Time and Distance formula with examples for ssc and bank exams

Time and Distance Formula for Competitive Exams

Solved Example 

Ex.1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr ?

Solution:

Aditya’s speed = 20 km/hr $$  =   \left(20×{5\over18}\right)m/sec={50\over9} mm/sec$$

∴ Time taken to cover 400 m $$  =   \left(400×{9\over50}\right)sec=72\ sec=1{12\over60} min =1{1\over5}min$$

Ex.2. A cyclist covers a distance of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist ?

Solution:

$$ Speed = \left({750\over150}\right)m/sec =5m/sec=\left(5×{18\over5}\right)km/hr=18 km/hr $$

Ex.3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.

Solution:

Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare be y.

$$Then, 3x = 4y → x={4\over 3}y→4x={16\over 3}y$$

∴ Ratio of speeds of dog and hare = Ratio of distance covered by them in the same time.

   $$ = 4x: 5y={16\over 3}y:5y={16\over 3}:5=16:15$$

Ex.4. An aeroplane flies along the four sides of a square at the speeds of 200, 400, 600 and 800 km/hr. Find the average speed of the plane around the field.

Solution:

Let the side of the square be x km and let the average speed of the plane around the field be y km/hr. 

$$ Then, {x\over200}+{x\over400}+{x\over600}+{x\over800}={4x\over y} $$

$$↔{25x\over2400}={4x\over y}↔y=\left({2400×4\over25}\right)=384.$$

∴ Average speed = 384 km/hr.

Ex.5. A goods train leaves a station at a certain time and at a fixed speed. After 6 hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the good train in 4 hours. Find the speed of the goods train.

Solution:

Let the speed of the goods train be x kmph.

Distance covered by goods train in 10 hours = Distance covered by express train in 4 hours

∴ 10x = 4×90 or x = 36.

So, speed of goods train = 36 kmph.

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