Two buses start from a bus terminal with a speed of 20 km/h at interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8 minutes?
A. 3 km/h
B. 4 km/h
C. 5 km/h
D. 7 km/h
Answer: Option C
Solution(By Examveda Team)
Let Speed of the man is x kmph. Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.Or, $$20 \times \frac{{10}}{{60}} = \frac{8}{60}\times\left(20+x\right) $$
Or, 200 = 160 + 8x
Or, 8x = 40
Hence, x = 5kmph. Detailed Explanation: A _____________M_______________B A = Bus Terminal. B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well. Distance Covered by Bus in 10 min = AB = $$\frac{{20}}{{60}} \times 10$$ = $$\frac{{10}}{3}$$ km. Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes. Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each other. Let Speed of the man = x kmph. Relative speed = 20 + x To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 + x) kmph = distance covered by bus in 10 minuted with speed 20 kmph.
Join The Discussion
Comments ( 27 )
Related Questions on Speed Time and Distance
A. 48 min.
B. 60 min.
C. 42 min.
D. 62 min.
E. 66 min.
A. 262.4 km
B. 260 km
C. 283.33 km
D. 275 km
E. None of these
A. 4 hours
B. 4 hours 30 min.
C. 4 hours 45 min.
D. 5 hours
Distance travelled by man in 8 mins and distance travelled by bus in 2 mins must be equal.
So, Distance travelled by Bus in 2 mins would be 20 x 2/60 = 2/3
In a same way, Distance travelled by man in 8 mins will become = X x 8/60
Now, 2/3 = 8X/60 (because both speeds should be same)
24x = 120 (by Cross Multiplying)
x = 120/24
x = 5 Answer
1) find distance b/w both buses in their interval of 10 min = speed*interval =(20km0min) *10min =3.33km
(interval distance of 3.33km b/w busses is distance covered by man and 2nd bus in their combine speed)
2) find combine speed of man and 2nd bus = (3.33km/8min)*60min =24.9 km/hr
3) find man's speed from combine speed = 24.9- bus speed = 24.9-20 = 4.9 ~ 5 km/hr
10/3
Speed(X+Y)= 10×60/3×8= 25
25-20= 5
Good explanation in the commens
If i have to shiw in the diagram the how can i show the relative speed with yhe help og the diagram
Total distance travelled by both (BUS+MAN) to meet each other => total distance covered by BUS (only) at the speed of 20 k/h in 10 mins. So, Distance between them = 20*(10/60) = 20/6 km.
ATQ, BUS and MAN travelling in opposite direction and meet after 8 min.
So, as we know S = D/T, It should be like = (20/6) / (20+x) = 8/60 (Let, speed of the man is x km/h)
=> x = 5 km/hr.
Hope this will help you to understand and visualize the concept of this question.
It can be solved easily in 3-4 lines using frame of reference concept.
1st bus starts and after 10min 2nd bus starts.
Man is moving in opposite direction. If man doesn't move he would have met the second bus exactly at 10th min
As man is moving with some speed in opposite direction definitely he will be meeting the 2nd bus taking time less than 10min i.e 8min as mentioned in the question
After man meeting the 1st bus consider the relative speed between man and the 2nd bus.
Let the speed of man be 'x' km/hr and speed of 2nd bus is 20km/hr(given)
The distance travelled by the man and 2nd bus in 8min is the distance travelled by the 2nd bus in 10min
Distance travelled by the man and 2nd bus in 8min is (x+20)8
Now (x+20)8=20*10
By solving x=5km/hr (speed of man)
A bus starts from X to Y at 40 kmph. Half an hour later another bus leaves Y for X travelling at 45 kmph. If the distance from X to Y is 105 km, how far from X will the buses meet? *
Time 10 min se 8 min krna h... 2 km hora 10 m 1/5 se time km hora to 1/4 se speed bdhani hogi... 20*1/4=5kmph
Time. 10:8
Speed. 4:5
4 20kmph k brabr h to 5 25 kmph k brabr hoga...
There should be a correction
(20+x) should be written as numerator
I still don't understand the concepts
Distance travelled by bus1 in 10 minutes= 20 km/hr / (10/60) =3.33 km
bus2 and bus 1 have constant 3.33 km difference due to their constant speed and now man travels 3.33km in 8 minutes from bus1 to 2 . so his speed is 3.33 km / (8/60) =25km/hr
relative speed in opposite shuld be plus or minus??= 25-20 = 5km/hr
Correction needed, 20+x would be multiply with 8 as like below
20*10/60=(20+x)*8/60
I m not clear.can anyone expain with a graphical explanation?
M:b
t 8:2
s 2:8
According to question,
8u=20
Then
1u=5/2
Then 2u= 5/2×2=5
Is it correct to say that point 'B' should come in b/w points 'A' & M' ?
M:B
t. 8:2
S 2:8
20÷8×2=5 km/h
Which is speed of man.
another better way to solve the problem
what annual payment will discharge a debt of rs.6550 due in 4 years at 5% per annum simple interest?
solution will be
20*10/60 = 8(20+x)/60
Or, 200 = 160+8x
Or, 8x = 40
Hence, x = 5kmph.
but on the above " {8/60(20+x)} " was done. but the right process is " {8(20+x)/60} "
The relative distance covered by Bus and Man in 8 minutes should be equal to distance covered by bus in 10 minutes. I.e
Distance covered by bus and man together with relative speed 20 + x kmph = distance covered by bus in 10 minutes.
Sir e question ka answer kese aya.how a distance cover by a man and bus same hoga
i was not clear with the solution ,can anyone help me ?
How can you equal these distances. Give me a logic
Where its mentioned the distance covered by bus and man are same ???
very good