Time and Distance Questions Answers:-
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Time and Distance Formula :-
Time and Distance Tricks 1:- If a person travels half a distance in a journey at x km/hr and remaining half at the speed of y km/hr, the average speed of the whole journey is given by:
A person travels half of journey at the speed of 30 km/hr and the next half at a speed of 15 km/hr. What is the average speed of the person during the whole journey?
Here; x= 30 and y= 15 Putting the in above formula the Answer will be= 20 km/hr
Time and Distance Tricks 2:- If a person travels three equal distances in a journey at a speed of x km/hr, y km/hr and z km/hr. The average speed is given by:
Rajesh starts from Faizabad to Lucknow a distance of about 300 kms. He divided his journey into a distance of thee equal parts in terms of distance and covered them with the speed of 30 km/hr, 60 km/hr and 90 km/hr, calculate his average speed during the journey:
Here: x= 30, y= 60, z= 90. Putting the values we get: Answer= 49.09 km/hr
Time and Distance Tricks 3:- If a person travels Mth part of a distance at x km/hr, Nth part at y km/hr and the remaining Pth part at z km/hr, Then his average speed in km/hr is given by:
If instead of proportion, the parts of the distance are given in percentage i.e. M%, N% and P% respectively, then the formula becomes:
Hari travels 20% of a distance at an average speed of 20 km/hr, next 30% distance at an average speed of 30 km/hr and the remaining 50% of distance at the average speed of 50 km/hr. Find his average speed during the whole journey.
Sol: Here M=20, N=30 and P= 50 also x= 20, y= 30 and z= 50
Putting the values we get: Answer =33.33 km/hr
Time and Distance Tricks 4:- If a car does a journey in ‘T’ hrs, the first half at ‘a’ km/hr and the second half at ‘b’ km/hr. The total distance covered by the car:
(2 x Time x a x b ) / (a + b).
Example:- A motorcar does a journey in 10 hrs, the first half at 21 kmph and the second half at 24 kmph. Find the distance?
Sol: Distance = (2 x 10 x 21 x 24) / (21+24)
= 10080 / 45
Answer = 224 km.
Time and Distance Tricks 5 .If a person goes from ‘A’ to ‘B’ at a speed of ‘a’ kmph and returns at a speed of ‘b’ kmph and takes ‘T’ hours in all, then the distance between the A and B:
= Total time taken x (Product of the two Speeds / Addition of the two speeds)
Example: A boy goes to school at a speed of 3 kmph and returns to the village at a speed of 2 kmph. If he takes 5 hrs in all, what is the distance between the village and the school?
Sol: Let the required distance be x km.
Then time taken during the first journey = x/3 hr.
and time taken during the second journey = x/2 hr.
x/3 + x/2 = 5 => (2x + 3x) / 6 = 5
=> 5x = 30.
=> x = 6
required distance = 6 km.
Time And Distance Solved Example
Ex .An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field.
Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,
x/200+x/400+x/600+x/800=4x/y=25x/2500=4x/y=y=(2400*4/25)=384 hence average speed =384 km/hr
Ex. If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.
Sol. Let the required distance be x km
Difference in the time taken at two speeds=1 min =1/2 hr
Hence x/5-x/6=1/5<=>6x-5x=6 =x=6
Hence, the required distance is 6 km
Ex. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and travels towards B at 65 kmph. Another train starts from B at 11 a.m. and travels towards A at 35 kmph. At what time do they meet?
Sol. Suppose they meet x hours after 10 a.m. Then,
(Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]=390.
65x + 35(x-1) = 390 => 100x = 425 => x = 17/4
So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.
Ex. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?
Sol. Let the distance be x km. Then,
( Time taken to walk x km) + (time taken to ride x km) =37 min.
( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.
But, the time taken to walk 2x km = 55 min.
Time taken to ride 2x km = (74-55)min =19 min.
Time and Distance Quiz
0 of 30 questions completed
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Question 1 of 30
1. Question1 pointsCategory: math
Every day a cyclist meets a train at a particular crossing. The road is straight before the crossing and both are traveling in the same direction. The cyclist travels with a speed of 10 Kmph. One day the cyclist comes late by 25 min. and meets the train 5km before the crossing. What is the speed of the train?Correct
Cyclist speed is 10kmph and he’ll take 30 min to cover 5 kms.
If Cyclist come 25 min early he would be 5 min before crossing.
So it means train takes 5 min to reach crossing.
We have speed of train calculated as
⇒ Speed= distance/time
⇒ Speed = 5km/5 min
⇒ Speed = 1 km/min
Or 60 km/hr
Question 2 of 30
2. Question1 pointsCategory: math
Two trains from the points A and B moving in opposite direction, at the point they meet the second train travels 120 kms more than the first. The speeds are 50kmph and 60kmph respectively find the distance between A and B?Correct
We know that
Distance = time × speed
Suppose first train travelled the distance x kms from A.
Then second train will travel the distance (x + 120) kms from B.
Let they met each other after t time
∴ (x + 120)/60 = (x/50)
⇒ x = 600 kms
But the distances between two point is (x + x + 120)
= 2x +120
= 2× 600 + 120
= 1320 km
Question 3 of 30
3. Question1 pointsCategory: math
One monkey climbs a pole at the rate of 6m/min for a minute and fells down 3m in the alternate minute. Length of the pole is 60 m, how much time it will take to reach the top?Correct
One monkey climbs a pole at the rate of 6m/min for a minute
And fells down 3m in the alternate minute.
∴ He climbs 3 m in 2 mins.
∴ He will climb (60 – 6) = 54 m in time t and last 6 m in 1 min.
∴ Time (t) taken in 54 m climb = (2/3) × 54 = 36 min
∴ The total time taken to climb 60 m pole = 36 + 1 = 37 min
Question 4 of 30
4. Question1 pointsCategory: math
A train B speeding with 120 kmph crosses another train C running in the same direction, in 2 minutes. If the lengths of the trains B and C be 100 m and 200 m respectively, what is the speed (in kmph) of the train C?Correct
Speed of train B = 120 kmph= 120 × (5/18) = 100/3 m/s
Let Speed of C = C m/s
Since, both the trains are travelling in same direction, so relative speed = ([100/3) – C] m/s
To cross the train C, B will have to travel the distance equal to length of B + length of C
Time = 2 mins = 2 × 60 second
Since, Time = Distance / Relative Speed
⇒ 4000 – 120C = 300
⇒ C = 370/12 m/s
⇒ C = (370/12) × (18/5) = 111 kmph
Question 5 of 30
5. Question1 pointsCategory: math
River is running at 2 kmph. It took a man twice as long to row up as to row down the river. The rate (in km/hr) of the man in still water is:Correct
Let the speed man of in still water=u kmph
Speed of river= v kmph = 2 kmph
Speed of downstream = u + v
Speed of upstream= u – v
Let the distance covered by a man = y km
According to the question,
Time taken in Upstream = 2 × time taken in downstream
⇒ 2 × (u – v) = (u + v)
⇒ 2 × (u – 2) = (u + 2) (∵ v = 2 kmph)
⇒ u = 6 kmph
So the correct option is (4).
Question 6 of 30
6. Question1 pointsCategory: math
If Rachit drives a car four times a lap @10, 20 30, 60 kmph. What is his average speed?Correct
Let 60 km be the length of the lap.
Then time taken by Rachit to complete the lap at 10 km/hr = Distance/speed= 60/10
= 6 hours
Then time taken by Rachit to complete the lap at 20 km/hr = Distance/speed= 60/20
= 3 hours
Then time taken by Rachit to complete the lap at 30 km/hr = Distance/speed= 60/30
= 2 hours
Then time taken by Rachit to complete the lap at 60 km/hr = Distance/speed= 60/60
= 1 hour
Total distance covered = 60 + 60 + 60 + 60 = 240 kms
Total time taken = 6 + 3 + 2 + 1 = 12 hours
Average speed = Total distance/Total time
= 20 km/hr
Question 7 of 30
7. Question1 pointsCategory: math
A truck covers a distance of 640 km in 10 hrs. A car covers the same distance in 8 hrs. What is the respective ratio between the speed of the truck and the car?Correct
Both truck and car covers a distance = 640 km
Time taken by truck is 10 hrs.
∴ Speed of truck = 640/10 = 64 km/hr
Time taken by car is 8 hrs.
∴ Speed of car = 640/8 = 80 km/hr
∴ Ratio between speed of truck and car =
Speed of truck : speed of car = 64 : 80 = 4 : 5
So the correct option is (5).
Question 8 of 30
8. Question1 pointsCategory: math
A 280 meters long train crosses a platform thrice its length in 6 minutes 40 seconds. What is the speed of the train?Correct
For crossing the platform train have to cross (total distance = platform length + train length).
According to the question
Train length = 280 m
Platform length = 3 × train length = 3 × 280
= 840 m
∴ Total length = 280 + 840 = 1120 m
And time taken to cross platform = 6 min 40 sec = 360 sec + 40 sec
= 400 sec.
∴ Speed of train = total length/time = 1120/400
= 2.8 sec
Question 9 of 30
9. Question1 pointsCategory: math
A man travelled a certain distance at the rate of 15 miles an hour and came back at the rate of 10 miles an hour. What is his average speed?Correct
Let the man covers m miles distance.
Then in forward journey, time taken by man = Distance/Speed
Time taken by man in backward journey = m/10
Hence, total distance = 2m.
Total time taken = m/10 + m/15
= m( 1/10 + 1/15)
= m (5/30)
Hence, Average speed = Total Distance/Total time
Average speed = 2m ÷ m/6
= 2m × 6/m
Question 10 of 30
10. Question1 pointsCategory: math
A pig takes 4 leaps for every 5 leaps of a goat but 3 leaps of a pig are equal to 4 leaps of the goat. Compare their speeds.Correct
It is given that 3 leaps of pig = 4 leaps of goat.
Therefore, 1 leap of pig = 4/3 leap of goat.
Pig took 4 leaps, that is 4 × 4/3 = 16/3 leaps of goat while at the same time, goat took 5 leaps.
Hence, the ratio of speed of pig to goat is 16/3 : 5
= (16/3) × (1/5)
That is, the ratio is 16 : 15
Question 11 of 30
11. Question1 pointsCategory: math
While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance left to cover by him was 5 times of the covered distance. What was his speed in meters per second?Correct
we have been given that the man walked for 1 hour 40 min , ie 1 + 2/3 = 5/3 hours.
Let us take that the distance covered by him is m
Then distance left = 5m
Hence total distance = m + 5m
⇒ 6m = 24
⇒ m = 24/6
⇒ m = 4
Hence, we walk 4 km in 5/3 hours.
⇒ 4 km = 4000 meter
⇒ 5/3 hours = 3600 × 5/3
= 6000 sec.
Hence speed = distance / time
Speed = 4000 meter / 6000 sec
= 2/3 m/s
Question 12 of 30
12. Question1 pointsCategory: math
Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, 11/5 seconds after he sees it strike the tree. How far is the man from the wood chopper?Correct
Distance covered = Speed × Time
⇒Distance = 1100 × 11/5
= 1100 × 2.2
Hence distance is 2420 feets.
Question 13 of 30
13. Question1 pointsCategory: math
Rajdhani Express travels at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point?Correct
Since, Rajdhani Exp. Travels at 100 km/hr,
Time taken by it to cover 600 km = distance/speed
= 6 hours
Since it takes break after every 75 km, number of breaks taken = 600/75 – 1
= 8 – 1
We subtracted one because after reaching at 600 km, it won’t take a break
Hence total time spent in breaks = 3 × 7
= 21 minutes
Total time = 6 hours and 21 minutes
Question 14 of 30
14. Question1 pointsCategory: math
Two cars are 15 kms apart. One is moving at a speed of 50kmph and the other at 40kmph in the same direction. How much time will it take for the two cars to meet?Correct
Assuming the car with speed 40 kmph is 15 Km ahead of Car with speed 50kmph
Relative speed = 50kmph – 40kmph = 10kmph
Distance = 15 kms
Time = x Hrs
Distance = (Speed) (Time)
15 = 10x
⇒ x = 15 / 10
⇒ x = 3 / 2 hrs
Question 15 of 30
15. Question1 pointsCategory: math
A car travels a certain distance taking 7 hrs in forward journey. During the return journey, the speed is increased by 12km/hr and the car takes 5 hrs to reach the destination. What is the distance travelled in one way?Correct
For Forward Journey :
Speed = S kmph
Time = 7 Hrs
Distance = D kms
Distance = (Speed ) × (Time)
D = 7S ………………. (i)
For Return Journey
Speed = ( S + 12 ) kmph
Time = 5 hrs
Distance = D kms
Distance = (Speed ) × (Time)
D = (S+12) × 5 ………………. (ii)
Distance travelled is same,
So, 7S = 5S + 60
S = 30 kmph
Distance = 30 × 7 = 210 kms
Question 16 of 30
16. Question1 pointsCategory: math
A train crosses a platform 100 meters long in 60 seconds at a speed of 45 km per hour. The time taken by the train to cross an electric pole is:Correct
Say L is the length of the train.
Speed of the train = 45 km/hr = 45 × (5/18) m/s = 12.5 m/s
Time taken by the train to cross the platform i.e. travel the distance (L + 100) m is 60 seconds.
⇒ (L + 100)/12.5 = 60 [Distance/speed = Time]
⇒ (L + 100) = 60 × 12.5 = 750
⇒ L = 650 m
∴ Time taken to cross an electric pole = (650/12.5) seconds = 52 second
Question 17 of 30
17. Question1 pointsCategory: math
A train 50 meters long passes a platform 100 meters long in 10 seconds The speed of the train is:Correct
Distance covered by train in crossing the platform of 100 meters is (100 + 50) = 150 meters.
Time taken to cross the platform = 10 seconds.
Speed = Distance/Time
⇒ Speed of train = 150/10 = 15 m/s
∴ Speed of train = 15 m/s
= 15 × (18/5) km/hr
= 54 km/hr
Question 18 of 30
18. Question1 pointsCategory: math
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:Correct
Let the speed of trains be A m/s and B m/s respectively.
Length of First train = 27 × A meters
Length of Second train = 17 × B meters
Speed of first train relative to second train = (A + B) m/s
First train has to cover (27A + 17B) meters in 23 seconds with speed (A+B) m/s
⇒ (27A + 17B) ÷ (A + B) = 23
⇒ 27A + 17B = 23 × (A + B)
⇒ 4A = 6B
⇒ A/B = 6/4 = 3/2
⇒ A : B = 3 : 2Incorrect
Question 19 of 30
19. Question1 pointsCategory: math
The length of the bridge, on which a train 130 meters long and travelling at 45 km/hr can cross in 30 seconds, is:Correct
Let L is the length of Bridge
Speed of the train = 45 km/hr = 45 × (5/18) m/s = 12.5 m/s
Time taken by train to travel (130 + L) meters at 12.5 m/s is 30 seconds.
⇒ (130 + L)/12.5 = 30
⇒ (130 + L) = 12.5 × 30
⇒ 130 + L = 375
⇒ L = 245 meters
Question 20 of 30
20. Question1 pointsCategory: math
2 hours after a freight train leaves Delhi, a passenger train leaves the same station travelling in the same direction at an average speed of 60 km/hr. After travelling 4 hrs the passenger train overtakes the freight train. The average speed of the freight train was?Correct
Let the Distance Covered by Freight train = Passenger Train = D kms
Speed of Passenger Train = 60 km/hr
Time taken by Passenger Train to Travel distance D = 4 Hrs
Time taken by Freight Train to Travel Distance D = ( 4 + 2 ) = 6 Hrs
Distance = Speed × Time
Distance Freight = Speed Freight × Time Freight …….. (i)
Distance Passenger = Speed Passenger × Time Passenger …….. (ii)
Distance Freight = Distance Passenger
⇒ Speed Freight × Time Freight = Speed Passenger × Time Passenger
⇒ Speed Freight × 6 = 60 × 4
⇒ Speed Freight = 40 km/hr
Question 21 of 30
21. Question1 pointsCategory: math
If the total distance of a journey is 120 km. If one goes by 60 kmph and comes back at 40kmph what is the average speed during the journey?Correct
For Forward Journey
Distance =120 km
Speed = 60kmph
Time Taken = (Distance Travelled) / (Speed) = 120/60 = 2 hrs
For Backward Journey
Distance =120 km
Speed = 40kmph
Time Taken = (Distance Travelled) / (Speed) = 120/40 = 3 hrs
Total Distance = 120 + 120 = 240
Total Time taken = 3 + 2 = 5
Avg. Speed = (Total Distance Travelled) / (Total Time)
Avg. Speed = 240/5 = 48kmph
Question 22 of 30
22. Question1 pointsCategory: math
An aeroplane flies twice faster than a train which covers 60 miles in 80 minutes. What distance will the aeroplane cover in 20 minutes?Correct
Since train covers 60 miles in 80 minutes,
The distance covered by train in 20 minutes = (20/80) × 60
= 15 miles
Now we know that speed of aero plane is 2 times the speed of the train
Hence, it will cover twice the distance as covered by the train
= 2 × 15
= 30 miles
Question 23 of 30
23. Question1 pointsCategory: math
A train 50 metres long passes a platform 100 metres long in 10 seconds. The speed of the train is:Correct
We have the total distance covered by the train = length of the platform + length of the train
= 100+ 50 metres
= 150 metres
Total time taken = 10 sec
Using the formula
Speed of the train = 150/10 m/s
= 15 metres/sec
= 15 × (18/5) = 54 km/hr
Question 24 of 30
24. Question1 pointsCategory: math
A boat travels 20 kms upstream in 6 hrs and 18 kms downstream in 4 hrs. Find the speed of the boat in still water.Correct
Let speed of boat in still water = a kmph
Let speed of water current = b kmph
Speed of boat in upstream = a-b kmph
Speed of boat in downstream = a+b km/hr
Speed = Distance / Time
Now we have
⇒ a – b = 20/6
⇒ a – b = 3.333 —-
⇒a + b = 18/4
⇒a + b = 4.5 ——
Adding 1 and 2 equations, we get
⇒ 2a = 7.833 or 47/6
⇒ a = 47/12 km/hr
Hence, speed of boat in still water = 47/12 km/hr
Question 25 of 30
25. Question1 pointsCategory: math
A bus covers a distance of 2,924 kms. in 43 hours. What is the average speed of the bus?Correct
Given, distance covered = 2924 km and time taken = 43 hours
∵ Speed = Distance/time
⇒ Speed of the bus = 2924/43 kmph = 68 kmph
Question 26 of 30
26. Question1 pointsCategory: math
A train passes a station platform in 36 sec. and a man standing on the platform in 20 sec. If the speed of the train is 54 Km/hr, what is the length of the train?Correct
(Note): When a train passes a man/object, the distance travelled by the train while passing that object, would be equal to the length of the train.
(Note): When a train passes a platform or a long object, the distance travelled by the train, while crossing that object, would be equal to the sum of the lengths of the train and length of that object.
Let the lengths of the train and the platform are ‘x’ metres and ‘y’ metres respectively, then
We know that, formula-
Speed = Distance/Time
When the train crosses the platform,
⇒ x + y = 15 × 36 = 540 metres …………………………………..(1)
When the train crosses the manm
⇒ x = 300 metres ………………………………………………………..(2)
Solving equations (1) and (2) we get,
X = 300 metres and Y = 240 metres.
Hence the length of the train is 300 metres.
Question 27 of 30
27. Question1 pointsCategory: math
Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?Correct
Say D is the total distance in km.
Time = Distance / Speed
Time taken to cover (2/3) of D at 4 kmph = (2/3) × D/4 = D/6 hours.
Time taken to cover remaining (1/3) of D at 5 kmph = (1/3) × D /5 = D/15 hours.
1 hour 24 minutes = 1 hour + (24/60) hours = 1.40 hours.
∴ Total time = 1.40 hours
So, (D/6) + (D/15) = 1.40
⇒ (7/30) × D = 1.40
⇒ D = 1.40 × (30/7) = 6 km
Question 28 of 30
28. Question1 pointsCategory: math
Ferrari is leading car manufacturer. Ferrari S.P.A. is an Italian sports car. It has enjoyed great success. If Mohan’s Ferrari is 3 times faster than his old MERCEDES which gave him 35 km/h if Mohan travelled 490 km in his Ferrari the how much time in hours he took?Correct
Given, speed of Mohan’s old Mercedes = 35 km/h
Speed of Ferrari is 3 times speed of Mercedes
⇒ Speed of Ferrari = 35 × 3 = 105 km/h
∵ Time taken = Distance/Speed
∴ Time taken to travel 490 km at 105 km/h = 490/105 = 4.67 hours
Question 29 of 30
29. Question1 pointsCategory: math
A bus starts at 6:00 pm. from starting point at the speed of 18m/s, reached its destination and waited for 40 minutes. And again returned back at the speed of 28m/s. If the time taken in forward journey is same as time taken in reverse and waiting time.
What will be the time when it reaches again at its starting point?Correct
Say distance is d m.
Time taken = Distance/Speed
Time taken in forward journey = d/18 seconds.
Time taken in reverse journey = d/28 seconds.
Given that, d/18 = d/28 + (40 × 60) [40 minutes = 40 × 60 seconds]
⇒ d × (5/252) = 2400
⇒ d = 2400 × 252/5 = 120960 m
∴ Total time taken = (d/18) + (d/28) + 2400 = 13440 seconds
⇒ Time taken = 13440/60 minutes = 224 minutes = 3 hours and 44 minutes
∴ The bus reaches back at 9:44 PM
Question 30 of 30
30. Question1 pointsCategory: math
There is a circular track of length 400 mts. If A and B starts at the same point but in opposite direction with a speeds of 8 m/sec and 12 m/s respectively. When will they meet for the first time at the starting point?Correct
n order to meet them at starting point for the first time
The distance covered by A = 400 × n (where n = no of rounds taken by A) (∴ n is an integer)
The distance covered by B = 400 × m (where m = no of rounds taken by B) (∴ m is an integer)
Let the time after they meet at the starting point for the 1st time = t seconds
Distance = Speed × time
For A ⇒ 400 × n = 8 × t…………….. (i)
For B ⇒ 400 × m = 12 × t…………….. (ii)
Dividing Equation (i) & (ii)
⇒ n/m = 8/12
⇒ n/m = 2/3………… (iii)
As n and m are no. of rounds taken by A & B respectively to meet at the starting point
⇒ n, m has to be integers
Lowest value for the (n, m) according to Eq. (iii) is (2, 3)
⇒ n = 2 & m = 3
Putting Value of n or m in equation (i) or (ii) respectively
⇒ 400 × 2 = 8 × t OR 400 × 3 = 12 × t
⇒ t = 100 seconds
⇒ 1 min 40 seconds