Circular Races: Solved Examples

Example 1: In a km race A beats B by 50 metres or 10 seconds. Find the time taken by A to complete the race.

Solution: When A covered the distance of 1 km = 1000 m, B must be 50 m behind A which B can cover in 10 seconds
⇒ The speed of B = 50/10 = 5 m/s ⇒ time taken by B to cover 1000 m = 1000/5 = 200 sec
⇒ Time taken by A to complete the race of 1000 m = 200 – 10 = 190 sec

Example 2: A is 1(4/5) times as fast as B. If A gives a start of 80 metres, how long should the racecourse be such that race ends in a dead heat?

Solution: Speed of A = 9/5 Speed of B. When B covers 5 m, A covers 9 m ⇒ A can give a start of (9 – 5) = 4 m to B in a 9 m race. ⇒ 80 m (4 × 20) start in 9 × 20 = 180 m, which should be the length of the racecourse.

Example 3: In a race of 800 m, A can beat B by 80 m and in a race of 600 m, B can beat C by 60 m. By how many metres will A beat C in a race of 400 m?

Solution: When A covers distance of 800 m, B covers distance of (800 – 80) = 720 m ⇒ Speed of A : Speed of B = 800:720 = 10:9 Also, when B covers distance of 600 m, C covers distance of (600 – 60) = 540 m ⇒ Speed of B : Speed of C = 600:540 = 10:9 ⇒ Speed A : Speed B : Speed C = 100 : 90 : 81 ⇒ in 100 m race, A can beat C by (100 – 81) = 19 m ⇒ in 400 m race, A can beat C by 19 × 4 = 76 m

Example 4: A beats B by 37 m and C by 23 m in a race of 200 m. By how many metres will C beat B in a race of 354 m?

Solution: When A covers a distance of 200 m, B covers (200 – 37) = 163 m and C covers (200 – 23) = 177 m. ⇒ Speed A : Speed B : Speed C = 200:163:177 ⇒ in 177 m race, C can beat B by (177 – 163) = 14 m ⇒ in 354 m race, C can beat B by (14/177)x354 = 28 m

Example 5: In a game of 80 points, A can give B a lead 20 points and C 29 points, then how many points of lead can B give to C in a game of 180 points?

Solution: This question is no different from the previous ones and involves the concept of races only. A can give B 20 points means when A scores 80 points, B scores (80 – 20) = 60 points ⇒ A:B = 80:60 Likewise, when A scores 80 points, C scores (80 – 29) = 51 points ⇒ A:C = 80:51 ⇒ A:B:C = 80:60:51 ⇒ in a game of 60 points, B can give C (60 – 51) = 9 points ⇒ in a game of 180 (60 × 3) points, B can give C (9 × 3) = 27 points

Example 6: Two people A and B started running from a same point on a circular track of length 400 m in opposite directions with initial speeds of 10 m/s and 40 m/s respectively. Whenever they meet, A’s speed doubles and B’s speed halves. After what time from the start will they meet for the third time?

Solution: Time taken to meet for the 1st time = 400/10+40 = 8 sec, since they are moving in opposite direction

Now A’s speed = 20 m/s and B’s speed = 20 m/s ⇒ Time taken to meet for the 2nd time = 400/40+10 = 10 sec

Now A’s speed = 10 m/s and B’s speed = 40 m/s ⇒ Time taken to meet for the 3rd time = 400/10+10 = 8 sec

⇒ Total time to meet for the third time = (8 + 10 + 8) = 26 sec

Example 7: Three persons started running on a circular track of length 36 km from the same starting point with speeds 7 km/h, 11km/h and 15 km/h respectively in the same direction. How many points are there where they will meet on the track, including the starting point?

Solution: Time taken to meet for the 1st time at starting point = LCM (36/7,36/11,36/15) = 36 hours

Time taken to meet for the 1st time = LCM (36/(11-7),36/(15-7)) = LCM(36,36)/HCF(4,8) = 36/4 = 9 hours

⇒ Number of meeting points = 36/9 = 4 points

[Shortcut: Since they are moving in same direction, one may directly solve by taking HCF {(11 – 7), (15 – 7)} = 4 points, but this is applicable only when HCF of given speeds is 1]

Example 8: A and B start running around a circular track of circumference 240 m at speeds of 3 m/s and 7 m/s, starting from the same point at the same time in the same direction. After how much time will they meet at a point which is diametrically opposite the starting point?

Solution: Time taken to meet for the 1st time at starting point = LCM(240/3,240/7) = 240 sec

Time taken to meet for the 1st time = LCM(240/(7-3)) = 60 sec

⇒ Number of meeting points = 240/60 = 4 points which would be symmetrical on the track.

⇒ They will meet after every 60 seconds at four different points including starting point.
⇒ They will meet at a point which is diametrically opposite the starting point after 120 sec.