IMPORTANT FACTS AND
FORMULAE
Distance Distance
1. Speed = Time
, Time= Speed
, Distance = (Speed *
Time)
2. x km / hr
= x
* 5
18
3. x m/sec
= (x * 18/5) km /hr
4. 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.
5. 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
18
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
150
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
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
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
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.
ð 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.
