10/02/2019 2:44 pm
TSD (Time speed and distance)
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A left pune at noon sharp. Two hours later, H startesd from pune in same direction. B overtook A at 8pm. Find the average speed of two trains over this journey if the sum of their average speeds is 70km/hr.
Gaurav Jain 12/02/2019 4:13 pm
This post was modified 6 years ago by Gaurav Jain
from 12 to 8
A and H covered same distance in 8 and 6 hours respectively
so the ratio of their speed will be 6:8 (s1/s2=t2/t1)
=>
SA/SH=3/4
let's say their speeds are 3x and 4x respectively
sum of their speed is 7x = 70 (given)
x=10
hence speeds are 30 and 40
average of speed is 35 kmph
1
10/02/2019 2:56 pm
Two trains, A and B , start at the same time from stations P And Q respectively towards each other. After passing each other , they take 12 hours and 3 hours too reach Q and P respectively. If A is moving at speed of 48km/hr, then what is speed of B?
Gaurav Jain 12/02/2019 4:07 pm
Let's say they meet at a point x, after t hours
speed of A is a and speed of B is b.
so distance PX = at
distance QX=bt
now after meeting at x train A still needs to cover 'bt' distance and
and train B needs to cover 'at' distance
ratio of time taken by both trains to cover respective distance = (bt/a) to (at/b) = 12/3
so
(b/a)2= (12/3)
a=48 (given)
so b= 96 kmph
1
10/02/2019 3:04 pm
Walking at 3/4 of hus normal speed , a man takes 2.5 hours more than the normal time. Find the normal time.
aniket prajapati 23/02/2019 9:52 pm
Let earlier speed is 4 now it becomes 3
So time ratio becomes 3:4, difference 1x=>2.5
Actual i.e,3x=7.5
tothepoint 15/07/2019 11:46 am
Speed is reducing by 1/4th, hence the time will be increased by 1/3rd. This increase is given to be 2.5 hours. => 1/3 of the actual time = 2.5 hours => Actual time =7.5 hours.
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06/03/2018 9:37 pm
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 = 39 miles/hr
0
10/03/2018 6:18 pm
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
Anonymous 10/03/2018 6:34 pm
Distance = 540 km
Let speed of P is Sp and Q is Sq
(540)/(Sp + Sq) = 2
=> Sp + Sq =270 …………….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.
Sq x 2 = Sq x ( t-3) ……………….2)
Sq x t = Sp x 2……………………….3)
2) Divided by 3)
2/3 = (t-3)/2
=> 4 = t2 – 3t
t2 – 3t – 4 = 0
t =4
Put t = 4 in equation 3)
Sq x 4 = Sp x 2
2Sq =Sp
Now from eq. 1)
2Sq + Sq =270 => Sq =90 km/hr
0
10/03/2018 6:20 pm
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.
Anonymous 10/03/2018 6:37 pm
Deepak travels in 30 min what the bus travels in 30 + 30 = 60 min --> Speed of Deepak = twice that of the bus.
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 3/4 × 50 = 37.5 min.
Total time taken by Deepak = 30 + 37.5 = 67.5. Therefore, total time taken by the bus = 2 × 67.5 = 135 min = 2.25 hr
Speed = 80/2.25 = 35.5 km/h
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16/03/2018 5:34 pm
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:
0
16/03/2018 6:11 pm
Hello Tina,
Please find the solution,
t1 -t2 = 20 minutes
(d/12) -(d/15) =(20/60)
d/60 = 20/60
d=20 km
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09/07/2018 1:51 am
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?
TG Team 09/07/2018 4:46 pm
Let the distance between his home and office is 'd' kms .
Then , d/40 - 20/60 = d/60
d = 40kms.
0
09/07/2018 1:54 am
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?
TG Team 09/07/2018 4:51 pm
Hi Nilesh 🙂
Let the normal speed = s = D/t
and new speed = s1 = 3D/4t
s : s1 = D/t : 3d/4t = 4 : 3 .
0
12/07/2018 6:42 pm
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?
TG Team 13/07/2018 1:57 pm
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.
ratio of speeds of A and B = 1000 / ( 960 - 19v )..... (1)
and when A gives B a headstart of 30s B wins by 40 m , therefore when A travels 960m B travels 1000 - 30v m
Ratio of speeds of A and B = 960/ ( 1000 - 30v) ..... (2)
from (1) and (2) v = 6.66 m/s
Hence time taken by B to travel 1000m = 1000/6.66 = 150s
and time taken by A to travel 1000m = 125 s
0
14/07/2018 8:54 pm
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?
0
20/07/2018 5:10 pm
A man can cover a certain distance in 3 hr36 min if he walk at the rate of 5km/hr. If he cover the same distance on bike at the rate of 24km/hr, then what will be time taken by him?
TG Team 20/07/2018 6:17 pm
Hello Nancy!
Distance travelled by the man in 3hr 36 min ( i.e 18/5 hours ) = (18/5) x 5 = 18 km .
Hence , to travel 18km with the speed of 24km/hr , it takes 18/24 = 3/4 = 45 minutes .
TG Team 20/07/2018 6:23 pm
Alternate Approach : The distance is constant in the case . Now the speed of 24km/hr is 24/5 times of his walking speed. Therefore , the time taken will be 5/24 times of 3 hr 36 min i.e = (5/24) x (18/5) = 45 minutes
0
20/07/2018 5:47 pm
If a boy walks from his house to school at the rate of 4 km/hr, he reaches school 10 min earlier. However if he walks at the rate of 3km/hr he reach 10 min late . find the distance of his school from his house?
TG Team 20/07/2018 6:49 pm
Let the time taken to travel the distance at the speed of 4km/hr be t hr .
Now the speed of 3km is 3/4 of its previous speed , therefore the time taken to travel the distance with the speed of 3km/ hr will be 4/3 of t .
Now , 4t/3 - t = 20min
t/3 = 1/3 hr
t = 1 hr .
Hence, the distance between his house to the school = 1 x 4 = 4km.
TG Team 31/07/2018 12:17 pm
Alternate Approach :
House __________________D __________________School
Let the original time taken be t then ,
D/4 = ( t - 10)/ 60 or t - 10 = 15 D .... (1)
and (t + 10)/60 = D/3 or t + 10 = 20D .... (2)
Solving (1) and (2)
We get D = 4km .
0
30/07/2018 2:49 pm
At the end of each completed minute between 12:00 and 23:59 we measured in degrees the angle between the hour and the minutes hand of a regular wall clock. What is the measurement of the smallest recorded angle.
A. 0.25
B. 0.5
C. 2.5
D. 3