→1·2 km is 1200 metres. A start of 70 metres means A begins 70 metres up the track, so A has 1200 − 70 = 1130 metres of running to do while B has the full 1200.

In equal time, distances stand in the ratio of the speeds→5 : 6. So in the time B covers 1200 metres, A covers five-sixths of that — 1000 metres.

Where A stands when B breasts the tape→70 + 1000 = 1070 metres from the start line. B has won, and A is 1200 − 1070 = 130 metres short of the finish.

The same result in one line→Over 1200 metres B’s extra speed is worth 1200 x (6 − 5) / 6 = 200 metres. The start returns 70 of those to A, and 200 − 70 = 130 metres is what is left of B’s advantage.

The dead-heat threshold, worth memorising→For speeds m : n with n the larger, over a distance D, the start that ties the race is D x (n − m) / n — here 1200 x 1/6 = 200 metres. More than 200 and A wins; exactly 200 and it is a dead heat; less and A loses. Seventy metres is barely a third of the requirement, which is why both ‘A wins by 30 m’ and ‘dead heat’ fail: a ratio is not a fixed handicap, it scales with the distance.

Why 200 is on the page→It is B’s winning margin from a LEVEL start — this question with its one complication removed, the advantage computed and then never spent. On this race 200 metres carries two meanings at once, which is what makes it the best-built distractor in the set. Note also that ‘A beats B by x metres’ and ‘A gives B a start of x metres’ are different statements and are set up differently.

Study next

Races expressed as a start in seconds rather than metres, and how to convert between the twoCircular-track meeting and overtaking problems, where the same ratio idea gives the number of meeting pointsTime, speed and distance with a fixed distance, where speed and time vary inverselyRelative speed for two bodies moving in the same and in opposite directionsWork-and-time problems, which are the same proportionality dressed as rates of output

Common traps

Leaving 1·2 km as 1·2 and comparing it with a start measured in metres.
Giving the start to the wrong runner, so that B is made to run 1270 metres instead of A running 1130.
Applying the ratio to A's shortened 1130 metres rather than to the full 1200 that B runs.
Computing B's advantage of 200 metres and reporting it as the answer without spending the 70-metre start.
Assuming a close ratio implies a close race. The handicap scales with the distance, so 5 : 6 over 1200 metres is worth 200 metres.

The quantitative block of an EPFO EO/AO paper usually carries one time-speed-distance item, and races with a start or a beat are the favoured dressing because they let a single ratio be tested through a wordy stem. Options are short phrases naming a winner and a margin, so a candidate who gets the winner right and the margin wrong still scores nothing — the arithmetic has to be completed, not estimated.

Related PYQs

EPFO_EOAO_2020_Q83Open & attempt →

‘M’ is 60 years old. ‘R’ is 5 years junior to ‘M’ and 4 years senior to ‘V’. The youngest brother of ‘V’ is ‘B’ and he is 6 years junior to ‘V’. What is the age difference between ‘M’ and ‘B’ ?

(a) 18 years
(b) 15 years
(c) 13 years
(d) 11 years

Answer(b) 15 years

The next item of this quantitative block, and the same discipline from a different angle: a chain of comparative statements that has to be turned into numbers in the right direction before any subtraction is attempted. On both, the arithmetic is trivial and the reading is the question.

Practice

practice — not a real PYQ
The ratio of the speeds of ‘A’ and ‘B’ is 3 : 4. What start must ‘B’ give ‘A’ in a 1 km race so that the race ends in a dead heat ?
- (a)200 m
- (b)250 m
- (c)300 m
- (d)400 m
Answer(b) 250 m
practice — not a real PYQ
In a 500 metre race, ‘A’ beats ‘B’ by 50 metres. What is the ratio of the speed of ‘A’ to the speed of ‘B’ ?
- (a)9 : 10
- (b)10 : 9
- (c)11 : 10
- (d)5 : 4
Answer(b) 10 : 9

Why