Two cars (A and B) start at the starting line at the same time on a 3-mile long circular track, going in opposite directions. As they drive around the course, they pass each other many times. Exactly one hour after starting, they pass each other for the 33rd time. At this point car A has completed exactly 20 laps. What is the average speed of Car B?

A motorcyclist has to cover a distance of 200 km to reach city b from city a. After travelling a certain distance, his motorcycle develops a problem and travels at 3/4th of its original speed, thereby he reached b 1 hour late. Had the problem developed 30 km earlier, he wud hav reached b 12 minutes later. Find theÂ initialÂ distance itÂ travelled without a problem and the speed over that part of the journey

Difference of 12 minutes comes because of 30 km.Â

--> Difference of an hour comes because of 150 km.Â

Therefore, the problem happened afterÂ **200 - 150 = 50 km**.Â

Let the original time taken for this 150 km track was t.Â

With 3/4^{th}Â speed, new time will be 4t/3.

Extra time = 4t/3 - t = 1 h (given) --> t = 3h. The train takes 3h to cover 150 km. Therefore, itsÂ **normal speed is 50 kmph**. Speed after the problem = 3/4^{th}Â of the speed =Â **37.5 kmph**

Appu and Gappu both start running around a circular racetrack in the same direction, each of them going at a constant speed. Appu is on a bicycle and Gappu is on foot. Appu goes all the way around the track and catches up with Gappu. The moment he meets Gappu, he turns around and heads back the starting point while Gappu keeps on going on his path. When Appu reaches the starting point he again meets Gappu who is just finishing his first round. What is the ratio of the speeds of Appu and Gappu?

Hi Nilesh !Â

Let A is the starting point and B is the point where Appu and Gappu meet the first time .Â

Let the smaller distance between A and B is y and larger one is x .Â

x/ ( 2x + y ) = y/xÂ

x^2 = 2xy + y^2Â

dividing the equation by y^2 ,Â

(x/y)^2 = 2x/y + 1Â

on solving we get x/y = 1 + sqrt{2} or 1 -Â Â sqrt{2}Â

Hence ,theÂ ratio of speeds of appu and Gappu isÂ 1 + sqrt{2} : 1 .Â

Ram starts on his Honda Civic from CL center (A) towards Workshop arena (B). Ramesh starts on foot from the same point, A towards point B at the same time. After some time, Ramesh calls up Ram, just when Ram has traveled half the distance. Ramesh continues towards B while Ram turns back. He picks up Ramesh and together they turn back again and reach B, taking 1.5 times the time Ram would normally take while Ramesh saves 15 minutes. If AB = 3 kms, the the speed of Ramesh is

Â

Hi Kinshuk,

Whenever both cars meet at any point, it means that they have covered a lap (3 miles).

So, total no. of laps=33.

Now, if car A has completed 20 laps, then car B would have covered 33-20=13 laps.

avg. speed of car B= 13*3 miles / 1 hr =Â Â 39 miles/hr

Two cars P and Q start simultaneously from points A and B respectively at different speeds and travel towards each other. They meet each other in two hours. After that, P takes 3 hours less to reach B than the time Q takes to reach A. If the distance between A and B is 540km, find the speed of the car Q (in kmph)

- 90
- 120
- 60
- 180

Distance = 540 km

Let speed of P is S_{p} and Q is S_{q}

(540)/(S_{p} + S_{q}) = 2

=> S_{p} + S_{q} =270 â€¦â€¦â€¦â€¦â€¦.1)

Distance travel by P in 2 hours is equal to distance travel by Q in t hours.

And distance travel by Q in 2 hours is equal to distance travel by P in ( t-3) hours.

S_{q} x 2 = S_{q} x ( t-3) â€¦â€¦â€¦â€¦â€¦â€¦.2)

S_{q} x t = S_{p} x 2â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.3)

2) Divided by 3)

2/3 = (t-3)/2

=> 4 = t^{2} â€“ 3t

t^{2} â€“ 3t â€“ 4 = 0

t =4

Put t = 4 in equation 3)

S_{q} x 4 = S_{p} x 2

2S_{q} =S_{p}

Now from eq. 1)

2S_{q} + S_{q} =270 => S_{q} =90 km/hr

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The distance between two towns Aurangabad & Jalna is 80KM. A bus left Aurangabad & traveled with constant speed. Thirty minutes later, Deepak left Aurangabad for Jalna in his car. He overtook the bus in 30 minutes and continued on his way to Jalana. Without stopping at Jalna, he turned back & again encountered the bus 80 minutes after he left Aurangabad. Determine the speed of the bus.

Deepak travels in 30 min what the bus travels in 30 + 30 = 60 min --> Speed of Deepak = twice that of the bus.Â

Now he encounters the bus the second time 50 min after the first encounter. Let the remaining distance from the first meet to the finish be d. Now the total distance traveled by Deepak and bus in 50 min = 2d. The bus will travel 1/3rd of that i.e. 2d/3 and Deepak will travel 2/3rd i.e. 4d/3. So Deepak will travel d distance in 3/4Â Ã— 50 = 37.5 min.Â

Total time taken by Deepak = 30 + 37.5 = 67.5. Therefore, total time taken by the bus = 2Â Ã— 67.5 = 135 min = 2.25 hr

Speed = 80/2.25 = 35.5 km/h

Please solve.

A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead, he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is:

Hello Tina,Â

Please find the solution,Â

t1 -t2 = 20 minutesÂ

(d/12) -(d/15) =(20/60)

d/60 = 20/60

d=20 km

If Rajesh drives his bike at a speed of 40km/hr from his office, then he reaches him home 10 min late. If he drives at 60km/hr he reaches 10 min early. Find the distance between his office and home?

Let the distance between his home and office is 'd' kms .Â

Â

Then , d/40 - 20/60 = d/60Â

Â

d = 40kms.Â

A distance is covered at a certain speed in a certain time. If the triple of this distance is covered in four times the time, then what is the ratio of the two speeds?

Hi Nilesh ðŸ™‚Â

Â

Let the normal speed = s = D/tÂ

and new speed = s1 = 3D/4tÂ

Â

s : s1 = D/t : 3d/4t = 4 : 3 .Â

In a kilometer race, if A gives B a head start of 40m, then A wins by 19 seconds. If A gives B a head start of 30 seconds, then B wins by 40m. The time taken by each of them to run a kilometer race (in seconds) are?Â

Hello Samyak!

When A gives B a head start of 40 m he wins by 19s , therefore when A travels 1000m B travels 1000 - 40 - 19v m, where v is the speed of B.Â

ratio of speeds of A and B = 1000 / ( 960 - 19v )..... (1)

and when A gives B a headstart of 30sÂ B wins by 40 m , therefore when A travels 960m B travels 1000 - 30v mÂ

Ratio of speeds of A and B = 960/ ( 1000 - 30v)Â ..... (2)Â Â

from (1) and (2) v = 6.66 m/sÂ

Hence time taken by B to travel 1000m = 1000/6.66 = 150sÂ

and time taken by A to travel 1000m =Â 125 sÂ

A, B, C start from the same point on a linear track and travel in the same direction at constant speeds. The ratio of speeds of A,B and C is 3:4:6. The ratio of time taken by B and C to overtake A from the time after they start is 2:3. B and C started X and Y hours respectively after A's start. What is the ratio X:Y?Â

2:9 ?

Yes