A 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 ?

Correct — D, (d) 7·4 km/h.

THE STEP THAT DECIDES THIS QUESTION is to convert the two advantages — a 50 metre head start and a 15 second margin — into a distance and a time for B, and then divide.

WHAT A DOES. The race is 1 km and A runs it at 8 km/h, so

A's time = 1000 m ÷ 8000 m per hour = 1/8 hour = 7·5 minutes = 450 seconds.

WHAT B DOES. "A gives B a start of 50 m" means B begins 50 metres up the course, so B has only 950 metres to cover. "Still beats him by 15 seconds" means that when A crosses the line, B is still 15 seconds from finishing, so

B's time = 450 + 15 = 465 seconds for 950 metres.

B'S SPEED. Distance over time, converted to kilometres per hour:

950 m in 465 s → 950 × 3600 / 465 metres per hour = 7354·8 m/h ≈ 7·35 km/h

which the option list rounds to 7·4 km/h. A handy form of the same calculation: for a distance of 950 metres, the speed in km/h is 3420 divided by the time in seconds, since 0·95 km × 3600 = 3420. Here 3420 ÷ 465 = 7·355.

THE ESTIMATE THAT ANSWERS THIS IN FIVE SECONDS, and which is the most useful thing on this card. A takes 450 seconds. B is only 15 seconds behind, and he was given a 50 metre head start into the bargain. So B is a LITTLE slower than A, not a lot slower — his time is 465 seconds against 450, a difference of about three per cent, and his head start makes up part even of that. A speed a little below 8 km/h is the only sensible answer, and exactly one option lies within a kilometre per hour of 8. The four options are spaced a full km/h apart — 4·4, 5·4, 6·4, 7·4 — so no fine judgement is needed at all.

BACK-SUBSTITUTION, which turns the estimate into a proof. For a speed v in km/h, B's time over 950 metres is 3420/v seconds, and his losing margin is that minus A's 450 seconds:

v = 4·4 → 777 s → beaten by about 5 minutes 27 seconds v = 5·4 → 633 s → beaten by about 3 minutes 3 seconds v = 6·4 → 534 s → beaten by about 1 minute 24 seconds v = 7·4 → 462 s → beaten by about 12 seconds

Only the last is anywhere near the 15 seconds the stem states. The small shortfall — 12 seconds rather than 15 — is simply the effect of rounding: the exact speed is 7·355 km/h, and the paper offers figures to one decimal place, so 7·4 is the nearest available.

A SECOND ROUTE, by scaling B up to the full distance. B covers 950 metres in 465 seconds, so he would cover the full kilometre in 465 × 1000/950 = 489·5 seconds. Comparing like with like, the two men's times over 1 km are 450 and 489·5 seconds, and since speed is inversely proportional to time over a fixed distance, B's speed is 8 × 450/489·5 = 7·35 km/h. The same answer, reached without ever converting to metres per second.

The decimals in all four options are printed with a raised middle dot, and the unit is set as km/h with a solidus and no spaces, as this booklet does throughout.

Why the others are wrong

• (a)4·4 km/h — At 4·4 km/h B would take 3420 ÷ 4·4 = about 777 seconds to cover his 950 metres, against A's 450 seconds — he would be beaten by more than five and a half minutes, not by fifteen seconds. The figure is barely more than half A's speed, which is the pace of a brisk walk against a run, and no head start of fifty metres in a race of a thousand could disguise a gap that size. The option is worth dwelling on only because it shows how far an answer can drift when the two adjustments are applied in the wrong direction. A candidate who adds the fifty metres to B's distance instead of subtracting it, and who adds the fifteen seconds to A's time as well, compounds two errors that both work the same way. The guard against it is the estimate: settle first whether B is a little slower or much slower than A, and only then compute. Fifteen seconds lost out of four hundred and fifty is a small margin, and small margins mean speeds that are close together.
• (b)5·4 km/h — At 5·4 km/h B's 950 metres would take about 633 seconds, so he would trail A by more than three minutes. Like the lowest option this is not a near miss but a different order of result, and it can be rejected without arithmetic by anyone who has read the stem carefully. The phrase that fixes the scale is "still beats him by 15 seconds": fifteen seconds is a margin at the finish of a close race, and the whole race lasts seven and a half minutes for the winner. A loser who is two-thirds as fast as the winner does not finish fifteen seconds behind; he finishes when the winner has long gone. It is worth noticing that the option list here is an evenly spaced ladder rising by exactly one kilometre per hour at each step, which means a candidate who has merely estimated the answer as "a bit under eight" can pick it out with complete confidence, while a candidate who has computed a wrong figure will find a home for it somewhere on the ladder.
• (c)6·4 km/h — At 6·4 km/h B would need about 534 seconds for his 950 metres and would lose by roughly a minute and twenty-four seconds — nearly six times the stated margin. This is the nearest of the three wrong options and therefore the one most worth checking properly. Six point four is exactly eight-tenths of A's speed, so B would be twenty per cent slower; over a race in which A takes 450 seconds, a twenty per cent deficit costs about ninety seconds even before the head start is allowed for, and the head start is worth only about twenty-three seconds at B's pace. The arithmetic never comes close to fifteen. A candidate might arrive here by forgetting the head start altogether and dividing 1000 metres by 465 seconds, but that gives about 7·7 km/h, which is nearer the answer, not further from it — so this option is not the product of any single natural slip, and back-substitution disposes of it at once.

RACES ARE PROBLEMS ABOUT TWO RUNNERS COMPARED AT THE SAME MOMENT, and the whole topic depends on translating the language of racing into distances and times.

THE VOCABULARY, which is where most errors begin:

"A BEATS B BY x METRES" — when A finishes, B is x metres short of the line. So in the time A covers the full course, B covers (course − x). This is a comparison at a fixed TIME. "A BEATS B BY t SECONDS" — B crosses the line t seconds after A. This is a comparison at a fixed DISTANCE. "A GIVES B A START OF x METRES" — B begins x metres up the course and has only (course − x) to run. "A GIVES B A START OF t SECONDS" — B sets off t seconds before A. "A DEAD HEAT" — both finish together.

This item combines a metre start with a second margin, which is what makes it worth more than a routine sum: B's DISTANCE is reduced by the head start and his TIME is increased by the margin, and the two adjustments are made to different quantities. Handling them in the same breath is where candidates go wrong, so it is worth writing B's distance and B's time on separate lines before dividing.

THE UNDERLYING RELATION is the familiar one, speed = distance ÷ time, together with the fact that over a FIXED distance speed is inversely proportional to time. That second fact gives the scaling route used above: bring both runners onto the same distance, compare their times, and invert the ratio to get the ratio of speeds.

UNIT DISCIPLINE deserves its own note, because this problem mixes kilometres per hour with metres and seconds. Two conversions cover almost everything: 1 km/h = 5/18 m/s, and 1 m/s = 18/5 = 3·6 km/h. A useful derived shortcut for a fixed distance d in kilometres: the speed in km/h equals 3600 d divided by the time in seconds. With d = 0·95 that is 3420 divided by the time, which is the form used here.

FINALLY, ESTIMATION AS A METHOD RATHER THAN A LAST RESORT. In a race problem the two speeds are usually close together, because a race in which one runner is twice as fast as the other is not a race. So the answer is nearly always near the given speed, and an option list spread over several kilometres per hour is answering itself. Making the estimate first also protects against the direction errors — adding when one should subtract — that produce wildly wrong figures rather than near misses.

Key facts

• A covers 1 km at 8 km/h, so A's time is 1/8 hour, that is 450 seconds — the reference against which everything else is measured.
• A start of 50 m means B runs only 950 m; a margin of 15 seconds means B's time is 450 + 15 = 465 seconds.
• B's speed is therefore 950 metres in 465 seconds, which is 950 × 3600/465 = about 7354 metres per hour, or 7·35 km/h, rounded on the option list to 7·4.
• For a fixed distance d in kilometres, the speed in km/h is 3600d divided by the time in seconds; with d = 0·95 that is 3420 divided by the time.
• The estimate settles the item on its own: a loser only 15 seconds behind over a 450 second race must be only slightly slower, and just one option lies near 8 km/h.
• Back-substitution confirms it — the four options imply losing margins of about 327, 183, 84 and 12 seconds respectively, and only the last is near the stated 15.
• A start in METRES reduces the loser's distance while a margin in SECONDS increases his time, and the two adjustments apply to different quantities.
• Scaling B up to the full kilometre gives 489·5 seconds against A's 450, and since speed is inversely proportional to time, B's speed is 8 × 450/489·5 = 7·35 km/h.
• The conversions that carry this topic are 1 km/h = 5/18 m/s and 1 m/s = 3·6 km/h.

Common traps

• Giving B the full 1000 metres. A start of 50 m means he runs 950, and forgetting it inflates his computed speed.
• Applying the head start to B's time or the margin to B's distance. The metres change the distance and the seconds change the time.
• Abandoning a computed value of 7·35 because no option matches exactly. The options are a full km/h apart and the nearest is intended.
• Skipping the estimate. A margin of 15 seconds in a race lasting 450 seconds means the speeds must be close, which alone identifies the answer.
• Mixing units mid-calculation. Either work throughout in metres and seconds and convert once at the end, or work throughout in kilometres and hours.

• practice — not a real PYQ

  In a race of 1 km, A beats B by 100 metres. If A runs at a steady 10 km/h throughout, what is B's speed in the same race ?

  • (a)8 km/h
  • (b)9 km/h
  • (c)10 km/h
  • (d)11 km/h

  Answer(b) 9 km/h — beating by a distance is a comparison at the same instant, so in the time A covers 1000 metres B covers 900. Their speeds are therefore in the ratio 1000 : 900, and nine-tenths of 10 km/h is 9 km/h.

• practice — not a real PYQ

  In a race of 500 metres, A finishes the course in 100 seconds and beats B by 20 seconds. What is B's speed over that race ?

  • (a)12 km/h
  • (b)15 km/h
  • (c)18 km/h
  • (d)20 km/h

  Answer(b) 15 km/h — beating by a time is a comparison at the same distance, so B covers the full 500 metres in 120 seconds. That is 500/120 metres per second, which multiplied by 3·6 gives exactly 15 km/h.