SPEED, TIME AND DISTANCE
Concept of speed, time and distance is based on the formula Speed × Time = Distance
Speed (s) :
The rate at which any moving body covers a particular distance is called its speed.
Speed Distance Time
= ;
Time (t) :
It is the time duration over which the movement has occured. The unit used for measuring time is synchronous with denominator of the unit used for measuring speed. Thus, if the speed is measured in terms of km/h then time is measured in hours.
Time =Distance Speed ;
Unit : SI unit of speed is metre per second (mps). It is also mea-sured in kilometers per hour (kmph) or miles per hour (mph). Conversion of units :
(i) 1 hour = 60 minutes = 60 × 60 seconds. 1 km = 1000 m
1 km = 0.625 mile
Þ 1 mile = 1.60 km, i.e. 8 km = 5 miles 1 yard = 3 feet 1 foot = 12 inches 1 km/h =185 m/sec, 1 m/sec 18 5 = km/h 1 miles/hr 22 15 = ft/sec.
* Average speed Total Distance Total time =
* While travelling a certain distance d, if a man changes his speed in the ratio m : n, then the ratio of time taken becomes n : m.
* If a certain distance (d), say from A to B, is covered at ‘a’ km/hr and the same distance is covered again say from B to A in ‘b’ km/hr, then the average speed during the whole journey is given by :
Average speed = æçèa b2ab+ ö÷ø km/h
... (which is the harmonic means of a and b
Also, if t1 and t2 is time taken to travel from A to B and B to A, respectively, the distance ‘d’ from A to B is given by :
d = (t1 + t2) æça bab+ ö÷ è ø d = (t1 – t2) æçèb aab- ö÷ø d = (a – b) 1 2 2 1 t t t t æ ö ç - ÷ è ø
* If a body travels a distance ‘d’ from A to B with speed ‘a’ in time t1 and travels back from B to A i.e., the same distance with m
n of the usual speed ‘a’, then the change in time taken to cover the same distance is given by :
Change in time = æçmn-1ö÷
è ø × t1; for n > m = æç1-mn ö÷
è ø × t1; for m > n
* If first part of the distance is covered at the rate of v1 in time t1 and the second part of the distance is covered at the rate of v2 in time t2, then the average speed is 1 1 2 2
1 2 v t v t t t æ + ö ç + ÷ è ø
Relative Speed : When two bodies are moving in same direction with speeds S1 and S2 respectively, their relative speed is the difference of their speeds.
i.e. Relative Speed = S1 – S2
When two bodies are moving in opposite direction with speeds S1 and S2 respectively, their relative speed is the sum of their speeds.
i.e. Relative Speed = S1 + S2
Example 1 :
The driver of a maruti car driving at the speed of 68 km/h locates a bus 40 metres ahead of him. After 10 seconds, the bus is 60 metres behind. Find the speed of the bus.
Solution :
Let speed of Bus = SB km/h.
Now, in 10 sec., car covers the relative distance = (60 + 40) m = 100 m \ Relative speed of car =100
10 = 10 m/s = 10 18 36 km / h
5
´ =
SPEED, TIME & DISTANCE (TRAINS, BOAT & STREAM)
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\ 68 – SB=36 B
S 32km / h
Þ =
Example 2 :
If a person goes around an equilateral triangle shaped field at speed of 10, 20 and 40 kmph on the first, second and third side respectively and reaches back to the starting point, then find his average speed during the journey.
Solution :
Let the measure of each side of triangle is D km. The person travelled the distance from A to B with 10 kmph, B to C with 20 kmph and C to A with 40 kmph.
D
B C
A
D D
If TAB = Time taken by the person to travel from A to B, TBC = Time taken by the person to travel from B to C and TCA = Time taken by the person to travel from C to A. Then total time = TAB + TBC + TCA
D D D 10 20 40 = + + D 8 4 2 7D 80 40 + + æ ö = ç ÷= è ø
Total distance travelled = D + D + D = 3D Hence, average speed
3D 120 17 kmph.1 7D 7 7 40 = = = Example 3 :
Two guns were fired from the same place at an interval of 15 min, but a person in a bus approaching the place hears the second report 14 min 30 sec after the first. Find the speed of the bus, supposing that sound travels 330 m per sec.
Solution :
Distance travelled by the bus in 14 min 30 sec could be travelled by sound in (15 min – 14 min 30 sec) = 30 second. \ Bus travels = 330 × 30 in 141
2 min. \ Speed of the bus per hour
330 30 2 60 99 12 29 1000 29 ´ ´ ´ ´ = = ´ 1188 28 40 km / hr 29 29 = = Example 4 :
A hare sees a dog 100 m away from her and scuds off in the opposite direction at a speed of 12 km/h. A minute later the dog perceives her and gives chase at a speed of 16 km/h. How soon will the dog overtake the hare and at what distance from the spot where the hare took flight ?
Solution :
Suppose the hare at H sees the dog at D.
D H K
\ DH = 100 m
Let K be the position of the hare where the dog sees her. \ HK = the distance gone by the hare in 1 min
12 1000 1m 60
´
= ´ = 200 m
\ DK = 100 + 200 = 300 m The hare thus has a start of 300 m. Now the dog gains (16 – 12) or 4 km/h. \ The dog will gain 300 m in 60 300
4 1000 ´ ´ min or 1 4 2 min. Again, the distance gone by the hare in 41
2 min 12 1000 1 4 900m 60 2 ´ = ´ =
\ Distance of the place where the hare is caught from the spot H where the hare took flight = 200 + 900 = 1100 m * If two persons (or vehicles or trains) start at the same time in
opposite directions from two points A and B, and after crossing each other they take x and y hours respectively to complete the journey, then
Speed of first y Speed of second= x
Example 5 :
A train starts from A to B and another from B to A at the same time. After crossing each other they complete their journey in 31
2 and 4 2
7 hours respectively. If the speed of the first is 60 km/h, then find the speed of the second train.
Solution : 4 2 1st train 's speed y 7 1 2nd train 's speed x 3 2 = = 18 2 6 7 7 7 = ´ = \ 60 6 2nd train 's speed =7 Þ 2nd train’s speed = 70 km/h. * If new speed is a
b of usual speed, then Usual time Change in time
b 1 a = æ - ö ç ÷ è ø Example 6 :
A boy walking at 35 of his usual speed, reaches his school 14 min late. Find his usual time to reach the school.
Usual time = 514 14 3 21 min 2 1 3 ´ = = -Example 7 :
A train after travelling 50 km, meets with an accident and then proceeds at 45 of its former rate and arrives at the terminal 45 minutes late. Had the accident happened 20 km further on, it would have arrived 12 minutes sooner. Find the speed of the train and the distance.
Solution :
Let A be the starting place, B the terminal, C and D the places where the accidents to be placed.
A C D B
By travelling at 45 of its original rate the train would take 5 4 of its usual time, i.e.,1
4 of its original time more. \ 14 of the usual time taken to travel the distance
CB = 45 min. ...(i)
and 14 of the usual time taken to travel the distance DB = (45 – 12) min ...(ii)
Subtracting (ii) from (i), 1
4 of the usual time taken to travel the distance CD = 12 min. \ Usual time taken on travel 20 km = 48 min.
\ Speed of the train per hour 20 60 48
= ´ or 25 km/h. From (i), we have
Time taken to travel CB = 45 × 4 min = 3 hrs. \ The distance CB = 25 × 3 or 75 km.
Hence the distance AB = the distance (AC + CB) = 50 + 75 or 125 km.
* A man covers a certain distance D. If he moves S1 speed faster, he would have taken t time less and if he moves S2 speed slower, he would have taken t time more. The original speed is given by
(
1 2)
2 1 2 S S S S ´ ´ -Example 8 :A man covers a certain distance on scooter. Had he moved 3 km/h faster, he would have taken 20 min less. If he had moved 2 km/h slower, he would have taken 20 min more. Find the original speed.
Solution :
Speed 2
(
3 2)
12km / hr. 3 – 2´ ´
= =
* If a person with two different speeds U & V cover the same distance, then required distance
U V
Difference between arrival time U V
´
= ´
-Also, required distance otal time taken U V U V
´
= T ´
+
Example 9 :
A boy walking at a speed of 10 km/h reaches his school 12 min late. Next time at a speed of 15 km/h reaches his school 7 min late. Find the distance of his school from his house?
Solution :
Difference between the time = 12 – 7 = 5 min 5 1hr 60 12 = = Required distance = 12 1 10 15 10 15 ´ -´ 150 1 2.5 km 5 12 = ´ =
* A man leaves a point A at t1 and reaches the point B at t2. Another man leaves the point B at t3 and reaches the point A at t4, then they will meet at
(
)(
)
(
2 1)
(
4 1)
1 2 1 4 3 t – t t – t t t – t t – t + + Example 10 :A bus leaves Ludhiana at 5 am and reaches Delhi at 12 noon. Another bus leaves Delhi at 8 am and reaches Ludhiana at 3 pm. At what time do the buses meet?
Solution :
Converting all the times into 24 hour clock time, we get 5 am = 500, 12 noon = 1200, 8 am = 800 and 3 pm = 1500 Required time 500 (1200 – 500)(1500 – 500) (1200 – 500) (1500 – 800) = + + 00 700 1000 700 700 ´ = 5 + + = 1000 = 10 am.
* Relation between time taken with two different modes of transport : t2x + t2y = 2(tx + ty)
where,
tx = time when mode of transport x is used single way. ty = time when mode of transport y is used single way. t2x = time when mode of transport x is used both ways. t2y = time when mode of transport y is used both ways.
Example 11 :
A man takes 6 hours 30 min. in going by a cycle and coming back by scooter. He would have lost 2 hours 10 min by going on cycle both ways. How long would it take him to go by scooter both ways?
Solution :
Clearly, time taken by him to go by scooter both way = 6h.30m – 2h.10m = 4h.20m =
3 1 4 h
Example 12 :
A man travels 120 km by ship, 450 km by rail and 60 km by horse taking altogether 13 hrs 30 min. The speed of the train is 3 times that of the horse and 1½ times that of the ship. Find the speed of the train.
Solution :
If the speed of the horse is x km/hr; that of the train is 3x and
that of the ship is 1½3x = 2x km/hr
\ 1202x + 4503x + 60 x =
27 2
\ 60x + 150 x + 60 x = 27 2 \ 270 x = 27 2 \ x = 20 \ Speed of the train = 60 km/hr..
Example 13 :
Rajesh travelled from the city A to city B covering as much distance in the second part as he did in the first part of his journey. His speed during the second part was twice his speed during the first part of the journey. What is his average speed of journey during the entire travel ?
(1) His average speed is the harmonic mean of the individual speed for the two parts.
(2) His average speed is the arithmetic mean of the individual speed for the two parts.
(3) His average speed is the geometric mean of the individual speeds for the two parts.
(4) Cannot be determined.
Solution :
The first part is ½ of the total distance & the second part is ½ of the total distance. Suppose, he travels at a km/hr speed during the first half & b km/hr speed during the second half. When distance travelled is the same in both parts of the journey, the average speed is gives by the formula 2ab
a b+ i.e. the harmonic mean of the two speeds.
Example 14 :
Two friends X and Y walk from A to B at a distance of 39 km, at 3 km an hour and 31
2km an hour respectively. Y reaches B, returns immediately and meet x at C. Find the distance from A to C .
Solution :
When Y meets X at C, Y has walked the distance AB + BC and X has walked the distance AC.
X C Y A y B
So, both X and Y have walked together a distance = 2 × AB = 2 × 39 = 78 km . The ratio of the speeds of X and Y is 3 : 31
2 i.e, 6 7 Hence , the distance travelled by X = AC = 6
6 7+ × 78 = 36 km
Example 15 :
A man rides one-third of the distance from A to B at the rate of ‘a’ kmph and the remainder at the rate of ‘2b’ kmph. If he had travelled at the uniform rate of 3c kmph, he could have rode from A to B and back again in the same time. Find a relationship between a, b and c.
Solution :
Let the distance between A and B is X km and T1 and T2 be the time taken, then
1 X T 3a = , T2 2X X 6b 3b = = , T1+T2=X a b3éê ab+ ùú ë û
Let T3 be the time taken in third case, then T3= 2X3c
Þ 2X X (a b) 3c =3ab + Þ 2 a b 2ab c c ab a b + = Þ = + Example 16 :
Two cyclists start from the same place to ride in the same direction. A starts at noon at 8 kmph and B at 1.30 pm at 10kmph. How far will A have ridden before he is overtaken by B ? Find also at what times A and B will be 5 km apart.
Solution :
If A rides for X hours before he is overtaken, then B rides for (X – 15) hrs.
Þ 8X = 10 (X – 1.5) Þ X = 7.5
Þ A will have ridden 8 × 7.5 km or 60 km.
For the second part, if Y = the required number of hours after noon, then
8X = 10 (X – 1.5) ± 5
Þ X = 10 or 5 according as B is ahead or behind A. Þ The required times are 5 p.m. and 10 p.m.
Example 17 :
Two men A and B start from a place P walking at 3 kmph and 3½ kmph respectively. How many km apart will they be at the end of 2½ hours ?
(i) If they walk in opposite directions ? (ii) If they walk in the same direction ?
(iii) What time will they take to be 16 km apart if. (a) they walk in opposite directions ? (b) in the same direction ?
Solution :
(i) When they walk in opposite directions, they will be
1 1 3 3 6 km 2 2 æ + ö= ç ÷ è ø apart in 1 hour.. \ ln 21
2 hours they will be
1 5 1
6 16 km
2 2´ = 4 apart. (ii) If they walk in the same direction, they will be
1 1
3 3 km
2- =2 apart in 1 hour.. Þ ln 21
2 hours they will be
1 5 1
1 km 2 2´ = 4 apart. (iii) Time to be 16 km apart while walking in opposite directions = 16 2 6 1 13 3 3 2 = + hours.
time = 16 32 hours 1 3 3 3 = -TRAINS
A train is said to have crossed an object (stationary or moving) only when the last coach of the train crosses the said object completely. It implies that the total length of the train has crossed the total length of the object.
* Time taken by a train to cross a pole/a standing man =Length of trainSpeed of train .
* Time taken by a train to cross platform/bridge etc. (i.e. a stationary object with some length)
length of train + length of platform/bridge etc. . speed of train
=
* When two trains with lengths L1 and L2 and with speeds S1 and S2 respectively, then
(a) When they are moving in the same direction, time taken by the faster train to cross the slower train
=difference of their speedsL1+L2 .
(b) When they are moving in the opposite direction, time taken by the trains to cross each other
=sum of their speedsL1+L2 .
* Suppose two trains or two bodies are moving in the same direction at u km/hr and v km/hr respectively such that u > v, then
their relative speed = (u – v) km/hr.
If their lengths be x km and y km respectively, then time taken by the faster train to cross the slower train (moving in the same direction) = æçèx yu v+- ö÷øhrs.
* Suppose two trains or two bodies are moving in opposite directions at u km/hr and v km/hr, then their relative speed = (u + v) km/hr.
If their lengths be x km & y km, then : time taken to cross each other = æçèx yu v++ ö÷ø hrs.
* If a man is running at a speed of u m/sec in the same direction in which a train of length L meters is running at a speed v m/sec, then (v – u) m/sec is called the speed of the train relative to man. Then the time taken by the train to cross the man = 1
v u- seconds
* If a man is running at a speed of u m/sec in a direction opposite to that in which a train of length L meters is running with a speed v m/sec, then (u + v) is called the speed of the train relative to man.
Then the time taken by the train to cross the man = v u1+ seconds.
* If two trains start at the same time from two points A and B towards each other and after crossing, they take a and b hours in reaching B and A respectively. Then,
A’s speed : B’s speed =
(
b : a)
.Example 18 :
How long does a train 90 m long running at the rate of 54 km/h take to cross –
(a) a Mahatma Gandhi’s statue ? (b) a platform 120 m long ?
(c) another train 150m long, standing on another parallel track ?
(d) another train 160 m long running at 36 km/h in same direction ?
(e) another train 160 m long running at 36 km/h in opposite direction ?
(f) a man running at 6 km/h in same direction ? (g) a man running at 6 km/h in opposite direction ?
Solution :
(a) The statue is a stationary object, so time taken by train is same as time taken by train to cover a distance equal to its own length.
Now, 54 km/h 54 5 15m / s 18 = ´ = \ Required time 90 15 = = 6 sec.
(b) The platform is stationary of length = 120 m. Length to be covered
= Length of the train + Length of the platform = 90 + 120 = 210 m
\ Required time 210 15
= = 14 sec. (c) Length to be covered
= Length of the train + length of the other train = 90 + 150 = 240 m.
\ Required time 240 15
= = 16 sec. (d) Another train is moving in same direction.
Length to be covered
= Length of the train + length of the other train = 90 + 160 = 250 m
Relative speed = 54 – 36 = 18 kmph. \ Required time 250 50 sec.
5 18
18
= =
´
(e) Another train is moving in opposite direction. Length to be covered
= Length of the train + length of the other train = 90 + 160 = 250 m
Relative speed = 54 + 36 = 90 kmph \ Required speed 5250 10 sec
90 18
= =
´ .
(f) The man is moving in same direction,
so Length to be covered = Length of the train, and relative speed = speed of train – speed of man
\ Required time 90 5 (54 – 6) 18 = ´ 90 3 27 63 40 4 4 = ´ = = sec. (g) The man is moving in opposite direction, so
Length to be covered = Length of the train, and relative speed = speed of train + speed of man \ Required time 90 5 27 5 sec.2
5 5 (54 6) 18 = = = + ´ Example 19 :
Two trains of equal lengths are running on parallel tracks in the same direction at 46 km/h and 36 km/h, respectively. The faster train passes the slower train in 36 sec. Find the length of each train is ?
Solution :
Let the length of each train be x metres. Then, the total distance covered = (x + x) = 2x m
Relative speed = (46 – 36) = 10 km/h = 1018´5m/s Now, 50 18 x 2 36= ´ or x = 50 m Example 20 :
A train 110 m in length travels at 60 km/h. How much time does the train take in passing a man walking at 6 km/h against the train ?
Solution :
Relative speeds of the train and the man
= (60 + 6) = 66 km/h = m/s 18
5 66´ Distance = 110 m
Therefore, time taken in passing the men
= 6s 5 66 18 110 = ´ ´ Example 21 :
Two trains 137 metres and 163 metres in length are running towards each other on parallel lines, one at the rate of 42 kmph and another at 48 kmph. In what time will they be clear of each other from the moment they meet?
Solution :
Relative speed of the trains
= (42 + 48) kmph = 90 kmph
= æçè90´185ö÷ø m/sec = 25 m/sec. Time taken by the trains to pass each other
= Time taken to cover (137 + 163) m at 25 m/sec
= æçè30025 ö÷ø sec = 12 seconds.
BOAT & STREAM
Stream : It implies that the water in the river is moving or flowing. Upstream : Going against the flow of the river.
Downstream : Going with the flow of the river.
Still water : It implies that the speed of water is zero (generally, in a lake).
Flow of water Boat
Boat Down stream Up stream
Let the speed of a boat (or man) in still water be X m/sec and the speed of the stream (or current) be Y m/sec. Then,
* Speed of boat with the stream (or downstream or D/S) = (X + Y) m/sec.
* Speed of boat against the stream (or upstream or U/S) = (X – Y) m/sec.
* Speed of boat in still water is (X Y) (X – Y) X 2 + + = Downstream Upstream 2 + =
* Speed of the stream or current is Y (X Y) – (X – Y) 2 + = Downstream Upstream 2 + = Example 22 :
A boat is rowed down a river 28 km in 4 hours and up a river 12 km in 6 hours. Find the speed of the boat and the river.
Solution : Downstream speed is 28 7 kmph 4 = Upstream speed is 126 = 2 kmph Speed of Boat 1 2
= (Downstream + Upstream Speed)
1[7 2] 4.5 2 = + = kmph Speed of current 1 2 = (Downstream–Upstream speed) 1(7 – 2) 2.5 kmph 2 = = Example 23 :
P, Q, and R are the three towns on a river which flows uniformly. Q is equidistant from P and R. I row from P to Q and back in 10 hours and I can row from P to R in 4 hours. Compare the speed of my boat in still water with that of the river.
Solution :
Let the speed of the boat be v1 and the speed of the current be v2 and d be the distance between the cities.
Now, 4 v v d 2 1 = + and 1 2 d 6 v -v =
Þ 4 6 v v v v 2 1 2 1 = -+ or 1 2 2v 10 2v = 2 or 1 2 v 5 :1 v = Required ratio = (5 + 1) : 5 = 6 : 5
* A man can row X km/h in still water. If in a stream which is flowing of Y km/h, it takes him Z hours to row to a place and back, the distance between the two places is Z(X – Y )2 2
2X Example 24 :
A man can row 6 km/h in still water. When the river is running at 1.2 km/h, it takes him 1 hour to row to a place and back. How far is the place ?
Solution :
Man’s rate downstream = (6 + 1.2) = 7.2 km/h. Man’s rate upstream = (6 – 1.2) km/h = 4.8 km/h. Let the required distance be x km.
Then 7.2x +4.8x = 1 or 4.8x + 7.2x = 7.2 × 4.8 Þ x 7.2 4.8 2.88km 12 ´ = = By direct formula : Required distance
(
)
2 2 1 6 – (1.2) 2 6 ´ = ´ 36 1.44 34.56 2.88 12 12 -= = = km* A man rows a certain distance downstream in X hours and returns the same distance in Y hours. If the stream flows at the rate of Z km/h, then the speed of the man in still water is given by
Z(X Y) km / hr Y – X
+
* And if speed of man in still water is Z km/h then the speed of stream is given by
Z(Y – X)km / hr X Y+ Example 25 :
Vikas can row a certain distance downstream in 6 hours and return the same distance in 9 hours. If the stream flows at the rate of 3 km/h, find the speed of Vikas in still water.
Solution :
By the formula,
Vikas’s speed in still water 3 9 6
(
)
9 – 6+
= = 15 km/h
* If a man capable of rowing at the speed u of m/sec in still water, rows the same distance up and down a stream flowing at a rate of v m/sec, then his average speed through the journey is
= UpstreamMan 's rate in still water´Downstream = (u v) (u v) u
- +
Example 26 :
Two ferries start at the same time from opposite sides of a
river, travelling across the water on routes at right angles to the shores. Each boat travels at a constant speed though their speeds are different. They pass each other at a point 720m from the nearer shore. Both boats remain at their sides for 10 minutes before starting back. On the return trip they meet at 400m from the other shore. Find the width of the river.
Solution :
Let the width of the river be x. Let a, b be the speeds of the ferries.
720 (x 720) a b -= ...(i) (x 720) 400 720 (x 400) 10 10 a a b b - -+ + = + + ... (ii)
(Time for ferry 1 to reach other shore + 10 minute wait + time to cover 400m)
= Time for freely 2 to cover 720m to other shore + 10 minute wait + Time to cover (x – 400m))
Using (i), we get ab =(x 720)720
-Using (ii), (x 320)-a =(x 320)+a Þ =ab (x 320)(x 320)-+ On, solving we get, x = 1760m
Example 27 :
A man rows 27 km with the stream and 15 km against the stream taking 4 hours each time. Find this rate per hour in still water and the rate at which the stream flows.
Solution :
Speed with the stream = 27 63 4 = 4 kmph \ Speed against the stream = 15 33
4 = 4kmph.
\ Speed of the man in still water = 1 63 33 5 kmph1
2 4 4 4
æ + ö=
ç ÷
è ø
\ Speed of the stream = 1 63 33 1.5 kmph
2 4 4
æ - ö =
ç ÷
è ø
Example 28 :
On a river, B is between A and C and is also equidistant from A and C. A boat goes from A to B and back in 5 hours 15 minutes and from A to C in 7 hours. How long will it take to go from C to A if the river flows from A to C ?
Solution :
If the speed in still water is x kmph and speed of the river is y kmph, speed down the river = x + y and speed up the river x – y. \ d d 51 x y x y+ + - = 4 ... (1) 2d 7 x y+ = ... (2) Multiplying (1) by 2, we get x y x y2d+ + 2d- =1012
1. An aeroplane flies along the four sides of a square at the speeds of 200, 400, 600 and 800 km/h. Find the average speed of the plane around the field.
(a) 384 km/h (b) 370 km/h (c) 368 km/h (d) None of these
2. A monkey ascends a greased pole 12 metres high. He ascends 2 metres in first minute and slips down 1 metre in the alternate minute. In which minute, he reaches the top ?
(a) 21st (b) 22nd
(c) 23rd (d) 24th
3. A man walks a certain distance and rides back in 6 h.1 4 He can walk both ways in 7 h.3
4 How long it would take to ride both ways ?
(a) 5 hours (b) 4 hours1 2 (c) 4 hours3
4 (d) 6 hours
4. There are 20 poles with a constant distance between each pole. A car takes 24 seconds to reach the 12th pole . How much time will it take to reach the last pole?
(a) 25.25 s (b) 17.45 s (c) 35.75 s (d) 41.45 s
5. A man is walking at a speed of 10 km per hour. After every kilometre, he takes rest for 5 minutes. How much time will he take to cover a distance of 5 kilometres?
(a) 48 min. (b) 50 min. (c) 45 min. (d) 55 min.
6. On a journey across Bombay, a tourist bus averages 10 km/h for 20% of the distance, 30 km/h for 60% of it and 20 km/h for the remainder. The average speed for the whole journey was
(a) 10 km/h (b) 30 km/h (c) 5 km/h (d) 20 km/h
7. In a 800 m race around a stadium having the circumference of 200 m, the top runner meets the last runner on the 5th minute of the race. If the top runner runs at twice the speed of the last runner, what is the time taken by the top runner to finish the race ?
(a) 20 min (b) 15 min
(c) 10 min (d) 5 min
8. A man walks half of the journey at 4 km/h by cycle does one third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hours and 12 minutes. The length of the journey is
(a) 36 km (b) 1332km
67
(c) 40 km (d) 28 km
9. A train does a journey without stoppage in 8 hours, if it had travelled 5 km/h faster, it would have done the journey in 6 hours 40 minutes. Find its original speed.
(a) 25 km/h (b) 40 km/h (c) 45 km/h (d) 36.5 km/h
10. A train leaves station X at 5 a.m. and reaches station Y at 9 a.m. Another train leaves station Y at 7 a.m. and reaches station X at 10: 30 a.m. At what time do the two trains cross each other ?
(a) 7 : 36 am (b) 7 : 56 am (c) 8 : 36 am (d) 8 : 56 am
11. Cars C1 and C2 travel to a place at a speed of 30 km/h and 45 km/h respectively. If car C2 takes 21
2 hours less time than C1 for the journey, the distance of the place is
(a) 300 km (b) 400 km (c) 350 km (d) 225 km
12. A man covers a certain distance on a toy train. If the train moved 4 km/h faster, it would take 30 minutes less. If it moved 2 km/h slower, it would have taken 20 minutes more. Find the distance.
(a) 60 km (b) 58 km
(c) 55 km (d) 50 km
13. 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 goods train in 4 hours. Find the speed of the goods train.
(a) 36 kmph (b) 40 kmph (c) 30 kmph (d) 42 kmph
14. Without stoppages, a train travels certain distance with an average speed of 80 km/h, and with stoppages, it covers the same distance with an average speed of 60 km/h. How many minutes per hour the train stops ?
(a) 15 (b) 18
(c) 10 (d) None of these
15. If a man walks to his office at 3/4 of his usual rate, he reaches office 1/3 of an hour later than usual. What is his usual time to reach office.
(a) 1hr
2 (b) 1 hr
(c) 3hr
4 (d) None of these
16. A train running between two stations A and B arrives at its destination 10 minutes late when its speed is 50 km/h and 50 minutes late when its speed is 30km/h. What is the distance between the stations A and B ?
(a) 40 km (b) 50 km
(c) 60 km (d) 70 km
17. A thief goes away with a Maruti car at a speed of 40 km/h. The theft has been discovered after half an hour and the owner sets off in another car at 50 km/h. When will the owner overtake the thief from the start.
(a) 21
2hours (b) 2 hr 20 min
(c) 1 hr 45 min (d) cannot be determined 18. A starts 3 min after B for a place 4.5 km away. B on reaching
his destination, immediately returns back and after walking a km meets A. If A walks 1 km in 18 minutes then what is B’s speed ?
(a) 5 km/h (b) 4 km/h (c) 6 km/h (d) 3.5 km/h
19. A long distance runner runs 9 laps of a 400 metres track everyday. His timings (in minutes) for four consecutive days are 88, 96, 89 and 87 resplectively. On an average, how many metres/minute does the runner cover ?
(a) 40 m/min (b) 45 m/min (c) 38 m/min (d) 49 m/min
20. 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 speed of the car is :
(a) 4 : 3 (b) 3 : 4
(c) 3 : 2 (d) 2 : 3
21. 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:
(a) 1 hours (b) 2 hours (c) 3 hours (d) 4 hours
22. Points A and B are 70 km apart on a highway. One car starts form A and the another one from B at the same time. If they travel in the same direction, they meet in 7 hours. But if they travel towards each other, they meet in one hour. The speeds of the two cars are, respectively.
(a) 45 and 25 km/h (b) 70 and 10 km/h (c) 40 and 30 km/h (d) 60 and 40 km/h
23. Anil calculated that it will take 45 minutes to cover a distance of 60 km by his car. How long will it take to cover the same distance if the speed of his car is reduced by 15 km/hr? (a) 36 min (b) 55.38 min
(c) 48 min (d) 40 min
24. The jogging track in a sports complex is 726 metres in circumference. Pradeep and his wife start from the same point and walk in opposite directions at 4.5 km/h and 3.75 km/h, respectively. They will meet for the first time in : (a) 5.5 min (b) 6.0 min
(c) 5.28 min (d) 4.9 min
25. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water (in litres) will fall into the sea in a minute?
(a) 4,00,000 (b) 40,00,000
(c) 40,000 (d) 4,000
26. The speed of a boat in still water is 15 km/h and the rate of stream is 5 km/h. The distance travelled downstream in 24 minutes is
(a) 4 km (b) 8 km
(c) 6 km (d) 16 km
27. A person can swim in still water at 4 km/h. If the speed of water is 2 km/h, how many hours will the man take to swim back against the current for 6 km.
(a) 3 (b) 4
(c) 41
2 (d) Insufficient data
28. A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water? (a) 4 km/h (b) 6 km/h
(c) 8 km/h (d) Data inadequate
29. A boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30km upstream and 21 km downstream in 6 hours and 30 minutes. The speed of the boat in still water is : (a) 10 km/h (b) 4 km/h
(c) 14 km/h (d) 6km/h
30. A man who can swim 48 m/min in still water swims 200 m against the current and 200 m with the current. If the difference between those two times is 10 minutes, find the speed of the current.
(a) 30 m/min (b) 29 m/min (c) 31 m/min (d) 32 m/min
31. A circular running path is 726 metres in circumference. Two men start from the same point and walk in opposite directions at 3.75 km/h and 4.5 km/h, respectively. When will they meet for the first time ?
(a) After 5.5 min (b) After 6.0 min (c) After 5.28 min (d) After 4.9 min
32. A train after travelling 150 km meets with an accident and then proceeds with 3/5 of its former speed and arrives at its destination 8 h late. Had the accident occurred 360 km further, it would have reached the destination 4 h late. What is the total distance travelled by the train?
(a) 840 km (b) 960 km (c) 870 km (d) 1100 km 33. A man swimming in a steam which flows 11
2km/hr., finds that in a given time he can swim twice as far with the stream as he can against it. At what rate does he swim?
(a) 51 2km/hr (b) 1 4 2 km/hr (c) 71 2 km/hr (d) None of these
34. Two persons start from the opposite ends of a 90 km straight track and run to and fro between the two ends. The speed of first person is 30 m/s and the speed of other is 125/6 m/s. They continue their motion for 10 hours. How many times they pass each other?
(a) 10 (b) 9
(c) 12 (d) None of these
35. A man starts from B to K, another from K to B at the same time. After passing each other they complete their journeys in 313 and 445 hours, respectively. Find the speed of the second man if the speed of the first is 12 km/hr.
(a) 12.5 kmph (b) 10 kmph (c) 12.66 kmph (d) 20 kmph
36. A train 100 metres long moving at a speed of 50 km/hr. crosses a train 120 metres long coming from opposite direction in 6 sec. The speed of the second train is
(a) 60 km/hr. (b) 82 km/hr. (c) 70 km/hr. (d) 74 km/hr.
37. A passenger sitting in a train of length 100 m, which is running with speed of 60 km/h passing through two bridges, notices that he crosses the first bridge and the second bridge in time intervals which are in the ratio of 7 : 4 respectively. If the length of first bridge be 280 m, then the length of second bridge is:
(a) 490 m (b) 220 m
(c) 160 m (d) Can’t be determined 38. A man can row a certain distance against the stream in six
hours. However, he would take two hours less to cover the same distance with the current. If the speed of the current is 2 kmph, then what is the rowing speed in still water? (a) 10 kmph (b) 12 kmph
(c) 14 kmph (d) 8 kmph
39. A boat, while going downstream in a river covered a dis-tance of 50 mile at an average speed of 60 miles per hour. While returning, because of the water resistance, it took one hour fifteen minutes to cover the same distance . What was the average speed of the boat during the whole jour-ney?
(a) 40 mph (b) 48 mph
(c) 50 mph (d) 55 mph
40. Two trains, 130 m and 110 m long, are going in the same direction. The faster train takes one minute to pass the other completely. If they are moving in opposite directions, they pass each other completely in 3 seconds. Find the speed of each train.
(a) 38 m/sec, 36 m/sec (b) 42 m/sec, 38 m/sec (c) 36 m/sec, 42 m/sec (d) None of these
41. A passenger sitting in a train of length 100 m, which is running with speed of 60 km/h passing through two bridges, notices that he crosses the first bridge and the second bridge in time intervals which are in the ratio of 7 : 4 respectively. If the length of first bridge be 280 m, then the length of second bridge is:
(a) 490 m (b) 220 m
(c) 160 m (d) Can’t be determined
42. If a train runs at 70 km/hour, it reaches its destination late by 12 minutes. But if it runs at 80 km/hour, it is late by 3 minutes. The correct time to cover the journey is
(a) 58 minutes (b) 2 hours (c) 1 hour (d) 59 minutes
43. A car covers four successive 6 km stretches at speeds of 25 kmph, 50 kmph, 75 kmph and 150 kmph respectively. Its average speed over this distance is
(a) 25 kmph (b) 50 kmph (c) 75 kmph (d) 150 kmph
44. A is faster than B. A and B each walk 24 km. The sum of their speed is 7 km/hr and the sum of time taken is 14 hours. A’s speed is
(a) 4 km/hr (b) 3 km/hr (c) 5 km/hr (d) 7 km/hr
45. In a 1 km race A beats B by 40 meters or 7 seconds. Find A’s time over the course.
(a) 172 sec (b) 150 sec (c) 160 sec (d) 168 sec
46. A hare sees a dog 100 meters away from her and scuds off in the opposite direction at the speed of 12 km/hr. A minute later the dog sees her and chases her at a speed of 16 km/hr. How soon will the dog overtake her ?
(a) 240 sec (b) 360 sec (c) 270 sec (d) 180 sec
47. On reducing my speed to 3 km/hr. I reach office 10 minutes late. I usually travel at 133% of this speed and reach early by the same margin. How far is my office ?
(a) 4 km (b) 6 km
(c) 3.5 km (d) 4.5 km
48. A monkey climbs a slippery pole 12 m height rises 1 m in every 1 min and slips ½ metre in every next minute. Find how soon it will reach the top.
(a) 45 min (b) 40 min (c) 35 min (d) 48 min
49. It takes 8 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 down by train and the rest by car. The ratio of the speed of the train to that of the car is
(a) 2 : 3 (b) 3 : 2 (c) 3 : 4 (d) 4 : 3 1 (a) 11 (d) 21 (a) 31 (c) 41 (c) 2 (a) 12 (a) 22 (c) 32 (c) 42 (c) 3 (c) 13 (a) 23 (b) 33 (b) 43 (b) 4 (d) 14 (a) 24 (c) 34 (c) 44 (b) 5 (b) 15 (b) 25 (b) 35 (b) 45 (a) 6 (d) 16 (b) 26 (b) 36 (b) 46 (d) 7 (c) 17 (a) 27 (a) 37 (c) 47 (c)
8 (a) 18 (a) 28 (b) 38 (a) 48 (a)
9 (a) 19 (a) 29 (a) 39 (b) 49 (a)
10 (b) 20 (b) 30 (d) 40 (b)
1. (a) Let each side of the square be x km and let the average speed of the plane around the field be y km/h. Then, x x x x 4x 200 400 600 800+ + + = y 25x 4x 2400 4 y 384. 2400 y 25 ´ æ ö Þ = Þ =ç ÷= è ø \ Average speed = 384 km/h. 2. (a) In 2 minutes, he ascends = 1 metre
\ 10 metres, he ascends in 20 minutes. \ He reaches the top in 21st minute.
3. (c) We know that, the relation in time taken with two different modes of transport is
twalk both + tride both = 2 (twalk + t ride)
ride both
31 t 2 25
4 + = ´ 4
Þ tride both =25 31 192 – 4 = 4 =4 hrs34 4. (d) Let the distance between each pole be x m.
Then, the distance up to 12th pole = 11 x m Speed = 11x
24 m/s
Time taken to covers the total distance of 19x = 41.45s x 11 24 x 19 ´ =
5. (b) Rest time = Number of rest × Time for each rest = 4 × 5 = 20 minutes
Total time to cover 5 km 5
60 minutes + 20 minutes = 50 minutes. 10
æ ö
=ç ´ ÷
è ø
6. (d) Let the average speed be x km/h. and total distance be y km. Then, 0.2 0.6 0.2 y y y y 10 +30 + 20 =x 1 x 20km / h 0.05 Þ = =
7. (c) After 5 minutes (before meeting), the top runner covers 2 rounds i.e., 400 m and the last runner covers 1 round i.e., 200 m.
\ Top runner covers 800 m race in 10 minutes. 8. (a) Let the length of the journey =x km.
\ Journey rides by horse cart x 1 1 1 2 3
æ ö
= ç - - ÷
è ø
=16x km.
Then, total time taken to complete journey 31hr 5 = 1 2 3 31 t t t 5 Þ + + = x 1 x 1 x 31 2 4 3 12 6 9 5 Þ ´ + ´ + = ´ 31 216 x 36.2km 36km 5 37 Þ = ´ = »
9. (a) Let original speed = S km/h
Here, distance to be covered is constant 20 S 8 (S 5) 3 æ ö \ ´ = + ç ÷ è ø 20 100 8S S 3 3 Þ - = S 100 25 km / h 4 Þ = =
10. (b) Let the distance between X and Y be x km. Then, the speed of A is 4 x km/h and that of B is 7 x 2 km/h. X x km Y4km/h x h / km 7 x 2
Relative speeds of the trains
= x 2x 15x km / h
4 7 28
æ + ö=
ç ÷
è ø
Therefore the distance between the trains at 7 a.m. = km 2 x 2 x x- =
Hence, time taken to cross each other
= 60 56min 15 14 x 15 28 2 x 28 x 152 x = ´ = ´ =
Thus, both of them meet at 7 : 56 a.m. 11. (d) Let C1 takes t hrs. Then,
Q Distance is same. \ 30t 45 t 5 2 æ ö = ç - ÷ è ø 15 t hrs 2 Þ = 15 Distance = 30× 225 km 2 \ =
12. (a) Let the distance be x km and speed of train be y km/h. Then by question, we have
x x 30
y 4+ = -y 60 ...(i)
and x x 20
y 2- = +y 60 ...(ii) On solving (i) and (ii), we get x = 3y Put x = 3y in (i) we get
3y 1
3
y 4+ = -2 Þ =y 20 Hence, distance = 20 × 3 = 60 km.
13. (a) 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. 14. (a) Due to stoppages, it covers 20 km less .
Time taken to cover 20km 20 = h 1h 80 =4 1 60 min 4 = ´ = 15 min 15. (b) If new speed is a b of original speed. Then,
usual time × b–1 change in time a æ ö = ç ÷ è ø \ usual time × 4 1 1 3 3 æ - =ö ç ÷ è ø 1 usual time = 3 1 hr 3 Þ ´ =
16. (b) Let the distance between the two stations be x km. Then, x 10 x 50 50- 6 =30- 6 x 1 x 5 50 6 30 6 Þ - = -or 3 2 50 x 30 x - = or x=50km
Thus distance between the station A and B = 50 km 17. (a) Distance to be covered by the thief and by the owner
is same.
Let after time 't', owner catches the thief. 1 40 t 50 t – 2 æ ö \ ´ = ç ÷ è ø 5 1 10t 25 t hr 2 hr 2 2 Þ = Þ = =
18. (a) A covers 3.5 km before he meets B in (18 × 3.5 + 3) = 66 min = h
10 11 60 66 =
Now, B covers a distance of 5.5 km in 10 11 hours Þ B’s speed = 5km/h 11 10 2 11´ =
19. (a) Average speed =Total distance Total time 400 4 9 400 4 9 88 96 89 87 360 ´ ´ ´ ´ = = + + + = 40 metres /minute
20. (b) Let the speed of the train and the car be x km/h and y km/h, respectively. Now, 8 y 480 x 120+ = …(i) and 3 25 y 400 x 200+ = …(ii) From (i),
120y + 480x = 8xy and …(iii)
From (ii), 200y + 400x = xy 3 25
…(iv) From (iii) and (iv),
25 ) x 400 y 200 ( 3 8 x 480 y 120 + = + or 15y + 60x = 24y + 48x or 12x = 9y or 4 3 y x =
21. (a) Let the duration of the flight be x hours. Then,
600 600 600 1200 200 200 1 x x x 2x 1 2 - = Þ - = + + Þ x (2x + 1) = 3 Þ 2x2 + x – 3 = 0 Þ (2x + 3) (x – 1) = 0 Þ x = 1 hr. [neglecting the –ve value of x]. 22. (c) Let the speed of the cars be x km/h and y km/h
respectively.
Their relative speed when they are moving in same direction = (x – y) km/h.
Their relative speed when they are in opposite directions = (x + y) km/h. Now, 1 y x70 =+ or x + y = 70 ... (i) and 7 ) y x ( 70 = - or x – y = 10 ... (ii) Solving (i) and (ii), we get
x = 40 km/h and y = 30 km/h. 23. (b) D = S × T 60 = S´ç ÷æ4560öhr è ø S = 60 60 45 ´ Þ 80km/hr
Now, new speed = 80 – 15 = 65 km/hr. \ Time Distance 60hr. Speed 65 = = or 60 60 min 55.38min. 65´ =
Hence, Time to taken by car to travel same distance is 55.38 min.
24. (c) Let the husband and the wife meet after x minutes. 4500 metres are covered by Pradeep in 60 minutes. In x minutes, he will cover x
60 4500
metres. Similarily,
In x minutes, his wife will cover x 60 3750 m. Now, x 726 60 3750 x 60 4500 = + Þ 5.28min 8250 60 726 x= ´ =
25. (b) Volume of water flowed in an hour
= 2000 × 40 × 3 cubic metre = 240000 cubic metre \ volume of water flowed in 1 minute =240000 4000
60 = cubic metre = 40,00,000 litre 26. (b) Downstream speed = 15 + 5 = 20 km/h.
\ Required distance 20 24 8km. 60
= ´ =
27. (a) Man's speed in upstream = 4 – 2 = 2 km/h. \ Required time 6 3 km / h 2 = = 28. (b) Rate downstream = æçè162 ö÷ø kmph = 8 kmph; Rate upstream = æç164 ö÷ è ø kmph = 4 kmph. \ Speed in still water = 1
2 (8 + 4) = 6 km/h.
29. (a) Let speed of the boat in still water be x km/h and speed of the current be y km/h.
Then, upstream speed = (x – y) km/h and downstream speed = (x + y) km/h
Now, 6 ) y x ( 28 ) y x ( 24 = + + - …(1) and 2 13 ) y x ( 21 ) y x ( 30 = + + - …(2)
Solving (1) and (2), we get x = 10 km/h and y = 4 km/h
30. (d) Let vm = velocity of man = 48 m/min Let vc = velocity of current
then t1= time taken to travel 200 m against the current.
i.e., 1 200 – m c t v v = ....(1)
and t2 time taken to travel 200 m with the current
i.e., 2 200 m c t v v = + ....(2) Given : t1 – t2 = 10 min \ 200 200 10 – m c m c v v -v +v = Þ 2 2 40 m c c v -v = v Þ 2 40 (48)2 0 c c v + v - = Þ vc =32, 72
-Hence, speed of the current = 32 (Qvc ¹ -72). 31. (c) Their relative speeds = (4.5 + 3.75) = 8.25 km/h
Distance = 726 metres = 0.726km 1000726 = Required time = 60 5.28 25 . 8 726 . 0 = ´ min
32. (c) Let the total distance to be travelled = x km Speed of train = v km/h
and time taken = t hr.
(
)
150 150 8 3 5 x t v v -+ = + æ ö ç ÷ è ø ...(1)(
)
510 510 4 3 5 x t v v -+ = + ...(2) Eq (2) – Eq (1) 510 150 510 150 4 3 3 5 5 x x v v v v - -- + - = -360 -360 5 4 3 v v ´ - = - Þ v = 60 km/hr.. t = 60x Put in eqn (1) 150 150 8 3 60 60 60 5 x- æ x ö + =ç + ÷ ´ è ø 5 150 8 2 36 60 x- x + = + 150 5 11 8 36 60 2 2 x- - x = - = 10 1500 6 11 360 2 x- - x= Þ 4x– 1500 = 360 11 1980 2 ´ = Þ 4x = 3480 x = 3480km 870 km 4 =33. (b) Let the speed of swimmer be x km/hr When he swim with the flow
then speed = ( x + 3/2) km/h. 1 3 2 S æx ö t \ =ç + ÷´ è ø
When he swim against the flow of stream then speed –3 2 x t æ ö = çè ÷ø 2 –32 S æx öt \ = ç ÷ è ø
According to the ques
S1 = 2S2. 3 3 2 – 2 2 x t x t æ + ö = æ ö ç ÷ ç ÷ è ø è ø 3 2 – 3 2 2 2 x x t t æ + ö = æ ö ç ÷ ç ÷ è ø è ø 2 3 2 – 3 2 x+ x æ ö Þç ÷= è ø Þ 2x + 3 = 4x – 6 Þ 9 = 2x Þ 9 2 x= = 4 km / hr1 2 34. (c) The speeds of two persons is 108 km/h and 75 km/h.
The first person covers 1080 km in 10 hours and thus he makes 12 rounds. Thus, he will pass over another person 12 times in any one of the direction.
35. (b) 2nd man 's speed1st man 's speed = b a = b a = 4 4 5 1 3 3 = 24 3 5 ´10 = 36 25 = 6 5 \ 2nd man 's speed12 = 65 \ 2nd man’s speed = 60 6 = 10 km/hr.. 36. (b) Let speed of the second train = x km/hr.
Relative speed of trains = (50 + x) km/hr.
Distance travelled by trains = (100 + 120) = 220 metres Distance = Speed × Time
220 6 km (50 ) km/hr. hr 1000 3600 æ ö = + ´æ ö ç ÷ ç ÷ è ø x è ø 50 + x = 220 36001000 6´ ´ 50 + x = 132 x = 132 – 50 = 82 km/hr
37. (c) Note here the length of the train in which passenger is travelling is not considered since we are concerned with the passenger instead of train. So, the length of the bridge will be directly proportional to the time taken
by the passenger respectively. t ® Time l ® Length of bridge Therefore. 1 1 2 2 t l t =l 7 280 4 = 2 Þ x = 160 m
38. (a) If the rowing speed in still water be x kmph, and the distance by y km, then 6 – 2 y x = Þ y = 6 (x – 2) ...(1) and, 4 2 y x+ = Þ y = 4 (x + 2) ...(2) Þ 6 (x – 2) = 4 (x + 2) Þ x = 10 kmph
39. (b) Time taken by the boat during downstream journey = h
6 5 60 50 =
Time taken by the boat in upstream journey = h 4 5 Average speed = 2 50 100 24 48 5 5 50 6 4 ´ ´ = = + mph
40. (b) Let the Speed of faster train be x and speed of slower train be y.
Now, when both the train move in same direction their relative speed = x – y
Now, total distance covered = 130 + 110 = 240 Now, distance = speed × time
\ 240 = ( x– y) × 60 ( 1min 60sec)Q =
Þ x – y = 4 …(1)
When the trains move in opposite direction then their relative speed = x + y
\ 240 = ( x + y) × 3
Þ 80 = x + y …(2)
on solving eqs (1) and (2), we get x = 42 m/sec and y = 38 m/sec
41. (c) Note here the length of the train in which passenger is travelling is not considered since we are concerned with the passenger instead of train. So, the length of the bridge will be directly proportional to the time taken by the passenger respectively.
t ® Time l ® Length of bridge Therefore. 1 1 2 2 t l t =l 7 280 4= 2 Þ x = 160 m
42. (c) Let correct time to cover journey be t hours 12 70 t 60 æ + ö ç ÷ è ø = 3 80 t 60 æ + ö ç ÷ è ø 70t + 14 = 80t+ 4 10t = 10 t = 1 hour
43. (b) Average Speed =Total Distance Covered Total Time Taken
6 6+6+6+6 24 = 6 6 6 6 1 1 1 1 25 50 75 150 25 50 75 150 Þ é ù + + + ê + + + ú ë û 24 300 50 km/hr 6 24 ´ = Þ ´
44. (a) Let speed of A and B are S1 and S2 respectively. S1 + S2 = 7 ... (1) 1 2 24 24 14 S +S = or 1 2 1 1 14 S +S = 24 ... (2) 1 2 1 2 S S 14 S S 24 + = Þ 1 2 7 14 S S =24 S1S2 = 12 S1(7 – S1) = 12 S12 – 7S 1 + 12 = 0 (S1 – 4)(S1 – 3) = 0 S1 = 4, 3 Corresponding values of S2 = 3, 4 As, S1 > S2 A’s speed is 4 km/h 45. (d) B covers 40 m in 7 seconds Speed of B = 40 7 m/s
Time taken by B to cover 1 km 1000 7 175 S 40
´
= =
A’s time over the course = (175 – 7) = 168 S
46. (c) Let at time ‘t’ dog overtook hare distance travelled by dog = 16 t
distance between dog and hare =12 tæçè +601 ö÷ø+0.1 m
1 12 t 0.1 16 t 60 æ + ö+ = ç ÷ è ø 1 1 12t 16 t 5 10 + + = 3 4 t 10= 3 t h 40 = 3 t 3600 270 40 = ´ = seconds 47. (a)
48. (a) The monkey 1 meter in 4 min. This pattern will go on till he reaches 11 meters. i.e., 11 × 4 = 44 mins. After that he would have climb 1 meter and he will reach the pole. So, the total time taken = 44 + 1 = 45 min.
49. (c) Let T be the speed of train and C be the speed of car. 120 480 T + C = 8 Þ 1 4 1 T C+ =15 ...(1) 200 400 T + C = 20 8 60 + Þ 1 2 1 T C+ =24 ...(2) Subtracting (2) from (1) 2 1 1 (2 1) C - =15 24 -2 C = 1 40 Þ C = 80 1 T = 1 15 – 4 80 1 T = 1 60 Þ T = 60 Required ratio = 60 : 80 = 3 : 4
