DISTANCE = SPEED * TIME
Distance will remain constant. Hence if there are two sets of speed and time as (S1,T1) and (S2,T2) in covering the same distance D
Rahul normally leaves his home at 8am and reaches his office at 9am. But today he increased his speed 1.5 times. What time did he reach the office?
Given data:
T1 = 1 hour S2/S1 = 1.5 = 3/2
Solution:
Using the formula T1/T2 = S2/S1
T2 = T1 * S1/S2 = 1 hour * 2/3 = 2/3 hours = 40 mins.
Hence he will reach his office at 8am + 40 mins = 8:40 am
Malini drives from her home to office and reaches in 1 hour. But today she travelled half the distance at 2/3 rd of the normal average speed. At what average speed should she cover the remaining distance so that the overall time taken still remains 1 hour.
As Malini takes 1 hour for the entire distance, for half of the distance it is 30 minutes at normal speed.
Using the formula T1/T2 = S2/S1, time taken for the first half at 2/3 rd speed,
T2 = T1S1/S2 = 30 mins * 3/2 = 45 minutes.
Hence Malini has to cover the remaining half of the distance in 1hour – 45 mins = 15 minutes. Hence using the formula T1/T2 = S2/S1 for the remaining second half,
S2 = T1S1/T2 = (30mins/15mins) * S1 = 2*S1
Hence Malini has to double the original speed for the second half of the distance
Karthik cycles to his school at a certain average speed in 2 hours. On a given day, he covered first half at 2/3rd of his normal speed and the second half at 3/2 nd of his normal speed. What is the difference in time taken to reach the school on that day?
As Karthik, takes 2 hours for the entire distance, for half of the distance it is 1 hour at normal speed.
For the first half time taken T2 = T1S1/S2 = 60 mins * 3/2 = 90 mins
For the second half time taken T2 = T1S1/S2 = 60 mins * 2/3 = 40 mins
Total time taken = 90+40 = 130 mins which is 10 mins more than normal time (of 2 hours or 120 mins). So he is late to school by 10 mins.
AVERAGE SPEED FOR ONWARD AND RETURN JOURNEY = 2xy/(x+y)
x – onward journey speed
y – return journey speed
It is WRONG to calculate the average speed as (x+y)/2
Vijay travelled to and from from Chennai to Bangalore. He covered a total distance of 720 kms (including onward and return) in 12 hours. If his onward journey speed was 45 kmph, find the return journey speed.
Average speed for onward and return journey = 720kms/12 hrs = 60 kmph
Using the formula, 60 = 2xy(x+y).
As x=45 kmph, 60 = 2*45*y(45+y)
90y = 60*45 + 60y
y = 60*45/30 = 90 kmph
Return journey speed = 90 kmph.
Miranda drove for 3 hours at a rate of 40 miles per hour and for 2 hours at 50 miles per hour. What was her average speed for the whole journey?
Average speed for the whole journey= Total Distance/ Total Time Taken
Total Distance = 3*40 + 2*50 = 220 miles.
Total Time Taken = 3 + 2 = 5 hours.
Average speed for the whole journey= 220/5 = 44 mph
A van whose speed is 100 kmph starts from point A to B and a truck whose speed is 80 kmph starts from point B to A simultaneously. If the distance between A and B is 540 kms, after how many hours will they meet each other?
Time Taken To Meet = Difference In Distance / Relative Speed
= 540/(100+80) = 3 hours
A truck starts from Bangalore to Delhi at 6am. A bus starts from Bangalore to Delhi at 8am and travels in the same path as the truck. If the speed of truck is 60kmph and that of the bus is 80kmph, at what time will the bus overtake the truck?
Difference in distance = Distance travelled by truck from 6am to 8am = 2 * 60 = 120kms
Relative speed = 80 – 60 = 20 kmph
Time taken to meet (overtake) = 120/20 = 6 hrs.
Hence the bus will overtake the truck at 8am + 6hrs = 2pm.
Two trains start at the same time, from stations A and B towards each other. After meeting each other on the way they take 5 hours and 20 hours respectively to arrive at their destinations B and A. If the speed of the train that started from A is 50 kmph, what is the speed of the other train?
Let the time taken from the time of start to meet each other be m hours
A 200 m long train travelling at 72 kmph crosses a platform in 20 seconds. Find the length of the platform.
Note: The last compartment of the train also should cross the platform to consider that the train has crossed the platform
Speed of train = 72 kmph = 72 * 5/18 m/sec= 20 m/sec.
Let the length of the platform be x metres. The train crosses the platform in 20 seconds.
From the diagram, the train covers its length plus the platform length in 20 seconds.
Hence Distance covered = 20 sec * 20 m/sec = 200m + x
x = 400m-200m = 200m.
The length of the platform is 200 metres.
Balu would reach this office 15 minutes early if he walked at 4 kmph from his house. He would reach it 45 minutes late if he walked at 3 kmph from his house. Find the distance between his house and office.
Let time taken at 4 kmph be x hours. So T1 = x and S1=4
When the speed is 3kmph, T2 = x hours + 60 minutes (15 minutes early + 45 minutes late) = x + 1 hours. So T2=x+1, S2=3
Distance = T1*S1 = T2*S2
4x = 3 (x+1), Solving x=3 hours.
Distance = T1S1 = 4*3 = 12 kms.
In a 100 m race, A beats B by 10 m and C by 13 m. What is the distance B will beat C in a race of 180 m?
SC = SA – 13 = 100-13 = 87 metres.
From the diagram, B beats C by 3 metres when B has covered 90 metres. (BC = SB-SC = 90-87)
Let B beat C by x metres in 180 metres race.
90 : 3 = 180 : x
Hence x= 180*3/90 = 6 metres.
In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. What is the distance A can beat C in the same race?
When A covers 100m, B covers 75m. Hence A/B = 100/75
When B covers 100m, C covers 96m. Hence B/C = 100/96
A/C = A/B * B/C = 100*100/(75*96) = 4*25/(3*24) = 25/18.
That is when A covers 25m, C covers 18m.
Hence when A covers 100m, C covers 100*18/25 = 72m.
So A beats C by 100-72 = 28m
A and B are running in a circular track in a direction opposite to C. C is running at twice the speed of A and thrice the speed of B. They start running from the same point. If the average speed of A is 3m/s and the track is 120 metres in circumference, When will A,B,C meet at the starting point from the time of their start?
Irrespective of the direction of running,
A takes 120/3 = 40 seconds to complete one round. B takes 120/2 = 60 seconds. C takes 120/6 = 20 seconds.
L.C.M of 40,60,20 = 120 seconds.
Hence they will meet after 120 seconds or 2 minutes after the start.
