Time Speed Distance

Speed Time And Distance

Question 1
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 cars is:
A
3 : 4
B
4 : 3
C
2 : 3
D
3 : 2
Question 1 Explanation: 
\begin{align} & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}} \cr & {\text{and}}\,{\text{that}}\,{\text{of}}\,{\text{the}}\,{\text{car}}\,{\text{be}}\,y\,{\text{km/hr}} \cr & {\text{Then}},\,\frac{{120}}{x} + \frac{{480}}{y} = 8 \cr & \Rightarrow \frac{1}{x} + \frac{4}{y} = \frac{1}{{15}}\,.....\left( i \right) \cr & {\text{and}},\,\frac{{200}}{x} + \frac{{400}}{y} = \frac{{25}}{3} \cr & \Rightarrow \frac{1}{x} + \frac{2}{y} = \frac{1}{{24}}\,.....\left( {ii} \right) \cr & {\text{Solving}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right){\text{,}}\, \cr & {\text{we}}\,{\text{get}},\,x = 60\,{\text{and}}\,y = 80 \cr & \therefore {\text{Ratio}}\,{\text{of}}\,{\text{speeds}} \cr & = 60:80 = 3:4 \cr\end{align}






Question 2
A car after traveling 18 km from a point A developed some problem in the engine and the speed became 4/5th 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
B
35
C
20
D
30
Question 2 Explanation: 
He proceeds at 4/5 S where S is his usual speed means 1/5 decrease in speed which will lead to 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

Thus,

1/4 * T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km

T = 36 min = 36/60 hours = 0.6 hours.

Speed of the car = 12/0.6 = 20 kmph.

Question 3
A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
A
15 km
B
16 km
C
14 km
D
17 km
Question 3 Explanation: 
\begin{align} & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{foot}}\,{\text{be}}\,x\,km \cr & {\text{Then,}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{bicycle}} = \left( {61 - x} \right)km \cr & {\text{So}},\,\frac{x}{4} + \frac{{\left( {61 - x} \right)}}{9} = 9 \cr & \Rightarrow 9x + 4\left( {61 - x} \right) = 9 \times 36 \cr & \Rightarrow 5x = 80 \cr & \Rightarrow x = 16\,km \cr\end{align}






Question 4
Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
A
12 kmph
B
11 kmph
C
14 kmph
D
8 kmph
Question 4 Explanation: 
\begin{align} & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{by}}\,x\,{\text{km}} \cr & {\text{Then}},\,\frac{x}{{10}} - \frac{x}{{15}} = 2 \cr & \Rightarrow 3x - 2x = 60 \cr & \Rightarrow x = 60\,km \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{travel}}\,60\,km\,{\text{at}}\,10\,{\text{km/hr}} \cr & = \left( {\frac{{60}}{{10}}} \right)\,hrs = 6\,hrs \cr & {\text{So,}}\,{\text{Robert}}\,{\text{started}}\,{\text{6}}\,{\text{hours}}\,{\text{before}}\,2\,P.M.\,i.e.,\,at\,A.M. \cr & \therefore {\text{Required}}\,{\text{speed}} \cr & = \left( {\frac{{60}}{5}} \right)\,kmph = 12\,kmph \cr\end{align}
Question 5
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
624 km
B
600 km
C
584 km
D
684 km
Question 5 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 = X/30.

Time taken to cover X km at 40 km/h = X/40

Now, \begin{align} & \left( {\frac{X}{{30}}} \right) + \left( {\frac{X}{{40}}} \right) = 17.5 \cr & \left[ {\frac{{\left( {40X + 30X} \right)}}{{1200}}} \right] = 17.5 \cr & 70X = 17.5 \times 1200 \cr & X = 300\,km \cr & {\text{Total}}\,{\text{distance}},\,2x \cr & = 600\,km \cr\end{align}







Question 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
7 seconds
C
5 seconds
D
6 seconds
Question 6 Explanation: 
Length of the 1st train = 105 m;

Length of the 2nd train = 90 m.

Relative speed of the trains,

= 45+72 = 117kmph

= 117*5/18 = 32.5 m/sec;

Time taken to cross each other,

= (Length of 1st train + length of 2nd train) /relative speed of the trains.

Time taken = 195/32.5 = 6 secs.

Question 7
A man goes to the fair with his 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
1000 m
B
848 m
C
1675 m
D
800 m
Question 7 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 * 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,

= 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,

= 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 = 80/100 = 0.8 min.

Now remaining distance = 80 - (0.8 * 20) = 64 m.







Question 8
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
30 h
B
24 h
C
25 h
D
15 h
Question 8 Explanation: 

Let

Usual Speed = X kmph.

Now,

Speed of bus after increasing the speed = (X +14)kmph. -------------- (1)

A .....1680km .....B In first case,

Time taken to covered the distance 1680 km = 1680/X ----------- (2)

In Second Case,

Time Taken to covered the distance 1680 km = 1680/(X + 14).

Time difference = 6 Hours.

So,

\begin{align} & \Rightarrow \left[ {\frac{{1680}}{x} - \frac{{1680}}{{\left( {x + 14} \right)}}} \right] = 6 \cr & \Rightarrow 1680\left[ {\frac{{14}}{{\left( {{x^2} + 14x} \right)}}} \right] = 6 \cr & \Rightarrow 280 \times 14 = {x^2} + 14x \cr & \Rightarrow {x^2} + 70x - 56x - 3920 = 0 \cr & \Rightarrow x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & \Rightarrow 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\end{align}

Question 9
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
283.33 km
C
260 km
D
275 km
Question 9 Explanation: 
Difference in time of departure between two trains = 45 min. = 45/60 hour = 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

(t+3/4)

= ((4t+3)/4);

Now,

100 * (4t+3)/4 = 136t;

Or, 25(4t+3) = 136t;

Or, 100t+75 = 136t;

Or, 36t = 75;

Or, t = 75/36 = 2.08 hours;

Then, distance x km = 136 * 2.084 = 283.33 km.







Question 10
In a race of 1000 m, A can beat B by 100m. In a 400m, B beats C by 40 m. In a race of 500m. A will beat C by
A
60 m
B
45 m
C
95 m
D
50 m
Question 10 Explanation: 
When A runs 1000 m, B runs 900 m.

Hence, when A runs 500 m, B runs 450 m.

Again, when B runs 400 m, C runs 360 m.

And, when B runs 450 m, C runs = 360 * 450/400 = 405 m.

Required distance = 500 - 405 = 95 meter.

That means when A runs 500 meter then B can run 450m then C runs 405 m.

There are 10 questions to complete.

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