Chapter-26
Speed, Time and Distance Speed
The Relative Speed
If the location of an object changes with time then it is said to be in motion. A bus running on a road, an ant crawling on a floor, a monkey climbing up a greased pole etc — all are the examples of objects in motion because the locations of these objects keep on changing with respect to their surroundings. Let an object move from a point A to the point B through any path, then the actual length of the path followed by the object is called the distance travelled by the object. The rate at which any moving body covers a particular distance is called its speed.
Speed =
Distance travelled Time taken
If the distance is constant, Speed
Time =
1 Time
Distance Speed
If the time is constant, Distance Speed Distance = Time × Speed. If the speed is constant, Distance Time We can say that for constant distance travelled, speed is inversely proportional to the time taken. Thi s can be expl ain ed by a sim ple exa mpl e. To cover a distance of 100 kms, if a person goes at the speed of 25 kmph, he will require 4 hours to complete the journey and travelling at a speed of 50 kmph, 2 hours will be required.
(a) Uniform Speed If the object covers equal distance in equal time intervals, howsoever small the interval may be then its speed is called the uniform speed.
(b) Variable Speed If the object travels different distances in equal intervals of time, then its speed is called a variable speed. In this case the speed changes from instance to instance.
(i) Objects are moving in opposite directions The rel at iv e spee d of one objec t wit h respe ct to the other, will have magnitude greater than individual speed of each object. This is why, for example, a train A moving with speed 10 km/hr will cross another train B moving in opposite directions with speed 25 km/hr, with a relative speed of (10 + 25 =) 35 km/hr which is greater than the individual speed of either train.
(ii) Objects are moving in same direction The re lative speed of one o bject with respect to the other, will have magnitude either less than or greater than individual speed of each object. This is why, for example, a train A moving with the speed of 20 km/hr will cross the another train B moving in same direction with the speed of 15 km/hr, with a relative speed of (20 – 15 =) 5 km/hr which is less than the individual speed of either train. Take anot her exampl e, a tra in A movi ng wit h the speed of 60 km/hr will cross the another train B moving in same direction with the speed of 20 km/hr with a relative speed of (60 - 20 =) 40 km/hr which is less than the train A and greater than the train B.
Average Speed Average Speed =
Total Distance Covered
Total Time Taken For example, A person divides his total route of journey into three equal parts and decides to travel the three parts with speed of 40, 30 and 15 km/hr respectively. Find his average speed during the whole journey. Let the three equal parts of journey be x km. Time taken to travel first part of the journey
x hour 40
=
Time taken to t ravel second part of the journey
x hour 30
=
450
C o n c ep ep t o f A r i t h m e t i c
Time taken to travel third part of the journey
x hour 15
=
x x x hours Total time taken = 40 30 15 Total distance travelled = x + x + x = 3x km Average Speed =
Total Distance Travelled
An Important Result If two persons (or trains) A and B start at the same time in opposite directions from two points, and arrive at the point in a and b hours respectively after having met, then A’s rate : B’s rate =
b : b : a . Proof: Suppose A starts from P and B starts from Q and they meet at R.
Total Time Taken =
Let A’s rate be x km per hour, and B’s rate be y km per hour. QR = ax km, PR = by km. Now time taken by A in travelling the
x km/hr x x x 40 30 15
= 24 km/hr
distance by km is
by
hours. x And time taken by B in travelling the
Units of Measurement Time is measured in seconds (s), minutes (min) or hours (hr) Distance is usually measured in metres (m), kilometres (km), miles, yards or feet. Speed is usually measured in metres per second (mps), kilometres per hour (kmph or km/hr) or miles per hour (mph).
K KUNDAN ax
distance ax km is
Since both start at the same time and meet, those two times must be equal. Hence ax
Conversion of units 1 1 8 1
y
hour = 60 minutes = (60 × 60) seconds. kilometre = 1000 metres kilometres = 5 miles yard = 3 feet
1 kilometre per hour =
1 kilometre 1 hour
1000 metres 5 = (60 60) seconds 18 metre per second
y hours.
by
x
x 2 2
b
a
y
x
y
b
a
For example, A man sets out to cycle from Bhiwani to Rohtak, and at the same time another man starts from Rohtak to cycle to Bhiwani. After passing each other they complete their journeys
4 and 4 hours respectively. At what rate 3 5 does the second man cycles if the first cycles 8 km per hour? We proceed as follows: in 3
1
18 kilometres per hour = 5 metres per second [To convert kilometres per hour into metres per second we multiply the given speed in kilometres per hour with
5
. And to convert 18 metres per second into kilometres per hour we multiply the given speed in metres per second with
18 5
.]
or
8 Second man' s rate
second man’s rate =
1 kilometre per hour =
1 mile per hour =
First man' s rate Second man' s rate
22 15
5 8
mile per hour.
feet per second.
4 5 6 1 5 3 3 4
6 5
5 6
8 = 6
2 3
km per hour.
451
S p ee d , T i m e a n d D i s t a n c e
Solved Examples Ex. 1:
A man wal ks 22.5 km in 5 hour s. How m u c h h e w i l l w a l k i n 4 h o ur s ?
Soln:
We have, distance covered = 22.5 km, time taken = 5 hours.
Speed =
Now, 1 hour = 60 minutes = (60 × 60) seconds = 3600 seconds Distance covered by the goods train in one hour = (Speed × Time) = (4 × 3600) metres
Distance Time
= 4
22.5 km/hr = 4.5 km/hr or, Speed = 5
Ex. 2:
Now, Distance = Speed × Time or, Distance covered in 4 hours = (4.5 × 4) km = 18 km. Hence, the man will walk 18 km in 4 hours. A car travels at the speed of 72 km/hr. How many metres will it travel in one second? We have, speed of the car = 72 km/hr
=
3600
km
1000
72
km = 14.4 km 5 Thus, the dista nce covere d by the goods train in 1 hour = 14.4 km. Hence, the speed of the train is 14.4 km/hr.
K KUNDAN
Soln:
Time = 1 sec =
1
60
minute
Ex. 4:
A m a n t r a v e l s so m e d i s t a n c e a t a s p ee d o f 1 2 k m / h r a n d r e t u r n s a t 9 k m / h r . If the total tim e taken by him is 2 hours 2 0 m i n u t e s, f i n d t h e d i s t a n c e.
Soln:
Let the distance be x km. Then, time taken at 12 km/hr =
60 seconds 1minute 1 minute 1 sec 60
x 60 minutes = 5x minutes. 12
1 hour 60 60
Time taken at 9 km/hr =
60 minutes 1hour 1 minute 60 hour 1 1 minute hour 60 60 60 1
or,
or,
10 m = 20 m. 36 1
3
x 140
15x 20x 3 35x 3
140
140
or, 35x = 3 × 140 or, x =
Thus, the car will trav el 20 metr es in 1 sec.
In order to find the speed of the goods train in km/hr, we have to find the distance travelled (in km) by it in one hour. We have, Speed = 4 m/sec and Time = 1 hour.
20
5 x
[ 1 km = 1000 m]
Soln:
hour
Since the total time taken is 2 hours 20 minutes ie 140 minutes. Now, according to the question,
1 1000 m = 72 3600
T h e s p e ed o f a g o o d s t r a i n i s 4 m / s ec . W h a t i s i t s sp e ed i n k m / h r ?
9
20x x 60 minutes = minutes. 3 9
1 km = 72 3600
Ex. 3:
x
=
Now, Distance = Speed × Time Distance covered in one second = Speed × Time
hour
=
=
= 72
x
12
3 410 35
x = 12 Hence, the distance is 12 km. Ex. 5:
W a l k i n g a t 4 k m / h r , a p er s o n r e a ch e s h i s of f i c e 5 m i n u t e s l a t e . I f h e w a l k s at 5 k m/hr , he will be 4 mi nutes too ea r l y . F i n d t h e d i s t a n c e o f h i s o f f i c e f r o m h i s r e si d e n c e.
452 Soln:
Co n c ep t o f A r i t h m e t i c Let the required distance be x km. Time taken at 4 km/hr =
x 4
When the person walks at 5 km/hr, then
x
hours
5
x 60 minutes = 15x minutes 4
=
Time taken at 5 km/hr =
x
=
5
x 5
60
x
t
x
x
or,
1
....(ii) 5 15 Now, subtracting equation (ii) from equation (i), we have,
hours
60 minutes = 12x minutes
4
Since the difference between the two times taken is (5 + 4) = 9 minutes. 15x - 12x = 9 or, 3x = 9 or, x =
4
t
5
= t
1
=
or,
5 x 4x 20
1
1 t 12 15
12
1 15
=
54 60
=
9 60
9
60
9
= 3. 3 Hence, the required distance is 3 km. Alternative Method I: Let the distance to the office be 1 km. Then, time taken to cover 1 km at the rate of 4 km/hr
x
or,
9
K KUNDAN Distance
=
Speed
1
x =
1 60 = 15 minutes 4
Distance Speed
A gun is fired at a d istance of 3.32 km a w a y f r o m R oh i t . H e h e a r s t h e s o u n d 1 0 s e co n d s l a t e r . F i n d t h e s p ee d o f t h e sound.
Soln:
Since Rohit is at a distance of 3.32 km from the gun and he hears the sound 10 seconds later. This means that in 10 seconds sound covers a distance of 3.32 km. Distance covered = 3.32 km = (3.32 × 1000) metres = 3320 metres and, time taken = 10 seconds.
1
5 hour
1 60 = 12 minutes 5
=
Now, Speed =
Difference in time taken
= (15 – 12) = 3 minutes But the actual difference in time = (5 + 4) = 9 minutes Thus, when the differ ence in time taken is 3 minutes, the distance to office = 1 km If the difference in time is 9 minutes, 1 the distance to office = 9 = 3 km 3 Hence the distance to office = 3 km. Alternative Method II: Let x km be the distance of office of the person and t hours be the time required to reach the office by the person. When the person walks at 4 km/hr, then
x 4 or,
t
x 4
t
5
12
Speed =
Ex. 7:
Soln:
Distance Time
3320 10
Time taken by a man to reach exhibition from his house Distance =
Speed
=
18 15
hours =
6 5
hours
Time taken by a man to reach his house from the exhibition =
....(i)
m/sec = 332 m/sec.
A m a n t r a v e l s a d i s t a n c e of 1 8 k m f r o m his house to an exhibiti on by tanga at 1 5 k m / h r a n d r e t u r n s ba c k o n c y cl e a t 1 0 k m / h r . F i n d h i s a v er a g e sp e ed f o r t h e w h o l e j ou r n e y .
Distance
60 1
9 20
Ex. 6:
Time taken to cover 1 km at the rate of 5 km/hr =
60
= 3 60 Hence, the required distance is 3 km.
4 hour
=
20
Speed
=
18 10
hours =
9 5
hours
453
S p ee d , T i m e a n d D i s t a n c e Total time of journey
They are 16 km apart in
15 6 9 hours = hours = 3 hours. 5 5 5
16 32 6 hours. 2 1 13 13 6 2
=
Total dista nce cover ed = (18 + 18) km = 36 km Average speed for the whole journey =
Total Distance Covered Total Time Taken
=
Ex. 8:
36
(iv) From (ii), discussed above, 1 1 They are 3 3 km apart in 1 hour. 2 2 They are 16 km apart in 16 = 32 hours. 1 2
km/hr = 12 km/hr
3 Two men A and B start from a place P 1 walking at 3 km and 3 km an hour 2
Ex. 9:
respectively. How many km will they 1
be apart at the end of 2
Soln:
A t r a i n t r a v e l l i n g 2 5 k m a n h o u r l e a ves D el h i a t 9 a m a n d a n o t h er t r a i n t r a v el l i n g 3 5 k m a n h o u r s t a r t s a t 2 p m i n t h e s a m e d i r e ct i o n . H o w m a n y k m f r o m D el h i w i l l t h ey b e t o g et h er ?
Let the required distance be x km. A train leaves Delhi at 9 am and another train leaves Delhi at 2 pm in the same direction. ie difference of time = 5 hours. Distance travelled by the first train in 5 hours = (25 × 5) = 125 km Now, according to the question, Time taken by the train to cover (x – 125) km is equal to the time t aken by the second train to cover the distance of x km.
K KUNDAN 2
hours,
(i)
if t he y w al k in o pp os it e directions? (i i) if they walk in the same direction? What time will they take to be 16 km apart, (iii) if they walk in the opposite directions? (i v) if they walk in the same direction? Soln: (i) When they walk in opposite directions, Their rela tiv e veloc ity will be
1 1 3 3 6 km 2 2
ie, They will be 6
2
in
1
2
1
2
km apart in one hour.
hours
they
will
be
1 1 3 3 km 2 2
2
25 5
or 12
= 12
1
Ex. 10: W a l k i n g
km apart. (iii) From (i), discussed above,
They wil l be 3 3 one hour.
1 1 6 km apart in 2 2
2
km.
1
hours. 10 2 the required distance from Delhi
km apart in one hour.
1 1 1 in 2 hours they will be 2 1 2 2 4 2
1
Alternative Method : The first trai n has a start of 25 × 5 km and the second train gains (35 – 25) or 10 km per hour. the second train will gain 25 × 5 km in
(ii) When they walk in same direction, Their rela tiv e velo city will be
1
x 125
they will be together = 437
1 1 1 6 2 16 km apart. 2 4 2
ie, They will be
x
35 25 or, 25x = 35x – 35 × 125 35 125 875 1 437 x = 10 2 2 The distance from Delhi after which
3 4
1 2
35 = 437
1 2
km.
o f h i s u s u a l s p ee d , a p e r s on
1 i s 1 h o u r s l a t e t o h i s o f f i c e. F i n d h i s 2 u s u a l t i m e t o c o v er t h e d i s t a n c e .
Soln:
It is easy to see that if the speed of a train or man be changed in the ratio of a : b ,
454
Co n c ep t o f A r i t h m e t i c then the time required to travel a certain distance will be changed in the ratio of b : a . Since the man walks at
3 4
Soln:
To walk 1 km, I requi re
of his usual first case, and
4
rate, the time that he takes is
3
of his
4
of usual time = usual time + 1
3 1 3
of usual time = 1
usual time = 1
1
1 2
1 2
4
3
hour in the
hour in the second case.
1 1 hour in the first 3 4
hours.
1
case, and
4
hour in the second case.
1 1 hour or 5 minutes 3 4
Thefore, I save
hours.
3 hours = 4
1
1
Therefore, I save
usual time.
Suppose I have to walk 1 km.
in the second case. But , by the question, I save (40 + 30) or 70 minutes. Hence the required distance = (70 5 =) 14 km. Alternative Method : Let the required distance be D km. Time taken to cover D km at 3 km/hr
1
hours. 2 2 Alternative Method : Let the usual speed of the person be x km/ hr and the distance of his office = D km. His usual time to cover the distance D = hours x Now, according to the question,
K KUNDAN 3
Speed =
4
D hour 3
=
Time taken to cover D km at 4 km/hr
of his usual speed
D hour 4
=
3 x km/hr 4
=
Total diff eren ce in tim e = 40 minutes late + 30 minutes early
Time taken to cover the distance D km
70 7 hours 60 6
= 70 minutes =
D 4D = hours 3x 3x 4
Now, according to the question,
D 3
Again,
4D
3x
or,
or,
or,
D
x
1
1
2
hours =
3
2
D
or,
D
D =
x
7 6
1 1 7 3 4 6
3 1 x 3 2
D
4
or, D
hours
D 4
x
D
7 6
7 12
1
9
Ex. 12: T w o m e n A a n d B w a l k f r o m P t o Q a
2
d i s t a n c e of 2 1 k m , a t 3 a n d 4 k m a n h o u r r e s p ec t i v e l y . B r e a c h e s Q , r e t u r n s i m m e d i a t e l y a n d m eet s A a t R . F i n d t h e dista nce fr om P to R.
3
3
12
= 14 km 6 Hence the required distance = 14 km.
2
Usual time =
9 2
= 4
1 2
hours.
Ex. 11: I h a v e t o b e a t a c e r t a i n p l a c e a t a c er t a i n t i m e a n d f i n d t h a t I s h a l l b e 4 0 m i n u t e s t o o l a t e, i f I w a l k a t 3 k m a n h o u r a n d 3 0 m i n u t e s t o o s oo n , i f I w a l k a t 4 k m a n h o u r . H ow f a r h a v e I to walk?
Soln:
When B meets A at R, B has walked the distance PQ + QR and A the distance PR. That is both of them have together walked twice the distance from P to Q, ie 42 km.
455
S p ee d , T i m e a n d D i s t a n c e Now the rates of A and B are 3 : 4 and they have walked 42 km. Hence the distance PR travelled by A =
distance PQ (= 55 km) and when they meet at S for the second time, they have together covered three times the distance PQ or 165 km.
3
of 42 km. = 18 km. 7 Alternative Method I: Now PR =
3
1 of PQ = 3 2 2
32 11
55 km
= 30 km. QP + PS is the distance covered by B when he meets A for the second time.
Let the required distance be x km. Now, according to the question, A and B both walk for the same distance Distance travelled by B = (21 + 21 – x ) = (42 – x ) km
2
QP + PS =
42 x hours 4
Time taken by B =
1 2
32
1 2
of 165 km = 75 km.
PS = 75 - QP = (75 - 55) km = 20 km. SR = PR - PS = (30 - 20) km = 10 km.
Distance travelled by A = x km
K KUNDAN x Time taken by A = hours 3
x
Ex. 14: Po i n t s A a n d B a r e 9 0 k m a p a r t f r o m ea c h o t h e r o n a h i g h w a y . A ca r s t a r t s f r o m A a n d a n o t h er f r o m B a t t h e s a m e t i m e. I f t h e y g o i n t h e sa m e d i r ec t i o n t h e y m e et i n 9 h o u r s a n d i f t h ey g o i n
42 x
3 4 or, 4x = 126 – 3x or, 7x = 126 or, x =
126
9
7
h o u r s . F i n d t h e i r s p e ed s .
= 18
7 required distance = 18 km Alternative Method II: A’s speed = 3 km B’s speed = 4 km Let us consider that A and B meets after t hours. Distance covered by A in t hours = 3 t km Distance covered by B in t hours = 4 t km Total distance covered by A and B = 3t + 4t = 7t km But the total distance covered by A and B is twice the distance between P and Q. So, 7t = 21 × 2
Soln:
Let X and Y be two cars starting from points A and B respectively. Let the speed of car X be x km/hr and that of car Y be y km/hr. Case I: When two cars move in the same direction:
a t R , r e a ch Q a n d P , r e t u r n i m m e d i a t e l y a n d m eet a g a i n a t S . F i n d t h e d i s t a n c e from R to S.
Suppose two cars meet at point Q. Then, distance travelled by car X = AQ, distance travelled by car Y = BQ. It is given that two cars meet in 9 hours. Distance travelled by car X in 9 hours = 9x km or AQ = 9x km Distance travelled by car y in 9 hours = 9y km BQ = 9y km Clearly, AQ - BQ = AB 9x - 9y = 90 [ AB = 90 km] or, x - y = 10 .... (i) Case II: When two cars move in opposite directions: Suppose two cars meet at point P. Then, distance travelled by car X = AP distance travelled by car Y = BP
When A and B meet at R for the first time, they have together covered the whole
In this case, two cars meet in
t = 2
21
7 t = 6 hours So, the distance between P and R = Distance travelled by A = 3 × 6 = 18 km. Ex. 13: A a n d B s t a r t a t t h e sa m e t i m e f r o m P a n d Q (5 5 k m a p a r t ) t o Q a n d P a t 3
1 a n d 2 k m a n h o u r r e sp e ct i v el y , m e et 2
Soln:
o p p o s i t e d i r ec t i o n s t h e y m e et i n
9 7
hours.
456
Co n c ep t o f A r i t h m e t i c Distance travelled by car X in 9 7
9
hours =
7
x km or AP =
Distance travelled by car Y in
=
9
y km
or BP =
9
7 7 Clearly, AP + BP = AB
9 7
x
9 7
9 7 9 7
x km
hours
y km
=
200
hours x Time taken to cover 400 km by train 400 =
y 90
9
(x y ) 90 7 or, x + y = 70 ... (ii) Solving (i) and (ii), we get x = 40 and y = 30. Hence, speed of car X is 40 km/hr and speed of car Y is 30 km/hr. or,
Case II: When he travels 200 km by train and the rest by car: If Ved trav els 200 km by train, then distance travelled by car is (600 - 200) km = 400 km. Now, time taken to cover 200 km by train
y hours
In this case the total time of journey is 8 hours 20 minutes
200
x
200
400 y = 8 hours 20 minutes 400
1
K KUNDAN or,
x
y
8
[ 8 hours 20 minutes
Ex. 15: V ed t r a v el s 6 0 0 k m t o h i s h o m e p a r t l y
by train and par tly by car. He tak es 8 h o u r s i f h e t r a v el s 1 2 0 k m b y t r a i n a n d t h e r es t b y ca r . H e t a k e s 2 0 m i n u t e s l o n g e r i f h e t r a v el s 2 0 0 k m b y t r a i n a n d t h e r es t b y c a r . F i n d t h e sp e ed o f t h e t r a i n a n d t h e ca r .
Soln:
Let the speed of the train be x km/hr and the speed of the car be y km/hr. Case I: When he travels 120 km by train and the rest by car : If Ved travels 120 km by train, then distance covered by car is (600 - 120) km = 480 km. Now, time taken to cover 120 km by train =
120 x
It is given that the total time of the journey is 8 hours. 120 x
480 y
8
15 60 8 x y
or, 8
15 or,
x 15
or,
x
200
or,
60
60
y
1
1 0 .... (i) y
x
400 y
20
hours = 8
60
1 3
hours]
25 3
8 16 25 3 x y
or, 25
8 16 1 or, x y 3 24
or,
x
24
Time taken to cover 480 km by car 480 = y hours
= 8
hours
Distance Time speed
3
or,
x
48
48
y
y
1
Putting
x
1
1 0
.... (ii)
1
u and
y
v in (i) and (ii),
we get 15u + 60v - 1 = 0 .... (iii) 24u + 48v - 1 = 0 .... (iv) On solving equations (iii) and (iv), we have u =
1 60
and v =
Now, u =
1 x
1 80 1 60
1 x
x = 60,
1 1 1 and v = y 80 y y = 80. Hence, speed of train = 60 km/hr and speed of car = 80 km/hr.
457
S p ee d , T i m e a n d D i s t a n c e Ex. 16: X t a k es 3 h o u r s m o r e t h a n Y t o w a l k 1 a h ea d o f Y b y 1 h o u r s . F i n d t h e i r 2
3
Now, u =
10
s p eed o f w a l k i n g .
Soln:
Let the speeds of X and Y be x km/hr and y km/hr respectively. Then, time taken by X to cover 30 km =
30
x = 1
1
and v =
5
10
Ex. 17: A f t e r c ov er i n g a d i s t a n c e o f 3 0 k m w i t h a u n i f o r m s p ee d t h er e i s so m e d e f ec t i n a t r a i n en g i n e a n d t h e r e f o r e, i t s
3
y
10
km/hr and 3 Y’s speed = 5 km/hr.
By the given conditions, we have 30
3
1
y hours
x
3
x
Hence, X’s speed =
30
30
10
1
10
y 5 y = 5.
hours
x And, time taken by Y to cover 30 km =
3
10u – 3 = 0 u =
3 0 k m . B u t , i f X d o u b l e s h i s p a c e, h e i s
K KUNDAN 10
or,
x
1
y
.... (i)
30
hours 2x Time taken by Y to cover 30 km
o f i t s o r i g i n a l
5
s p e ed . Co n s e q u en t l y , t h e t r a i n r e a c h e s i t s d e st i n a t i o n l a t e by 4 5 m i n u t e s . H a d i t h a p p en e d a f t e r c ov er i n g 1 8 k i l o m e t r es m o r e, t h e t r a i n w o u l d h a v e r e a c h ed 9 m i n u t e s e a r l i e r . Fi n d t h e s p ee d o f t h e t r a i n a n d t h e d i s t a n c e of j o u r n ey .
If X doubles his pace, then speed of X = 2x km/hr Time taken by X to cover 30 km =
4
s p ee d i s r e d u c ed t o
10
Soln:
Let the original speed of the train be x km/hr and the distance of the jou rney be
30
=
y hours
y km. Then, time taken =
According to the given conditions, we have 30 y
30
or,
y
10
or,
y
30
30
5
1
2x
2x
x
1
2
hours. x Case I: When defect in the engine occurs after covering 30 km. Speed for first 30 km = x km/hr and, speed for the remaining ( y - 30) km
3
=
2
1 x
4
5
x km/hr
1
Time taken to cover 30 km =
2
Time taken to cover (y - 30) km
10 20 or, x y 1 .... (ii) Putting
y
u and
1 y
v , equations (i)
and (ii) become: 10u – 10v = 1 .... (iii) –10u + 20v = 1 ... (iv) Adding (iii) and (iv), we get: 1 10v – 2 = 0 v . 5 Pu tting v
1 5
in (iii), we get:
=
30 x
hours
5 y 30 hours = (y - 30) hours 4x 4x 5
According to the given condition, we have
30 x or, or,
30 x
5 4x
(y 30)
5y 150 4x
120 5y 150
y
y x
x
45 60
3 4
4y 3x
4x 4x or, 5y – 30 = 4y + 3x or, 3x – y + 30 = 0
458
Co n c ep t o f A r i t h m e t i c Case II: When defect in the engine occurs after covering 48 km. Speed for first 48 km = x km/hr Speed for the remaining ( y – 48) km =
4x 5
Total time t aken by t he train if no accident
d hours x
happens =
Case I: Time taken by t he t rain t o cover the whole length of the trip
km/hr
Time taken to cover 48 km =
48 x
(d 3x ) = 3 1 hours 75 x 100
hours
Time taken to cover (y - 48) km
y 48 5(y 48) hour = hour = 4 x 4x 5
= 4
4(d 3x )
hours
3x
Now, according to the question,
According to the given condition, the train now reaches 9 minutes earlier ie it is 36 minutes late.
4
4(d 3x ) 3x
d x
4
K KUNDAN 48 x
or,
or,
or,
48 x
5(y 48) 4x
5y 240 4x
192 5y 240 4x
5y 48 4
y x
y
x
36 60 3 5
5y 3x 5 x
5y 3x 5
or, 25y - 240 = 20y + 12x or, 12x - 5y + 240 = 0 Thu s, we ha ve th e fo llo wi ng sy st em of simultaneous equations: 3x - y + 30 = 0 ... (i) 12x - 5y + 240 = 0 ... (ii) On solving equations (i) and (ii), we have, x = 30 and y = 120 Hence, the original speed of the train is 30 km/hr and the length of the journey is 120 km.
Ex. 18: A t r a i n m et w i t h a n a c ci d e n t 3 h o u r s a f t e r s t a r t i n g , w h i c h d et a i n s i t f o r o n e h o u r , a f t e r w h i c h i t p r o c eed a t 7 5 % of i t s o r i g i n a l s p e ed . I t a r r i v e s a t t h e d es t i n a t i o n 4 h o u r s l a t e. Ha d t h e a c c i d e n t t a k en p l a c e 1 5 0 k m f a r t h e r along the rai lway li ne, th e tra in would
or,
4d 12x
d
3x x or, 4d – 12x = 3d or, d = 12x .... (i) Case II: If the train had covered 150 km more before the accident then the distance of the accident = (3x + 150) km Remaining distance = (d – (3x + 15)) km Time taken to cover the whole lengt h of the trip 3x 150 d (3x 150) 1 75 x x 100
Now, according to the question,
3x 150 d (3x 150) d 7 1 3x x x 2 4
or, or,
3x 150 x
3x 150 x
4d 12x 600 3x
Soln:
Let the length of the trip be d km and the original speed of the train be x km/hr. As the accident takes place after 3 hours. distance covered in 3 hours by the train = (3 × x ) = 3x km Remaining distance = (d - 3 x ) km
d x
or,
or,
2
1
3x
or,
7
4 12x 12x 600
1 h a v e a r r i v ed o n l y 3 h o u r s l a t e . Fi n d 2 t h e l e n gt h o f t h e t r i p a n d t h e o r i g i n a l s p ee d of t h e t r a i n .
9x 450 36x 600 3x 3x 150 12x 200 x 3x 150 12x 200 x
29
29
29
2
2
2
12x x
5 2
d 12x
459
S p ee d , T i m e a n d D i s t a n c e
or,
15x 50
6D or, V(V 6) 4
29
x 2 or, 30x - 100 = 29x or, x = 100 Hence, speed = 100 km/hr and the length of the trip (d) = 12x = 12 × 100 = 1200 km
....(i) 6 When the person moves 6 km/hr slower, then
Ex. 19: A t r a i n c ov er e d a c er t a i n d i s t a n c e a t a u n i f o r m s p eed . I f t h e t r a i n w o u l d h a v e b een 6 k m / h r f a s t er , i t w o u l d h a v e t a k en 4 h o u r s l e ss t h a n t h e sc h e d u l e d t i m e. A n d , i f t h e t r a i n w er e s l ow e r b y 6 km /hr, it would have taken 6 hours m o r e t h a n t h e sc h ed u l e d t i m e. F i n d t h e l e n gt h o f t h e j o u r n e y .
Soln:
Let the actual speed of the train be x km/hr and the actual time taken be y hours. Then, Distance = (x y) km ... (i) [ Distance = speed × time] If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours ie when speed is ( x + 6) km/hr, time of journey is (y - 4) hours. Distance = (x + 6) (y - 4) km or, xy = (x + 6) (y – 4) [Using (i)] or, –4x + 6y – 24 = 0 or, –2x + 3y – 12 = 0 ... (ii) When the speed is reduced by 6 km/hr, then the time of journey is increased by 6 hours ie when speed is ( x - 6) km/hr, time of journey is (y + 6) hours Distance = (x – 6) (y + 6) or, xy = (x – 6) (y + 6) [Using (i)] or, 6x – 6y – 36 = 0 or, x – y – 6 = 0 .... (iii) On solving equations (ii) and (iii), we get x = 30 and y = 24. Putting the values of x and y in (i), we obtain Distance = (30 × 24) km = 720 km. Hence, the length of the journey is 720 km. Alternative Method: Let the original speed and distance be V km/hr and D km respectively. Time taken to complete the whole journey
4V(V 6)
D =
D V6 or,
D V 6
D
D
V
6
V
6
D(V V 6) or,
V(V 6)
6
or, D = V(V – 6) ....(ii) Combining equations (i) and (ii), we get
K KUNDAN =
D
hours V When the train moves 6 km/hr faster, then
D V6 or,
D V 6
D
D
V V
4 4
D(V V 6) or,
V(V 6)
4
4V(V 6)
V(V 6) 6 or, 4V + 24 = 6V – 36 or, 2V = 60 V =
60
= 30 2 Hence original speed = 30 km/hr Putting the value of V in equation (ii), we have D = 30 (30 – 6) = (30 × 24) = 720 km
Ex. 20: A h a r e s e es a d o g 1 0 0 m e t r e s a w a y f r o m h e r a n d s cu d s o f f i n t h e o p p o si t e d i r ec t i o n a t a s p eed o f 1 2 k m a n h o u r . A m i n u t e l a t e r t h e d o g p er c e i v es h e r a n d g i v es ch a s e a t a s p e ed o f 1 6 k m p e r h o u r . H o w s oo n w i l l t h e d o g o v er t a k e t h e h a r e , a n d a t what d i s t a n c e f r o m t h e sp o t w h e n ce t h e h a r e took flight?
Soln:
Suppose the hare at H sees the dog at D.
DH = 100 metres. Let K be the position of the hare where the dog sees her. HK = the distance gone by the hare in 1 minute.
12 1000
1 m = 200 m. 60 DK = 100 m + 200 m = 300 m The hare thus has a start of 300 m. Now the dog gains (16 - 12) or 4 km in an hour. =
the dog will gain 300 m in 4
1 2
minutes.
60 300 4 1000
or
460
Co n c ep t o f A r i t h m e t i c Again, the distance gone by the hare in
4
DH = 100 metres. Let K be the position when the dog sees hare.
1 minutes. 2 =
12 1000
4
HK = distance gone by hare in
1
m = 900 m. 60 2 distance of the place where the hare is caught from the spot H whence the hare took flight = (200 + 900) m = 1100 m.
=
12 1000
1 2
min
1
= 100 metres 60 2 DK = 100 metres + 100 metres = 200 metres The hare thus has a strat of 200 metres. Now the dog gains (16 – 12) or 4 km in an hour Distance gained by dog in 1 min
Ex. 21: A h a r e , p u r s u e d b y a g r e y -h o u n d , i s 5 0 o f h e r o w n l e a p s a h ea d o f h i m . W h i l e t h e h a r e t a k es 4 l e a p s t h e g r e y -h o u n d t a k e s 3 l e a p s . I n o n e l e a p t h e h a r e g o es
=
3 3 1 m e t r es a n d t h e g r e y -h o u n d 2 4 4
Now
m e t r e s. I n h o w m a n y l e a p s w i l l t h e gr e y - h o u n d o v e r t a k e t h e h a r e?
200 3
4 1000 60
200 3
metres
metres is covered in time
K KUNDAN
Soln:
50 leaps of the hare = 50 1
3 4
the grey-hound should gain
m =
175
175 2
2
m
4 3
hare goes
4
3
1
3 4
3
4
m over
Ex. 23: T w o g u n s w e r e f i r e d f r o m t h e s a m e p l a c e a t a n i n t er v a l o f 1 3 m i n u t es bu t a p er s o n i n a t r a i n a p p r o a c h i n g t h e p l a c e h e a r s t h e s ec o n d r e p o r t 1 2 m i n u t e s 3 0 s ec o n d s a f t er t h e f i r s t . F i n d t h e s p ee d o f t h e t r a i n , s u p p o s i n g t h a t s o u n d t r a v e l s 3 3 0 m e t r e s p e r s ec o n d .
Soln:
m whilst the
m
5
grey-hound
=
gains
175 2
m
=
in
175 12 210 leaps. 5 2 w h en a d o g i s 1 0 0 m e t r e s o f f . A f t e r h a l f a m i n u t e t h e d og s ees h a r e a n d p u r s u e s a t 1 6 k m p er h o u r . H ow s o o n w i l l h e ca t c h h a r e ?
Suppose the hare H sees the dog at D.
330 30 2 60 25 1000
1188 25
or 47
13 25
km per hour
Ex. 24: A c a r r i a g e d r i v i n g i n a f o g p a s sed a m a n w h o w a s w a l k i n g a t t h e r a t e of 3 k m a n h o u r i n t h e s a m e d i r e ct i o n . H e c o u l d s ee t h e c a r r i a g e f o r 4 m i n u t e s a n d i t w a s v i si b l e t o h i m u p t o a d i s t a n c e o f 1 0 0 m . W h a t w a s t h e s p ee d o f t h e c a r r i a g e?
Ex. 22: A h a r e st a r t s t o r u n a t 1 2 k m p er h o u r
Soln:
1
min. 2 the speed of the train per hour
m in one leap.
the
It is easy to see that the distance travelled by the train in 12 min 30 seconds could be travelled by sound in (13 min - 12 min 30 seconds) or 30 seconds. the train travels 330 × 30 metres in
12
3 3 4 the grey-hound gains 2 1 or 4 3 4
12
1 3
200 = 3 min. 200 Hence, dog will catch hare in 3 minutes.
leaps.
the grey-hound goes 2
= 1 min metres is covered in time =
the hare. Now the grey-hound takes 3 leaps whilst the hare takes 4 leaps. the grey-hound takes 1 leap whilst the hare takes
200
Soln:
Th e di st an ce t ra v el le d by ma n in 4 minutes =
3 1000 60
4 metres = 200 metres.
461
S p ee d , T i m e a n d D i s t a n c e distance travelled by
Distance covered in 1 hour x = km 50 15
carriage in 4 minutes = (200 + 100) or 300 metres. speed of carriage
300
=
4 1
= 4
2
60 1000
In the second case: Distance to be covered = 2x km
km per hour Speed = 2
x
km/hr 50 15 25 15 Number of hours he walks per day = (24 – 2 × 9 =) 6 hours Distance covered in 1 day x = 6 km 25 15
km per hour.
Ex. 25: T w o t r a i n s s t a r t a t t h e sa m e t i m e , on e from A t o B and the other from B to A. I f t h e y a r r i v e a t B a n d A r e s p ec t i v e l y 5 h o u r s a n d 2 0 h o u r s a f t e r t h e y p a s sed ea c h o t h e r . S h o w t h a t o n e t r a v el s t w i c e as fast as the other.
Soln:
x
6x
Now
km is covered in 1 day 25 15 2x km shall be covered in
Let the two trains be P and Q.
=
2x 25 15
= 125 days.
K KUNDAN Let the train P starts from A and Q starts from B and they meet at C. Let P’s speed be x km per hour and Q’s speed be y km per hour. BC = 5x km; AC = 20y km Now time taken by P to travel a distance
Ex. 27: A p er s o n w a l k s f r o m A t o B a t t h e r a t e o f 3 k m p h a n d a f t er t r a n s a ct i n g s o m e b u s i n es s w h i c h o c cu p i e s h i m a n h o u r , returns to A by tr amw ay at th e rate of 5 k m p h . H e t h en f i n d s t h a t h e h a s b een a b s en t f o r 2 h o u r s 2 0 m i n u t e s. F i n d th e distna ce fr om A to B.
20y
20y =
hours x And time taken by Q to travel a distance
Soln:
2 3 5 3.75 kmph 3 5
y hours
The time for which he tra vels
Since both start at t he same time and meet, those two times must be equal.
20y x
1 1 1 1 hours 3 3
= 2
5 x
1 1 Distance = 3.75 1 2 3
y
2
2
or, 5x = 20y x 2
or,
y 2
or,
y
x
= 4
multiplied 3.75 1
2
x 50
km
Since he rests for 9 hours, Distance covered in (24 – 9 =) 15 hours
x 50
km
1
by . Because we 3 2
away. He walk s at th e rate of 4 km /hr f o r t h e f i r s t 1 6 k m a n d t h en t r a v el s i n a s co o t e r f o r t h e r e st o f t h e j ou r n e y . H o w e ve r , i f h e h a d t r a v e l l e d b y s co o t er f o r t h e 1 6 k m a n d c ov er e d t h e r em a i n i n g d i s t a n c e o n f o ot a t 4 k m / h r , h e w o u l d h a v e t a k en a n h o u r l o n g er t o co m p l e t e t h e j o u r n e y . F i n d t h e sp e e d o f t h e scooter.
Suppose initially he covers x km in 50 days
Distance covered in 1 day =
1
will get twice the distance otherwise. Ex. 28: A p e r s o n h a s t o r e a c h a p l a c e 4 0 k m
1
d a y s w h en h e r e s t s 9 h o u r s ea c h d a y , how long it will tak e for him to walk t w i ce a s f a r i f h e w a l k s t w i c e a s f a s t a n d r e st s t w i c e a s l o n g e a c h d a y ?
=
= 2.5 km
Note: In calculating distance we have
Ex. 26: A c a n w a l k a c er t a i n d i s t a n c e i n 5 0
Soln:
We have average speed =
5x
5x =
6x
Soln:
Total distance covered = 40 km Distance covered on foot = 16 km Distance covered in a scooter = 40 km – 16 km = 24 km Suppose speed of the scooter = x km per hour
462
Co n c ep t o f A r i t h m e t i c Now, according to the question, speed of the man = 4 km per hour Then time taken in the first case =
16 4
km at a speed of 24 km/hr. 2 Th en , ti me t ak en to tr av el t he w ho le
24 4 hours x x
24
x x journey = hours 2 21 2 24
Also time taken in the second case =
Now, according to the question,
16 6 hours x 4 x
16
24
x 2 21
Since the time taken in the second case is one hour longer or,
24 16 6 4 1 x x or,
16 x 16
64
24
24 x
x 2 24
24x 21x 21 24 2
x = = 1
x
km/hr and
= 10
= 10
2 10 21 24 21 24
224 km.
Ex. 31: A m o n k e y t r i e s t o a s ce n d a g r e a s ed p o l e 1 4 m et r e s h i g h . H e a s c en d s 2 m e t r e s i n f i r s t m i n u t e a n d sl i p s d ow n 1 m e t r e i n t h e a l t e r n a t e m i n u t e . I f h e co n t i n u e s t o a s cen d i n t h i s f a s h i o n , h o w l o n g d o es h e t a k e t o r ea c h t h e t o p ?
K KUNDAN or,
or,
or,
x
8 x
x
= 1 – 6 + 4
1
Soln:
8
=1 x or, x = 8 Hence, speed of the scooter = 8 km per hour.
Ex. 29: A b o y g o e s t o s c h o o l a t a s p e e d o f 3
k m / h r a n d r et u r n s t o t h e v i l l a g e a t a s p ee d of 2 k m / h r . I f h e t a k e s 5 h o u r s i n a l l , w h a t i s t h e d i s t a n c e b et w een t h e v i l l a g e a n d t h e sc h o o l ?
Soln:
x
a t t h e r a t e o f 1 5 k m a n d 1 6 k m p er h o u r r es p e ct i v e l y . F i n d t h e d i s t a n c e t r a v el l e d w h en o n e t a k e s 1 6 m i n u t e s l o n g er t h a n t h e o t h e r .
Soln:
or,
x 3
x 2
6
5
Ex. 30: A m o t o r c a r d o es a j o u r n e y i n 1 0 h o u r s , t h e f i r st h a l f a t 2 1 k m / h r a n d t h e s ec o n d h a l f a t 2 4 k m / h r . F i n d t h e distance.
Let the distance be x km.
2
x
15
hours
Time taken by the second runner
x hours 16 Now, according to the question, x
5
or, 5x 30 x = 6 required distance = 6 km
x
Time taken by the first runner =
=
2x 3x
Then
Let the distance be x km.
hour.
3 And time taken during the second journey x = hour. 2
Soln:
Ex. 32: T w o r u n n e r s c o ve r t h e s a m e d i s t a n c e
Let the required distance be x km. Then time taken during the first journey =
In every 2 minutes he is able to ascend (2 – 1 =) 1 metre. This way he ascends upto 12 metres because when he reaches at the top, he does not slip down. Thus, upto 12 metres he takes 12 × 2 = 24 minutes and for the last 2 metres he takes 1 minute. Therefore, he takes (24 + 1 =) 25 minutes to reach the top. That is, in 26th minute he reaches the top.
km is travelled at a speed of 21
15 or,
x 16
x (16 15 ) 15 16
x=
16 60
16 60
16 60
15 16 = 64 km.
Ex. 33: W i t h o u t a n y s t o p p a g e a p e r so n t r a v el s a c er t a i n d i s t a n c e a t a n a v e r a g e s p ee d o f 8 0 k m / h r a n d w i t h s t op p a g es h e c o ve r s t h e s a m e d i s t a n c e a t a n a v e r a g e s p eed o f 6 0 k m / h r . H o w m a n y m i n u t es per hour does he stop?
463
S p ee d , T i m e a n d D i s t a n c e Soln:
Let the total distance be x km. Time taken at the speed of 80 km/hr
Remaining distance = 1
x
=
total distance
hours. 80 Time taken at the speed of 60 km/hr
x
=
60
hours.
20x
60 80 240 his rest per hour =
x
240
x 60
5
of the total distance= 390
5
2
260 65 km/hr. 4
distance =
Ex. 36: W h e n a m a n t r a v el s eq u a l d i s t a n c e a t
x
2
3 5 = 260 km. average speed for the remaining
x x hours he rested for 60 80 =
3 2 of the 5 5
s p e e d s V 1 a n d V 2 k m / h r , h i s a v er a g e s p eed i s 4 k m / h r . B u t w h e n h e t r a v e l s a t t h e s e s p ee d s f o r e q u a l t i m e h i s a v e r a g e s p ee d i s 4 . 5 k m / h r . Fi n d t h e d i f f er e n c e o f t h e t w o s p e ed s a n d a l s o f i n d t h e v a l u es o f V1 a n d V2 .
hours
x 240
60 x
K KUNDAN 1
=
Soln:
hours = 15 minutes.
4 Ex. 34: A m a n r o d e o u t a c er t a i n d i s t a n c e by train at the rate w a l k ed b a ck a t hour. The whole a n d 4 8 m i n u t es . ride?
Soln:
D D are V hours and V hours respectively. 1 2
of 25 km an hour and t h e r a t e o f 4 k m p er j ou r n ey t o o k 5 h o u r s What distance did h e
average speed =
Let the distance be x km. Then time spent in journey by train
x
=
hours. 25 And time spent in journey by walking =
x 4
29x
100
x
25
5
100 5
48 60
x 4
Now,
5 hours 48 minutes.
29
V1 V2 3 km/hr
5
V1 + V2 = 9 km/hr .. ... (ii) On solving equations (i) and (ii), we have V1 = 6 km/hr and V 2 = 3 km/hr.
20 km
i t , h e f i n d s t h a t h e h a s c ov er e d
3 5
o f
Total distance covered in (3 + 4.5) hours = 3 × 40 + 4.5 × 60 = 390 km. Now, according to the question, of the distance = 390 km
.....(i) and
Ex. 37: A m a n t a k e s 8 h o u r s t o w a l k t o a
t h e t o t a l d i s t a n c e. A t w h a t a v e r a g e s p ee d s h o u l d h e t r a v e l t o c o v er t h e remaini ng distance in 4 hours?
5
V1 V2 2 V1 V2 2 4V1V2 = 81 – 72 = 9
c er t a i n p l a c e a n d r i d e b a c k . H o w e v er , h e co u l d h a v e g a i n e d 2 h o u r s , i f h e h a d c ov er e d b o t h w a y s b y r i d i n g . H o w l o n g w o u l d h e h a v e t a k en t o w a l k b o t h ways?
s p eed o f 4 0 k m / h r a n d f o r 4 . 5 h o u r s a t t h e sp e ed o f 6 0 k m / h r . A t t h e en d o f
3
V1 V2
4.5 km/hr 2 or, V1 + V2 = 9 and V 1 V2 = 18
Ex. 35: A p e r s o n t r a v e l s f o r 3 h o u r s a t t h e
Soln:
Total time
In second case,
hours. x
Total distance
2D 2V1V2 4 km/hr D D V1 V2 V1 V2
average speed =
Therefore,
or,
Suppose the equal distance = D km Then ti me ta ken wit h V 1 and V 2 speeds
Soln:
Walking time + Riding time = 8 hours .... (1) 2 Riding time = (8 – 2 =) 6 hours .... (2) 2 × (1) – (2) gives the result 2 × walking time = (2 × 8 – 6 =) 10 hours. both ways walking will take 10 hours.
Ex. 38: A p er s o n t r a v el l e d 1 2 0 k m b y s t e a m e r , 4 5 0 k m b y t r a i n a n d 6 0 k m b y h o r s e. I t t o o k 1 3 h o u r s 3 0 m i n u t e s. If t h e r a t e of the train i s 3 times tha t of the horse
464
Co n c ep t o f A r i t h m e t i c a n d 1 . 5 t i m es t h a t o f t h e st e a m er , f i n d t h e r a t e of h o r s e , t r a i n a n d s t e a m e r p er h o u r .
Soln:
Suppose the speed of horse = x km/hr. Then speed of the train = 3x km/hr and speed of the steamer = 2x km/hr Now, according to the question, 120 450 60 13.5 hours 2x 3x x (Since 13 hours 30 minutes = 13.5 hours) or,
360 900 360
Ex. 40: A p e r s o n c o v e r s a d i s t a n c e i n
40 m i n u t e s i f h e r u n s a t a s p eed o f 4 5 k m p e r h o u r o n a n a v er a g e. F i n d t h e s p ee d a t w h i c h h e m u s t r u n t o r ed u c e t h e t i m e o f j o u r n e y t o 3 0 m i n u t es .
Soln:
D 45
20 6 13.5 Hence, speed of horse = 20 km/hr Speed of train = 3x = 3 × 20 = 60 km/hr Speed of steamer = 2x = 2 × 20 = 40 km/hr
40 60
2 3
2
45 = 30 km 3 Let the required speed be x km per hour. Now, according to the question,
1620
x
= 40 minutes =
D =
13.5
6x
Let the distance be D km.
x
1 2
= 30 1 30 minutes 2 hour
K KUNDAN
Ex. 39: A m a n c o v e r s a c e r t a i n d i s t a n c e o n
x = 30 × 2 = 60 km/hr. Ex. 41: T h e d i s t a n c e b e t w e e n t w o s t a t i o n s ,
s co o t er . H a d h e m o v ed 3 k m / h r f a s t er , h e w o u l d h a v e t a k en 4 0 m i n u t e s l ess . I f h e h a d m o v e d 2 k m / h r s l o w er , h e w o u l d h a v e t a k en 4 0 m i n u t e s m o r e . F i n d t h e d i s t a n c e (i n k m ) a n d o r i g i n a l speed.
Soln:
D el h i a n d A m r i t s a r , i s 4 5 0 k m . A t r a i n s t a r t s a t 4 p m f r o m D el h i a n d m o v es t o w a r d s A m r i t s a r a t a n a v e r a g e s p eed o f 6 0 k m / h r . A no t h er t r a i n s t a r t s f r o m A m r i t s a r a t 3 . 2 0 p m a n d m o ves t o w a r d s D el h i a t a n a v er a g e s p ee d o f 8 0 k m / h r . Ho w f a r f r o m D el h i w i l l t h e t w o t r a i n s m eet a n d a t w h a t t i m e ?
Suppose the distance is D km and the initial speed is x km/hr.
D
Then, we hav e
x 3
D
x 2
or,
D
x
D
x 3
3D
D
x 2
x
D
x
40
60
Soln:
and
40 60
2
Suppose the trains meet at a distance of x km from Delhi. Let the trains from Delhi and Amritsar be A and B respectively. Th en , [Time taken by B to cover (450 – x ) km] – [Time t aken by A to cover x km]
3
=
2
or, x (x 3) 3 and
D
D
x
2D
.... (1)
or,
2
..... (see note)
450 x 80
x
60
40 60
or, 7x = 1190 or, x = 170. Thus, the trains meet at a distance of 170 km from Delhi. Time taken by A to cover 170 km
2 .... (2)
From (1) and (2), we have 3D 2D x (x 3)
60
3(450 – x ) – 4x = 160
3
or, x (x 2) 3
40
170 hours = 2 hours 50 min. 60
x (x 2)
=
or, 3(x 2) 2(x 3) or, 3x 6 2x 6
Note:
So, the trains meet at 6.50 pm RHS = 4 : 00 pm – 3.20 pm
x 12 km/hr Now, if we put this value in (1), we get D =
2
12 15
= 40 km. 3 3 Hence, the distance is 40 km and the original speed is 12 km/hr.
= 40 minutes =
40
hour 60 LHS comes from the fact that the train from Amritsar took 40 minutes more to travel up to the meeting point because it had started its journey at 3.20 pm whereas
465
S p ee d , T i m e a n d D i s t a n c e the train from Delhi had started its journey at 4 pm and the meeting time is the same for both the trains. Ex. 42: A man leaves a point P at 6 am and reaches the point Q at 10 am. Another man leaves the point Q at 8 am and reaches the point P at 12 noon. At what time do they meet? Soln: Let the distance PQ be A km. And they meet x hours after the first man starts. Average speed of first man =
A
A
A
A
or,
12 8
4
x 2 5x
2
or, 2x 2 + 10x = 1500 or, x 2 + 5x - 750 = 0 or, x 2 + 30x - 25x - 750 = 0 or, x (x + 30) - 25 ( x + 30) = 0 or, (x - 25) (x + 30) = 0 or, x = 25 or x = -30. Negative value of x should be rejected. Hence, the original speed of the train is 25 km/hr. Ex. 44: A f a s t t r a i n t a k e s 3 h o u r s l e ss t h a n a s l ow t r a i n f o r a t h e s p ee d o f t h e l e ss t h a n t h a t o f s p ee d s of t h e t w
km/hr. 10 6 4 Average speed of second man =
300x 1500 300x
Soln:
km/hr.
journey of 600 km . If slow tra in is 10 k m/hr t h e f a st t r a i n , f i n d t h e o tra ins.
Let the speed of the slow train be x km/hr. Then, speed of the fast train is (x + 10) km/hr. Time taken by the slow train to cover 600
K KUNDAN Distance travelled by first man =
Ax
km 4 The y me et x hours after the first man starts. The second man, as he starts 2 hours late, meets after ( x - 2) hours from his start. Therefore, the distance travelled by the second man =
Ax
Now,
A(x 2) 4
A(x 2)
hours. x Time taken by the fast train to cover 600
600 hours. x 10 Now, according to the question, km =
km
600 x
km = A
4 4 or, 2x – 2 = 4 x = 3 hours. They meet at 6 am + 3 hours = 9 am
or,
a t c o n s t a n t s p ee d . I f t h e s p ee d o f t h e t r a i n i s i n c r e a sed b y 5 k m a n h o u r , t h e j ou r n ey w o u l d h a v e t a k e n 2 h o u r s l e s s . F i n d t h e o r i g i n a l s p e ed o f t h e train.
300
x
300 x 5
or,
x (x 5)
x 10x
3
Ex. 45: A p l a n e l ef t 3 0 m i n u t e s l a t er t h a n t h e s ch e d u l e t i m e a n d i n o r d e r t o r e a ch i t s d es t i n a t i o n 1 5 0 0 k m a w a y i n t i m e i t h a s t o i n c r e a s e i t s sp e ed b y 2 5 0 k m / h r f r o m i t s u s u a l s p ee d . F i n d i t s u s u a l speed.
Soln:
2
6000
2
3
or, (x + 50) (x - 40) = 0 or, x = –50 or x = 40 or, x = 40 [ x cannot be negative] or, Hence, the speeds of two trains are 40 km/hr and 50 km/hr.
2
300(x 5) 300x
x (x 10)
or, x x 50 40 x 50 0
300
300
3
or, x 2 50x 40x 2000 0
hours.
hours. x 5 It is given that the time to cover 300 km is reduced by 2 hours.
x 10
or, x 2 10x 2000 0
x Tim e ta ken to cov er 300 km wh en th e speed is increased by 5 km/hr =
600
or, 3(x 2 10x ) 6000
Let x km/hr be the constant speed of the train. Then, time taken to cover 300 km =
600(x 10 ) 600 x
or,
Ex. 43: A t r a i n t r a v el s a d i s t a n c e o f 3 0 0 k m
Soln:
600
km =
Let the usual speed of the plane be x km/hr. Then,
466
Co n c ep t o f A r i t h m e t i c Tim e ta ken to cov er 1500 km wi th th e 1500 usual speed = hours x Tim e ta ken to cov er 1500 km wi th th e speed of (x + 250) km/hr =
or,
1500 x
1500 x
1500 x 250
1500 x 250
or,
x 250
2
x 250 x
600 hour x
Hence, duration of flight =
1 2
600 hour = 1 hour. 600
1
=
2
x (x 250) 1500 250
[ x cannot be negative] So, the original speed of the aircraft was 600 km/hr.
150
1500 x 1500 250 1500 x or,
or, x = 600 or, x = -400 or, x = 600
1
Ex. 47: T w o t r a i n s l ea v e a r a i l w a y s t a t i o n a t t h e sa m e t i m e. T h e f i r s t t r a i n t r a v el s d u e w e s t a n d t h e s e co n d t r a i n d u e n o r t h . T h e f i r s t t r a i n t r a v el s 5 k m / h r f a s t e r t h a n t h e s ec on d t r a i n . I f a f t e r two h ours, they are 50 km ap art, find t h e a v er a g e s p eed o f e a c h t r a i n .
2
1
2
or, 750000 = x 2 + 250x or, x 2 + 250x – 750000 = 0 or, x 2 + 1000x – 750x – 750000 = 0 or, x (x + 1000) - 750 ( x + 1000) = 0 or, (x + 1000) (x - 750) = 0 or, x = -1000 or x = 750 or, x = 750 [ speed cannot be negative] Hence, the usual speed of the plane is 750 km/hr.
K KUNDAN
Ex. 46: I n a f l i g h t o f 6 0 0 k m , a n a i r cr a f t w a s
s l o w ed d o w n d u e t o b a d w e a t h e r . I t s a v e r a g e s p ee d f o r t h e t r i p w a s r e d u c ed b y 2 0 0 k m / h r a n d t h e t i m e of f l i g h t i n c r e a s ed b y 3 0 m i n u t es . F i n d t h e duration of flight.
Soln:
Soln:
Let the speed of the second train be x km/hr. Then, the speed of the first train is (x + 5) km/hr. Let O be the position of the railway station from which the two trains leave. Distance travelled by the first train in 2 hours = OA = speed × time = 2(x + 5) km Distance travelled by the second train in 2 hours = OB = speed × time = 2x km By Pythagoras Theorem, AB2 = OA2 + OB2
Let the original speed of the aircraft be x km/hr. Then, new speed = ( x – 200) km/hr Duration of flight at original speed
600 hour x
=
Duration of flight at reduced speed
600 hour x 200
=
600 x 200
600 x
1 2
600x 600(x 200) or,
or,
x (x 200) 120000 2
x 200x 2
1 2
1 2
or, x – 200x – 240000 = 0 or, x 2 - 600x + 400x - 240000 = 0 or, x (x - 600) + 400 (x - 600) = 0 or, (x - 600) (x + 400) = 0 or, x - 600 = 0 or x + 400 = 0
or, 502 = [2(x + 5)]2 + {2x }2 2 or, 2500 = 4(x + 5) + 4x 2 or, 8x 2 + 40x – 2400 = 0 or, x 2 + 5x – 300 = 0 or, x 2 + 20x – 15x – 300 = 0 or, x (x + 20) – 15 (x + 20) = 0 or, (x + 20) (x – 15) = 0 or, x = –20 or x = 15 [ x cannot be negative] Hence, the speed of the second train is 15 km/hr and the speed of the first train is 20 km/hr. Ex. 48: A w a l k s h a l f a k m p h f a s t er t h a n B a n d t h r ee q u a r t e r s o f a k m p h f a s t er t h a n C . T o w a l k a c er t a i n d i s t a n c e C t a k e s
467
S p ee d , T i m e a n d D i s t a n c e t h r ee qu a r t er s o f a n h o u r m o r e t h a n B and tw o hours more tha n A. Find the d i s t a n c e c o v er e d a n d t h e t i m e t a k en b y B . A l s o f i n d h i s sp e ed .
Soln:
or, 4y 2 17y 4 0 or, y 4,
Let the speed of B be y kmph
speed of C = y
1
T = 10 hours. (y =
4
is not possible).
1
Ex. 49: A m a n s t a r t e d f r o m h i s h o u se t o h i s
2
workpl ace 8 k m away at th e rate of 4 k m p h s o a s t o r e a ch j u s t i n t i m e. A f t er 5 m i n u t es h e r e a l i s ed t h a t h e h a d l e f t s om e i m p o r t a n t d o c u m e n t s a t h o m e , s o h e t u r n ed b a c k , a n d n o w w a l k i n g a t a n i n c r e a s ed s p e e d , st i l l s u c ce ed e d in reaching his workp lace in ti me. What w a s h i s i n c r e a s ed s p e ed ?
kmph and
1 3 1 y kmph 2 4 4
Let the distance travelled be d km and time for A is T.
d 2d 1 2y 1 y 2
Now, for A, T =
4
Now from (5), d = 45 and from (3),
1 Speed of A = y kmph and 2 speed of B = y
1
... (1)
Soln:
K KUNDAN d 4d 1 4y 1 y 4
For C, T + 2 =
4d
or, T
4y 1
2
4d 8y 2 4y 1
... (2)
and for B, T
5
4
or, T
d
d 5 T or, y y 4
4d 54
... (3)
4y
Let A be the house and B be the workplace. Th e wo rk pl ac e is 8 km aw ay . Ha d he walked all the wa y at 4 kmph he would have taken 2 hours to reach his workplace, which would be just in time. Therefore, total time = 2 hours Referring to the above diagram, the total time is made up (i) walking from A to C for 5 minutes
From (1) and (2), 2d
2y 1
5
=
60 12 now, 4 km in 1 hour
4d 8y 2 4y 1
1
1
or, d (4y - 1) = (2y + 1) (2d - 4y + 1) or, 4dy - d = 4dy - 8y 2 + 2y + 2d - 4y + 1 or, 8y 2 + 2y - 3d - 1 = 0
8y 2 2y 1 or, d = 3 and from (2) and (3)
Distance AC =
4d 8y 2 4y 1
.... (4)
3
AC
AB
= .... (5)
2
or, 32y 8y 4 36y 9y
1
x
hours =
1
8 x
hours
Now, according to the question,
2
8y 2 2y 1 12y 2 3y 3 4
hours
x hours x 3 (iii) Walking from A to B time taken
or, 12y 2 3y 4d 0 12y 2 3y 4 From (4) and (5),
12
hours
km 3 (ii) If x is the increased speed, time for walking from C to A =
4y
or, 16dy 32y 2 8y 16dy 20y 2 4d 5y
2
km in
4d 5y
or, d
1
hours =
1 12
x = 4
1 3x 8
23
8 x
kmph
468
Co n c ep t o f A r i t h m e t i c
Ex. 50: D i n k y i s p i c k e d u p b y h i s f a t h er b y c a r
by travelling partly by foot and partly by car, he takes 45 minutes longer than if he would have travelled the whole distance by car. Therefore time taken to walk the distance BC = t w = (7.5 + 45 =) 52.5 minutes. t w = 52.5 minutes But for a constant distance BC, s d t w
f r o m s ch o o l e v er y d a y a n d t h e y r e a c h h o m e a t 5 . 0 0 p m . O n e d a y , s i n c e s ch o o l g o t o ver a n h o u r e a r l i e r t h a n u s u a l , h e started walk ing towards home at 3 km ph. He met his father on the way a n d t h e y r ea c h ed h o m e 1 5 m i n u t es ea r l i e r t h a n t h e i r u s u a l t i m e. W h a t i s t h e sp e ed o f t h e c a r ?
Soln:
s w
s d
t d
52.5
3 7.5 s d = 21 kmph
Let the speed of car be sd and the speed of Dinky be s w s w = 3 kmph Referring to the above diagram, if A represents the home and B represents the school, the father starts at his usual time but meets his son on the way at C. So, in going from A to C and back to A he saves the time he would have used commuting from C to B and back to C. Since they reach back 15 minutes earlier than usual, ie the time saved = 15 minutes. Therefore time taken to drive from C to B and back is 15 minutes. Therefore time taken to
Ex. 51: T w o p l a c es P a n d Q a r e 1 6 2 k m a p a r t . A t r a i n l ea v es P f o r Q a n d a t t h e sa m e t i m e a n o t h e r t r a i n l e a v es Q f o r P . B o t h t h e t r a i n s m e et 6 h o u r s a f t e r t h e y st a r t m o v i n g . I f t h e t r a i n t r a v el l i n g f r o m P t o Q t r a v el s 8 k m / h r f a s t e r t h a n t h e o t h e r t r a i n , f i n d t h e s p eed o f t h e t w o trains.
K KUNDAN drive the distance BC = t d =
Soln:
15
Suppose the speeds of the two trains are x km/hr and y km/hr respectively. Now, Total distance travelled by both the trains in 6 hours = (6x + 6y ) km Now, according to the question, (6x + 6y ) = 162 or, x y
162
27 ....(i) 2 and x – y = 8 ....(ii) Solving equations (i) and (ii), we have x = 17.5 km/hr and y = 9.5 km/hr. Hence, speed of the two trains are 17.5 km/hr and 9.5 km/hr.
2
t d = 7.5 minutes
Dinky starts 1 hour earlier than usual. Had he moved at driving speed, ie speed of the car, he would have reached 1 hour earlier. But he reaches only 15 minutes earlier. Therefore he loses 45 minutes, ie
Practice Exercise
1.
2.
3.
For a journey, walking
6
of his usual speed, 7 a man becomes late by 25 minutes. What is his usual time taken for the journey? A motorist covers a distance from A to B at a speed of 20 km/hr and return journey from B to A at a speed of 30 km/hr. If he takes 5 hours for the whole journey, find the distance from A to B. Shivangi starts from her house for her school at a certain fixed time. If she walks at the rate of 5 km/hr, she is late by 7 minutes. However, if she walks at the rate of 6 km/hr, she reaches the school 5 minutes earlier than
4.
5.
6.
the scheduled time. What is the distance of the school from her house? I will reach my destination 40 minutes late if I walk at the rate of 3 km/hr. However, I will reach 30 minutes before time if I walk at the rate of 4 km/hr. Find the distance of my destination from the starting point. A student travels to his school at a speed of 4 km/hr and reaches the school 15 minutes late. On travelling at a speed of 6 km/hr, he reaches the school 5 minutes early. At what speed must he travel to reach the school just in time? A person travels a certain distance on a bicycle at a certain speed. Had he moved 3 km/hr
469
S p ee d , T i m e a n d D i s t a n c e
7.
8.
9.
faster, he would have taken 40 minutes less. Had he moved 2 km/hr slower, he would have taken 40 minutes more. Find the distance and original speed of the person. Two pla ces A and B are 80 km apar t fr om each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet each other in 8 hours. If they move in opposite directions towards each other, they meet in 1 hour 20 minutes. Determine the speeds of the cars. A train covers a distance between stations A and B in 45 minutes. If the speed is reduced by 5 km/hr, it will cover the same distance in 48 minutes. What is the distance between the two stations A and B (in km)? Also, find the speed of the train. A car covering half of a distance of 100 km develops some engine trouble and later travels at half of its original speed. As a result, it arrives 2 hours later than its normal time. What was the original speed of the car?
station B meets with an accident one hour after starting. After stopping there for 30 4 minutes, it proceeds at of its usual speed 5 and arrives at B 2 hours late. Had the train covered 80 km more before the accident, it would have been just one hour late. Determine the original speed of the train and the distance between A and B. 17. A train after travelling 50 km meets with an 3 accident and then proceeds at of its former 4 speed and arrives at its destination 25 minutes late. Had the accident occurred 24 km behind, it would have reached the destination only 35 minutes late. Find the speed of the train and the distance travelled by the train. 18. Ravi can walk a certain distance in 40 days, when he rests 9 hours a day. How long will he take to walk twice the distance, twice as fast and rest twice as long each day? 19. Two men set out at the same tim e to wal k towards each other from two points A and B, 72 km apart. The first man walks at the rate of 4 km/hr. The second man walks 2 km in
K KUNDAN
10 . A train covers a distance of 193
1 3
km in 4
1
4
hours with one stoppage of 10 minutes, two of 5 minutes and one of 3 minutes on the way. Find the average speed of the train. 11. Distance between two places X and Y is 60 km. Two persons A and B start from X towards Y at the same time. Speed of B is 4 km/hr less than the speed of A. A reaches Y, returns at once and meets B at a distance of 12 km from Y. What is the speed of B? 12 . An army bomb squad man set a fuse for blasting a rock to take place after one minute. He ran away from the site at the speed of 13 m/s. Sound travels at the speed of 325 m/s. Upto what distance could the army man run, before he heard the sound of blast? 13. On a particular day a person starts walking from a place X at 2 am and reaches place Y at 5 am. A second person starts walking from a place Y at 4 am and reaches place X at 9 am on the same day. At what time do they cross each other? 14. A covers some distance in 50 days when he rests 9 hours a day. In how many days will he cover the double distance by resting twice as before? 15. A man travelled a total distance of 3990 km, part of it by air, part by water and the rest by land. The time he spent in travelling by air, water and land was in ratio 1 : 16 : 2 respectively and the average speed of each mode of travel was in the ratio 20 : 1 : 3 respectively. If his overall average speed was 42 km/hr, find the distance covered by water. 16. A goods train travelling from station A to
the first hour,
20.
21 .
22.
23 .
2
1
km in the second hour, 3 2 km in the third hour and so on. Find the time after which the two men will meet. Two trains start out towards each other from points 650 km apart. If they start out at the same time, they will meet in 10 hours, but if one of them starts out 4 hours and 20 minutes after the other, they will pass each other 8 hours following the departure of the latter. Determine the average speed of each other. Distance between two stations X and Y is 220 km. Trains P and Q leave station X at 8 am and 9.51 am respectively at the speed of 25 km/hr and 20 km/hr respectively for journey towards Y. A train R leaves station Y at 11.30 am at a speed of 30 km/hr for journey towards X. When and where will P be at equal distance from Q and R. Two places P and Q are 3 36 km apart. A t rain leaves P for Q and at the same time another train leaves Q for P. Both trains meet at the end of the 6 hours. If one train travels 8 km/hr faster than the other, find the speeds of the other trains. On a 2-km road, a total number of 201 trees are planted on each side of the road at equal distances. How many such trees in all will be planted on both sides of a 50-km road such that the distance between two consecutive trees is the same as that of the consecutive trees on the 2-km road?
470
Co n c ep t o f A r i t h m e t i c
Answers and explanations 1.
Let the usual speed of the person be x km/hr and the distance of his journey be D km. D His usual time to cover the distance = hour x Now, according to the question,
6
Speed =
7
Since the difference between the two times taken is (7 + 5 =) 12 minutes 12x – 10x = 12 or, 2x = 12 12 = 6 x = 2 Hence, the required distance is 6 km. Alternative Method: Let x km be the distance between her house and school and t hours be the time required to reach the school from her house. When Shivangi walks at 5 km/hr, then x 7 t ....(i) 5 60 When Shivangi walks at 6 km/hr, then x 5 t 6 60
6 x km/hr 7
of his usual speed =
Time taken to cover the distance D km
D 7D = hour 6x 6x 7 Again, 7D
K KUNDAN 6x
or, or,
25 x 60
D
5 1 x 6 12
or
5 5 6 hours 12 2
D
x
Usual time = 2
2.
1
hours. 2 Let the distance from A and B is x km. Time taken to cover the distance from A to B at 20 km/hr =
x
x hours. 30 Total time taken = 5 hours. (Given)
20
x
30
x
30
4.
1
12
7
60
57 60
=
70 60
hours =
7
6
5
x 3
hours
x
hours 4 According to the question,
= 60 km 5 Let the required distance be x km.
x 3 x 5
hours
x = 60 minutes = 12x minutes 5 Time taken to walk at 6 km/hr =
1
hours
Time taken at 4 km/hr =
300
Time taken to walk at 5 km/hr =
60
= 6 5 Hence the required distance = 6 km Let the required distance be x km. Difference of time taken at different speeds = (40 + 30) minutes = 70 minutes
Time taken at 3 km/hr =
5 60 or, 5x = 60 × 5 = 300
12
30
5
3x 2x
or, x =
x
x =
to A at 30 km/hr =
x
or,
hours.
20 And time taken to cover the distance from B
3.
1 t ....(ii) 6 12 Subtracting equation (ii) from equation (i), we get x x 7 1 t t 5 6 60 12
D 7
x 6
or,
or,
4
7 6
4x 3x 12
x 12
hours
x = 60 minutes = 10x minutes 6
x
7 6
7 6
7
12 = 14 6 Distance of the destination = 14 km
or, x =
471
S p ee d , T i m e a n d D i s t a n c e 5.
Let the distance of the school be x km. Time taken in first case =
x 4
or, D
....(i) 9 When the person moves 2km/hr slower, then
hours
But this time is 15 minutes late or
15
1
D V 2
60 4 hours late Actual time for reaching the school in time
or,
x 1 should be hours 4 4 Time taken in second case =
or,
x 6
2V V 3
hours
D V 2 D V 2
D
D
D
V
V
V
40
2
2
60
3
3
DV DV 2D or,
5 1 But this time is 5 minutes early or 60 12
V(V 2)
2 3
2D 2 or, V(V 2) 3
hours early. Actual time for reaching the school in time
K KUNDAN or, D
From the above, we have
2V(V 3)
x 1 x 1 4 4 6 12
or,
x 1 4
12
or, 12x 12 8x 8
or, 4x 20 or, x = 5 km The distance of the school be 5 km and
6.
(x 1)
4 = 1 hour The required speed is 5 km/hr. Let the original speed and distance be V km/hr and D km respectively. Time taken to complete the whole journey D = hours. V When the person moves 3 km/hr faster, then D V3 or,
or,
D V3
D V
D
D
V
V
D V3
40 60
2
2
V V 3
3D 2 or, V V 3 3
V(V 2)
we get D =
7.
12 10
= 40 km. 3 Case I: When the cars are moving in the same direction.
Let A and B be two places and C be the place of meeting. Let the speed of car starting from A be x km/hr and the car starting form B be y km/hr. Relative speed = (x – y ) km/hr According to the question, (x – y ) × 8 = 80 or, x – y = 10 ...(i) Case II: When the cars are moving in the opposite directions and they meet at point C.
3 Relative speed = (x + y ) km/hr Time taken = 1 hour 20 minutes
3
DV 3D DV or,
9 3 or, 2(V + 3) = 3(V – 2) or, 2V + 6 = 3V – 6 or, 3V – 2V = 6 + 6 or, V = 12 km/hr Putting the value of V in equation (ii),
2x 1
actual time to reach school in time =
V(V 2)
....(ii) 3 Combining equations (i) and (ii), we get
x 1 hours should be 6 12
= 1
2 3
1 4 hours 3 3
Again, according to the question,
(x y )
4 3
80
472
8.
Co n c ep t o f A r i t h m e t i c
or, x + y = 60 ...(ii) Solving equations (i) and (ii), we have x = 35 and y = 25 Speeds of the cars = 35 km/hr and 25 km/hr. Suppose the distance is x km and the speed of the train is y km/hr. Thus we hav e two rel ati onsh ips: (1)
(2)
x
45
x
y
60
y 5
3 4
48 60
3
x 4
4
=
580 15 58 3
= 50 km/hr
11.
y 4
x
5
Average speed of the train =
580 3 km/hr 58 15
5
(y 5)
From (1) and (2), we have
3 4
4
y
5
(y 5 )
4 3 4 5 4
Let A and B meet after t hours. Let the speed of B be x km/hr. Speed of A = (x + 4) km/hr Distance covered by A in t hours = 60 = 72 Distance covered by B in t hours = 60 = 48 Now, according to the question, xt = 48 ....(i) (x + 4)t = 72 ....(ii) On dividing equation (ii) by equation have
+ 12 km – 12 km
K KUNDAN or, y
4 20
or, y =
16 15
= 80 km/hr
Theref ore speed = 80 km/hr and distan ce x =
9.
3
80 = 60 km
4 Half of the original speed means double the normal time. It means that the car should have covered half of the distance of 100 km, ie 50 km, in 2 hours. Hence, the original speed of the car
50 = 25 km/hr 2
=
10 . Distance covered by train = 193
1
km =
580
km 3 3 Time taken by the train to cover this distance
17 hours = hours 4 4 Total stoppage during the journey = 10 × 1 + 5 × 2 + 3 × 1 = 4
1
= 23 minutes =
23 60
hours
Actual time taken by the train to cover the above distance =
=
=
17 4
23 60
17 15 23
60
232 60
=
58 15
hours
72
3
x 48 2 or, 2x + 8 = 3x or, x = 8 Speed of A = 8 km/hr 12 . Time after which the bomb is set to explode = 1 minute = 60 seconds Speed of the man = 13 m/sec Distance covered by man in 60 sec = 13 × 60 = 780 metres So, distance to be travelled by sound before it catches up with army man = 780 metres Speed of the sound = 325 m/sec (given) Since the man and sound are travelling in the same direction, the relative speed of sound = (325 – 13 =) 312 m/sec Time taken by sound to travel 780 metres =
780
= 2.5 sec 312 Now, during this time man would have travelled further. So, distance covered by man in 2.5 seconds = 2.5 × 13 = 32.5 m The total distance trave lled by man = 780 + 32.5 = 812.5 metres. 13. X P Y l
60 255 23
x 4
(i), we
l
l
Let the speed of the person who starts from X be x km/hr and speed of the person who starts from Y be y km/hr. Time taken by the person who starts from X = 5 am – 2 am = 3 hours
473
S p ee d , T i m e a n d D i s t a n c e Time taken by the person who starts from Y = 9 am – 4 am = 5 hours Again, let the distance between X and Y be D km . Now, according to the question,
D
x km/hr =
3
and
y km/hr =
14.
Let the distance for A be x km Number of hours A walks daily = (24 – 9 =) 15 hours Number of days = 50 days
Speed (in km/hr) =
D 5
x 50 15
..... (1)
In second situation Let the number of days be Y Distance = 2x Number of hours for which A walks daily = 6 hours Speed in second case (in km/hr)
If the person starting from X reaches the meeting point after t hours, person starting from Y will reach the meeting point after (t – 2) hours. Since the person starting from X starts moving at 2 am while the person starting from Y starts moving at 4 am. And the difference of time = (4 am – 2 am)= 2 hours Distance (XP) travelled by the person
=
Distance
2x
..... (2) Time Y6 In both the cases, the speed remains the same
D t km 3
starting from X =
2x Y6
2x 50 15
K KUNDAN and the distance (YP) travelled by the person
or, Y × 6 = 50 × 15
D starting from Y = (t 2) km 5
Total distance t ravelled by both before meeting = Distance travelled by person from X + Distance travelled by person from Y
D D t (t 2) D 3 5
=
t t 2 D 5 3
or, D
or,
t
3
t 2 5
= 3990
16.
5t 3t 6
or, t =
21
2
5
hours 8 8 Converting this in hours, minutes and seconds, we get 2 hours 37 minutes and 30 seconds. [2
5 8
= 125 days 6 15. Total distance trave lled = 3990 km Ratio of time spent in travelling by air, water and land = 1 : 16 : 2 Ratio of respective speeds = 20 : 1 : 3 From the given fact, the ratio of respective distances will be 20 : 16 : 6 = 10 : 8 : 3 Sum of the ratios = 10 + 8 + 3 = 21 Distance travelled by steamer will be
1
1 15 or, 8t – 6 = 15 or, 8t = 15 + 6 = 21 or,
50 15
or, Y =
5 60 minutes 8
hours = 2 hours +
1 75 37 minutes 2 2
= 2 hours +
= 2 hours + 37 minutes +
1 2
1 60 2
30 seconds = 2 hours 37 minutes and 30 seconds]
= 1520 km
Let the distance between station A and station B be d km. Again, let the initial speed of the goods train be x km/hr. As the accident takes place after 1 hour distance covered in 1 hour by the goods train = x km Remaining distance = (d – x ) km Total tim e taken, if no accident happened
d hours x
=
Case I: Time taken by the goods tra in to cove r the distance = 1
minutes
= 2 hours + 37 minutes +
8
21
30 d x 4x 60 5
1 5(d x ) = 1 hours 2 4x Now, according to the question, 1
1 2
5(d x ) 4x
d x
2
474
Co n c ep t o f A r i t h m e t i c
or,
or,
or,
or,
(d x ) 5
d 1
d
4x (d x ) 5 4x
x
x
4x d 5x
2 1
5d 5x 4d
17.
Let the distance be D km and speed be the x km/hr From the question, we have 50 (D 50) 4 D 25 D 5
x
2
1
or,
2 or,
1
4x 2 or, 2d – 10x = 4x or, 2d = 14x or, d = 7x ....(i) Case II: If the goods train had covered 80 km more before the accident, then the distance of site of the accident = (x + 80) km Remaining distance = [d – (x + 80)] km Time taken to cover the whole of the distance
3x
150 4D 200
3x 4D 50
3x
x
60
x
12
12D 5x 12x
12D 5 x 12x
or, 16D 200 12D 5x ... (i) and 4D – 5x = 200
50 24 x
(D 26) 4
3x
D x
26
4D 104
35
60
D
7
x 12
12D 7x 12x
12D 7 x
K KUNDAN ( x 80) 30 d (x 80) hours = 4x x 60 5
According to the question,
x 80 30 d ( x 80) d 1 4x x 60 x 5
or, 1
or,
or,
or,
or,
80 x
80 x
1
2
1 2
5[d ( x 80]
4x
5[d (x 80)] 4x
320 5d 5x 400 4x
5d 5x 80 4x
1 2
1
2
1 2
d x
1
d x
d x
d x
4d 5d 5x 80 4x
or, 2x = 5x – d + 80 Putting the value of d from equation (i), we have 2x = 5x – 7x + 80 or, 4x = 80
x =
or, or, or,
x
3x
78 4D 104 3x
4D 26 3x
12x
12D 7x 12x
12D 7x 12x
or, 4D 7x 104 .... (ii) Now, subtracting equation (ii) from equation (i), we have 2x = 96 x = 48 km/hr Put the value of x in equation (i) and find the distance (D) or, 4D – 5 × 48 = 200 or, 4D = 200 + 240 = 440
D =
440
= 110 km. 4 18. Time for work per day in first condition = (24 – 9 =) 15 hours Time for work per day in second conditi on = (24 – 9 × 2 =) 6 hours Here we have four quantities : Speed, Distance, Work and Days. We have to calculate number of days. Hence, Days will be in the last column. Here following relationships exist: More speed , less days (Inverse) More distance , more days (Direct) Less hours of work , more days (Inverse)
80
= 20 4 Hence original speed of the train = 20 km/hr. Distance between the stations A and B = d = 7x (From i) = (7 × 20) = 140 km.
Hence,
2 : 1 1 : 2 :: 40 : x 6 : 15 or, 2 × 1 × 6 : 1 × 2 × 15 :: 40 : x (Compounding the ratios)
475
S p ee d , T i m e a n d D i s t a n c e or, 2 × 1 × 6 × x = 1 × 2 × 15 × 40 (Product of extreme terms = Product of mean terms)
x = 19.
1 2 15 40 2 1 6
Hence the required time = 100 days. Let A starts from point X, B starts from point Y and they meet after t hours. A
B
X
Y
P
Sum of an AP =
2a n 1d
2 where n = number of terms, a = first term and d = common difference
hours
13x
=
13x
8x + 8y = 650
YP = 2 + 2.5 + 3 + .... t terms This is an AP.
3
3 3 Both the trains meet 8 hours after train A leaves P. Now if they meet at C 1 then PC1 = 8 × x = 8x km BC1 = 8 × y = 8y km According to the question,
XP = 4 × t = 4t km
n
13
= AP = x ×
= 100
13
Distance covered by train A in
or, 8x y 650 or, 8 65 650
3
13x 3
13x 3
K KUNDAN YP =
t 1 t t 1 2 2 (t 1) 4 2 2 2 2 2
t 2 t 7t t 2 7t t 2 = 2t 4 4 4 4 4
But it is given that XY = 72 or XP + PY = 72
7t t 2 + 4t = 72 4
or, 7t + t 2 + 16t = 288 or, t 2 + 23 t - 288 = 0 or, t 2 + 32 t – 9t - 288 = 0 or, t (t + 32) – 9 (t + 32) = 0 or, (t + 32) )(t – 9) = 0 + 32 = 0 t or, t = –32 (Not possible) t – 9 = 0 or, t = 9 They meet after 9 hours. 2 0 . Let the trains A and B travel at speed of x and y km/hr respectively and meets 10 hours after departure.
From the figure it can be seen that AC = (x × 10) km BC = (y × 10) km AC + BC = x × 10 + y × 10 or, 650 = 10(x + y ) or, x + y = 65 In the second situation when the other train starts after 4 hours and 20 minutes
4 hours and 20 minutes = 4
20 60
4
1 3
13 3
hours
or,
13x 3
or, x =
650 520 130
130 3
= 30 km/hr 13 Speed of train A = 30 km/hr Speed of train B = (65 – 30) km/hr = 35 km/hr 21 . As given, speed of the train P = 25 km/hr Speed of the train Q = 20 km/hr Speed of the train R = 30 km/hr Q
20 t
P
B
1
A
P
Q
X 33 km
25 t
1
1
R
R
30 t
Y
87.5 km
Distance travelled by train P between 8:00 to 11:30
1
7
hours =
25
175
87.5 km 2 2 2 Distance travelled by train Q between 9 : 51 to 11 : 30 ie. in ie in 3
33 20 20 60 20 = 33 km Assume that trains P and Q are at A and B respectively at 11 : 30 am. Also assume that t minutes after 11 : 30 am, train P was equidistant from train Q and train R. At the equidistant position train P, Q and R were at P1, Q1 and R 1. XP1 = XA + AP1 = (87.5 + 25 t ) km XQ1 = XB + BQ1 = (33 + 20 t ) km P1 Q1 = XP1 – XQ1 = (87.5 + 25 t ) - (33 + 20 t ) = (54.5 + 5 t ) km Distance RR1 = 30 t km 1 hour 39 minutes = 1
39
476
Co n c ep t o f A r i t h m e t i c P1R1 = Total distance - XP 1 - RR 1 = 220 - (87.5 + 25 t ) - 30 t = (132.5 - 55 t ) km 1 1 = P1R 1 P Q 5t + 54.5 = 132.5 - 55 t 78 60 minutes or, 60 t = 78 or t = 60 or, t = 78 minutes So 78 minutes after 11 : 30 am ie at 12 : 48 pm train P will be equidistant from train Q and R. 78 XP1 = 87.5 + 25 t = 87.5 + 25 × 60 = 87.5 + 32.5 XP1 = 120 km At 120 km away from station X, trains would be at equal distances.
Let the speed of train from P = x km/hr and that from Q = (x + 8) km/hr Both trains meet after 6 hours (x × 6) + (x + 8) × 6 = 336 or, 6x + 6x + 48 = 336 or, 12x = 336 – 48 = 288 or, x =
288
= 44 12 Speed of one train = 24 km/hr Speed of the other train = (24 + 8 =)32 km/hr 23 . Distance between 2 trees on a 2-km road
2 1000 = 10 m 201 1
=
Number of trees planted on both sides of a 50-km road
50 1000 1 = 10002 10
= 2
22.
K KUNDAN Let R be the meeting point.