1. 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
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
2. 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
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.
3. 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
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
4. 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
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
5. 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
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)
$$\eqalign{ & \frac{{ {\frac{{3S}}{2}} }}{{\left( S \right)}} = \frac{{ {T + {\frac{5}{{12}}} } }}{T} \cr & 3T = 2T + \frac{{10}}{{12}} \cr & T = \frac{{10}}{{12}}{\text{hours}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{car}} \cr & \frac{{3S}}{2} = \frac{{100}}{{ {\frac{{10}}{{12}}} }} \cr & \frac{{3S}}{2} = 120 \cr & S = 80\,kmph \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{bike}} = 80\,kmph \cr} $$
6. 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
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
7. 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
Explanation: Let length of each train be X m, then
$$\eqalign{ & {S_1} = \frac{X}{4} \cr & {S_2} = \frac{X}{5} \cr & {\text{Required time to cross each other,}} \cr & = {\frac{{2X}}{{ { {\frac{X}{4}} + {\frac{X}{5}} } }}} \cr & = \frac{{40}}{9}{\text{seconds}} \cr} $$
8. 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
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
9. 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
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
10. 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
Explanation: let required speed be S km/h
50 × 30 = S × 20
S = 75 km/h