# TCS Questions on Time, speed, distance with Solutions – 1

1.Joke is faster than Paul, Joke and Paul each walk 24 KM. The sum of their speed is 7 Km per hour. And the sum of times taken by them is 14 hours. Then, Joke speed is
a. 3 KM/Hr
b. 4 KM/Hr
c. 5 KM/Hr
d.7 KM/Hr
Answer:a
Explaination :Speed = Distance/Time
let the speed of joke x then speed of paul will be 7-x
24/x + 24/(7−)x =14
Try to plugin the values from the options. If Joke speed is 4 the paul is 3.

2.George while driving along the highway saw road markers which are at equal distances from each other. He crosses the markers every 20 seconds. If he increases his speed by x meters per second, he crosses the markers at every 15 seconds. But if he increases his speed by y meters per second, he crosses the marker at every 10th second. If y-x = 40 meters per second, then what is the distance between two markers.
Answer:1200
Explaination: Let speed be =z m/s then Distance= 20z m
(z+x)15=20z; (z+y)10=20z
Also given that y – x = 40
solving we get 20z=1200

3.Suresh Raina and Gautam Gambhir after a scintillating IPL match decide to travel by cycle to their respective villages. Both of them start their journey travelling in opposite directions. Each of their speeds is 6 miles per hour. When they are at a distance of 50 miles, a housefly starts flying from Suresh Raina’s cycle towards Gautam Gambhir at a relative speed of 17 miles per hour with respect to Raina’s speed. What will be the time taken by housefly to reach Gambhir?
a. 10 hrs
b. 15 hrs
c. 20 hrs
d. 25 hrs
Answer:10 hrs
Explaination:
Fly speed is 17 kmph w.r.t to suresh as fly is moving in opposite direction to suresh, its actual speed is 17 – 6 = 11.
Now relative speed of fly and gambhir = 11 – 6 = 5 kmph
So fly takes =
50/(11−6) = 10 Hrs

4.Car A leaves city C at 5pm and is driven at a speed of 40kmph. 2 hours later another car B leaves city C and is driven in the same direction as car A. In how much time will car B be 9 kms ahead of car A if the speed of car is 60kmph
Answer:267 mins
Explaination :Relative speed = 60 – 40 = 20 kmph
Initial gap as car B leaves after 2 hours = 40 x 2 = 80 kms
Car B should be 9 km ahead of the A at a required time so it must be 89 km away
Time = 89 / 20 = 4.45 hrs or 267 mins

5.Two cars start from the same point at the same time towards the same destination which is 420 km away. The first and second car travel at respective speeds of 60 kmph and 90 kmph. After travelling for sometime the speeds of the two cars get interchanged. Finally the second car reaches the destination one hour earlier than the first. Find the time after which the speeds get interchanged?
Answer: 4
Explaination :
Let the total time taken by the cars be a and b
Let the time after which the speed is interchanged be t
For car A, 60t+90(a-t) = 420, 90a – 30t = 420 …….(1)
For car B, 90t + 60(b-t) = 420, 60b + 30t = 420 ….(2)
Using both (1) and (2), we get 90a + 60b = 840
But as a – b =1, 90a + 60(a-1) = 840.
Solving a = 6.
Substituting in equation 1, we get t = 4

6.A and B start from house at 10am. They travel fro their house on the MG road at 20kmph and 40 kmph. there is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm.
Answer: 160 km
Explaination :
Distnace between House and T junction = 20 x 2 = 40.
ie., B reached T at 11 am.
B continued to right after 11 am and travelled upto 2. So distance covered by him = 3 x 40 = 120
A reached T at 12 noon and travelled upto 2 So distanced travelled by him = 2 x 20 = 40
So total distance between them = 120 + 40 = 160 km

7.At 12.00 hours, J starts to walk from his house at 6 kmph. At 13.30, P follows him from J’s house on his bicycle at 8 kmph. When will J be 3 km behind P?
Anwer: 19.30 mins
Explaination:
By the time P starts J is 1.5 hr x 6 = 9 km away from his house.
J is 3 km behind when P is 3 km ahead of him. ie., P has to cover 12 km. So he takes 12 / (8 – 6) = 6 hrs after 13.30. So the required time is 19.30Hrs

8.J is faster than P. J and P each walk 24 km. Sum of the speeds of J and P is 7 kmph. Sum of time taken by them is 14 hours. Then J speed is equal to
a) 7 kmph
b) 3 kmph
c) 5 kmph
d) 4 kmph
Answer:
Explaination :
Given J > P
J + P = 7, only options are (6, 1), (5, 2), (4, 3)
From the given options, If J = 4 the P = 3. Times taken by them =
24/4 + 24/3= 14