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Races and Games for CUET Applied Maths

About 5 min to learn · 73 questions

Races and games are part of Section B2 of CUET Maths, the Applied Mathematics section you answer if you choose Applied Mathematics instead of Section B1. Every question compares two players by distance, time or speed, so one habit, reading the margin correctly, settles most of them. Learn each rule here, from beating a rival by a distance or a time to giving a start, dead heats, games of points and runners on a circular track, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area
Section B2: Applied Mathematics
On this page · 9 sections
  1. 01Beating by a distance or a time
  2. 02Giving a start and dead heats
  3. 03Games of points
  4. 04Circular tracks
  5. 05How CUET asks this chapter
  6. 06Exam shortcuts
  7. 07Common mistakes
  8. 08Quick revision
  9. 09Check your understanding

Beating by a distance or a time

At a school sports day, Meera wins the 500 m race and Tara is 25 m behind her at the line. That one sentence tells us how their speeds compare, and from it how a longer race, or a race with a head start, would end.

What does "A beats B by x metres" tell you?

It means that while A runs the whole race of dd metres, B runs only d−xd - x metres in the same time. Equal times make the speeds proportional to the distances, so A's speed : B's speed =d:(d−x)= d : (d - x).

Example 1 · Meera and Tara

In a 500 m race Meera beats Tara by 25 m. What is the ratio of their speeds?

  1. While Meera runs 500 m, Tara runs 500−25=475500 - 25 = 475 m.

  2. Speeds: 500:475=20:19500 : 475 = 20 : 19.

Answer

Meera : Tara =20:19= 20 : 19.

"A beats B by tt seconds" means B reaches the line tt seconds after A. When a margin is given both ways, "by 25 m or 5 seconds", the loser covers that distance in that time, which gives the loser's speed at once.

Watch out

Two margins do not simply add. If A beats B by 15 m and B beats C by 20 m in 200 m races, then while A runs 200 m, B runs 185 m and C runs 180200×185=166.5\frac{180}{200} \times 185 = 166.5 m, so A beats C by 33.5 m, not 35.

Giving a start and dead heats

What is a dead heat, and how much start gives one?

A dead heat is a race in which both runners reach the line together. The faster runner A gives the slower runner B a start of exactly the distance B would lose by over the full race: while A runs dd, B runs d×B’s speedA’s speedd \times \frac{\text{B's speed}}{\text{A's speed}}, and the rest is the start.

Example 2 · A start for a dead heat

Tara's speed is 45\frac{4}{5} of Meera's. How much start should Meera give her in a 1,000 m race for a dead heat?

  1. While Meera runs 1,000 m, Tara runs 45×1,000=800\frac{4}{5} \times 1{,}000 = 800 m.

  2. So Tara should start 200 m ahead.

Answer

A start of 200 m.

Games of points

What does "A can give B 10 points in a game of 50" mean?

It means that when A scores 50 points, B scores 50−10=4050 - 10 = 40 in the same time, so their scoring rates are 50:40=5:450 : 40 = 5 : 4. Games of points are races in which the distance is counted in points.

With three players, chain the ratios: if A : B and B : C are known, A : C is their product, and the points one can give another follow from it.

Circular tracks

When do two runners on a circular track meet?

Running the same way, they meet each time the faster has gained one full lap, so the time between meetings is lap ÷ (difference of speeds). Running opposite ways, they meet each time they have covered one lap between them: lap ÷ (sum of speeds).

Example 3 · A 300 m track

Two runners start together on a 300 m track at 6 m/s and 4 m/s. When do they first meet, running the same way, and running opposite ways?

  1. Same way: 3006−4=150\frac{300}{6 - 4} = 150 seconds.

  2. Opposite ways: 3006+4=30\frac{300}{6 + 4} = 30 seconds.

Answer

150 seconds and 30 seconds.

To find when both are back at the starting point together, take the LCM of their lap times.

How CUET asks this chapter

  • Margins: a speed ratio from a margin, a margin from two times, a time from a margin given both ways.
  • Starts: the start for a dead heat, starts between three runners, who wins with a given start.
  • Games: points one player can give another, rates from points, chaining two games.
  • Tracks: first meeting the same way or opposite ways, meeting at the start, how many passes in a given time.

Exam shortcuts

  • A beats B by xx in dd: speeds d:(d−x)d : (d - x).
  • Margin given as distance and time: loser's speed = distance ÷ time.
  • Dead-heat start = the distance B loses by over the full race.
  • Track: same way, lap ÷ difference; opposite ways, lap ÷ sum.

Common mistakes

  • Adding margins from two different races.
  • Turning a speed ratio upside down.
  • Using a sum of speeds for runners going the same way.
  • Counting a meeting that falls after the time asked about.

Quick revision

Races

  • Equal times: speeds ∝ distances
  • Start for a dead heat = margin over the full race

Games and tracks

  • Points: same as metres in a race
  • Track: difference of speeds (same way), sum (opposite)

Next in Section B2: comparing quantities with numerical inequalities.

Next · 08Quick revision

Check your understanding

73 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Beating by a distance or a timeQ1 of 73Q1

Beating by a distance or a time

26 questions

In a 100 m race, A beats B by 20 m. What is the ratio of A's speed to B's?

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AnswerB · 5 : 4

  1. While A runs 100 m, B runs 100−20=80100 - 20 = 80 m in the same time.

  2. In equal times, speeds are in the ratio of distances: 100:80=5:4100 : 80 = 5 : 4.

Worth noting

4 : 5 is B's speed to A's, the ratio turned round.

A runs 1.5 times as fast as B. By how many metres does A beat B in a 300 m race?

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AnswerA · 100 m

  1. While A runs 300 m, B runs 3001.5=200\frac{300}{1.5} = 200 m.

  2. A wins by 300−200=100300 - 200 = 100 m.

Worth noting

150 m takes half of 300, as if A were twice as fast.

A runs 65\frac{6}{5} times as fast as B. By how many metres does A beat B in a 600 m race?

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AnswerB · 100 m

  1. When A runs 600 m, B runs 600×56=500600 \times \frac{5}{6} = 500 m.

  2. A wins by 600−500=100600 - 500 = 100 m.

Worth noting

120 m takes 15\frac{1}{5} of the race; the margin is the 16\frac{1}{6} of A's 600 m that B has not covered.

In a 1,000 m race A beats B by 100 m, and in another 1,000 m race B beats C by 50 m. The runners keep the same speeds in every race.

By how many metres would B beat C in a 400 m race?

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AnswerB · 20 m

  1. C runs 950 m for every 1,000 m that B runs, a margin of 5%.

  2. 5% of 400 m is 20 m.

Worth noting

50 m keeps the 1,000 m margin; it scales with the length of the race.

In a 200 m race, A finishes in 25 seconds and beats B by 40 m.

What is A's speed in this 200 m race?

Options for question 5

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AnswerD · 8 m/s

  1. 200 m in 25 seconds.

  2. 20025=8\frac{200}{25} = 8 m/s.

Worth noting

6.4 m/s is B's speed.

In a 1,000 m race A beats B by 100 m, and in another 1,000 m race B beats C by 50 m. The runners keep the same speeds in every race.

What is the ratio of A's speed to B's in these races?

Options for question 6

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AnswerD · 10 : 9

  1. While A runs 1,000 m, B runs 900 m.

  2. 1,000:900=10:91{,}000 : 900 = 10 : 9.

Worth noting

9:109 : 10 is B's speed to A's.

In a 400 m race, A beats B by 50 m. What is the ratio of A's speed to B's?

Options for question 7

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AnswerB · 8 : 7

  1. While A runs 400 m, B runs 400−50=350400 - 50 = 350 m.

  2. Speeds are in the ratio of distances in the same time: 8:78 : 7.

Worth noting

7 : 8 is B's speed to A's.

In a 1,000 m race, A beats B by 100 m. What is the ratio of A's speed to B's?

Options for question 8

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AnswerD · 10 : 9

  1. While A runs 1,000 m, B runs 1,000−100=9001{,}000 - 100 = 900 m.

  2. Speeds are in the ratio of distances in the same time: 10:910 : 9.

Worth noting

9 : 10 is B's speed to A's.

A runs 1 km in 3 minutes 10 seconds and B runs it in 3 minutes 20 seconds. By how many metres does A beat B?

Options for question 9

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AnswerC · 50 m

  1. A finishes 10 seconds before B. B's speed is 1,000200=5\frac{1{,}000}{200} = 5 m/s.

  2. In those last 10 seconds B still had 5×10=505 \times 10 = 50 m to run.

Worth noting

1,00019\frac{1{,}000}{19} m uses A's speed for those 10 seconds; it is B who is still running.

In a 200 m race, A beats B by 25 m or by 5 seconds. How long does A take to run the race?

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AnswerA · 35 seconds

  1. B covers the 25 m in 5 seconds, so B runs at 5 m/s and takes 2005=40\frac{200}{5} = 40 seconds for the race.

  2. A finishes 5 seconds earlier: 35 seconds.

Worth noting

40 seconds is B's time, not A's.

In a 100 m race, A beats B by 10 m, and in another 100 m race B beats C by 10 m. By how many metres would A beat C in a 100 m race?

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AnswerB · 19 m

  1. When A runs 100 m, B runs 90 m. When B runs 90 m, C runs 90100×90=81\frac{90}{100} \times 90 = 81 m.

  2. So A beats C by 100−81=19100 - 81 = 19 m.

Worth noting

20 m adds the two margins, but B's 10 m lead over C was over a full 100 m, and B only runs 90 m here.

In a 100 m race, A beats B by 10 m, and A runs the race in 9 seconds. By how many seconds does A beat B?

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AnswerB · 1 second

  1. In 9 seconds B runs 90 m, so B's speed is 10 m/s and B takes 10 seconds for 100 m.

  2. A beats B by 10−9=110 - 9 = 1 second.

Worth noting

0.9 seconds uses A's speed for B's last 10 m.

In a 300 m race A beats B by 30 m. Running at the same speeds, by how many metres does A beat B in a 1 km race?

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AnswerD · 100 m

  1. B runs 270300=910\frac{270}{300} = \frac{9}{10} of A's distance in the same time.

  2. When A runs 1,000 m, B runs 900 m: A wins by 100 m.

Worth noting

30 m keeps the old margin; it grows with the length of the race.

Read the two statements and choose the correct option.

I. If A beats B by xx m in a dd m race, A's speed : B's speed =d:(d−x)= d : (d - x).

II. If A beats B by xx m or by tt seconds, B's speed is xt\frac{x}{t} m/s.

Options for question 14

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AnswerC · Both I and II

  1. I: in the same time A runs dd and B runs d−xd - x. True.

  2. II: B covers the last xx m in the tt seconds after A finishes. True.

A runs 54\frac{5}{4} times as fast as B. By how many metres does A beat B in a 800 m race?

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AnswerA · 160 m

  1. When A runs 800 m, B runs 800×45=640800 \times \frac{4}{5} = 640 m.

  2. A wins by 800−640=160800 - 640 = 160 m.

Worth noting

200 m takes 14\frac{1}{4} of the race; the margin is the 15\frac{1}{5} of A's 800 m that B has not covered.

A runs 1 km in 4 minutes and B runs it in 4 minutes 10 seconds. By how many metres does A beat B?

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AnswerC · 40 m

  1. A finishes 10 seconds ahead. B's speed is 1,000250=4\frac{1{,}000}{250} = 4 m/s.

  2. In those 10 seconds B still had 4×10=404 \times 10 = 40 m to go.

Worth noting

1253\frac{125}{3} m uses A's speed for the last 10 seconds; it is B who is still running.

In a 400 m race, A beats B by 40 m or by 8 seconds. How long does A take to run the race?

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AnswerA · 72 seconds

  1. B runs the last 40 m in 8 seconds, so B's speed is 5 m/s and B takes 4005=80\frac{400}{5} = 80 seconds.

  2. A finishes 8 seconds earlier: 72 seconds.

Worth noting

80 seconds is B's time.

In a 100 m race A beats B by 20 m, and in a 100 m race B beats C by 25 m. By how many metres does A beat C in a 100 m race?

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AnswerC · 40 m

  1. When A runs 100 m, B runs 8080 m.

  2. C runs 75100\frac{75}{100} of what B runs: 80×75100=6080 \times \frac{75}{100} = 60 m, so A wins by 4040 m.

Worth noting

45 m adds the margins, but B's margin over C shrinks because B runs less than the full race.

In a 200 m race A beats B by 20 m, and in a 200 m race B beats C by 20 m. By how many metres does A beat C in a 200 m race?

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AnswerB · 38 m

  1. When A runs 200 m, B runs 180180 m.

  2. C runs 180200\frac{180}{200} of what B runs: 180×180200=162180 \times \frac{180}{200} = 162 m, so A wins by 3838 m.

Worth noting

40 m adds the margins, but B's margin over C shrinks because B runs less than the full race.

In a 200 m race, A finishes in 25 seconds and beats B by 40 m.

What is B's speed in this 200 m race?

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AnswerC · 6.4 m/s

  1. When A finishes, B has run 200−40=160200 - 40 = 160 m in the same 25 seconds.

  2. 16025=6.4\frac{160}{25} = 6.4 m/s.

Worth noting

8 m/s is A's speed.

In a 200 m race, A finishes in 25 seconds and beats B by 40 m.

How long does B take to run the 200 m?

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AnswerD · 31.25 seconds

  1. B runs at 6.4 m/s.

  2. 2006.4=31.25\frac{200}{6.4} = 31.25 seconds.

Worth noting

30 seconds adds 5 seconds for the 40 m at A's speed; B covers it more slowly.

In a 200 m race, A finishes in 25 seconds and beats B by 40 m.

By how many seconds does A beat B?

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AnswerB · 6.25 seconds

  1. B takes 31.25 seconds and A 25 seconds.

  2. A wins by 6.25 seconds.

Worth noting

5 seconds is the time A needs for 40 m; B, slower, needs 6.25 seconds for it.

In a 1,000 m race A beats B by 100 m, and in another 1,000 m race B beats C by 50 m. The runners keep the same speeds in every race.

By how many metres does A beat C in a 1,000 m race between them?

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AnswerC · 145 m

  1. When A runs 1,000 m, B runs 900 m; C runs 9501,000\frac{950}{1{,}000} of B's distance: 900×0.95=855900 \times 0.95 = 855 m.

  2. A wins by 1,000−855=1451{,}000 - 855 = 145 m.

Worth noting

150 m adds the two margins, but B runs only 900 m in A's race, so B's lead over C shrinks to 45 m.

In a 250 m race, A beats B by 10 m. What is the ratio of A's speed to B's?

Options for question 24

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AnswerA · 25 : 24

  1. While A runs 250 m, B runs 250−10=240250 - 10 = 240 m.

  2. Speeds are in the ratio of distances in the same time: 25:2425 : 24.

Worth noting

24 : 25 is B's speed to A's.

In a 1 km race A beats B by 50 m, and in another 1 km race B beats C by 40 m. By how many metres does A beat C in a 1 km race?

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AnswerA · 88 m

  1. When A runs 1,000 m, B runs 950 m, and C runs 950×9601,000=912950 \times \frac{960}{1{,}000} = 912 m.

  2. A beats C by 1,000−912=881{,}000 - 912 = 88 m.

Worth noting

90 m adds the two margins.

In a 1,000 m race A beats B by 100 m, and in another 1,000 m race B beats C by 50 m. The runners keep the same speeds in every race.

A gives C a 100 m start in a 1,000 m race. By how many metres does A still win?

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AnswerA · 45 m

  1. C has 900 m to run. In the time A runs 1,000 m, C runs 855 m.

  2. C is still 900−855=45900 - 855 = 45 m short when A finishes.

Worth noting

50 m takes the start from the added margins, 150−100150 - 100; the true margin is 145 m.

Giving a start and dead heats

13 questions

A runs 100 m in 15 seconds. He gives B a start of 10 seconds, and they finish together. How long does B take to run 100 m?

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AnswerD · 25 seconds

  1. B sets off 10 seconds before A and they reach the line together.

  2. So B runs for 15+10=2515 + 10 = 25 seconds.

Worth noting

5 seconds subtracts the start; B is the slower runner and takes longer.

A runs at 8 m/s and B at 6 m/s. How many seconds of start should A give B in a 240 m race for a dead heat?

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AnswerC · 10 seconds

  1. A takes 2408=30\frac{240}{8} = 30 seconds and B takes 2406=40\frac{240}{6} = 40 seconds.

  2. B must start 10 seconds earlier.

Worth noting

40 seconds is B's whole time, not the difference.

A runs 43\frac{4}{3} times as fast as B. How much of a start should A give B in a 400 m race so that the race ends in a dead heat?

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AnswerC · 100 m

  1. A dead heat means both finish together. While A runs 400 m, B runs 34×400=300\frac{3}{4} \times 400 = 300 m.

  2. So B should start 400−300=100400 - 300 = 100 m ahead.

Worth noting

300 m is how far B runs, not the start.

In a 100 m race, A can give B a start of 20 m and C a start of 30 m. How much of a start can B give C in a 100 m race?

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AnswerD · 12.5 m

  1. While A runs 100 m, B runs 80 m and C runs 70 m, so B : C = 80 : 70.

  2. While B runs 100 m, C runs 7080×100=87.5\frac{70}{80} \times 100 = 87.5 m, so B can give C a start of 12.5 m.

Worth noting

10 m subtracts the starts, but they were measured against A's 100 m, not B's.

Read the two statements and choose the correct option.

I. In a dead heat, both runners reach the finishing line at the same moment.

II. If A beats B by xx m in a dd m race, then a start of xx m to B over dd m gives a dead heat.

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AnswerC · Both I and II

  1. I is the meaning of a dead heat.

  2. II: while A runs dd m, B runs d−xd - x m; with an xx m start, that is exactly what B needs. True.

A runs 240 m in 30 seconds. He gives B a start of 40 m in a 240 m race, and they finish together. How long does B take to run 240 m?

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AnswerD · 36 seconds

  1. B runs 200 m in A's 30 seconds, so B's speed is 20030=203\frac{200}{30} = \frac{20}{3} m/s.

  2. 240 m takes B 240×320=36240 \times \frac{3}{20} = 36 seconds.

Worth noting

30 seconds is the time B took for 200 m, with the start.

A runs 400 m in 80 seconds and B runs it in 100 seconds. How much of a start, in metres, should A give B for a dead heat in a 400 m race?

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AnswerD · 80 m

  1. B's speed is 400100=4\frac{400}{100} = 4 m/s. In A's 80 seconds, B runs 4×80=3204 \times 80 = 320 m.

  2. So B needs a start of 400−320=80400 - 320 = 80 m.

Worth noting

20 is the start in seconds, not metres.

In a 100 m race, A can give B a start of 10 m and C a start of 28 m. How much of a start can B give C in a 100 m race?

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AnswerD · 20 m

  1. While A runs 100 m, B runs 90 m and C runs 72 m.

  2. So when B runs 100 m, C runs 72×10090=8072 \times \frac{100}{90} = 80 m: B can give C 20 m.

Worth noting

18 m subtracts the starts, 28−1028 - 10, forgetting that B runs only 90 m in A's race.

Assertion (A): If A gives B a start of 30 m in a 300 m race and they dead-heat, then in a 300 m race with no start A beats B by 30 m.

Reason (R): In the time A runs 300 m, B runs 270 m.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. R: the dead heat means B covered 270 m while A covered 300 m. True.

  2. So with no start, when A finishes B is 30 m short: A is true, and R is exactly why.

A and B run at speeds in the ratio 3 : 2. How much of a start should A give B in a 600 m race for a dead heat?

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AnswerB · 200 m

  1. When A runs 600 m, B runs 600×23=400600 \times \frac{2}{3} = 400 m.

  2. B must start 600−400=200600 - 400 = 200 m ahead.

Worth noting

400 m is how far B runs, not the start.

In a 500 m race, the speeds of A and B are in the ratio 3 : 4. A is given a start of 140 m. Who wins, and by how much?

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AnswerD · A, by 20 m

  1. A has 500−140=360500 - 140 = 360 m to run. In that time B runs 43×360=480\frac{4}{3} \times 360 = 480 m.

  2. B is still 20 m short when A finishes, so A wins by 20 m.

Worth noting

B is faster, but not by enough to make up 140 m over the distance A still has to run.

In a 1 km race, A beats B by 40 m and C by 64 m. By how many metres does B beat C in a 1 km race?

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AnswerB · 25 m

  1. While A runs 1,000 m, B runs 960 m and C runs 936 m.

  2. While B runs 1,000 m, C runs 936960×1,000=975\frac{936}{960} \times 1{,}000 = 975 m, so B wins by 25 m.

Worth noting

24 m is 64−4064 - 40, measured over A's kilometre, not B's.

In a 1 km race the speeds of A and B are in the ratio 5 : 4, and A gives B a start of 150 m. By how many metres does A win?

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AnswerA · 50 m

  1. When A runs 1,000 m, B runs 1,000×45=8001{,}000 \times \frac{4}{5} = 800 m.

  2. B had 850 m to run, so A finishes with B still 50 m short.

Worth noting

200 m is A's margin with no start; the 150 m start takes most of it back.

Games of points

16 questions

'A can give B 15 points in a game of 60' means that when A scores 60 points, B scores how many?

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AnswerC · 45 points

  1. Giving 15 points means B needs only 60−15=4560 - 15 = 45 points to keep level.

  2. So when A reaches 60, B has 45.

Worth noting

75 adds the points given instead of taking them away.

In a game of 80, A can give B 16 points. What is the ratio of A's scoring rate to B's?

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AnswerB · 5 : 4

  1. When A scores 80, B scores 80−16=6480 - 16 = 64.

  2. 80:64=5:480 : 64 = 5 : 4.

Worth noting

5:15 : 1 compares 80 with the 16 points given, not with B's score.

Read the two statements and choose the correct option.

I. 'A can give B 20 points in a game of 100' means A's and B's scoring rates are in the ratio 5 : 4.

II. Then A can give B 20 points in a game of 200 as well.

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AnswerA · Only I

  1. I: A scores 100 while B scores 80, and 100:80=5:4100 : 80 = 5 : 4. True.

  2. II: in a game of 200, B scores 160, so A gives 40 points, not 20. False.

In a game of 100 points, A can give B 20 points and C 25 points.

What is the ratio of A's scoring rate to B's in this game?

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AnswerD · 5 : 4

  1. When A scores 100, B scores 80.

  2. 100:80=5:4100 : 80 = 5 : 4.

Worth noting

5:15 : 1 compares 100 with the 20 points given.

In a game of 100 points, A can give B 20 points and C 25 points.

How many points can A give C in a game of 60?

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AnswerB · 15 points

  1. C scores 75100=34\frac{75}{100} = \frac{3}{4} of A's points.

  2. When A scores 60, C scores 45: A gives 15 points.

Worth noting

25 points keeps the figure for a game of 100.

In a game of 90, A can give B 15 points. What is the ratio of A's scoring rate to B's?

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AnswerA · 6 : 5

  1. When A scores 90, B scores 90−15=7590 - 15 = 75 in the same time.

  2. 90:75=6:590 : 75 = 6 : 5.

Worth noting

5 : 6 is B's rate to A's.

In a game of 100 points, A can give B 20 points and C 28 points. How many points can B give C in a game of 100?

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AnswerB · 10 points

  1. When A scores 100, B scores 80 and C scores 72.

  2. When B scores 100, C scores 7280×100=90\frac{72}{80} \times 100 = 90, so B can give C 10 points.

Worth noting

8 subtracts the points given, which were measured against A's 100, not B's.

Assertion (A): If A can give B 20 points in a game of 100, then A can give B 10 points in a game of 50.

Reason (R): With fixed scoring rates, the points given are proportional to the length of the game.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. A : B = 100 : 80. When A scores 50, B scores 40, so A gives 10 points. A is true.

  2. R is the reason: halving the game halves the points given. Both are true and R explains A.

A scores 3 points for every 2 that B scores. How many points can A give B in a game of 90?

Options for question 48

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AnswerA · 30 points

  1. When A scores 90, B scores 90×23=6090 \times \frac{2}{3} = 60.

  2. So A can give B 90−60=3090 - 60 = 30 points.

Worth noting

60 is B's score, not the points given.

In a game of 100, A can give B 10 points and C 19 points. How many points can B give C in a game of 100?

Options for question 49

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AnswerC · 10 points

  1. When A scores 100, B scores 90 and C scores 81.

  2. When B scores 100, C scores 81×10090=9081 \times \frac{100}{90} = 90: B gives C 10 points.

Worth noting

9 subtracts the points, 19−1019 - 10, which is B's lead in A's game of 100, not in B's own game of 100.

In a game of 100, A can give B 20 points. How many points can A give B in a game of 150?

Options for question 50

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AnswerC · 30 points

  1. B scores 80100=45\frac{80}{100} = \frac{4}{5} of A's points.

  2. When A scores 150, B scores 120: A can give 30 points.

Worth noting

20 points keeps the old number; it grows in proportion to the game.

In a game of 100 points, A can give B 20 points and C 25 points.

What is the ratio of B's scoring rate to C's in this game?

Options for question 51

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AnswerA · 16 : 15

  1. When A scores 100, B scores 80 and C scores 75.

  2. 80:75=16:1580 : 75 = 16 : 15.

Worth noting

21:2021 : 20 treats B as giving C 5 points in a game of 100; it is 5 points in a game of 80.

In a game of 100 points, A can give B 20 points and C 25 points.

How many points can B give C in a game of 80?

Options for question 52

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AnswerC · 5 points

  1. C scores 1516\frac{15}{16} of B's points.

  2. When B scores 80, C scores 75: B gives 5 points.

Worth noting

4 points scales 5 down to a game of 80, but the 5 points already belong to a game of 80.

In a game of 60, A can give B 10 points, and in a game of 75, B can give C 15 points. How many points can A give C in a game of 90?

Options for question 53

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AnswerD · 30 points

  1. A : B = 60 : 50 = 6 : 5 and B : C = 75 : 60 = 5 : 4, so A : C = 6 : 4 = 3 : 2.

  2. When A scores 90, C scores 23×90=60\frac{2}{3} \times 90 = 60: A can give C 30 points.

Worth noting

25 adds the 10 and 15 points given, which belong to games of different lengths.

In a game of 60, A can give B 12 points and C 18 points. How many points can B give C in a game of 80?

Options for question 54

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AnswerD · 10 points

  1. When A scores 60, B scores 48 and C scores 42, so C scores 4248=78\frac{42}{48} = \frac{7}{8} of B's points.

  2. In a game of 80, C scores 80×78=7080 \times \frac{7}{8} = 70: B gives C 10 points.

Worth noting

6 is the gap in A's game of 60; the game here is 80, and B's own pace sets it.

In a game of 50, A can give B 5 points, and in a game of 60, B can give C 12 points. How many points can A give C in a game of 50?

Options for question 55

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AnswerB · 14 points

  1. When A scores 50, B scores 45. C scores 4860=45\frac{48}{60} = \frac{4}{5} of B's points.

  2. So when A scores 50, C scores 45×45=3645 \times \frac{4}{5} = 36: A gives C 14 points.

Worth noting

17 adds the margins, but B's 12-point lead is over a game of 60, and B scores only 45 in A's game.

Circular tracks

18 questions

Two cyclists start together from the same point on a 600 m circular track and ride in opposite directions at 4 m/s and 6 m/s. After how many seconds do they first meet?

Options for question 56

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AnswerA · 60 seconds

  1. Riding towards each other, they close the gap at 4+6=104 + 6 = 10 m/s.

  2. They meet when together they have covered one lap: 60010=60\frac{600}{10} = 60 seconds.

Worth noting

300 seconds subtracts the speeds, which is for riding the same way.

Assertion (A): Two runners going the same way round a track from the same point first meet when the faster has run exactly one lap more than the slower.

Reason (R): Their relative speed is the difference of their speeds.

Options for question 57

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. R: in the same direction, the gap between them grows at the difference of the speeds. True.

  2. They meet again when that gap reaches one whole lap, so A is true, and R explains how the gap builds up.

Two runners start together from the same point on a 360 m circular track and run in opposite directions at 3 m/s and 5 m/s. After how many seconds do they first meet?

Options for question 58

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AnswerB · 45 seconds

  1. Running towards each other, they close the gap at 88 m/s.

  2. One lap between them takes 3608=45\frac{360}{8} = 45 seconds.

Worth noting

180 seconds uses the relative speed for the same directions.

A and B run round a 400 m circular track from the same point at the same moment, A at 5 m/s and B at 3 m/s.

If they run in opposite directions round the track, when do they first meet?

Options for question 59

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AnswerA · 50 seconds

  1. They close the gap at 5+3=85 + 3 = 8 m/s.

  2. 4008=50\frac{400}{8} = 50 seconds.

Worth noting

200 seconds is for running the same way.

A and B start together from the same point on a 400 m circular track and run in the same direction at 5 m/s and 3 m/s. After how many seconds do they first meet again?

Options for question 60

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AnswerD · 200 seconds

  1. Running the same way, A gains on B at 5−3=25 - 3 = 2 m/s.

  2. They meet when A has gained one full lap: 4002=200\frac{400}{2} = 200 seconds.

Worth noting

50 seconds adds the speeds, which is right only when they run towards each other.

Read the two statements and choose the correct option.

I. Running in opposite directions from the same point, two runners first meet when together they have covered one lap.

II. Running in the same direction from the same point, the faster runner first meets the slower one when he has run one more lap than him.

Options for question 61

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AnswerC · Both I and II

  1. I: their distances add up to the lap when they meet. True.

  2. II: their distances differ by one lap when the faster catches up. True.

A completes a lap in 40 seconds and B in 56 seconds. They start together from the starting point. After how many seconds are they next at the starting point together?

Options for question 62

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AnswerC · 280 seconds

  1. Each is at the start after a whole number of its laps: A at multiples of 40, B at multiples of 56.

  2. The first common time is LCM(40,56)=280\text{LCM}(40, 56) = 280 seconds.

Worth noting

2,240 is 40×5640 \times 56, a common multiple but not the least.

A completes a lap in 6 minutes and B in 9 minutes, running the same way from the same point. After how many minutes does A first overtake B?

Options for question 63

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AnswerC · 18 minutes

  1. A gains 16−19=118\frac{1}{6} - \frac{1}{9} = \frac{1}{18} of a lap a minute.

  2. A full lap's gain takes 18 minutes.

Worth noting

185\frac{18}{5} minutes adds the rates, which is for running opposite ways.

Two runners start together from the same point on a 500 m circular track and run in the same direction at 7 m/s and 5 m/s. After how many seconds do they first meet?

Options for question 64

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AnswerD · 250 seconds

  1. Running the same way, the faster gains at 22 m/s.

  2. One lap between them takes 5002=250\frac{500}{2} = 250 seconds.

Worth noting

1253\frac{125}{3} seconds uses the relative speed for opposite directions.

Three runners take 30, 40 and 60 seconds to complete a lap. They start together from the starting point. After how many seconds are all three next at the start together?

Options for question 65

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AnswerB · 120 seconds

  1. They are all at the start at a common multiple of 30, 40 and 60.

  2. LCM(30,40,60)=120\text{LCM}(30, 40, 60) = 120 seconds.

Worth noting

240 is a common multiple, but 120 comes first.

A completes a lap in 4 minutes and B in 6 minutes. They start together from the same point and run in opposite directions. After how many minutes do they first meet?

Options for question 66

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AnswerB · 2.4 minutes

  1. Together they cover 14+16=512\frac{1}{4} + \frac{1}{6} = \frac{5}{12} of a lap a minute.

  2. One lap between them takes 125=2.4\frac{12}{5} = 2.4 minutes.

Worth noting

12 minutes is when both are next at the start, LCM(4, 6), not their first meeting.

Two cyclists ride the same way round a 2 km circular track from the same point, at 18 km/h and 12 km/h. After how many minutes do they first meet?

Options for question 67

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AnswerD · 20 minutes

  1. The faster gains 18−12=618 - 12 = 6 km/h on the slower.

  2. A lap of 2 km takes 26=13\frac{2}{6} = \frac{1}{3} hour, that is 20 minutes.

Worth noting

4 minutes adds the speeds, which is for riding towards each other.

A and B run round a 400 m circular track from the same point at the same moment, A at 5 m/s and B at 3 m/s.

If they run the same way round the track, when do they first meet?

Options for question 68

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AnswerD · 200 seconds

  1. A gains on B at 5−3=25 - 3 = 2 m/s.

  2. A full lap's gain: 4002=200\frac{400}{2} = 200 seconds.

Worth noting

50 seconds adds the speeds, which is for running towards each other.

A completes a lap of a circular track in 60 seconds and B in 84 seconds. They start together from the starting point. After how many seconds are they next at the starting point together?

Options for question 69

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AnswerC · 420 seconds

  1. A is at the start every 60 seconds and B every 84 seconds, so they are there together at a common multiple.

  2. LCM(60,84)=420(60, 84) = 420 seconds, which is 7 minutes.

Worth noting

5,040 multiplies 60 by 84; it is a common multiple, but not the least.

On a circular track, A completes a lap in 5 minutes and B in 8 minutes, running the same way from the same point. How many times does A pass B in the first hour?

Options for question 70

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AnswerB · 4 times

  1. A gains 15−18=340\frac{1}{5} - \frac{1}{8} = \frac{3}{40} of a lap on B every minute, so a full lap every 403≈13.3\frac{40}{3} \approx 13.3 minutes.

  2. Each full lap gained is one pass. In 60 minutes A gains 60×340=4.560 \times \frac{3}{40} = 4.5 laps: 4 passes, the fifth coming only after about 66.7 minutes.

Worth noting

5 rounds 4.5 up, but half a lap gained is not a pass.

A and B run the same way round a circular track from the same point, with speeds in the ratio 5 : 3. At how many different points of the track do they meet?

Options for question 71

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AnswerA · 22

  1. With speeds 5k5k and 3k3k on a track of length LL, they meet every L2k\frac{L}{2k}, when A has run 5L2\frac{5L}{2}: 2.5 laps.

  2. So the meetings fall at half a lap and at the start, alternately: 2 points, which is 5−35 - 3.

Worth noting

8 is 5+35 + 3, the count for running in opposite directions.

A and B run round a 400 m circular track from the same point at the same moment, A at 5 m/s and B at 3 m/s.

Running the same way, at how many different points of the track do they meet?

Options for question 72

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AnswerB · 22

  1. They meet every 200 s, when A has run 1,000 m, two and a half laps.

  2. So the meetings alternate between the far side and the start: 2 points, which is 5−35 - 3.

Worth noting

8 is 5+35 + 3, the count for opposite directions.

A and B run round a 400 m circular track from the same point at the same moment, A at 5 m/s and B at 3 m/s.

After how many seconds are both runners next at the starting point together?

Options for question 73

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AnswerC · 400 seconds

  1. A laps in 4005=80\frac{400}{5} = 80 s and B in 4003\frac{400}{3} s; both are back at multiples of these.

  2. In thirds of a second: LCM(240,400)=1,200\text{LCM}(240, 400) = 1{,}200, so 400 seconds (A 5 laps, B 3 laps).

Worth noting

200 seconds is their first meeting anywhere on the track, not at the start.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.