Linear Races and Circular Races: Questions, Examples

Linear races and Circular races questions play a very significant role in competitive examination.

Circular races are those types of races that happen on an oval-shaped or circular race track with the same starting and finishing point. Circular race questions cover concepts like speed, distance, time, and relative positions on the track. Resources such as circular race concept, circular race PDF, circular race track diameter, circular races CAT questions, and circular race questions SSC CGL PDF help understand and solve problems related to circular races.

Linear races are those types of races that run in a straight line and have different starting and finishing lines. Linear race questions mainly focus on speed, distance, and time calculations. Linear chart examples and line chart race Python can visually represent competitors' progress in such races.

Linear Races: Concept

In a linear race, competitors run from the starting line to the ending line in a straight path or track. The primary object of linear race is to reach the end line in the shortest possible time. It primarily involves calculating speed, time, and distance.

There are some important terms used in linear races.

Starting point/line: From where the race starts.

Endine point/line: Where the race ends.

Head start: If a competitor, x has started the race after another competitor y reaches 500m, then we say that x gives y a head start of 500m.

If a competitor, x has started the race 30 seconds after another competitor y starts in the hope that he/she can still beat him. Then we can say that x gives y a head start of 30 seconds.

Dead heat: A dead heat race means all participants in the race finished the race at the same time and the same point.

Concepts and Tricks Required to Solve the Problems of Linear Races

There are many concepts to solving linear race questions. Important ones are:

• Speed, time, and distance: The fundamental concept of (Distance = Speed × Time) is used to solve linear race problems.

• Relative speed: Most of the time in a linear track competitors will run in the same direction. So their relative speed will be (Faster speed - Slower speed).

• Conversion of units: Speed may need to be converted from km/hr to m/s and vice versa.

• Average speed: Average speed can be calculated by dividing the total distance travelled by the total time taken.

Types of Problems of Linear Races

Many types of Linear races come in Competitive exams. Some of the important ones are discussed below.

A gives B a head start of x metre:

If there are two participants, A and B in the race. A’s speed is faster than B's. So A starts the race after B races to x metres in the hope that he/she can still catch him/her. Then we can say that A gives B a head start of x metres.

Example:

Ashok runs $2 \frac{2}{3}$ times as fast as Bharat. If Ashok gives Bharat a head start of 160 m, then how far must the winning post be so that Ashok and Bharat can reach it at the same time?

Sol: The ratio of speeds of Ashok and Bharat = $2\frac{2}{3}:1= 8:3$

That means in a race of 8 m, Ashok gains 5 m over Bharat.

⇒ 160 m will be gained by Ashok in a race of $\frac{8}{5}\times 160 = 256 \ \mathrm{m}$

Hence, the correct answer is 256 m.

A gives B a head start of t seconds:

If there are two participants, A and B in the race. A’s speed is faster than B's. So A starts the race t seconds after B starts in the hope that he/she can still catch him/her. Then we can say that A gives B a head start of t seconds.

Example:

Savitha and Parnika decided to run a 2250 m long race on a track as long as the length of the race. But Savitha gave Parnika a 396 m head start and also allowed her to start 9 seconds before Savitha started running. Parnika ran at 6 m per second and Savitha caught up with Parnika 250 seconds after Savitha started running. What was Savitha's speed (in m/s) during the race?

Sol: Let us denote Savitha's speed as $s$ m/s.

Since Parnika started 9 seconds before Savitha, the time Parnika ran is $(t+9)$ seconds.

The distance Parnika ran = $6×(t+9)$ metres.

Savitha started 250 seconds after Parnika and caught up with her.

⇒ The distance Savitha ran = $s\times t$ metres

Parnika had a head start of 396 metres.

According to the question,

$6×(t+9)+396=s×t$

⇒ $6×(t+9)+396=2250$

⇒ $6t+54+396=2250$

⇒ $6t=1800$

$\therefore t=300$ seconds

Now, Speed $=\frac{\text{Distance}}{\text{Time}}=\frac{2250}{300}= 7.5$ m/s

Hence, the correct answer is 7.5.

A beats B by x metres:

If there are two participants, A and B in the race. When participant A reaches the finish line, participant B is still x metres away from the finish line at the same time. Then we can say that A beats B by x metres.

Example:

In a 1500 m race, X beats Y by 100 m and X beats Z by 240 m. By what distance does Y beat Z in the same race?

Sol: Given: The length of the race = 1500 m,

X beats Y by 100 m

And X beats Z by 240 m.

When X is at 1500 m,

Y will be at 1500 – 100 = 1400 m

Z will be at 1500 – 240 = 1260 m

The ratio of distance covered by Y and Z = 1400 : 1260 = 10 : 9

$\therefore$ Distance by Y beats Z = $\frac{10-9}{10}×1500$ = 150 m

A beats B t seconds:

If there are two participants, A and B in the race. Participant A reaches the finish line first and Participant B reaches the finish line after t seconds. Then we can say A beats B by t seconds.

Example:

In a race of 1200 m, Ram can beat Shyam by 200 m or by 20 sec. What must be the speed of Ram?

Sol: Ram beats Shyam by 200 m or 20 sec.

This means Shyam covers a 200 m distance in 20 sec.

$\therefore$ Speed of Shyam = $\frac{200}{20}$ = 10 m/sec

Also, in the 1200 m race, Ram beats Shyam by 200 m.

Distance covered by Ram = 1200 m

Distance covered by Shyam in same time = 1000 m

Now, the 1000 m distance is covered by Shyam with a speed of 10 m/sec in $\frac{1000}{10}$ = 100 seconds.

So, 1200 m distance is covered by Ram in 100 seconds.

$\therefore$ Speed of Ram = $\frac{1200}{100}$ = 12 m/sec

Hence, the correct answer is 12 m/sec.

Dead heat:

Dead heat race means all participants in the race finished the race at the same time and at the same point. There will be no winner or loser in the dead heat race.

Suppose A and B started and finished the race together. Then that race between A and B finished in a dead heat.

Example:

A can run 250 m in 25 sec and B in 30 sec. How many metres start can A give to B in a 1 km race so that the race may end in a dead heat?

Sol: A can run 250 m in 25 seconds.

⇒ A can run 1000 m in $\frac{25}{250} \times 1000$ = 100 seconds

B can run 250 m in 30 seconds.

⇒ B can run 1000 m in $\frac{30}{250} \times 1000$ = 120 seconds

Distance covered by B in an extra 20 seconds = $\frac{20}{120} \times 1000 = \frac{500}{3}$ = 166.67 m

Therefore, A beats B by 166.67 m.

For a dead-heat race, A must give B a start of 166.67 m.

Hence, the correct answer is 166.67 m.

Circular Races: Concept

Circular races are races that run on circular tracks and competitors have the same starting and ending line. In this race, competitors can meet each other or lap each other multiple times. The primary factors to consider in circular races are the track length, the speeds of the competitors, and the number of laps completed.

There are some important terms used in circular races.

Track length: The total distance around the circular track is called Track length.

Number of laps: The number of times a competitor covers the circumference of the track. If a competitor covers two times, then it is considered 2 laps.

Starting point/Ending point: In general, in a circular race the point where the race starts is the point where the race usually ends. That point is called the starting or ending point.

Concepts and Tricks required to solve the problems of Circular Races

There are many concepts to solve problems of circular races. Important ones are:

• Relative speed: When two participants move in the same direction on a circular track, their relative speed is the difference between the speeds. (Higher speed - Lower speed)

When two participants move in the opposite direction on a circular track, their relative speed is the sum of their speeds. (Higher speed + Lower speed)

• Least Common Multiple(LCM): To determine when competitors on a circular track meet for the first time at the starting point, we need the LCM of the time to complete one lap.

• Meeting points: To calculate when the faster participant meets the slower participant on the circular track, we need to divide the circumference of the track by the relative speed of the participants.

• Ratios: Use the ratios to compare the speeds and determine relative distances covered in time.

Types of Problems on Circular Tracks/races

Many types of circular races come in Competitive exams. Some of the important ones are discussed below.

When will they meet at the starting point for the first time?

To determine when two participants meet at the starting point for the first time, we need to calculate the Least Common Multiple(LCM) of their time completing one lap.

Suppose A completes a 200-metre circular track in 20 seconds and B in 30 seconds. If they start together, they will first meet at the starting point in (LCM of 20, 30 =) 60 seconds.

Example:

A circular running path is 1452 m in circumference. Two men start from the same point and walk in opposite directions at speeds of 7.5 km/h and 9 km/h, respectively. When will they meet for the first time?

Sol: Relative speed = (9 + 7.5) = 16.5 km/h

Distance = $\frac{1452}{1000}$ = 1.452 km

Time = $\frac{\text{Distance}}{\text{Speed}}$

= $\frac{1.452}{16.5}$ hr

= $\frac{1.452}{16.5}×60$ min

= 5.28 min

They will meet for the first time after 5.28 minutes.

When will they meet for the first time on the circular track? (not necessarily at the starting point)

To determine when two participants meet the first time on the track not necessarily the starting point, we need to calculate the relative speed of those participants. Then divide the circumference of the track with the relative speed to get the meeting time for the first time.

Example:

Two cities P and Q are 181 km apart on a straight road. One man starts from P at 8:30 a.m. and travels toward Q at 30 km/hr. If another man starts from Q at 8:54 a.m. and travels towards P at a speed of 35 km/hr, then at what time will they meet?

Sol: Distance between P and Q at 8:30 a.m. = 181 km

Distance covered by P until Q starts = $30 \times \frac{24}{60}$ = 12 km

Distance between P and Q at 8:54 a.m. = 181 − 12 = 169 km

Let the time taken to cover 169 km be $t$ hours.

So, now both persons have to travel 169 km combined

$⇒30t + 35t = 169$

$⇒65t = 169$

$⇒t = \frac{169}{65}= 2 \frac{39}{65} =$ 2 hours 36 minutes

Therefore, the time of the meeting = 8:54 a.m. + 2 hours 36 minutes = 11:30 a.m.

At how many points will they meet while moving in the opposite direction?

If the ratio of speeds of two persons moving in opposite directions on a circular track is a : b, then the number of distinct meeting points will be a + b.

Example:

Two persons started running on a circular track at a speed of 20 m/s and 30m/s in opposite directions. If the circumference of the circular track is 100 m, find how many distinct points they will cross each other.

Sol: The ratio of the speed of both of the runners = 20 : 30 = 2 : 3

So, they will meet (2 + 3) 5 times. (sum of the ratio of their speeds)

At how many points will they meet while moving in the same direction?

If the ratio of speeds of two persons moving in the same directions on a circular track is a : b, then the number of distinct meeting points will be a - b where a > b.

Example:

Two runners, Sony and Mony, start running on a circular track of length 200 m at speeds of 18 km/h and 24 km/h, respectively, in the same direction. Find how many distinct points they will cross each other.

Sol: We know,

Sony and Mony, running on a circular track in the same direction of length 200 m at speeds of 18 km/h and 24 km/h.

Speed of Sony = 18 km/h = 18 × $\frac{5}{18}$ m/s = 5 m/s

Speed of Mony = 24 km/h = 24 × $\frac{5}{18}$ m/s = $\frac{20}{3}$ m/s

Ratio of their speeds = $5: \frac{20}{3}$ = 15 : 20 = 3 : 4

So, they will meet (4 - 3) = 1 time.

Summary of All Formulas Used in Circular Tracks/Races

• Distance = Speed × Time

• To determine 1 competitor's lap time, we have to divide the lap length by that competitor's speed.

• When two participants move in the same direction on a circular track, their relative speed is the difference between the speeds. (Higher speed - Lower speed)

When two participants move in the opposite direction on a circular track, their relative speed is the sum of their speeds. (Higher speed + Lower speed)

• If the ratio of speeds of two persons moving in opposite directions on a circular track is a : b, then the number of distinct meeting points will be a + b.

• If the ratio of speeds of two persons moving in the same directions on a circular track is a : b, then the number of distinct meeting points will be a - b where a > b.

• To calculate when the faster participant meets the slower participant on the circular track, we need to divide the circumference of the track by the relative speed of the participants.

• To calculate when the faster participant meets the slower participant on the circular track, we need to divide the circumference of the track by the relative speed of the participants.

Tips and Tricks

• When two participants move in the same direction on a circular track or linear track, their relative speed is the difference between the speeds. (Higher speed - Lower speed)

• To determine when competitors on a circular track meet for the first time at the starting point, we need the LCM of the time to complete one lap.

• $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

• If the ratio of their speeds is $V_1:V_2$, the distinct meeting points will be $V_1-V_2$.

• To determine when competitors on a circular track meet for the first time at the starting point, we need the LCM of the time to complete one lap.

• Average speed can be calculated by dividing the total distance travelled by the total time taken.

Practice Questions

Q1. Two people A and B started running from the 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?

1. 30 sec

2. 24 sec

3. 26 sec

4. 28 sec

Hint: Find the relative speed for each point till they meet for the third time. Calculate Time after every time they meet using $\frac{\text{distance}}{\text{speed}}$ and then calculate the total time.

Answer:

Given: Total length of the track = 400 m

Speed of A = 10 m/sec

Speed of B = 40 m/sec

They are running in the opposite direction

⇒ Relative speed = 10 + 40 = 50 m/sec

We know,

Speed = $\frac{\text{Distance}}{\text{Time}}$

Time when they meet first time = $\frac{400}{50}$ = 8 sec

Now, the speed of A will get doubled and that of B will get half.

⇒ The new speed of A = 20 m/sec

And, the new speed of B = 20 m/sec

⇒ Relative speed = 20 + 20 = 40 m/sec

Time when they meet second time = $\frac{400}{40}$ = 10 sec

Now, again the speed of A will get doubled and that of B will get half.

⇒ New speed of A = 40 m/sec

And, New speed of B = 10 m/sec

⇒ Relative speed = 40 + 10 = 50 m/sec

⇒ Time when they meet third time = $\frac{400}{50}$ = 8 sec

⇒ At the third time, they will meet after 8 + 10 + 8 = 26 sec

Hence, the correct answer is 26 sec.

Q2. In a circular race of 1600 m, A and B start from the same point and at the same time at speeds of 27 km/hr and 45 km/hr, respectively. After how long will they meet again for the first time on the racetrack, when they are running in the same direction?

1. 90 seconds

2. 320 seconds

3. 240 seconds

4. 180 seconds

Hint: The time they meet again on the track for the first time when they are both running in the same direction

$=\frac{\text{Length of the track}}{\text{The relative speed of A with B in the same direction}}$

Answer:

Given: Length of the track = 1600 m

The relative speed of A and B with both going in the same direction is 45 – 27 = 18 km/hr

⇒ $18 \times \frac{5}{18} = 5$ $\text{m/s}$

The time they meet again on the track for the first time when they are both running in the same direction

$=\frac{\text{Length of the track}}{\text{The relative speed of A with B in the same direction}}=\frac{1600}{5} = 320$ $\text{seconds}$

Hence, the correct answer is 320 seconds.

Q3. In a circular race of 4225 m, X and Y start from the same point and at the same time, at speeds of 54 km/hr and 63 km/hr, respectively. When will they meet again for the first time on the track, when they are running in opposite directions?

1. 140 seconds

2. 150 seconds

3. 130 seconds

4. 120 seconds

Hint: When two persons are moving in opposite directions, the relative speed is equal to the sum of their speeds.

Answer:

Relative speed = $54 + 63 = 117$ km/hr

Time is taken for both to meet for the first time on the track

$=\frac{\text{Distance}}{\text{Speed}} = \frac{4.225}{117} = 0.036 \text{ hours} = 0.036 \times 3600$ seconds = 130 seconds

Hence, the correct answer is 130 seconds.

Q4. In a 200-metre linear race, if A gives B a start of 25 m, A wins the race by 10 seconds. Alternatively, if A gives B a start of 45 m, the race ends in a dead heat. How long does A take to run 200 m?

1. 78 seconds

2. 77 seconds

3. 78.5 seconds

4. 77.5 seconds

Hint: Find the total time A takes to run 200 m using the formula, $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

Answer:

Given: In a 200 m linear race, if A gives B a start of 25 m, A wins by 10 seconds.

Alternatively, if A gives B a start of 45 m, the race ends in a dead heat.

So, the additional 20 m start given to B compensates for the 10 seconds.

i.e., B runs 20 m in 10 seconds.

Hence, B will take 100 seconds to run 200 m.

Also, A gives B a start of 45 m.

Since B runs 20 m in 10 seconds, 45 m will be covered in 22.5 seconds.

A will take 22.5 seconds less than B, or, 100 – 22.5 = 77.5 seconds, to

complete the race.

Hence, the correct answer is 77.5 seconds.

Q5. A is twice as fast as B and B is thrice as fast as C is. The journey covered by C in $1\frac{1}{2}$ hours will be covered by A in:

1. 15 minutes

2. 20 minutes

3. 30 minutes

4. 1 hour

Hint: Use the formula, to find the time required for A to cover the journey.

$\text{Time}=\frac{\text{Distance}}{{\text{Speed}}}$

Answer:

Given: That A is twice as fast as B and B is thrice as fast as C, it means that A is six times (2 × 3) as fast as C.

$\text{Time}=\frac{\text{Distance}}{{\text{Speed}}}$

If C covers a journey in $1\frac{1}{2}$ hours or $\frac{3}{2}$ hours, then A, being six times faster, will cover the same journey in $\frac{3}{2 × 6} = \frac{1}{4}$ hours.

So, A will cover the journey in $\frac{1}{4}$ hours or $\frac{1}{4}$ × 60 minutes = 15 minutes

Hence, the correct answer is 15 minutes.

Q6. In a 500-metre race, the ratio of speeds of two runners P and Q is $3:5$. P has a start of 200 metres, then the distance between P and Q at the finish of the race is:

1. P wins by 100 metres

2. Both reach at the same time

3. Q wins by 100 metres

4. Q wins by 50 metres

Hint: Use the formula: Distance = Speed × Time

Answer:

Given: In a 500-metre race, the ratio of speeds of two runners P and Q is $3:5$ and P has a start of 200 metres.

Distance to be covered by A = (500 – 200) = 300 metres

Time taken by A to cover 300 metres = $\frac{300}{3}=100$ time units

Distance covered by B in 100 time units,

⇒ (100 × 5) = 500 metres

Hence, the correct answer is 'Both reach at the same time'.

Q7. In a 1 km linear race, P beats Q by 120 metres or 30 sec. What is the time taken by P to cover the race?

1. 220 sec

2. 250 sec

3. 235 sec

4. 240 sec

Hint: P beats Q by 120 metres or 30 sec, which means Q covers 120 m in 30 sec, and use the formula: Distance = Speed × Time.

Answer:

P beats Q by 120 metres or 30 sec, which means Q covers 120 m in 30 sec.

Speed of Q = $\frac{120}{30}=4$ m/s

Time taken to cover 1000 m = $\frac{1000}{4} = 250$ sec

P beats Q by 30 seconds, which means P reached 30 seconds early.

Time taken by P to complete the race = 250 – 30 = 220 sec.

Hence, the correct answer is 220 sec.

Q8. Two friends, P and Q, start running around a circular track from the same point. They run in the same direction. P runs at 6 m/sec and Q runs at b m/sec. If they cross each other at exactly two points on the circular track and b is a natural number less than 6, how many values can b take?

1. 2

2. 1

3. 4

4. 3

Hint: If the ratio of their speeds is $V_1:V_2$, the distinct meeting points will be $V_1-V_2$.

Answer:

If the ratio of their speeds is $V_1:V_2$, the distinct meeting points will be $V_1-V_2$, with both running in the same direction. Try different values of $b$ from 1 to 5 and check which satisfies $V_1-V_2 = 2$

$\frac{V_1}{V_2}=\frac{6}{b}$; number of points is 2

Given: b is a natural number less than 6

Checking for $b$ = 1: $\frac{V_1}{V_2}=\frac{6}{1}$, the difference is 6 – 1 = 5 (not satisfied).

Checking for $b$ = 2: $\frac{V_1}{V_2}=\frac{6}{2}$$=\frac{3}{1}$, the difference is 3 – 1 = 2 (satisfied).

Checking for $b$ = 3: $\frac{V_1}{V_2}=\frac{6}{3}$$=\frac{2}{1}$, the difference is 2 – 1 = 1 (not satisfied).

Checking for $b$ = 4: $\frac{V_1}{V_2}=\frac{6}{4}$ $=\frac{3}{2}$, the difference is 3 – 2 = 1 (not satisfied).

Checking for $b$ = 5: $\frac{V_1}{V_2}=\frac{6}{5}$, the difference is 6 – 5 = 1 (not satisfied).

The condition is satisfied only when b = 2. So, b can take only one value.

Hence, the correct answer is 1.

Q9. A, B, and C run simultaneously, starting from a point, around a circular track of length 1200 m, at respective speeds of 2 m/s, 4 m/s, and 6 m/s. A and B run in the same direction, while C runs in the opposite direction to the other two. After how much time will they meet for the first time?

1. 10 minutes

2. 9 minutes

3. 12 minutes

4. 11 minutes

Hint: First, find the time when A will meet B and C separately. Use this information to solve the problem.

Answer:

Circular Track length = 1200 m

Speed of A = 2 m/s

Speed of B = 4 m/s

Speed of C = 6 m/s

The relative speed of A and B = 4 – 2 = 6 m/s

The relative speed of A and C = 2 + 6 = 8 m/s

A and B will meet on the track after every = $\frac{1200}{2}$ = 600 s

A and C will meet on the track after every = $\frac{1200}{8}$ = 150 s

Thus, A, B, and C will meet together for the first time on the track after = LCM (600, 150) = 600 seconds

$\therefore$ They will meet for the first time after = $\frac{600}{60}$ = 10 min

Hence the correct answer is 10 minutes.

Q10. Having started from the same point simultaneously, two runners P and Q are running around a circular track of length 500 m in opposite directions with speeds of 6 m/s and 10 m/s, respectively. If they exchange their speeds after meeting for the first time, who will reach the starting point first?

1. Q

2. P

3. Both P and Q will reach at the same time

4. No one of the P and Q

Hint: Calculate the relative speed of P and Q first, then find the time and distance at which they meet for the first time using Distance = Speed$\times$Time.

Answer:

Two runners P and Q run in a circular track of length with speeds 6 m/s and 10 m/s.

The time they meet for the first time $=\frac{\text{Length of Circular Track}}{\text{Relative Speed}}$

$= \frac{500}{10 + 6} = \frac{500}{16} = \frac{125}{4}$ sec

Now, Distance covered by P $= 6\times \frac{125}{4}=\frac{375}{2}$ m

And distance covered by Q $=10\times \frac{125}{4}=\frac{625}{2}$ m

After interchanging their speeds,

Speed of P = 10 m/s

Speed of Q = 6 m/s

The remaining distance for P and Q will be equal to (500 – Distance covered respectively)

Time taken by P to reach the starting point $= (500-\frac{375}{2})\times\frac{1}{10}$

$= \frac{625}{20}=\frac{125}{4}$ sec

Time taken by Q to reach the starting point $=(500-\frac{625}{2})\times\frac{1}{6}$

$= \frac{375}{12} = \frac{125}{4}$ sec

$\therefore$ They will reach the starting point at the same time.

Hence, the correct answer is “both P and Q will reach at the same time”.