Let the distance between the 2 cars be D km.

Let the speeds of the two cars be ‘a’ and ‘b’ respectively and a > b.

Case I)

Cars are moving in the opposite direction (towards each other)

Relative speed = a + b

Time taken = 1.5 hrs

Case II)

Cars are moving in the same direction (Car A chasing Car B)

Relative speed = a – b

Time taken = 10.5 hrs

In both the cases the distance between the cars is the same, ‘D’.

But the time taken is in the ratio 1.5 : 10.5 or 1 : 7

Therefore, the speeds will be in the ratio 7 : 1

a + b = 7(a – b)

8b = 6a

3a = 4b

Substituting b = 60 kmph,

We get, a = 80 kmph.

D = (a + b) * 1.5 = 210km

When they move towards each other the distance covered by them is in the ratio 4 : 3 and the total distance covered by them together is 210 km.

The slower car, B, covers 3/7 th of this 210 km which is 90 km.

 Let us denote the port by P,

the slower ship by S.

the faster ship by F.

 Triangle APB is right-angled, We know that angle APB is 60 degrees, because triangle SFA is equilateral. The right angle must be at point B, because angle PAB is less than 60 degrees.

sin(30) = 1/2=BP/AP

BP = 0.5(AP) = 8

BF = FP – BP = 24 – 8 = 16

That mean, the faster ship is twice as fast as the slower one.

So, when the faster ship reaches the Port covering 24 km, the slower ship covers only 12 km and has remaining 12 km left to cover.

Let the meeting point of the two trains between stations X and Y be M.

Let the time taken to reach M from X by train A be ‘t’ minutes. Since the two trains start simultaneously, the time taken by train B to reach M from Y will also be ‘t’ minutes.

We know that train A completes the entire journey in 10 minutes, so the time taken by train A to travel from M to Y will be ‘10 – t’ minutes. We are told that train B takes 9 minutes to reach X after it meets train A, which means, train B takes 9 minutes to travel from M to X.

When two bodies travel equal distances at constant speeds, the ratio of time taken by them to travel those distances will be the same.

The time taken by A and B to travel the distance between A and M is ‘t’ minutes and 9 minutes respectively. Similarly, the time taken by them to travel the distance between M and B is ‘10 – t’ and ‘t’ minutes.

The ratio of the times taken should be equal.

t/9=10−t/t

t^2=9(10−t)

t^2=90−9t

t^2+9t−90=0

t^2+15t−6t−90=0

t(t+15)−6(t+15)=0

(t+15)(t−6)=0

t = – 15 or t = 6

t can’t be negative, therefore, t is 6.

The total time taken by train B to reach station X from station Y is 6 + 9 =

15 minutes.

 Two trains A and B were moving in opposite directions, their speeds being in the ratio of 5 : 3.

The speed of train A is 5x.

The speed of train B is 3x.

Let’s assume length of train A and train B are La and Lb

The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.

So, the distance travelled is Lb. For this, the time taken is 46 sec.

Trains took another 69 seconds for the rear ends of the trains to cross each other.

So, the distance travelled is La. For this, the time taken is 69 sec.

La/5x+3x/Lb/5x+3x

 =69/46

The relative speed is (5x + 3x), since they are moving opposite to each other.

La/Lb

 = 3/2

Now both the trains A and B have 90 – 60 = 30 km to cover between them.

The relative speed of the two trains will be 40 + 20 = 60kmph.

With a relative speed of 60kmph, the two trains cover 30km in half an hour.

Hence the two trains will meet at 11: 00 AM

 Let the length of the race track be ‘x’ km.

By the time A finishes the race, B lags by 45km.

That is, in the same time, while A runs x kms, B runs (x-45) kms

Ratio of Speeds of A and B = x/x−45

By the time B finishes the race, C lags by 50km.

That is, in the same time, while B runs x kms, C runs (x-50) kms

Ratio of Speeds of B and C = x/x−50

By the time A finishes the race, C lags by 90km.

That is, in the same time, while A runs x kms, C runs (x-90) kms

Ratio of Speeds of A and C = x/x−90

(Ratio of Speeds of A and B) × (Ratio of Speeds of B and C) = (Ratio of Speeds of A and C)

x/x−45× /xx−50 = x/x−90

x/x−45 = x−50/x−90

If ab = c/d

; then a/b = c−a/d−b

Threfore, x/x−45 = x−50−x/x−90−(x−45)

x/x−45 = −50−45= 109

x/x−45

 = 10/9

9x = 10x – 450

x = 450.

or

Ratio of Speeds of A and B = 10/9

By the A covers x, B covers 0.9x

Since A beats B by 45km, 0.1x = 45

x = 450.

 R1 = 100m and R2 = 20m

Speed of Ram = 15km/hr and Rahim’s speed = 5km/hr

Time taken by Ram = 2π(100)/15

Time taken by Rahim = 2π(20)/5

Ratio of the time taken by Ram : Rahim = 5 : 3

After 15 units of time Ram and Rahim will be at starting point and meeting

That time ram would have done 3 rounds

 Let’s take speed of the train to be x and the time taken be t

Since speed is reduced to 1/3

 rd,

New speed = x/3

Since the speed is one-third, time taken will be tripled. T = 3t

This 3t is after the scheduled time, So extra 2t = 30 mintues

t = 15 minutes

Train travels at x km/hr takes 15 minutes and

Train travels at x/3

 km/hr takes 45 minutes

So, the train usually takes 15 minutes to cover the distance.

It travels 5 minutes at the usual speed. That is, it travels 1/3

 rd of the time at the usual speed. So it covers 1/3

 rd of the distance in 5 minutes.

To reach it’s destination in the on time, the train has to travel the remaining 2/3

 rds of the distance in 10 minutes. Since the train halts for 4 minutes, it should now cover the 2/3

 rds of the distance in 6 (10 – 4) minutes.

In other words, the train has to cover the same distance in 6/10

 th of the usual time.

In order to do so, the speed must be 10/6

 ths of the usual speed. Or the increased speed will be 4/6

 ths or 2/3

 rds of the usual speed. Which is an increase of 66.66% or nearly 67%.

 Two person to be A and B

A = 2km/hr and B = 4 km/hr

Speed of the train = t km/hr

Given, Train crosses A in 90 seconds (Length of the train is the distance)

t – 2 in 90 seconds and t – 4 in 100 seconds

Ratio of speed is opposite to time

(t – 2) 90 = (t – 4) 100

Solving t = 22km/hr

For A, Speed = t – 2 = 20 km/hr in 90 seconds

So travelling at 22 km/hr —-> 90×20/22 ≅ 82

 Speed are 8 km/hr and 15 km/hr

So it takes 15t times and 8t times

Difference of times = 15t – 8t = 7t

7t = 35 minutes (10:15 min – 9:40 min)

t = 5 minutes

He takes 75 minutes by travelling at 8 km/hr and 40 minutes if travelled at 15 km/hr

He starts at 9 am.

So, 9:10 to 10:00 = 50 minutes

15 km/hr = 40 min

And at what speed he should travel, so as to reach there within 50 minutes

15×40/50

 = 12 km/hr

 It is given that the cyclist starts at 10:00 am from A and reaches B at 11:00 am

Now, Motorcyclists start every minute from 10:01 am, and 45 such motorcyclists reach B before 11:00 am

If they leave one by one every minute, the 45th motorcyclist would have left by 10:45 am to reach B at 11:00 am.

Thus, time taken by one motorcyclist to reach B from A = 15 minutes.

Now, the cyclist doubles his speed. This means, he reaches B at 10:30 am

So, the last motorcyclist should have left A by 10:15 am

Thus, 15 motorcyclists would have reached B by the time the cyclist reaches B

Let the track length for John be ‘a’ and for Mary be ‘b’

So, Distance travelled by John = 9a

Distance travelled by Mary = 5b

Now, Time taken by John = hours

Time taken by Mary = hours

We know that Time taken by John = Time taken by Mary

= 

a = b

Total track length = 325 meters

So, b + b = 325 meters

b = 325 meters

b = 225 meters

a = 100 meters

Mary jogs at 7.5 Kmph = 7.5 x 

So, time taken = = 48 seconds

By the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.

So, the speeds of A and b are in the ratio 60:40 or 3:2
Hence, the time they take to cover a particular distance will be in the ratio 2:3

We know that A covers 60% of the distance at 10:00 AM and covers 100% of the distance at 10:12 AM.

That means A takes 12 minutes to cover 40% of the track. So to cover the entire track he must have taken 12+12+6 = 30 minutes.
(because 40% + 40% + 20% = 100%)

Since the time taken by A and B to complete the track are in the ratio 2:3, the time taken by B to complete the track will be 45 minutes.

At 10:00 AM, B has covered 40% of the track. If we can find out what time does B take to complete the remaining 60% of the track, we can find the finish time of B.

Time required to complete 60% of the track = 60% of 45 = 27 minutes.
Hence, B complete one single round at 10:27 AM.

The minimum time travelled by Car 1 = 6 hours.
Minimum time travelled by Car 2 = 5 hours
Now, the time taken can be of any value -> (6,5) (7,6) …… (1000,999)
However, since we want to calculate the highest possible percentage by which speed of car 2 could exceed car 1, we consider the lowest time taken – 6 hours and 5 hours, respectively.
Let the distance between the start and finish be 30 Kms (LCM of 6 and 5)
So, Car 1 travels at 5 Kmph and Car 2 travels at 6 Kmph
Percentage increase in speed by Car 2 = x 100 = x 100 = 20%

A beats B by 11 meters. When B completes the 11 meters, there is a lead of 80 meters to C
So, C must have travelled only 90 – 80 = 10 meters
When B travels 11 meters, C travels only 10 meters
Ratio of distance travelled by second and third horse are11x and 10 x respectively
We know that the second horse beats the third horse by 80 meters.
So, Length of the track = Distance travelled by the second horse = 11 x 80 = 880 meters

Let the distance be = 60 Kms.
So, Bimal travels a distance of 20Kms in each mode and Amal travels 10, 20 and 30 Kms respectively in 1 hour each
Time taken by Bimal = Time taken to travel 20 kms in 10 Kmph + Time taken to travel 20 kms in 20 Kmph + Time taken to travel 20 kms in 30 Kmph
Time taken by Bimal = 2 + 1 + hours = 3 + hours
Extra time taken by Bimal = hour
Percentage increase in time = x 100 = x 100 = 22
So, Percentage increase in time = 22

Circumference of A and B are in the ratio 3: 4
So, Ratio of Distance travelled in one revolution by A and B = 3: 4
Since they travel the same distance,
Ratio of number of revolutions of A and B = 4: 3 —– (1)
We know that each wheel of A requires 5000 more revolutions than B
So, the Ratio of number of revolutions of A and B = (n + 5000): n ——(2)
So, comparing (1) and (2)
Number of revolutions of A and B are 20000 and 15000 respectively
So, we know Bike B does 15000 revolutions in 45 minutes
Distance travelled = 2 x π x r x Number of revolutions
Speed of Bike B = mph
Speed of Bike B = 2 x x x 5 x 4 Kmph
Speed of Bike B = 16 Kmph

It is given that points A, P, Q and B lie on the same line such that P, Q and B are 100 km, 200 km and 300 km away from A. Let us draw the diagram first. All of them are on the same straight line and P, Q lie between A and B.
Cars 1 and Cars 2 leave A at the same time and move towards B. Simultaneously, Car 3 leaves B and moves towards A. Let us add more details to the diagram. C1 and C2 move towards B and C3 moves towards A. C3 meets C1 at Q and C3 meets C2 at P.

Let us assume the speeds of Car 1, 2, and 3 to be C1, C2 and C3 respectively. Car 1 is travelling quicker than Car 2. When Car 1 travels 200 km to reach Q, Car 3 has travelled 100 km or ratio of their speeds, C1 : C3 is 2 : 1.C1 and C2 move towards B and C3 moves towards A. C3 meets C1 at Q and C3 meets C2 at P.
Let us assume the speeds of Car 1, 2, and 3 to be C1, C2 and C3 respectively.When Car 3 travels 200 km to reach P, Car 2 has travelled 100 km or ratio of their speeds, C3 : C2 is 2 : 1. Now we have all the ratios. We know that C1 : C3 is 2 : 1 and C3 : C2 is 2 : 1. We can see that C3 is the common link


So, taking LCM of C3, we get,

We can see that ratio of speeds of Car 2 to Car 1 is 1 : 4. Hence, C2 : C1 = 1 : 4

Given that A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time.
They meet after one hour or their relative velocity is (a + b)km/hr and their relative distance is 350kms
a + b = 35013501 which is equal to 350 kms/hr

If they both move towards east, then they meet in 7 hrs.
We have to find the difference between their speeds, in km per hour. B is catching upon A, so once again their relative distance will be 350kms.
Relative speed = b – a , B travels faster so that it can meet a.
⟹ 350b−a350�−� = 7
⟹ b – a = 50 km/hr
We know that b + a = 350 km/hr
By this we can find the values of a and b but they have asked for only the difference between their speeds which is b – a = 50km/hr

Given that points A and B are 150 km apart.
Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A.
Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km.
The Car 1 is 20km away from A and it travels at 100kmph and the time taken is 2010020100 = 1515hr
So car 2 is 12 minutes behind car 1.
We have to find the distance in km, between car 2 and B when car 1 reaches B.

If car 1 reaches B, car 2 will take 12 minutes to reach b.
Distance between car 2 and B is 25 × 1515 = 5 kms
The distance, in km, between car 2 and B when car 1 reaches B is 5 kms.

Let Narayanan take X hrs to reach B then Partha would take X + 4 hrs
Its given that Partha reaches the mid-point of A and B two hours before Narayan reaches B
X+42�+42 = X – 2 => X=8 hrs
So, Partha would take 8+4 = 12 hrs to travel 60 Kms at a speed of 60126012 Kmph
Speed of Partha = 5 Kmph