1. Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in:
a) 15.5 seconds
b) 14.4 seconds
c) 20.2 seconds
d) 18.8 seconds
Discussion
Explanation: Let length of each train be x meter.
Then, speed of 1st train = $$\frac{x}{{18}}$$ m/sec
Speed of 2nd train = $$\frac{x}{{12}}$$ m/sec
When both trains cross each other, time taken
$$\eqalign{ & = {\frac{{2x}}{{ { {\frac{x}{{18}}} + {\frac{x}{{12}}} } }}} \cr & = \frac{{2x}}{{ {\frac{{ {2x + 3x} }}{{36}}} }} \cr & = \frac{{2x \times 36}}{{5x}} \cr & = \frac{{72}}{5} \cr & = 14.4\,\,{\text{seconds}} \cr} $$
2. A train passes two persons walking in the same direction at a speed of 3 kmph and 5 kmph respectively in 10 seconds and 11 seconds respectively. The speed of the train is
a) 28 kmph
b) 27 kmph
c) 25 kmph
d) 24 kmph
Discussion
Explanation: Let the speed of the train be S. And length of the train be x.
When a train crosses a man, its travels its own distance.
$$\eqalign{ & {\text{According to question}}; \cr & \frac{x}{{ {\left( {s - 3} \right) \times {\frac{5}{{18}}} } }} = 10 \cr & {\text{or}},\,18x = 50 \times s - 150.....({\text{i}}) \cr & {\text{and}} \cr & \frac{x}{{ {\left( {x - 5} \right) \times {\frac{5}{{18}}} } }} = 11 \cr & 18x = 55 \times s - 275......({\text{ii}}) \cr & {\text{Equating equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 50 \times s - 150 = 55 \times s - 275 \cr & {\text{or}},\,5 \times s = 125 \cr & {\text{or}},\,s = 25\,{\text{kmph}} \cr} $$
3. A plane left half an hour later than the scheduled time and in order to reach its destination 1500 kilometers away in time, it had to increase its speed by 33.33 percent over its usual speed. Find its increased speed.
a) 500 kmph
b) 250 kmph
c) 750 kmph
d) 1000 kmph
Discussion
Explanation: By increasing the speed by 33.33%, it would be able to reduce the time taken for traveling by 25%.
But since this is able to overcome the time delay of 30 minutes, 30 minutes must be equivalent to 25% of the time originally taken.
Hence, the original time must have been 2 hours and the original speed would be 750 kmph.
Hence, the new speed would be 1000 kmph.
4. Two rifles are fired from the same place at a difference of 11 minutes 45 seconds. But a man who is coming towards the place in a train hears the second sound after 11 minutes. Find the speed of train. (speed of sound = 330 m/s)
a) 72 kmph
b) 36 kmph
c) 81 kmph
d) 108 kmph
Discussion
Explanation: Distance travelled by man-train in 11 minute = distance traveled by sound in (11 min 45 sec - 11 min) = 45 sec.
Let the speed of train be x kmph.
$$x \times \frac{{11}}{{60}} = \frac{{45}}{{3600}} \times 330 \times \frac{{18}}{5}$$
Or, x = 81 kmph.
5. In a 100m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 kmph, then the speed of Bimal is:
a) 15.4 kmph
b) 14 kmph
c) 14.5 kmph
d) 14.4 kmph
Discussion
Explanation:
$$\eqalign{ & {\text{Time taken by Kamal}} \cr & = \frac{{100}}{{ {\frac{{18 \times 5}}{{18}}} }} \cr & = 20\,{\text{seconds}} \cr & {\text{Hence, }} \cr & {\text{Time taken by Bimal}} \cr & 20 + 5 = 25\,{\text{seconds}} \cr & {\text{So, Bimal's speed}} \cr & = \frac{{100}}{{25}} \cr & = 4\ \cr & = \frac{{4 \times 18}}{5} \cr & = 14.4\,{\text{kmph}} \cr} $$
6. Two trains 105 meters and 90 meters long, run at the speeds of 45 kmph and 72 kmph respectively, in opposite directions on parallel tracks. The time which they take to cross each other, is:
a) 8 seconds
b) 6 seconds
c) 7 seconds
d) 5 seconds
Discussion
Explanation: Length of the 1st train = 105 m
Length of the 2nd train = 90 m
Relative speed of the trains,
= 45 + 72 = 117 kmph
= $$\frac{{117 \times 5}}{{18}}$$ = 32.5 m/sec
Time taken to cross each other,
= $$\frac{{{\text{Length}}\,{\text{of}}\,{1^{{\text{st}}}}\,{\text{train}} + {\text{length of}}\,{2^{{\text{nd}}}}\,{\text{train}}}}{{{\text{relative }}\,{\text{speed }}\,{\text{of}}\,{\text{the }}\,{\text{trains}}}}$$
Time taken = $$\frac{{195}}{{32.5}}$$ = 6 seconds.
7. A train travelling with a speed of 60 kmph catches another train travelling in the same direction and then leaves it 120 m behind in 18 seconds. The speed of the second train is
a) 36 kmph
b) 26 kmph
c) 63 kmph
d) 35 kmph
Discussion
Explanation: Let speed of the 2nd train is S m/sec.
60 kmph $$ = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3}$$ m/sec.
As trains are traveling in same distance, Then Relative distance,
$$\eqalign{ & = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3} \cr & \frac{{50}}{3} - S = \frac{{120}}{{18}} \cr & \Rightarrow \frac{{50 - 3S}}{3} = \frac{{20}}{3} \cr & \Rightarrow 50 - 3S = 20 \cr & \Rightarrow 3S = 50 - 20 \cr & \Rightarrow 3S = 30 \cr & \therefore S = 10\,\,{\text{m/sec}} \cr} $$
Or, Speed of the 2nd train = $${\text{10}} \times \frac{{18}}{5}$$ = 36 kmph.
8. A car driver, driving in a fog, passes a pedestrian who was walking at the rate of 2 km/h in the same direction. The pedestrian could see the car for 6 minutes and it was visible to him up to distance of 0.6 km. what was speed of the car?
a) 30 kmph
b) 15 kmph
c) 20 kmph
d) 8 kmph
Discussion
Explanation: In 6 minutes, the car goes ahead by 0.6 km.
Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph.
Hence, Net speed of the car is 8 kmph.
9. Between 5 am and 5 pm of a particular day for how many times are the minute and the hour hands together?
a) 22
b) 11
c) 44
d) 33
Discussion
Explanation: It will come 11 times together.
10. Two boats A and B start towards each other from two places, 108 km apart. Speed of the boats A and B in still water are 12 km/h and 15 km/h respectively. If A proceeds down and B up the stream, they will meet after:
a) 4.5 hours
b) 4 hours
c) 5.4 hours
d) 6 hours
Discussion
Explanation: Let the speed of the stream be x kmph and both the boats meet after t hour.
According to the question,
(12 + x) × t + (15 - x) × t = 108
12t + 15t = 108
27t = 108
∴ t = $$\frac{{108}}{{27}}$$ = 4 hours
11. A train passes two bridges of length 800 m and 400 m in 100 seconds and 60 seconds respectively. The length of the train is:
a) 80 m
b) 90 m
c) 200 m
d) 150 m
Discussion
Explanation: Let length of the train be x m and speed of the train is s kmph.
Speed, s = $$\frac{{x + 800}}{{100}}$$ . . . . . (i)
Speed, s = $$\frac{{x + 400}}{{60}}$$ . . . . . (ii)
Equating equation (i) and (ii), we get,
$$\frac{{x + 800}}{{100}}$$ = $$\frac{{x + 400}}{{60}}$$
5x + 200 = 3x + 2400
2x = 400
x = 200m
12. A cyclist moving on a circular track of radius 100 meters completes one revolution in 2 minutes. What is the average speed of cyclist (approx.)?
a) 314 m/min
b) 200 m/min
c) 300 m/min
d) 900 m/min
Discussion
Explanation: Distance covered in one complete revolution
$$\eqalign{ & = 2\pi r \cr & = \frac{{2 \times 100 \times 22}}{7} \cr & \approx 628\,m \cr & {\text{Average}}\,{\text{speed}} \cr & = \frac{{628}}{2} \cr & = 314\,m/\min \cr} $$
13. A train, 300m long, passed a man, walking along the line in the same direction at the rate of 3 kmph in 33 seconds. The speed of the train is:
a) 30 kmph
b) $$32\frac{8}{{11}}$$ kmph
c) $$35\frac{8}{{11}}$$ kmph
d) 32 kmph
Discussion
Explanation: Let the speed of train = x km/ hr
Length of train = 300 metres
Their relative speed in same direction
= (x - 3) km/hr
According to the question,
$$\frac{{\left( {300 + 0} \right)m}}{{\left( {x - 3} \right) \times \frac{5}{{18}}m/s}} = 33$$
[Here man's length is 0 metre]
$$\eqalign{ & \frac{{100 \times 18}}{{5x - 15}} = 11 \cr & 1800 = 55x - 165 \cr & 55x = 1965 \cr} $$
∴ Speed of the train = $$\frac{{1965}}{{55}}$$ = $$35\frac{8}{{11}}$$ kmph
14. In a race of 200 meters, B can gives a start of 10 meters to A, and C can gives a start of 20 meters to B. The starts that C can gives to A, in the same race is:
a) 30 meters
b) 25 meters
c) 29 meters
d) 27 meters
Discussion
Explanation: When B runs 200 meters, A runs 190 meters;
Hence, when B runs 180 meters,
A runs = $$\frac{{190 \times 180}}{{200}}$$ = 171 meters;
When C runs 200m, B runs 180 meters.
C will give a start to A by = 200 - 171
= 29 meters
15. A person can row $$7\frac{1}{2}$$ km an hour in still water. Finds that it takes twice the time to row upstream than the time to row downstream. The speed of the stream is:
a) 2 kmph
b) 2.5 kmph
c) 3 kmph
d) 4 kmph
Discussion
Explanation: Let the distance covered be x km and speed of stream = y kmph.
Speed downstream = $$\frac{{15}}{2} + y$$ kmph
Speed upstream = $$\frac{{15}}{2} - y$$ kmph
$$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\frac{{2x}}{{ { {\frac{{15}}{2}} + y} }}} = {\frac{x}{{ { {\frac{{15}}{2}} - y} }}} \cr & or,\,15 - 2y = {\frac{{15}}{2}} + y \cr & 3y = 15 - {\frac{{15}}{2}} = \frac{{15}}{2} \cr & y = \frac{{15}}{6} = 2.5\,{\text{kmph}} \cr} $$
16. A train travelling at 48 kmph crosses another train, having half of its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of railway platform is:
a) 200 m
b) 300 m
c) 350 m
d) 400 m
Discussion
Explanation: Let the length of the train traveling at 48 kmph be 2x meters.
And length of the platform is y meters.
$$\eqalign{ & {\text{Relative speed of train}} \cr & = \left( {48 + 42} \right)\,{\text{kmph}} \cr & = {\frac{{90 \times 5}}{{18}}} \cr & = 25\,m/\sec \cr & {\text{And}}\,{\text{48}}\,{\text{kmph}} \cr & = \frac{{48 \times 5}}{{18}} \cr & = \frac{{40}}{3}\,m/\sec \cr & \cr & {\text{According to the question}}, \cr & \frac{{ {2x + x} }}{{25}} = 12; \cr & \,3x = 12 \times 25 = 300 \cr & \,x = \frac{{300}}{3} = 100m \cr & {\text{Then, length of the train}} \cr & = 2x \cr & = 100 \times 2 = 200m \cr & \frac{{200 + y}}{{ {\frac{{40}}{3}} }} = 45 \cr & 600 + 3y = 40 \times 45 \cr & {\text{Or}},\,3y = 1800 - 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200 \cr & {\text{Or}},\,y = \frac{{1200}}{3} = 400m \cr & {\text{Length of the platform}} \cr & = 400m \cr} $$
17. A man goes downstream with a boat to some destination and returns upstream to his original place in 5 hours. If the speed of the boat in still water and the stream are 10 kmph and 4 kmph respectively, the distance of the destination from the starting place is:
a) 16 km
b) 18 km
c) 21 km
d) 25 km
Discussion
Explanation: Let the distance of the destination from the starting point = x km.
Speed downstream = (10 + 4) = 14 kmph
Speed upstream = (10 - 4) = 6 kmph
Total time taken = 5 hours
$$\eqalign{ & {\frac{x}{{14}}} + {\frac{x}{6}} = 5 \cr & \frac{{ {3x + 7x} }}{{42}} = 5 \cr & {\text{or}},\,10x = 42 \times 5 \cr & {\text{or}},\,x = \frac{{42 \times 5}}{{10}} \cr & {\text{So}},\,x = 21\,{\text{km}} \cr} $$
18. The speed of a motor-boat is that of the current of water as 36:5. The boat goes along with the current in 5 hours 10 minutes. It will come back in:
a) 5 hours 50 minutes
b) 6 hours 50 minutes
c) 6 hours
d) 12 hours 10 minutes
Discussion
Explanation: Let the speed of the motor boat = 36x kmph and speed of current = 5x kmph.
The boat goes along with the current in 5 hours 10 minutes = $$\frac{{31}}{6}$$ hour
$$\eqalign{ & {\text{Hence, Distance}} \cr & = {\frac{{31}}{6}} \times \left( {36x + 5x} \right) \cr & = \frac{{41x + 31}}{6}\,km \cr & {\text{Speed of boat upstream}} \cr & = 36x + 5x = 41x\,{\text{kmph}} \cr & {\text{Hence, time taken to come back}} \cr & = {\frac{{ {41x \times {\frac{{31}}{6}} } }}{{31x}}} \cr & = \frac{{41}}{6}\,{\text{hours}} \cr & = 6\,{\text{hours}}\,\,50\,{\text{minutes}} \cr} $$
19. Two trains started at the same time, one from A to B and other from B to A. if they arrived at B and A respectively in 4 hours and 9 hours after they passed each other, the ratio of the speeds of the two trains was:
a) 2 : 1
b) 3 : 2
c) 4 : 3
d) 5 : 4
Discussion
Explanation: A →____________________← B
Ratio of the speed of trains is given by;
$$\eqalign{ & \frac{{{\text{Speed}}\,{\text{of}}\,{\text{train}}\,A}}{{{\text{Speed}}\,{\text{train}}\,{\text{of}}\,B}} = \frac{{\sqrt B }}{{\sqrt A }} \cr & = \frac{{\sqrt 9 }}{{\sqrt 4 }} \cr & = \frac{3}{2} \cr & = 3:2 \cr} $$
20. In a 1-kilometre race, A can beat B by 30 meters, while in a 500-meter race B can beat C by 25 meters. By how many meters will A beats C in a 100-meter race?
a) 7.25
b) 7.15
c) 7.85
d) 7.03
Discussion
Explanation: When A runs 100 meters, B runs 970 meters.
Hence, when A runs 100 meter, b runs 97 meter.
When B runs 500 meter, C runs 475 meter.
When B runs 97 meter, C runs, $$\frac{{475 \times 97}}{{500}}$$ = 92.15 meter.
A will beat C by (100 - 92.15) = 7.85 meters.
21. A plan left 40 minutes late due to bad whether and order to reach its destination 1600 km away in time, it had increase its speed by 400 kmph from its usual speed. Find the usual speed of the plane?
a) 600 km/hour
b) 400 km/hour
c) 1000 km/hour
d) 800 km/hour
Discussion
Explanation: Let usual speed be X kmph, then new speed will be (x + 400) kmph.
Time taken to cover 1600 km with speed X kmph,
= $$\frac{{1600}}{{\text{x}}}$$
Time taken to cover 1600 km with Speed (x + 400) kmph,
= $$\frac{{1600}}{{{\text{x}} + 400}}$$
Time difference = 40 minutes.
$$\frac{{1600}}{{\text{x}}} - \frac{{1600}}{{{\text{x}} + 400}} = \frac{{40}}{{60}}$$ hours
x2 + 400x - 960000 = 0 x = -1200, 800
Speed cannot be negative, So usual speed will be 800 km/hour
22. A thief seeing a policeman at a distance of 150 metres starts running at 10 kmph and the policeman gives immediate chase at 12 kmph. When the thief is overtaken the thief has traveled a distance of:
a) 750 m
b) 900 m
c) 800 m
d) 1 m
Discussion
Explanation: P__150m___T______x m_______Q.
Let Policeman caught thief at a distance (x + 150)m. And Thief has traveled x m.
Speed of Policeman
$$\eqalign{ & = 12\,\,{\text{kmph}} \cr & = \frac{{12 \times 5}}{{18}} \cr & = \frac{{60}}{{18}}\,\,{\text{m/sec}} \cr} $$
Speed of thief
$$\eqalign{ & = 10\,\,{\text{km}} \cr & = \frac{{10 \times 5}}{{18}} \cr & = \frac{{50}}{{18}}\,\,{\text{m/sec}} \cr} $$
In this case time is constant means Policeman covered (x + 150)m in same time thief covered x m.
$$\eqalign{ & \frac{{{\text{Speed of the thief}}}}{{{\text{Speed of Policeman}}}} = \frac{x}{{150 + x}} \cr & \frac{{50}}{{60}} = \frac{x}{{150 + x}} \cr & 7500 + 50x = 60x \cr & 10x = 7500 \cr & x = 750\,{\text{m}} \cr} $$
So, Thief has traveled 750 m before the caught.
23. The ratio between the speed of a bus and train is 15 : 27 respectively. Also, a car covered a distance of 720 km in 9 hours. The speed of the Bus is three-fourth the speed of the car. How much distance will the train cover in 7 hours?
a) 760 km
b) 756 km
c) 740 km
d) None of these
Discussion
Explanation: Ratio of speed of Bus and Train = 15 : 27
Let speed of the bus is 15X and Speed of the Train is 27X
Car Covered 720 km in 9 hours.
So, Speed of the Car = $$\frac{{720}}{9}$$ = 80 kmph
Given, Speed of the bus is $$\frac{3}{4}$$ of Car, So speed of the Bus,
= $$\frac{{80 \times 3}}{4}$$ = 60 kmph
15X = 60
X = 4
So, Speed of the train = 27X = 27 × 4 = 108 kmph.
Hence, Train will cover distance in 7 hours,
= 108 × 7
= 756 km
24. A starts 3 Minutes after B for a place 4.5 km distant. B, on reaching his destination, immediately returns and after walking a km meets A. If A can walk 1 Km in 18 minutes,then what is B's speed?
a) 4 km/h
b) 6 km/h
c) 7 km/h
d) 5 km/h
Discussion
Explanation: P__________3.5Km__________M__1km____Q
Speed of A = 1 km in 18 min = $$\frac{{1000}}{{18}}$$ = 55.55 m/min.
When A travels 3.5 km. B already has been traveled (4.5 + 1) = 5.5 km
A will take time to travel 3.5 km = $$\frac{{3500}}{{55.55}}$$ = 63 min.
So, B will take 66 min to travel 5.5 km (As B has started 3 min before of A)
Speed of B = $$\frac{{5500}}{{66}}$$ = 83.33 m/min = 5 km/h
25. A car travels 50% faster than a bike. Both start at the same time from A to B. The car reaches 25 minutes earlier than the bike. If the distance from A to B is 100 km, find the speed of the bike:
a) 120 kmph
b) 100 kmph
c) 80 kmph
d) 75 kmph
Discussion
Explanation: P __________100km__________Q
Let car takes time T hours to reach destination.
So, Bike will take $$\left( {{\text{T}} + \frac{{25}}{{60}}} \right)$$
Let speed of the bike = S kmph
Speed of Car = S + 50% of S = $$\frac{{3{\text{S}}}}{2}$$ kmph
For the both the case distance is constant. And when distance remain constant then time is inversely proportional to speed (As ST + D)
$$\eqalign{ & \frac{{ {\frac{{3S}}{2}} }}{{\left( S \right)}} = \frac{{ {T + {\frac{5}{{12}}} } }}{T} \cr & 3T = 2T + \frac{{10}}{{12}} \cr & T = \frac{{10}}{{12}}{\text{hours}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{car}} \cr & \frac{{3S}}{2} = \frac{{100}}{{ {\frac{{10}}{{12}}} }} \cr & \frac{{3S}}{2} = 120 \cr & S = 80\,kmph \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{bike}} = 80\,kmph \cr} $$
26. A soldier fired two bullets at an interval of 335 seconds moving at a uniform speed V1. A terrorist who was running ahead of the soldier in same direction, hears the two shots at an interval of 330 seconds. If the speed of sound is 1188 km/h, then who is the faster and how much?
a) Soldier, 22 km/h
b) Terrorist, 25 km/h
c) Soldier, 18 km/h
d) Terrorist, 20 km/h
Discussion
Explanation: $$\frac{{{\text{Speed }}\,{\text{of }}\,{\text{sound}}}}{{{\text{Relative speed of soldier and terrorist}}}} = $$ $$\frac{{{\text{Time }}\,{\text{taken}}}}{{{\text{Difference }}\,{\text{in }}\,{\text{time}}}}$$
$$\frac{{1188}}{{\text{S}}} = \frac{{330}}{5}$$
Soldier = 18 km/h
27. A coolie standing on a railway platform observe that a train going in one direction takes 4 second to pass him. Another train of same length going in opposite direction takes 5 seconds to pass him. The time taken (in seconds) by two train to cross each other will be:
a) 35 seconds
b) 36.5 seconds
c) $$\frac{{40}}{9}$$ seconds
d) 36 seconds
Discussion
Explanation: Let length of each train be X m, then
$$\eqalign{ & {S_1} = \frac{X}{4} \cr & {S_2} = \frac{X}{5} \cr & {\text{Required time to cross each other,}} \cr & = {\frac{{2X}}{{ { {\frac{X}{4}} + {\frac{X}{5}} } }}} \cr & = \frac{{40}}{9}{\text{seconds}} \cr} $$
28. A dog start chasing to a cat 2 hour later. It takes 2 hours to dog to catch the cat. If the speed of the dog is 30 km/h. what is the speed of cat?
a) 10 km/h
b) 12 km/h
c) 15 km/h
d) 20 km/h
Discussion
Explanation: Let speed of the cat be S km/h.
Time taken = $$\frac{{{\text{Distanced}}\,{\text{ advanced}}}}{{{\text{Relative}}\,{\text{ speed}}}}$$
2 = $$\frac{{2 \times {\text{S}}}}{{30 - {\text{S}}}}$$
S = 15 km/h
29. A car traveled first 36 km at 6 km/h faster than the usual speed, but it returned the same distance at 6 km/h slower than usual speed. I the total time taken by car is 8 hours, for how many hours does it traveled at the faster speed ?
a) 4 hours
b) 3 hours
c) 2 hours
d) 1 hours
Discussion
Explanation: Let the original speed be S then,
Total time taken,
$$ {\frac{{36}}{{{\text{S}} - 6}}} + {\frac{{36}}{{{\text{S}} + 6}}} $$ = 8 hours
On solving the equation
S = 12, -3
Then possible value of S = 12 km/h.
Thus, time taken by car at faster speed = $$\frac{{36}}{{12 + 6}}$$ = 2 hours
30. Roorkee express normally reaches its destination at 50 km/h in 30 hours. Find the speed at which it travels to reduce the time by 10 hours?
a) 38 km/h
b) 76 km/h
c) 75 km/h
d) 60 km/h
Discussion
Explanation: let required speed be S km/h
50 × 30 = S × 20
S = 75 km/h
31. Mr A started half an hour later than usual for the marketplace. But by increasing his speed to $$\frac{3}{2}$$ times his usual speed he reached 10 minutes earlier than usual. What is his usual time for this journey?
a) 1 hours
b) 2 hours
c) 3 hours
d) 45 minutes
Discussion
Explanation: Let the usual speed be S.
increased speed = $$\frac{{3{\text{S}}}}{2}$$
Let usual time taken be T
Thus, S × T = D, and also $$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = D. (T - 40, Because he started 30 minutes later and reached 10 minutes earlier, means he saved 40 minutes)
$$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = ST
$$\frac{{3 \times {\text{ST}}}}{2}$$ - 60S = ST
$$\frac{1}{2}$$ × ST = 60S
T = 120 minutes = 2 hours
32. Page and Plant are running on a track AB of length 10 metres. They start running simultaneously from the ends A and B respectively. The moment they reach either of the ends, they turn around and continue running. Page and Plant run with constant speeds of 2m/s and 5m/s respectively. How far from A (in metres) are they, when they meet for the 23rd time?
a) 0 meters
b) 10 meters
c) 40 meters
d) $$\frac{{60}}{7}$$ meters
Discussion
Explanation: The ratio of speeds is 2 : 5. So when slower one completes 2 one way journeys (and reaches its starting point), faster one travels 5 one way journeys (and reaches the other end). So that means after the faster one has traveled 5 one way journeys, both of them have reached same end and in next 5 one-way journey of faster runner, both reach their starting position simultaneously. Now most important to observe is that FASTER one will always meet the SLOWER one EXACTLY ONCE in each of its one-way journey, except when both of them have started with the same starting point.
Once you reduce that for the 5th time, they'll meet at A, and the next 5 rounds will have 4 meetings.
Just add 5 + 4 + 5 + 4 + 5 = 23 or just go by the position of Plant after every 5 rounds : A, B, A, B, A. The cycle repeats.
So, total distance will be 0 meters.
33. Two cars start simultaneously from cities A and B, towards B and A respectively, on the same route. Once the two cars reach their destinations they turned around and move towards the other city without any loss of time. The two cars continue shuttling in this manner for exactly 20 hours. If the speed of the car starting from A is 60km/hr and the speed of the car starting from B is 40km/hr and the distance between the two cities is 120 km, find the number of times the two cars cross each other?
a) 8
b) 10
c) 12
d) 20
Discussion
Explanation: Suppose both moves with the same speed 60 km/h then they will meet max 10 times. So answer will be less than 10. Thus answer will be 8.
34. Two persons A and B start simultaneously from P and Q respectively. A meets B at a distance of 60m from P. After A reaches Q and B reaches P, they turn around and start walking in opposite direction, now B meets A at a distance of 40m from Q. Find distance between P and Q?
a) 160 m
b) 100 m
c) 140 m
d) 105 m
Discussion
Explanation: P_____60m___R__Xm____40m____Q
Total distance = 100 + X
Initially, A traveled 60m and B traveled (40 + X)m. Since time is same. So, Speed ∝ Distance.
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}}$$
Second time,
So, now total distance traveled by A = 100 + X + 40
Total distance traveled by B = 100 + X + 60 + X
These will be in the ratio of Speed of A and B
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}} = \frac{{100 + {\text{X}} + 40}}{{100 + {\text{X}} + 60 + {\text{X}}}}$$
X = 40
Total distance = 100 + X = 100 + 40 = 140 m
35. Rohan, Shikha's boyfriend, had to pick her from her home for a live concert on her 23rd birthday. The venue of the concert and Shikha's home were in opposite directions from Rohan's office. He got late because of some work at office and realised that if he goes to pick Shikha from her home, which was a 48-minute drive from his office, they would be late for the show by 16 minutes. He asked her to start from her home towards his office in an auto-rickshaw and himself started driving towards her home. Both of them started simultaneously, he picked her as soon as they met and they managed to reach the venue just in time for the concert. If Rohan drives at an average speed of 60 km/hr, find the speed (in km/hr) of the auto-rickshaw.
a) 16 km
b) 20 km
c) 12 km
d) 15 km
Discussion
Explanation: Let the speed of the rickshaw = S km/h. Let after time T they meet.
So, 60t + ST = $$\frac{{4 \times 60}}{5}$$ $$\left[ {{\text{Since,}}\,\,48\,{\text{min}} = \frac{4}{5}\,\,{\text{hour}}} \right]$$
Rohan saves 16 min by making Shikha move as well. So, she saves 16 min by moving for distance X (Let).
2X = $$\frac{{60 \times 16}}{5}$$
X = 8 km
Speed of Auto rickshaw = 12 km. $$\left( {{\text{ST}} = 8\,{\text{km,}}\,\,{\text{T}} = \frac{2}{3}} \right)$$
36. A train 350 m long is running at the speed of 36 km/h. If it crosses a tunnel in 1 minute, the length of the tunnel in metres:
a) 200 meters
b) 250 meters
c) 300 meters
d) 150 meters
Discussion
Explanation: Let the length of the tunnel be x m
Time = $$\frac{{{\text{Length of train}} + {\text{Length of tunnel}}}}{{{\text{Speed}}}}$$
60 = $$\frac{{350 + {\text{x}}}}{{10}}$$
$$ {{\text{Speed}} = \frac{{36 \times 5}}{{18}} = 10\,\,{\text{m/sec}}} $$
x = 250 meters
So, length of tunnel is 250 meters
37. If a 250 m long train crosses a platform of same length as that of the train in 25 seconds, then the speed of the train is:
a) 150 m/sec
b) 200 m/sec
c) 20 km/h
d) 72 km/h
Discussion
Explanation: Speed of train
= $$\frac{{{\text{length of the train}} + {\text{length of platform}}}}{{{\text{Time}}}}$$
= $$\frac{{250 + 250}}{{25}}$$
= 20 m/sec
= 72 km/h
38. A postman goes with a speed of 36 km/h what is the speed of postman in m/s?
a) 4.5 m/s
b) 6 m/s
c) 5 m/s
d) 10 m/s
Discussion
Explanation: Speed = 36 km/h = $$\frac{{36 \times 5}}{{18}}$$ = 10 m/s
39. Sabarmati express takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is :
a) 132 m
b) 100 m
c) 80 m
d) 90 m
Discussion
Explanation: Let length of the train be x m
Speed of train,
$$\frac{{{\text{x}} + 162}}{{18}} = \frac{{{\text{x}} + 120}}{{15}}$$
x = 90 m
40. If two incorrect watches are set at 12:00 noon at correct time, when will both watches show the correct time for the first time given that the first watch gains 1 min in one hour and second watch looses 4 min in two hours.
a) 6 PM, 25 days later
b) 12 noon, 30 days later
c) 12 noon, 15 days later
d) 6 AM, 45 days later
Discussion
Explanation: First watch:
It shows correct time when it creates difference of 12 hours.
So, to create difference of 12 hour, time required = $$\frac{{60 \times 12}}{{24}}$$ = 30 days
Second watch:
It shows correct time when it creates difference of 12 hours.
So, to create difference of 12 hour, time required = $$\frac{{30 \times 12}}{{24}}$$ = 15 days
LCM of 30 and 15 gives the time when they show correct time together.
Required time = 30 days, at same time.
41. Narayan walking at a speed of 20 km/h reaches his college 10 minutes late. Next time he increases his speed by 5 km/h, but finds that he is still late by 4 minutes. What is the distance of his college from his house?
a) 10 km
b) 6 km
c) 12 km
d) 15 km
Discussion
Explanation: By increasing his speed by 25%, he will reduce his time by 20%. (This corresponds to a 6 minutes drop in his time.)
Hence, his time originally must have been 30 minutes.
Required distance = 20 kmph × 0.5 hours = 10 km.
42. A motor car does a journey in 17.5 hours, covering the first half at 30 km/h and the second half at 40 km/h. Find the distance of the journey?
a) 684 km
b) 600 km
c) 624 km
d) 584 km
Discussion
Explanation: Let the total distance be 2X.
A_____X km_____M_____X km_____B
Total time taken in the journey = 17.5 hours
Time taken to cover X km at 30 km/h = $$\frac{{\text{X}}}{{30}}$$
Time taken to cover X km at 40 km/h = $$\frac{{\text{X}}}{{40}}$$
$$\eqalign{ & {\frac{X}{{30}}} + {\frac{X}{{40}}} = 17.5 \cr & {\frac{{ {40X + 30X} }}{{1200}}} = 17.5 \cr & 70X = 17.5 \times 1200 \cr & X = 300\,km \cr & {\text{Total}}\,{\text{distance}},\,2x \cr & = 600\,km \cr} $$
43. A car after traveling 18 km from a point A developed some problem in the engine and the speed became $$\frac{4}{5}$$ th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:
a) 25 kmph
b) 30 kmph
c) 20 kmph
d) 35 kmph
Discussion
Explanation: He proceeds at $$\frac{4}{5}$$ S where S is his usual speed means $$\frac{1}{5}$$ decrease in speed which will lead to $$\frac{1}{4}$$ increase in time.
Now the main difference comes in those 12km (30 - 18) and the change in difference of time = (45 - 36) min = 9 min.
$$\frac{1}{4}$$ × T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km
T = 36 min = $$\frac{36}{60}$$ hours = 0.6 hours.
Speed of the car = $$\frac{12}{0.6}$$ = 20 kmph.
44. A hunter fired two shots from the branch of a tree at an interval of 76 seconds. A tiger separating too fast hears the two shots at an interval of 83 seconds. If the velocity of the sound is 1195.2 km/h, then find the speed of the tiger.
a) 112.8 km/h
b) 100.8 km/h
c) 80.16 km/h
d) 85.16 km/h
Discussion
Explanation: In the case of increasing gap between two objects
$$\frac{{{\text{Speed }}\,{\text{of }}\,{\text{sound}}}}{{{\text{Speed }}\,{\text{of }}\,{\text{tiger}}}} = $$ $$\frac{{{\text{Time }}\,{\text{Utilized}}}}{{{\text{Difference }}\,{\text{in }}\,{\text{time}}}}$$
$$\frac{{1195.2}}{{\text{S}}} = \frac{{83}}{7}$$
S = 100.8 km/h.
45. A man can cross a downstream river by steamer in 40 minutes and same by boat in 1 hour. If the time of crossing Upstream by streamer is 50% more than downstream time by steamer and the time required by boat to cross same river by boat in upstream is 50% more than time required by in downstream. What is the time taken for the man to cross the river downstream by steamer and then return to same place by boat half the way and by steamer the rest of the way?.
a) 85 minutes
b) 115 minutes
c) 120 minutes
d) 125 minutes
Discussion
Explanation: Downstream (steamer) = 40 min
Downstream (boat) = 60 min
Upstream (steamer) = 60 min
Upstream (boat) = 90 min
Required time = 40 + 30 + 45 = 115 minutes
46. A man can row 15 km/h in still water and he finds that it takes him twice as much time to row up than as to row down the same distance in the river. The speed of current (in km/h) :
a) 6 km/h
b) 6.5 km/h
c) 4.5 km/h
d) 5 km/h
Discussion
Explanation: Let the speed of the current be x km/h.
$$\frac{{15 + x}}{{15 - x}} = \frac{2}{1}$$
15 + x = 30 - 2x
3x = 30 - 15
3x = 15
x = 5 km/h
47. A boat covers 48 km upstream and 72 km downstream in 12 hours, while it covers 72 km upstream and 48 km downstream in 13 hours The speed of stream is
a) 2 kmph
b) 2.2 kmph
c) 2.5 kmph
d) 4 kmph
Discussion
Explanation: Let downstream speed be D and Upstream speed be U. Then
$$\eqalign{ & {\frac{{48}}{U} + \frac{{72}}{D}} = 12\,.\,.\,.\,.\,.\left( 1 \right) \cr & {\frac{{72}}{U} + \frac{{48}}{D}} = 13\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$
On solving the equation (1) and (2),
D = 12 kmph
U = 8 kmph
Thus, speed of the current
$$ = \frac{{D - U}}{2} = \frac{4}{2} = 2\,{\text{kmph}}$$
48. A motor boat takes 2 hours to travel a distance of 9 km downstream and it takes 6 hours to travel the same distance against the current. The speed of the boat in still water and that of the current (in km/h) respectively are:
a) 6, 5 km/h
b) 3, 1.5 km/h
c) 8, 5 km/h
d) 9, 3 km/h
Discussion
Explanation: Downstream speed,
= $$\frac{9}{2}$$ = 4.5 kmph
Upstream speed,
= $$\frac{9}{6}$$ = 1.5 kmph
Speed of boat in still water,
= $$\frac{{4.5 + 1.5}}{2}$$ = 3 km/h
Speed of current = $$\frac{{4.5 - 1.5}}{2}$$ = 1.5 km/h
49. A train met with an accident 120 km from station A. It completed the remaining journey at $$\frac{5}{6}$$ of its previous speed and reached 2 hour late at station B. Had the accident taken place 300 km further, it would have been only 1 hour late. What is the speed of the train?
a) 100 km/h
b) 12 km/h
c) 60 km/h
d) 50 km/h
Discussion
Explanation: A ____100 km____ P1 ________300 km_____ P2 ____X______ B
If speed becomes $$\frac{5}{6}$$ then time taken will be $$\frac{6}{5}$$ of original time.
Thus, extra time = 2 hour
$$\frac{1}{5}$$ = 2 hour
Thus unusual time which is require to complete the journey = 5 × 2 = 10 hours.
It means It covers 300 km distance in 5 hours.
Speed = $$\frac{{300}}{5}$$ = 60 km/h
50. A runs $$\frac{7}{4}$$ times as fast as B. If A gives B a start of 300 m, how far must the winning post be if both A and B have to end the race at same time?
a) 1400 m
b) 700 m
c) 350 m
d) 210 m
Discussion
Explanation: If A runs 7/4 times as fast as B
A : B = 7 : 4
Let x be the constant ratio
A : B = 7x : 4x
The difference is 300m
7x - 4x = 300
3x = 300
x = 100m
The winning post is the distance A has to run
A = 7x = 7(100) = 700 m
51. Two trains starting at the same time from two-stations 300 km apart and going in opposite directions, cross each other at a distance of 160 km from one of them. The ratio of their speeds is :
a) 7 : 9
b) 16 : 20
c) 8 : 7
d) 8 : 12
Discussion
Explanation: A → ___160________Meeting____140_____← B
Ratio of their distance covered = $$\frac{{160}}{{140}}$$ = 8 : 7
Ratio of their speeds = 8 : 7 [When time is constant Speed is proportional to Distance covered]
52. Two runner start running together for a certain distance, one at 5 km/h and another at 8 km/h. The former arrives one and half an hour before the latter. The distance in Km is :
a) 12 km
b) 20 km
c) 25 km
d) 36 km
Discussion
Explanation:
$$\eqalign{ & \text{Let }x\text{ be the distance} \cr & \frac{x}{5} - \frac{x}{8} = \frac{3}{2} \cr & \frac{8x - 5x}{40} = \frac{3}{2} \cr & \frac{3x}{40} = \frac{3}{2} \cr & x = 20\,{\text{km}} \cr} $$
53. A railway passenger counts the telegraph poles on the rail road as he passes them. The telegraph poles are at a distance of 50 metres. What will be his count in 4 hours, if the speed of the train is 45 km per hour.
a) 600
b) 2500
c) 3600
d) 5000
Discussion
Explanation: In train he travels = 4 × 45 = 180 km = 180000 m
The no. of poles,
= $$\frac{{180000}}{{50}}$$ = 3600 [Poles are at the distance of 50 m.]
54. Without stoppage, a train travels a certain distance with an average speed of 60 km/h, and with stoppage it covers the same distance with an average speed of 40 km/h. On an average, how many minutes per hour does train stop during the journey ?
a) 20 min/hour
b) 15 min/hour
c) 10 min/hour
d) 12 min/hour
Discussion
Explanation: Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute.
If it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min/hour.
55. A train without stopping travels 60 km/h and with stoppage 40 km/h. What is the time taken for stoppage on a route 300 km?
a) 10 hours
b) 20 hours
c) 5 hours
d) 2.5 hours
Discussion
Explanation: Since, the train travels at 60 km/h, it's speed per minute is 1 km per minute. Hence, if it's speed with stoppage is 40 km/h, it will travel 40 minutes per hour i.e. train stops 20 min per hour.
Time taken to travel 300 km with stoppage
$$ = \frac{{300}}{{40}} = 7.5\,\,{\text{hours}}$$
Time taken for stoppage as it stops for 20 min per hour
= 7 × 20 + 10
= 140 + 10
= 150 minutes.
= $$\frac{{150}}{{60}}$$ hours = 2.5 hours.
56. A person can row a boat d km upstream and the same distance downstream in 5 hours 15 minutes. Also, he can row the boat 2d km upstream in 7 hours. How long will it take to row the same distance 2d km downstream?
a) $$\frac{{3}}{{2}}$$ hours
b) 7 hours
c) $$\frac{{29}}{{4}}$$ hours
d) $$\frac{{7}}{{2}}$$ hours
Discussion
Explanation: Let the speeds of boat and stream was $$s$$ and $$v$$ km/hr respectively
Actual Speed Downstream = $$\left(s + v\right)$$ km/hr
Actual Speed upstream = $$\left(s - v\right)$$ km/hr
$$\eqalign{ & \frac{d}{{s + v}} + \frac{d}{{s - v}} = 5\,{\text{hr}}{\text{.}}\,15\,{\text{min}}{\text{.}} \cr & \Rightarrow \frac{d}{{s + v}} + \frac{d}{{s - v}} = \frac{{21}}{4}\,.\,....\left( 1 \right) \cr & {\text{and}} \cr & \frac{{2d}}{{s - v}} = 7 \cr & \Rightarrow \frac{d}{{s - v}} = \frac{7}{2}\,.....\left( 2 \right) \cr & {\text{By equation }}\left( 1 \right) - \left( 2 \right), \cr & \frac{d}{{s + v}} = \frac{{21}}{4} - \frac{7}{2} \cr & \Rightarrow \frac{d}{{s + v}} = \frac{{21 - 14}}{4} \cr & \Rightarrow \frac{d}{{s + v}} = \frac{7}{4} \cr & \Rightarrow \frac{{2d}}{{s + v}} = \frac{7}{2} \cr & \cr} $$
Hence, he takes $$\frac{{7}}{{2}}$$ hours to row 2d km distance downstream
57. If a man runs at 6 kmph from his house, he misses the train at the station by 8 min. If he runs at 10 kmph, he reaches 7 min before the departure of the train. What is the distance of the station from his house? (in Km).
a) $$4\frac{3}{4}$$ km
b) $$3\frac{1}{2}$$ km
c) $$4\frac{1}{4}$$ km
d) $$3\frac{3}{4}$$ km
Discussion
Explanation: Let the distance of the station from the house of the person = x km
$$\eqalign{ & {\text{Difference}}\,{\text{of}}\,{\text{time}} \cr & = 8 + 7 \cr & = 15\,{\text{minutes}} \cr & = \frac{1}{4}\,hr \cr & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & \frac{x}{6} - \frac{x}{{10}} = \frac{1}{4} \cr & \frac{{10x - 6x}}{{60}} = \frac{1}{4} \cr & \frac{{2x}}{{30}} = \frac{1}{4} \cr & \Rightarrow x = \frac{{15}}{4} = 3\frac{3}{4}km \cr} $$
58. A train running at a speed of 54 km/hr crosses a platform in 30 seconds. The platform is renovated and its length is doubled. Now, the same train running at same speed crosses the platform in 46 seconds. Find the length of the train.
a) 180 metres
b) 200 metres
c) 210 metres
d) 240 metres
Discussion
Explanation: Let length of the Platform is X m and Train is Y m.
Speed of the train = 54 kmph = $$\frac{{54 \times 5}}{{18}}$$ = 15 m/sec.
To cross the platform, train needs to travel (X + Y) m in 30 sec.
$$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & 15 = \frac{{{\text{X}} + {\text{Y}}}}{{30}} \cr & {\text{X}} + {\text{Y}} = 450\,.\,.\,.\,.\,.\,.\,.\left( 1 \right) \cr} $$
Now Platform is renovated and its length is doubled. So, train need to travel (2X + Y) m to cross the platform.
$$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & 15 = \frac{{{\text{2X}} + {\text{Y}}}}{{46}} \cr & {\text{2X}} + {\text{Y}} = 690\,.\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$
Multiplying equation (1) by (2)
2X + 2Y = 900 ---------- (3)
Now, equation (2) - (3)
2X + Y - 2X - 2Y = 690 - 900
- Y = - 210
Y = 210
Length of the train = 210 metres
59. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
a) 3.6 km/hr
b) 7.2 km/hr
c) 8.4 km/hr
d) 10 km/hr
Discussion
Explanation:
$$\eqalign{ & {\text{Speed}} = {\frac{{600}}{{5 \times 60}}} {\text{ m/sec}} = 2{\text{m/sec}} \cr & {\text{Converting}}\,{\text{m/sec}}\,{\text{to}}\,{\text{km/hr}} \cr & = {2 \times \frac{{18}}{5}} {\text{ km/hr}} \cr & = 7.2\,{\text{ km/hr}} \cr} $$
60. An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in $$1\frac{2}{3}$$ hours, it must travel at a speed of:
a) 300 kmph
b) 360 kmph
c) 600 kmph
d) 720 kmph
Discussion
Explanation:
$$\eqalign{ & {\text{Distance}} = {240 \times 5} = 1200\,{\text{km}} \cr & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & {\text{Speed}} = \frac{{1200}}{{ {\frac{5}{3}} }}\,{\text{ km/hr}}\, \cr & \left[ {{\text{We}}\,{\text{can}}\,{\text{write}}\,1\frac{2}{3}\,{\text{hours}}\,{\text{as}}\,\frac{5}{3}\,{\text{hours}}} \right] \cr & {\text{Required}}\,{\text{speed}} \cr & = {1200 \times \frac{3}{5}} \,{\text{ km/hr}} \cr & = 720\,{\text{ km/hr}} \cr} $$
61. The wheel of an engine of 300 cm in circumference makes 10 revolutions in 6 seconds. What is the speed of the wheel (in km/h)?
a) 18
b) 20
c) 27
d) 36
Discussion
Explanation: Circumference = One revolution
Distance covered in 10 revolutions = 300 * 10 = 3000 cm = 30 m
30 meters are covered in 6 seconds.
Speed of the wheel = $$\frac{{30}}{{\text{6}}}$$ = 5 m/s
5 m/s = $$\frac{{5 \times 18}}{5}$$ = 18 km/h. [ To convert m/s to km/h, we use to multiply by factor $$\frac{{18}}{5}$$ ]
62. A thief sees a jeep at a distance of 250 m, coming towards him at 36 km/h. Thief takes 5 seconds to realize that there is nothing but the police is approaching him by the jeep and start running away from police at 54 km/h. But police realise after 10 second, when the thief starts running away, that he is actually a thief and gives chase at 72 km/h. How long after thief saw police did catchup with him and what is the distance police had travel to do so?
a) 50 s, 1000 m
b) 65 s, 1150 m
c) 65 s, 1300 m
d) 45 s, 1050 m
Discussion
Explanation: Initial speed of the police = 10 m/s
Increased speed of the Police = 20 m/s
Speed of the thief = 15 m/s
Initial difference between speed of thief and police = 250 m
After 5 seconds difference between thief and police = 200 + (5 × 10) = 250 m
The time required by police to catch the thief = $$\frac{{250}}{5}$$ = 50 s
Total time = 50 + 15 = 65 s
Total distance = 1000 + (15 × 10) = 1150 m
63. A person X starts from Lucknow and another person Y starts from Kanpur to meet each other. Speed of X is 25 km/h, while Y is 35 km/h. If the distance between Lucknow and Kanpur be 120 km and both X and Y start their journey at the same time, when they will meet ?
a) 1 hours later
b) 2 hours later
c) $$\frac{1}{2}$$ hours later
d) 3 hours later
Discussion
Explanation: Lucknow(X →)_____120 km _____(← Y)Kanpur
Relative speed = (25+35) = 60 km/h
Time taken to complete 120 km
= $$\frac{{120}}{{60}}$$ = 2 h
They will meet 2 hours later.
64. Two persons A and B started from two different places towards each other. If the ratio of their speed be 3 : 5, then what is the ratio of distance covered by A and B respectively till the point of meeting?
a) 1 : 2
b) 3 : 4
c) 3 : 5
d) 5 : 3
Discussion
Explanation: When time is constant the distance covered by A and B will be in the ratio of their speeds. i.e. ratio of distance is 3 : 5 as time is constant.
65. The ratio of speeds of A and B is 2 : 3 and therefore A takes 20 minutes more time than B. What is the ratio of time taken by A and B?
a) 2 : 3
b) 2 : 5
c) 3 : 2
d) 3 : 5
Discussion
Explanation: When distance is constant then Speed is inversely proportional to time.
ST = D
When distance constant,
S ∝ $$\frac{1}{{\text{T}}}$$
Ratio of time taken by A and B = 3 : 2
66. From two places, 60 km apart, A and B start towards each other at the same time and meet each other after 6 hour. If A traveled with $$\frac{2}{3}$$ of his speed and B traveled with double of his speed, they would have met after 5 hours. The speed of A is:
a) 4 km/h
b) 6 km/h
c) 10 km/h
d) 12 km/h
Discussion
Explanation: A →_______60Km_________← B
Let the speed of A = x kmph and that of B = y kmph
x × 6 + y × 6 = 60
x + y = 10 --------- (i)
$$\left( {\frac{{2{\text{x}}}}{3} \times 5} \right) + \left( {2{\text{y}} \times 5} \right) = 60$$
10x + 30y = 180
x + 3y = 18 ---------- (ii)
From equation (i) × 3 - (ii)
3x + 3y - x - 3y = 30 - 18
2x = 12
x = 6 kmph
67. A, B and C start together from the same place to walk round a circular path of length 12km. A walks at the rate of 4 km/h, B 3 km/h and C $$\frac{3}{2}$$ km/h. They will meet together at the starting place at the end of:
a) 10 hours
b) 12 hours
c) 15 hours
d) 24 hours
Discussion
Explanation: Time taken to complete the revolution:
A → $$\frac{{12}}{{4}}$$ = 3 hours
B → $$\frac{{12}}{{3}}$$ = 4 hours
C → 12 × $$\frac{{2}}{{3}}$$ = 8 hours
Required time
= LCM of 3, 4, 8
= 24 hours
68. Ravi and Ajay start simultaneously from a place A towards B 60 km apart. Ravi's speed is 4km/h less than that of Ajay. Ajay, after reaching B, turns back and meets Ravi at a places 12 km away from B. Ravi's speed is:
a) 12 km/h
b) 10 km/h
c) 8 km/h
d) 6 km/h
Discussion
Explanation:
Ajay → (x + 4) kmph.
A ________ 60 km _________ B
Ravi → x kmph.
Let the speed of Ravi be x kmph;
Hence, Ajay's speed = (x + 4) kmph;
Distance covered by Ajay = 60 + 12 = 72 km;
Distance covered by Ravi = 60 - 12 = 48 km.
$$\eqalign{ & \frac{{72}}{{x + 4}} = \frac{{48}}{x} \cr & \frac{3}{{x + 4}} = \frac{2}{x} \cr & 3x = 2x + 8 \cr & x = 8\,{\text{kmph}} \cr} $$
69. The speed of A and B are in the ratio 3 : 4. A takes 20 minutes more than B to reach a destination. Time in which A reach the destination?
a) $$1\frac{1}{3}$$ hours
b) 2 hours
c) $$2\frac{2}{3}$$ hours
d) $$1\frac{2}{3}$$ hours
Discussion
Explanation: Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 (As Speed ∝ $$\frac{1}{{{\text{Time}}}},$$ When distance remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.
4x - 3x = $$\frac{{20}}{{60}}$$
x = $$\frac{{1}}{{3}}$$
Hence, time taken by A = 4x
hours = 4 × $$\frac{1}{3}$$ = $$1\frac{1}{3}$$ hours.
70. A man covers half of his journey at 6 km/h and the remaining half at 3 km/h. His average speed is
a) 9 km/h
b) 4.5 km
c) 4 km/h
d) 3 km/h
Discussion
Explanation:
$$\eqalign{ & {\text{Average}}\,{\text{speed}} \cr & = \frac{{2xy}}{{ {x + y} }} \cr & = \frac{{2 \times 6 \times 3}}{{ {6 + 3} }} \cr & = \frac{{36}}{9} \cr & = 4\,{\text{kmph}} \cr} $$
71. An athlete runs 200 meters race in 24 seconds. His speed in km/h is
a) 20
b) 24
c) 28.5
d) 30
Discussion
Explanation:
$$\eqalign{ & {\text{Speed of athlete}} \cr & = \frac{{200\,{\text{m}}}}{{24\,{\text{secs}}}} \cr & = \frac{{200 \times 18}}{{5 \times 24}} \cr & = 30\,{\text{km/hour}} \cr} $$
72. If A travels to his school from his house at the speed of 3 km/h, then he reaches the school 5 minutes late. If he travels at the speed of 4 km/h, he reaches the school 5 minutes earlier than school time. The distance of his school from his house is:
a) 1 km
b) 2 km
c) 3 km
d) 4 km
Discussion
Explanation: Let the distance between school and home be x km.
The difference of time when A goes school to school with these two different speed is 10 min
$$\eqalign{ & = \frac{{10}}{{60}}\,{\text{hour}} \cr & {\frac{x}{3}} - {\frac{x}{4}} = \frac{{10}}{{60}} \cr & \frac{x}{{12}} = \frac{1}{6} \cr & x = \frac{{12}}{6} \cr & = 2\,{\text{km}} \cr} $$
73. A man starts climbing a 11 m high wall at 5 pm. In each minute he climbs up 1 m but slips down 50 cm. At what time will he climb the wall?
a) 5:30 pm
b) 5:21 pm
c) 5:25 pm
d) 5:27 pm
Discussion
Explanation: Man climbs 1m and slips down 50 cm (0.5m) in one minute
i.e. he climbs (1 - 0.5 = 0.5 m) in one minute.
But in the last minute he will be climbing 1m as he gets on the top so no slip.
Time taken to climb 11 meter = $$ {\frac{{10}}{{0.5}} + 1} $$ = 21 minutes.
He climbs the wall at 5:21 pm
74. A monkey climbs a 60 m high pole. In first minute he climbs 6 m and slips down 3 m in the next minute. How much time is required by it to reach the top?
a) 35 minutes
b) 33 minutes
c) 37 minutes
d) 40 minutes
Discussion
Explanation: Monkey climbs 6m in 1st minute and slips down 3 m in next minute
i.e. Monkey climbs 3 meter in 2 minute then he climbs in one minute,
= $$\frac{3}{2}$$ m.
But in the last minute he climbs 6 m as he gets on the top so there is no slip.
Time required = $$2 \times \frac{{54}}{3} + \frac{6}{6}$$ = 37 minutes
75. An ant climbing up a vertical pole ascends 12 meters and slips down 5 meters in every alternate hour. If the pole is 63 meters high how long will it take it to reach the top?
a) 18 hours
b) 17 hours
c) 16 hours 35 min
d) 16 hours 40 min
Discussion
Explanation: Since it climbs 7 meter in every 2 hr. At the end of 14 hours it would have climbed up to 49 meter. In 15th hour it will reach 61 meter. Then drops back to 56 meter in the 16th hour. It will take another 35 min to travel remaining 7 meter since it will be ascending in the 17th hour. Thus, 16 hr and 35 min.
76. Two buses start from a bus terminal with a speed of 20 km/h at interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8 minutes?
a) 3 km/h
b) 4 km/h
c) 5 km/h
d) 7 km/h
Discussion
Explanation: Let Speed of the man is x kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
$$20 \times \frac{{10}}{{60}} = \frac{8}{60}\times\left(20+x\right) $$
200 = 160 + 8x
8x = 40
x = 5kmph.
77. Walking $$\frac{3}{4}$$ of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office:
a) 48 min.
b) 60 min.
c) 42 min.
d) 62 min.
Discussion
Explanation: $$\frac{4}{3}$$ of usual time = Usual time + 16 minutes;
Hence, $$\frac{1}{3}{\text{rd}}$$ of usual time = 16 minutes;
Usual time = 16 × 3 = 48 minutes.
78. Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together:
a) 262.4 km
b) 260 km
c) 283.33 km
d) 275 km
Discussion
Explanation: Difference in time of departure between two trains = 45 min. = $$\frac{{45}}{{60}}$$ hour = $$\frac{{3}}{{4}}$$ hour.
Let the distance be x km from Delhi where the two trains will be together.
Time taken to cover x km with speed 136 kmph be t hour
and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be
$$ {{\text{t}} + \frac{3}{4}} $$
= $$ {\frac{{4{\text{t}} + 3}}{4}} $$
Now,
100 × $$ {\frac{{4{\text{t}} + 3}}{4}} $$ = 136t
25(4t + 3) = 136t
100t + 75 = 136t
36t = 75
t = $$\frac{{75}}{{36}}$$ = 2.083 hours
Distance x km = 136 × 2.083 ≈ 283.33 km.
79. A man takes 6 hours 15 minutes in walking a distance and riding back to starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride back both ways is:
a) 4 hours
b) 4 hours 30 min.
c) 4 hours 45 min.
d) 5 hours
Discussion
Explanation: Time taken in walking both the ways = 7 hours 45 minutes ----- (i)
Time taken in walking one way and riding back = 6 hours 15 minutes ----- (ii)
By the equation (ii) × 2 - (i),
Time taken by the man in riding both ways,
= 12 hours 30 minutes - 7 hours 45 minutes
= 4 hours 45 minutes
80. A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:
a) 25 km/h
b) 28 km/h
c) 30 km/h
d) 33 km/h
Discussion
Explanation:
$$\eqalign{ & {\text{Let the total distance be 100 km}}. \cr & {\text{Average speed}} \cr & = \frac{{{\text{total}}\,{\text{distance}}\,{\text{covered}}}}{{{\text{time}}\,{\text{taken}}}} \cr & = \frac{{100}}{{ { {\frac{{30}}{{20}}} + {\frac{{60}}{{40}}} + {\frac{{10}}{{10}}} } }} \cr & = \frac{{100}}{{ { {\frac{3}{2}} + {\frac{3}{2}} + 1 } }} \cr & = \frac{{100}}{{ {\frac{{ {3 + 3 + 2} }}{2}} }} \cr & = \frac{{ {100 \times 2} }}{8} \cr & = 25\,\text{kmph} \cr} $$
81. Two planes move along a circle of circumference 1.2 km with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction, one plane overtakes the other every 60 seconds. Find the speed of the slower plane.
a) 0.04 km/s
b) 0.03 km/s
c) 0.05 km/s
d) 0.02 km/s
Discussion
Explanation: Let their speeds be x m/sec and y m/sec respectively.
$$\eqalign{ & \frac{{1200}}{{x + y}} = 15 \cr & \Rightarrow x + y = 80.....(i) \cr} $$
$$\eqalign{ & \frac{{1200}}{{x - y}} = 60 \cr & \Rightarrow x - y = 20.....(ii) \cr} $$
Adding (i) and (ii),
2x = 100 or x = 50
Putting x = 50 in (i), we get : y = 30
Speed of slower plane :
= 30 m/sec
= 0.03 km/sec
82. Two joggers left Delhi for Noida simultaneously. The first jogger stopped 42 min later when he was 1 km short of Noida and the other one stopped 52 min later when he was 2 km short of Noida. If the first jogger jogged as many kilometers as the second, and the second as kilometers as first, the first one would need 17 min less than the second. Find the distance between Delhi and Noida?
a) 5 km
b) 15 km
c) 25 km
d) 35 km
Discussion
Explanation:
$$\eqalign{ & {\text{Speed of first Jogger}} \cr & = {\frac{{ {x - 1} }}{{42}}} \times 60\,{\text{kmph}} \cr & {\text{Speed of }}\,{2^{nd}}\,{\text{jogger}} \cr & = {\frac{{ {x - 2} }}{{52}}} \times 60\,{\text{kmph}} \cr & {\text{Then}}, \cr & {\frac{{x - 2}}{{{s_b}}}} - {\frac{{x - 1}}{{{s_b}}}} \cr & \cr} $$
Now, check option one by one which gives us that option (b) is correct.
83. An ant moved for several seconds and covered 3 mm in the first second and 4 mm more in each successive second than its predecessor. If the ant had covered 1 mm in the first second and 8 mm more in each successive second, then the difference between the path it would cover during the same time and actual path would be more than 6 mm but less than 30 mm. find the time for which the ant moved (in seconds).
a) 5s
b) 4s
c) 6s
d) 2s
Discussion
Explanation: 3 + 7 + 11 + 15 + . . . . . (1)
1 + 9 + 17 + 25 + . . . . . (2)
The condition is satisfied for the 4 seconds in 2nd journey, ant covered 17m more than 1st one.
84. Two trains start from the same point simultaneously and in the same direction. The first train travels at 40 km /h, and the speed of the second train is 25% more than the speed of first train. Thirty minutes later, a third train starts from same point and in the same direction. It over takes the second train 90 minutes later than it overtook the first train. What is the speed of the third train?
a) 20 km/h
b) 40 km/h
c) 60 km/h
d) 50 km/h
Discussion
Explanation: A _______ B ________ C
Let both the trains start from point A.
At point B third train overtook the first train.
To overtake the first train by third train, Third train needs to cover,
Distance covered by first train in $$1\frac{1}{2}$$ h = distance covered by third train in 1 h.
[as third train has started 30 minutes later]
In this situation distance is constant, then
s × t = d; we get, S α $$\frac{1}{{\text{t}}}$$
Now,
$$\eqalign{ & \frac{{{\text{t}} + \frac{1}{2}}}{{\text{t}}} = \frac{{\text{s}}}{{40}} \cr & \frac{{2{\text{t}} + 1}}{{\text{t}}} = \frac{{\text{s}}}{{40}}\,.\,.\,.\,.\,.\,.\,.\,.\left( 1 \right) \cr} $$
From equation (1),
It is clear that time is in the ratio 3 : 2 then speed will be in 2 : 3 ratios.
Speed of the Third train will be 60 km/h.
85. A racetrack is in the form of a right triangle. The longer of the legs of track is 2 km more than the shorter of the legs (both these legs being on a highway). The start and end points are also connected to each other through a side road. The escort vehicle for the race took the side road and rode with a speed of 30 km/h and then covered the two intervals along the highway during the same time with a speed of 42 km/h. find the length of the race track.
a) 14 km
b) 10 km
c) 24 km
d) 36 km
Discussion
Explanation: The given conditions are met on a Pythagoras triplet 6, 8, 10.
Since, the racetrack only consists of the legs of the right angle triangle the length must be,
= 6 + 8 = 14 km.
86. An individual is cycling at a speed of 25 km per hour. He catches his predecessor who had started earlier in two hours. What is the speed of his predecessor who had started 3 hours earlier ?
a) 15 kmph
b) 12 kmph
c) 10 kmph
d) 8 kmph
Discussion
Explanation: The distance covered in two hour,
= 2 × 25 = 50 km
Time taken by first individual = (3h + 2h) = 5h
So, the speed of predecessor
= $$\frac{{50}}{5}$$ = 10 kmph.
87. A 6 cm long cigarette burns up in 15 minutes if no puff is taken.For every puff, it burns three times as fast during the duration of the puff.If the cigarette burns itself in 13 minutes, then how many puffs has the smoker taken if the average puff lasted 3 seconds?
a) 17
b) 18
c) 20
d) 22
Discussion
Explanation: Let the number of puffs,
$$3x \times \left( {3 \times \frac{1}{{150}}} \right) + \left( {13 \times 60 - 3x} \right) \times \frac{1}{{150}} = 6$$
So, x = 20 puffs.
88. An old man driving bike at 80 km per hour. However being sugar patient, old man could not travel continuously. He takes small breaks each of 2 minutes for every 15 minute of his drive. How much distance the old man will cover in 90 minutes?
a) 112 Km
b) 104 km
c) 89 km
d) 118 km
Discussion
Explanation: For every 15 minutes he takes a rest of 2 minutes.
Hence, for 90 minutes of drive he would require 12 minutes of rest.
He will be traveling for 90 - 12 = 78 minutes.
In 60 minutes he covers 80 Km.
In 1 minute he would cover $$\frac{{80}}{{60}}$$ Km.
In 78 minutes he would cover $$\frac{{80}}{{60}} \times 78$$ = 104 Km.
89. Two trains start simultaneously from two stations Howrah and Delhi, respectively towards each other on the same track. The distance between the two stations is 560 km and the speeds of trains are 30 kmph and 40 kmph.
Simultaneously with the trains, a sparrow sitting on the top of one of the train starts towards the other and reverses its direction on reaching the other train and so on. If the speed of sparrow is 80 kmph then the distance that the sparrow lies before being crushed between the train is :
a) 70 km
b) 560 km
c) 640 km
d) 650 km
Discussion
Explanation: Relative speed of the trains = 40 + 30 = 70 kmph.
Time taken by trains to collide,
= $$\frac{{560}}{{70}}$$
= 8 hours
In 8 hours sparrow will cover,
= 8 × 80
= 640 km.
90. Due to the technical snag in the signal system two trains start approaching each other on the same track from two different stations, 240 km away each other. When the train starts a bird also starts moving to and fro between the two trains at 60 kmph touching each train each time. The bird initially sitting on the top of the engine of one of the trains and it moves so till these trains collide. If these trains collide one and half hour after start, then how many kilometers bird travels till the time of collision of trains?
a) 90 km
b) 130 km
c) 120 km
d) 95 km
Discussion
Explanation: Time taken to collide the trains,
= one and half hour = $$\frac{3}{2}$$ hours
So, in $$\frac{3}{2}$$ hours bird travels,
= $$\frac{{3 \times 60}}{2}$$
= 90 km.
91. The driver of an ambulance sees a school bus 40 m ahead of him after 20 seconds, the school bus is 60 meter behind. If the speed of the ambulance is 30 km/h, what is the speed of the school bus?
a) 10 kmph
b) 12 kmph
c) 15 kmph
d) 22 kmph
Discussion
Explanation: Relative Speed,
$$\eqalign{ & = \frac{{{\text{Total }}\,{\text{distance}}}}{{{\text{Total time}}}} \cr & = \frac{{60 + 40}}{{20}} \cr & = 5\,\,{\text{m/s}} \cr & = \frac{{5 \times 18}}{5} \cr & = 18\,\,{\text{kmph}} \cr} $$
Relative Speed = (speed of ambulance - speed of school bus)
Speed of school bus = speed of ambulance - relative speed
= 30 - 18
= 12 kmph
92. A minibus takes 6 hour less to cover 1680 km distance, if its speed is increased by 14 kmph ? What is the usual time of the minibus ?
a) 15 hours
b) 24 hours
c) 25 hours
d) 30 hours
Discussion
Explanation: Let Usual Speed = x kmph
Speed of bus after increasing the speed = (x + 14)kmph
A ____________1680km ____________B
In first case, Time taken to covered the distance 1680 km = $$\frac{{1680}}{x}$$ . . . . . (1)
In Second Case, Time Taken to covered the distance 1680 km = $$\frac{{1680}}{{x + 14}}$$ . . . . . (2)
Time difference = 6 Hours.
So,
$$\eqalign{ & {\frac{{1680}}{x} - \frac{{1680}}{{ {x + 14} }}} = 6 \cr & 1680 {\frac{{14}}{{ {{x^2} + 14x} }}} = 6 \cr & 280 \times 14 = {x^2} + 14x \cr & {x^2} + 70x - 56x - 3920 = 0 \cr & x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & x = - 70,56. \cr & {\text{hence,}}\,{\text{speed}}\,{\text{of}}\,{\text{minibus}}\,{\text{is}}\,56\,km/h \cr & {\text{put}}\,x = 56\,{\text{in}}\,{\text{equation}}\,(2) \cr & T = \frac{{1680}}{{56}} \cr & \,\,\,\,\,\, = 30\,{\text{hours}} \cr} $$
93. A man reduces his speed from 20 kmph to 18 kmph. So, he takes 10 minutes more than the normal time. what is the distance traveled by him.
a) 30 km
b) 25 km
c) 50 km
d) 36 km
Discussion
Explanation: Usual time = 9 × 10 = 90 min = $$\frac{3}{2}$$ h
Distance traveled = Speed × time = $$20 \times \frac{3}{2}$$ = 30 km
94. A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speed is :
a) 1 : 16
b) 4 : 1
c) 2 : 1
d) 1 : 4
Discussion
Explanation: Let the original speed be S1 and time t1 and distance be D.
$$\eqalign{ & \frac{{ {\frac{D}{2}} }}{{2{t_1}}} = {S_2} \cr & {S_2} = \frac{D}{{4{t_1}}}\,{\text{and}}\,{S_1} = \frac{D}{{{t_1}}} \cr & \frac{{{S_1}}}{{{S_2}}} = \frac{4}{1} = 4:1 \cr} $$
95. A is twice fast as B and B is thrice fast as C. The journey covered by C in 78 minutes will be covered by A in :
a) 13 min
b) 15.5 min
c) 17 min
d) 12 min
Discussion
Explanation: The ratio of speeds of A, B, C = 6 : 3 : 1
The ratio of time taken by A, B, C = $$\frac{1}{6}$$ : $$\frac{1}{3}$$ : 1
To simplify it, we will multiply it by LCM of ratio of speeds given.
Hence, the ratio of time taken by A, B, C = 1 : 2 : 6
[Speed is inversely proportional to time, means if speed increase time decreases. So, ratio of time would be reciprocal of the ratio of speed given. ]
Time taken by C to covered given distance = 78 = 6 × 13
The ratio of time of A and C = 1 : 6
Thus, time taken by A = 13 min.
96. A man goes to the fair with his son and dog. Unfortunately man misses his son which he realizes 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to son and comes back to the man to show him the direction of his son. He keeps moving to and fro at 60 m/min between son and father, till the man meets the son. What is the distance traveled by the dog in the direction of the son?
a) 800 m
b) 1675 m
c) 848 m
d) 1000 m
Discussion
Explanation: In 20 minutes the difference between man and son,
= 20 × 20
= 400 m.
Distance traveled by dog when he goes towards the child,
= $$400 \times \frac{{60}}{{40}}$$
= 600 m and time required is 10 minutes.
In 10 min remaining distance between man and child,
= 400 - (20 × 10)
= 200 m.
Time taken by dog to meet the man,
= $$\frac{{200}}{{100}}$$
= 2 min
(Relative speed of dog with child is 40 m/min and same with man is 100 m/min.)
Remaining distance in 2 min,
= 200 - (2 × 20),
= 160 m.
Now, the time taken by dog to meet the child again,
= $$\frac{{160}}{{40}}$$
= 4 min.
In 4 min he covers = 4 × 60 = 240 m.
Now, remaining distance in 4 min = 160 - (4 × 20) = 80 m.
Time required by dog to meet the man once again = $$\frac{{80}}{{100}}$$ = 0.8 min.
Now remaining distance = 80 - (0.8 × 20) = 64 m.
Now, time taken by dog to meet the child again = (64/40) x (8/5) min
Distance travelled by dog = (8/5) x 60 = 96 m
Thus, we can observe that every next time dog just go 2/5th of previous distance to meet the child in the direction of child.
So, we can calculate the total distance covered by dog in the direction of child with the help of GP formula.
Here, first term (a) = 600 and common ratio (r) = 2/5
Sum of the infinite GP = a/(1 - r)
= 600/(1 - 2/5) = (600 x 5)/3 = 1000 m
97. A tiger is 50 of its own leaps behinds a deer. The tiger takes 5 leaps and per minutes to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?
a) 600 m
b) 700 m
c) 800 m
d) 1000 m
Discussion
Explanation: Speed of tiger = 40 m/min
Speed of deer = 20 m/min.
Relative speed = 40 - 20 = 20 m/min.
Initial difference in distance = 50 × 8 = 400 m
Time take to catch = $$\frac{{400}}{{20}}$$ = 20 min.
Distance traveled in 20 min,
= 20 × 40
= 800 m
98. A candle of 6 cm long burns at the rate of 5 cm in 5 hour and an another candle 8 cm long burns at the rate of 6 cm in 4h. What is the time required to each candle to remain of equal lengths after burning for some hours, when they starts to burn simultaneously with uniform rate of burning?
a) 1 hours
b) 1.5 hours
c) 2 hours
d) 4 hours
Discussion
Explanation: (6 - x) = (8 - 1.5x)
x = 4 cm.
It will take 4 hours to burn it in such a way that they will remain equal in lengths.
99. A man walking at the speed of 4 km/hr,cross a square field diagonally in 3 minutes.The area of the field is?
a) 20000 sqm
b) 25000 sqm
c) 18000 sqm
d) 19000 sqm
Discussion
Explanation: Speed of man,
= 4 kmph = $$\frac{{4 \times 5}}{{18}}$$ m/sec
In 3 min (180 sec) man will go = $$\frac{{20 \times 180}}{8}$$ = 200 m.
That means the diagonal of the square field = 200 m.
Diagonal of square,
= Side of Square × $$\sqrt 2 $$ = 200 m.
Side of Square = $$\frac{{200}}{{\sqrt 2 }}$$
Area of Square = Side2
Area = $$\frac{{200 \times 200}}{2}$$ = 20000 sqm.
100. A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. $$\frac{5}{{12}}$$ of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?
a) 1 : 6
b) 3 : 5
c) 6 : 1
d) 5 : 4
Discussion
Explanation:Train(T)__________ A_____5k____CAT__________B
T-------x------------------12k--------------------------
Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then
$$\eqalign{ & \Rightarrow \frac{v}{u} = \frac{x}{{5k}} = \frac{{ {x + 12k} }}{{7k}} \cr & 7x = 5\left( {x + 12k} \right) \cr & \frac{x}{k} = \frac{{30}}{1} \cr & {\text{Thus}}, \cr & \Rightarrow \frac{u}{v} = \frac{{30}}{5} = \frac{6}{1} \cr} $$
so, 6 : 1