# RS Aggarwal Quantitative Aptitude PDF Free Download: TIME AND DISTANCE

**TIME AND DISTANCE **

** IMPORTANT FACTS AND FORMULAE**

Distance Distance

- Speed = Time , Time= Speed , Distance = (Speed * Time)

- x km / hr = x *
__5__

18

- x m/sec = (x * 18/5) km /hr

- If the ratio of the speeds of A and B is a:b , then the ratio of the times taken by them to cover the same distance is
__1__:__1__a b

or b:a.

- Suppose a man covers a certain distance at x km/ hr and an equal distance at y km / hr . Then , the average speed during the whole journey is
__2xy__km/ hr.

x+y

## **SOLVED EXAMPLES**

** **

**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?**

**Sol.** Aditya’s speed = 20 km/hr = {20 * __5__} m/sec = __50__ m/sec

- 9

\Time taken to cover 400 m= { 400 * __9__ } sec =72 sec = 1 __12__ min 1 __1__ min.

50 60 5

** **

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

**Sol.** Speed = { __750__ } m/sec =5 m/sec = { 5 * __18__ } km/hr =18km/hr

- 5

**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.**

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

Then , 3x = 4y => x = __4__ y => 4x = __16__ y.

- 3

\ Ratio of speeds of dog and hare = Ratio of distances covered by them in the same time

= 4x : 5y = __16__ y : 5y =__16 __ : 5 = 16:15

- 3

**Ex. 4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5 of the remaining distance. What was his speed in metres per second?**

** 7**

**Sol.** Let the speed be x km/hr.

Then, distance covered in 1 hr. 40 min. i.e., 1 __ 2 __ hrs = __5x__ km

- 3

Remaining distance = { 24 – __5x__ } km.

3

__5x__=__5__{ 24 –__5x__} ó__5x__=__5__{__72-5x__} ó 7x =72 –5x

3 7 3 3 7 3

ó 12x = 72 ó x=6

Hence speed = 6 km/hr ={ 6 * __5__ } m/sec = __5__ m/sec = 1 __2__

18 3 3

** **

**Ex. 5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.**

**Sol.** Let the total distance be x km . Then,

__2__ x __1__ x

__3 __ + __ 3 __ = __7 __ ó __x__ + __x __ = __7__ ó 7x = 42 ó x = 6

4 5 5 6 15 5

**Ex. 6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.**

Sol. Average speed = { __2xy __} km/hr ={ __2*25*4 __} km/hr = __200__ km/hr

x+y 25+4 29

Distance traveled in 5 hours 48 minutes i.e., 5 __4__ hrs. = { __200 __ * __29__ } km = 40 km

5 29 5

Distance of the post-office from the village ={ __40__ } = 20 km

2

**Ex. 7.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.**

**Sol. :**

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/yó25x/2500ó4x/yóy=(2400*4/25)=384

hence average speed =384 km/hr

** **

**Ex. 8.Walking at 5 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.**

** 7 **

** **

**Sol. :**New speed =5/6 of the usual speed

New time taken=6/5 of the usual time

So,( 6/5 of the usual time )-( usual time)=10 minutes.

=>1/5 of the usual time=10 minutes.

- usual time=10 minutes

** **

**Ex. 9.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

óx=6

Hence, the required distance is 6 km

**Ex. 10. 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. 11. A goods train leaves a station at a certain time and at a fixed speed. After ^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 goods train in 4 hours. Find the speed of the goods train.**

** Sol.** 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 x 90 or x =36.

So, speed of goods train = 36kmph.

**Ex. 12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?**

** Sol.** Relative speed of the policeman = (10-8) km/hr =2 km/hr.

Time taken by police man to cover 100m 100 x 1 hr = 1 hr.

1000 2 20

In 1 hrs, the thief covers a distance of 8 x 1 km = 2 km = 400 m

20 20 5

** **

**Ex.13. 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.