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This set contains Races and Circular Motion Questions with Solutions — Set 3 (Q21 to Q30) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.

Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.

✏️ Attempt each question on your own first — then check the solution below.

Races and Circular Motion Questions 21 to 30 with Solutions

21. A and B run a race of 2000m. First A gives B a start of 200m and beats him by 30 sec. Next, A gives B a start of 3 minutes and is beaten by 1000m. find the time in minutes in which A and B can run the race separately.

22. A and B run a race of 3 km. A gives B a heaed-start of 400m and beats him by 30 sec. While coming back, A gives B a leadw of 2.5 min and gets beaten by 500m. What is the difference between the times in minutes in which A and B can run the race for one side separately.

23. In a 150m race, A wins over B by 35m and in the same race C can give a start of 25m to B. By how much distance can A give a start to C, so that A beats C by 5m ?

24. In a 1000m race A gives B a start of 40m and beats him by 19 sec. If A gives a start of 30 sec then B beats A by 40 m. Find the ratio of speeds of A and B.

25. In a race of 100m A can give a start of 20m to B and a start of 40m to C. How much start can B give to C in a 100m race.

26. A can beat B in a 100 m race by 10 m. B can beat C in a 100 m race by 10 m. What is the ratio of time taken by A, B and C to complete the race.

27. In a game of 100 points, A can give B 10 points and C 20 points. Then how many points B can give to C ?

28. If the ratio of speed of A and B is 5 : 9 and A loses the race by 48 m, what is the length of race track ?

29. The ratio of time taken by A and B to run a certain distance is 2 : 3 and A wins by 100 m. Then find the length of the race track.

30. In a 480 m race, A gave B a head-start of 6 seconds and still won the race by 10 sec. The ratio of the speed of A & B are 3 : 1. How long will B take to cover a race of 1000 m ?

Solutions — Races and Circular Motion Questions 21 to 30

21. A and B run a race of 2000m. First A gives B a start of 200m and beats him by 30 sec. Next, A gives B a start of 3 minutes and is beaten by 1000m. find the time in minutes in which A and B can run the race separately.
Sol:

DistanceTimeSpeedA BA B2000 18001000 2000t t + 30×

make speed of B equal in both cases
\(\frac{{1800}}{{t + 30}} = \frac{{2000}}{{\frac{t}{2} + 180}}\)
\(\frac{9}{{t + 30}} = \frac{{20}}{{t + 360}}\)
20t + 600 = 9t + 3240
11t = 2640
t = 240 sec = 4 min ⟵ Time taken by A    Answer
& time taken by B = \(\frac{t}{2} + 180 + \frac{{240}}{2} + 180\) = 300 sec = 5 min     Answer

22. A and B run a race of 3 km. A gives B a heaed-start of 400m and beats him by 30 sec. While coming back, A gives B a leadw of 2.5 min and gets beaten by 500m. What is the difference between the times in minutes in which A and B can run the race for one side separately.
Sol:

DistanceTimeSpeedA BA B3000 26002500 3000t t + 30

∴ \(\frac{{2600}}{{t + 30}} = \frac{{3000}}{{\frac{{5t}}{6} + 150}}\)
\(\frac{{13}}{{t + 30}} = \frac{{90}}{{5t + 900}}\)
90t + 2700 = 65t + 11700
25t = 9000
t = 360 secc = 6 min ⟵ Time taken by A
& time taken by B = \(\frac{{5t}}{6} + 150\)
= \(\frac{{5 \times 360}}{6} + 150\)
= 450 sec
= 7.5 min
∴ required time difference = 7.5 – 6 = 1.54 min       Answer

23. In a 150m race, A wins over B by 35m and in the same race C can give a start of 25m to B. By how much distance can A give a start to C, so that A beats C by 5m ?
Sol:

A B C150 1152330235125 150561906913812

A beats C by 12m & since it is given that A beats C by 5m so A gives a start of 12 – 5 = 7m to C    Answer

24. In a 1000m race A gives B a start of 40m and beats him by 19 sec. If A gives a start of 30 sec then B beats A by 40 m. Find the ratio of speeds of A and B.
Sol:

DistanceTimeA BA B1000 960960 1000t t + 19

Speed of B = Speed of B

\(\frac{{960}}{{t + 19}} = \frac{{1000}}{{\frac{{24t}}{{25}} + 30}}\)
\(\frac{{24}}{{t + 19}} = \frac{{625}}{{24t + 750}}\)
625t + 19 × 625 = 576t + 24 × 750
49t = 6125t = 125 sec
∴ \(\frac{{Speed\;of\;A}}{{Speed\;of\;B}} = \frac{{\frac{{1000}}{t}}}{{\frac{{960}}{{t + 19}}}} = \frac{{t + 19}}{t} \times \frac{{25}}{{24}}\)
= \(\frac{{144}}{{125}} \times \frac{{25}}{{24}}\)
= 6 : 5     Answer

25. In a race of 100m A can give a start of 20m to B and a start of 40m to C. How much start can B give to C in a 100m race.
Sol:

B A C80 100 55 5 3100 6020 25 154 5 3

∴ B gives a start of 1m in 4m race
∴ in 100m race start will be \(\frac{1}{4} \times 100\) = 25m     Answer

26. A can beat B in a 100 m race by 10 m. B can beat C in a 100 m race by 10 m. What is the ratio of time taken by A, B and C to complete the race.
Sol:

B A C100 90 910 10 9100 90100 90 819Time90×81 100×81 100×90 81 : 90 : 100 Answer

27. In a game of 100 points, A can give B 10 points and C 20 points. Then how many points B can give to C ?
Sol:

A : B : C100 90 8010 Answer

28. If the ratio of speed of A and B is 5 : 9 and A loses the race by 48 m, what is the length of race track ?
Sol:
Since time is constant so ratio of distance will be same as ratio of speeds.

A BSpeed 5 : 9Distance 5 : 94108 Answer48×12×12Since B wins the race∴ B cover whole track

29. The ratio of time taken by A and B to run a certain distance is 2 : 3 and A wins by 100 m. Then find the length of the race track.
Sol:
Since distance is constant
∴ ratio of speed ∝ \(\frac{1}{{ratio\;of\;time}}\)

A BTime 2 : 3Speed 3 : 21100×100300 m Answer×100

30. In a 480 m race, A gave B a head-start of 6 seconds and still won the race by 10 sec. The ratio of the speed of A & B are 3 : 1. How long will B take to cover a race of 1000 m ?
Sol:
Since distance is constant
∴ Speed ∝ \(\frac{1}{{Time}}\)

A BSpeed 3 : 1Time 1 : 3t (t + 16)(t + 6 + 10)

⟹ \(\frac{1}{t} = \frac{3}{{t + 16}}\)
⟹ t = 8
∴ required time = \(\frac{{8 + 16}}{{480}} \times 1000\) = 50 sec      Answer

Well done on completing Set 3!

Continue practising with Races and Circular Motion Questions 31 to 35 → Set 4 or revisit the Races and Circular Motion Concept Page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering races and circular motion for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Circular Motion on Wikipedia before attempting the next set.

This page is part of our complete series of races and circular motion questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core races and circular motion concept before your exam day.