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## Time And Distance Problems

Candidates who are going to appear any competitive examination and want to prepare Time And Distance Problems in a proper manner then you can get Time And Distance Problems With Formulas from this page. Here are some Shortcuts to solve questions easily. Candidates can download these question and answer through online mode from this page, which is well structured by team of recruitmentresult.com.

## Time And Distance Problems With Formulas

1. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

Speed = (600/ 5 x 60) m/sec.

= 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

= (2 x 18/5) km/ hr

= 7.2 km/hr.

2. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 5/3 hours, it must travel at a speed of:

Distance = (240 x 5) = 1200 km.

Speed = Distance/ Time

Speed = 1200/ (5/3) km/hr.

Required speed = (1200 x 3/ 5) km/ hr = 720 km/hr.

3. A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.

(1/2) x/ 21 + (1/2)x/ 24 = 10

x/ 21 + x/ 24 = 20

15x = 168 x 20

x = 168 x 20/ 15 = 224 km.

4.  The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:

Let the speed of two trains be 7x and 8x km/hr.

Then, 8x = 400/ 4 = 100

x = 100/ 8 = 12.5

Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

5. A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:

Total time taken = (160/ 4 + 160/ 80) hrs. = 9/2 hrs.

Average speed = 320 x 2/9  km/hr = 71.11 km/hr.

6. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then,  120/x  + 480/y = 8

1/x + 4/y = 1/15      ….(i)

And,    200/x  + 400/y = 25/3

1/x + 2/y = 1/24 ….(ii)

Solving (i) and (ii), we get: x = 60 and y = 80.

Ratio of speeds = 60 : 80 = 3 : 4.

7. A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, x/4 + (61 -x)/ 9 = 9

9x + 4(61 -x) = 9 x 36

5x = 80

x = 16 km.

8.  A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

Let distance = x km and usual rate = y kmph.

Then,  x/y – x (y + 3) = 40/ 60

2y (y + 3) = 9x ….(i)

And,    x (y -2) – x/ y = 40/ 60

Y (y – 2) = 3x ….(ii)

On dividing (i) by (ii), we get: x = 40.

9. A car travelling with5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.

Time taken = 1 hr 40 min 48 sec = 1 hr 204/5 min = 126/75 hrs = 126/ 75 hrs.

Let the actual speed be x km/hr.

Then,  5/7 x x 126/ 75 = 42

x = (42 x 7 x 75/ 5 x 126)

= 35 km/hr.

10. If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:

Let the actual distance travelled be x km.

Then,  x/ 10   = (x + 20)/ 14

14x = 10x + 200

4x = 200

x = 50 km

11. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Let speed of the car be x kmph.

Then, speed of the train =   150/100 x =   3/2 x   kmph.

75/x – 75/ (3/2) x = 125/ 10 x 60

75/x – 50/x = 5/ 24

x = (25 x24/5) = 120 kmph.

12. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = (9/ 54x 60) min = 10 min.

13. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

Let the duration of the flight be x hours.

Then,  600/x  – 600/ (x + (1/2)) = 200

600/ x – 1200/ (2x + 1) = 200

X (2x + 1) = 3

2×2 + x – 3 = 0

(2x + 3)(x – 1) = 0

x = 1 hr.      [neglecting the -ve value of x]

14.  In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay’s speed is:

Let Abhay’s speed be x km/hr.

Then,  30/x – 30/ 2x = 3

6x = 30

x = 5 km/hr.

15. Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

Let the distance travelled by x km.

Then,  x/10 – x/15 = 2

3x – 2x = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60/ 10 hrs = 6 hrs.

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

Required speed = 60/5 kmph. = 12 kmph.

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