Time, Speed and Distance
Exam Duration: 45 Mins Total Questions : 25
A train running at \(\frac { 8 }{ 11 }\) of its own speed reached a place in \(5\frac { 1 }{ 2 } \) h. How much time could be saved, if the train would have run at its own speed?
- (a)
\( 2\frac { 1 }{ 2 } h\)
- (b)
2h
- (c)
\(1\frac { 1 }{ 2 } h\)
- (d)
1 h
New speed \(= \frac { 8 }{ 11 } \) of usual speed
\(\therefore \) New time = \(\frac { 11 }{ 8 } \) of usual time
So, \(\frac { 11 }{ 8 } \) of usual time = \(\frac { 11 }{ 2 } h\)
\(\Rightarrow \) Usual time = \(\left( \frac { 11\times 8 }{ 2\times 11 } \right) =4h\)
Hence, time saved = \(5\frac { 1 }{ 2 } -4=1\frac { 1 }{ 2 } h\)
Starting from his one day, a student walks at a speed of 3 Km/h and reaches his school 5 min late. Next day he increases his speed by 1 Km/h and reaches the school 3 min early. How far is the school from his house?
- (a)
1.6 km
- (b)
1.24 km
- (c)
1.36 km
- (d)
1.8 km
Let the distance be x km.
Difference in timings = 8 min = \(\frac { 8 }{ 60 } h=\frac { 2 }{ 15 } h\)
\(\therefore \quad \frac { x }{ 3 } -\frac { x }{ 4 } =\frac { 2 }{ 15 } \Rightarrow \frac { x }{ 12 } =\frac { 2 }{ 15 } \Rightarrow x=\frac { 2\times 12 }{ 15 } =1.6km\)
Excluding stoppages, the speed of a bus is 24 km/h and including stoppages, it is 20 km/h. For how many minutes does the bus stop per hour?
- (a)
8 min
- (b)
12 min
- (c)
10 min
- (d)
14 min
Due to stoppages, it covers 4 km less per hour.
Time taken to cover 4 km \(\left( \frac { 4 }{ 24 } \times 60 \right) min=10min\)
In a flight of 800 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/h and the time of flight increased by 20 min. The duration of the flight is
- (a)
\(\frac { 3 }{ 5 } h\)
- (b)
\(\frac { 1 }{ 2 } h\)
- (c)
1 h
- (d)
\(1\frac { 1 }{ 2 } h\)
Let, the duration of the flight be x hour.
Then, \(\frac { 800 }{ x } -\frac { 800 }{ x+\frac { 1 }{ 3 } } =200\)
\(\Rightarrow \quad \) \(\frac { 1 }{ x } -\frac { 3 }{ 3x+1 } =\frac { 200 }{ 800 } \quad \)
\(\Rightarrow \quad \) \(\frac { 1 }{ x(3x+1) } =\frac { 1 }{ 4 } \)
\(\Rightarrow \quad \) \(3{ x }^{ 2 }+x-4=0\)
\(\Rightarrow \quad \) \(3{ x }^{ 2 }+4x-3x-4=0\)
\(\Rightarrow \quad \) \(x(3x+4)-1(3x+4)=0\)
\(\Rightarrow \quad \) \((3x+4)(x-1)=0\)
\(\Rightarrow \quad \) x = 1h \([As\quad x=-\frac { 4 }{ 3 } is\quad not\quad possible]\)
Two cyclists start from the same place in opposite directions. One goes towards North at 12 km/h and the other goes towards South at 15 km/h. What time will they take to be 8.1 km apart?
- (a)
18 min
- (b)
17 min
- (c)
16 min
- (d)
15 min
Let, the required time be x h.
Relative speed = (12+15)km/h = 27 km/h
\(27=\frac { 8.1 }{ x } \Rightarrow x=\frac { 8.1 }{ 27 } h\\ =\left( \frac { 8.1\times 60 }{ 27 } \right) min=18min\)
A train X leaves P at 6 am and reaches Q at 10 am. Another train Y leaves Q at 8 am reaches P at 11.30 am. At what time do the two trains cross each other?
- (a)
8 : 24 am
- (b)
8 : 56 am
- (c)
9 am
- (d)
None of these
Let, the distance between P and Q be d km.
\(Speed\quad ofX=\frac { d }{ 4 } km/h\\ Speed\quad of\Upsilon =\frac { 2d }{ 7 } km/h\)|
By the time train \(\Upsilon \) leaves Q at 8 am towards P,
train X has covered \(\left( \frac { d }{ 4 } \times 2 \right) km=\frac { d }{ 2 } km\)
Remaining distance \(\left( d-\frac { d }{ 2 } \right) km=\frac { d }{ 2 } km\)
Relative speed = \(\left( \frac { d }{ 4 } +\frac { 2d }{ 7 } \right) km/h=\left( \frac { 15d }{ 28 } \right) km/h\)
Let t hours be the time after 8 am when the two trains cross each other. Then,
\(\frac { 15d }{ 28 } =\frac { { d }/{ 2 } }{ t } \)
\(\Rightarrow \quad \quad \quad \quad t=\left( \frac { d\times 28 }{ 2\times 15d } \right) h=\frac { 14 }{ 15 } h\\ \quad \quad \quad \quad \quad \quad \quad =\quad \left( \frac { 14 }{ 15 } \times 60 \right) min=56min\)
Hence, the two trains meet at 8 : 56 am.
A man takes 4 h 20 min in walking to a certain place and riding back. He would have gained 1 h 30 min by riding both ways. The time he would take to walk both ways is
- (a)
5 h 50 min
- (b)
4 h 45 min
- (c)
5h 30 min
- (d)
4 h 30 min
Let, the distance be x km. Then,
(Time taken to walk x km) + (Time taken to ride x km) = \(\frac { 13 }{ 3 } h\)
(Time taken to walk 2x km) + (Time taken to ride 2x km) = \(\frac { 26 }{ 3 } h\)
But, time taken to ride 2x km = \(\frac { 17 }{ 6 } h\)
\(\therefore \) Time taken to walk 2x km = \(\left( \frac { 26 }{ 3 } -\frac { 17 }{ 6 } \right) h=\frac { 35 }{ 6 } h\)
\(=\left( \frac { 35 }{ 6 } \times 60 \right) \) min -= 350 min = 5h 50 min
A goods train runs at the speed of 63 km/h and crosses a 250 m long bridge in 28 s, what is the length of the goods train?
- (a)
230 m
- (b)
210 m
- (c)
250 m
- (d)
240 m
Let the length of the train be x metres.
Speed of the train = \(\left( 63\times \frac { 5 }{ 18 } \right) =\frac { 35 }{ 2 } m/s\\ \)
Time taken to cross te tunnel
\(=\frac { Length\quad of\quad train+Length\quad of\quad tunnel }{ Speed\quad of\quad the\quad train } \)
\(28=\left( \frac { x+250 }{ 35/2 } \right) \\ 28=\frac { 2(x+250) }{ 35 } \Rightarrow x=240m\)
How many seconds will a 375 m long train take to cross a man walking with a speed of 2 km/h in the direction of the moving train, if the speed of the train is 56 km/h?
- (a)
25 s
- (b)
24 s
- (c)
28 s
- (d)
21 s
Speed of the train relative to the man
\(=(56-2)km/h=54km/h\\ =\left( 54\times \frac { 5 }{ 18 } \right) m/s=15m/s\)
Time taken by the train to pass the man
\(=\frac { Length\quad of\quad train }{ Relative\quad speed } =\left( \frac { 375 }{ 15 } \right) s=25\quad s\)
Two trains of length 250 m and 140 m are running on parallel lines in the same direction at 58 km/h and 32 km/h respectively. Find the time taken by the slower train to pass the driver of the faster one.
- (a)
58 s
- (b)
54 s
- (c)
52 s
- (d)
48 s
Relative speed = (58-32) km/h = 26 km/h
\(=\left( 26\times \frac { 5 }{ 18 } \right) m/s=\frac { 65 }{ 9 } m/s\)
Time taken to cross each other = \(\left[ \frac { (250+140) }{ 65/9 } \right] s\)
\(=\left( \frac { 390\times 9 }{ 65 } \right) s=54\quad s\)
Two trains travel in opposite directions at 32 km/h and 40 km/h and a man sitting in slower train passes the faster train in 6 s. The length of the faster train is
- (a)
120 m
- (b)
114 m
- (c)
124 m
- (d)
136 m
Relative speed = (32+40)m km/h = 72 km/h
\(=\left( 75\times \frac { 5 }{ 18 } \right) m/s=20m/s\)
Length of train = \((20\times 6)m=120m\)
A boat running downstream covers a distance of 12 km in 2 h while for covering the same distance upstream it takes 3 h. What is the speed of the boat in still water?
- (a)
8 km/h
- (b)
4 km/h
- (c)
5 km/h
- (d)
9 km/h
Downstream speed = \(\left( \frac { 12 }{ 2 } \right) km/h=6km/h\)
Upstream speed = \(\left( \frac { 12 }{ 3 } \right) km/h=4km/h\)
\(\therefore \) Speed in still water = \(\frac { 1 }{ 2 } (6+4)km/h=5km/h\)
A boatman goes 3 km against the current of the stream in 1 h and goes 1 km along the current in 15 min. How long will it take to go 7 km in stationary water?
- (a)
2 h 15 min
- (b)
2 h
- (c)
2 h 30 min
- (d)
2 h 45 min
Downstream speed = \(\left( \frac { 1 }{ 15 } \times 60 \right) km/h=4km/h\)
Upstream speed = 3 km/h
Speed in still water = \(\\ \frac { 1 }{ 2 } (4+3)km/h=3.5km/h\)
\(\therefore \) Required time = \(\left( \frac { 7 }{ 3.5 } \right) h=2h\)
A boat running upstream takes 5 h 12 min to cover a certain distance, while it takes 4 h to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?
- (a)
23 : 3
- (b)
21 : 2
- (c)
25 : 7
- (d)
28 : 9
Let the man's upstream speed be x km/h and that downstream be y km/h. Then,
Distance covered upstream in 5 h 12 min
= Distance covered downstream in 4 h.
\(\Rightarrow \) \(\left( x\times \frac { 26 }{ 5 } \right) =(y\times 4)\Rightarrow y=\frac { 13x }{ 10 } \)
\(\therefore \) Required ratio = \(\left( \frac { y+x }{ 2 } \right) :\left( \frac { y-x }{ 2 } \right) =(y+x):(y-x)\)
\(=\left( \frac { 13x }{ 10 } +x \right) :\left( \frac { 13x }{ 10 } -x \right) =\frac { 23 }{ 10 } :\frac { 3 }{ 10 } =23:3\)
A man can row \(9\frac { 1 }{ 2 } \) km/h in still water. If in a river which is running at \(2\frac { 1 }{ 2 } \) km/h, it takes him 1 h 35 min to row to a place and back, how far off is the place?
- (a)
8.6 km
- (b)
8 km
- (c)
8.4 km
- (d)
7 km
Downstream speed = (9.5+2.5) km/h = 12 km/h
Upstream speed = (9.5-2.5) km/h =7 km/h
Let the required distance be x km. Then,
\(\frac { x }{ 12 } +\frac { x }{ 7 } =\frac { 19 }{ 12 } \Rightarrow \frac { 19x }{ 84 } =\frac { 19 }{ 12 } \Rightarrow x=7\)
Hence, the required distance = 7 km
In a stream running at 2 km/h, a motorboat goes 5 km upstream and back again to the starting point in 1 h 20 min. Find the speed of te motor boat in still water.
- (a)
4 km/h
- (b)
8 km/h
- (c)
10 km/h
- (d)
6 km/h
A boat covers 25km upstream and 24 km downstream in 16 h while it covers 20 km upstream and 30 km downstream in \(15\frac { 1 }{ 2 } h\) . The speed of the current is
- (a)
\(\frac { 1 }{ 2 } km/h\)
- (b)
\(\frac { 3 }{ 4 } km/h\)
- (c)
\(\frac { 1 }{ 8 } km/h\)
- (d)
\(\frac { 2 }{ 3 } km/h\)
Let, the upstream speed = x km/h and
Downstream speed = y km/h
Then, \(\frac { 25 }{ x } +\frac { 24 }{ y } =16\) ...(i)
and \(\frac { 20 }{ x } +\frac { 30 }{ y } =\frac { 31 }{ 2 } \) ...(ii)
Multiplying Eq. (i) by 4 and Eq. (ii) by 5 and subtracting, we get
\(\frac { 54 }{ y } =\frac { 27 }{ 2 } \Rightarrow y=\left( \frac { 54\times 2 }{ 27 } \right) =4\)
Putting y=4 in Eq. (i) we get x = 2.5
Upstream speed = 4 km/h and downstream speed
= 2.5 km/h
Hence, speed of current = \(\frac { 1 }{ 2 } (4-2.5)=\frac { 3 }{ 4 } km/h\)
A worker may claim Rs 15 each km which he travels by taxi and Rs 5 for each km which he drives his own car. If in one week he claimed Rs 500 for travelling 80 km, how many km did he travel by taxi?
- (a)
10
- (b)
20
- (c)
30
- (d)
80
Let the distance covered by taxi = x km
Let the distance covered by car = y km
So, \(x+y\quad =\quad 80\) ...(i)
\(15x+5y\quad =\quad 500\) ...(ii)
Solving both the equations x = 10 km
A boat travels upstream from Q to P and downstream from p to Q in 3 h. If the distance between P to Q is 4 km and the speed of the current is 1 km/h , then what is the speed of the boat in still water?
- (a)
1.5 km/h
- (b)
5.2 km/h
- (c)
4.5 km/h
- (d)
3 km/h
Let speed of boat in still water = x km/h
Upstream \(=\frac { 4 }{ x-1 } \) and Downstream \(=\frac { 4 }{ x+1 } \)
So, \(\frac { 4 }{ x-1 } +\frac { 4 }{ x-+1 } =3\Rightarrow x=3km/h\)
A, B and C are three participants in a kilometer race. If A can give B a start of 40 m and B can give C a start of 25 m, how many meters of a start can A give to C?
- (a)
60 m
- (b)
64 m
- (c)
62 m
- (d)
66 m
Ratio of speed of A : B = 1000 : 960
Ratio of speed of B : C = 1000 : 975
\(\therefore \) If B travels 1000 m, C travels 975 m and if B travels 960 m, C will travels \(\frac { 975 }{ 1000 } \times 960=936m\)
\(\therefore \) A can give C start of (1000-936) = 64 m
A and B run a 5 km race on a round course of 400 m. If their speeds be in the ratio 5:4 , how often does the winner pass the other?
- (a)
\(4\frac { 1 }{ 2 } times\)
- (b)
\(2\frac { 3 }{ 4 } times\)
- (c)
\(3\frac { 1 }{ 2 } times\)
- (d)
\(2\frac { 1 }{ 2 } times\)
It is clear from the question that When A covers 500m, B covers 400m i.e., A takes a lead of 100 m in every 500 m of distance. Therefore, a lead of 400 m will be taken in travelling a distance of 200 m.
Or in other words A passes B every after 2000 m.
Hence, total number of such pass = \(\frac { 5000 }{ 2000 } =\frac { 5 }{ 2 } \)
\(=2\frac { 1 }{ 2 } lines\)
A ship 55 km from the shore springs a leak which admits 2 ton of water in 6 min; 80 ton would suffice to sink her, but the pumps can throws out 12 ton an hour. The average rate of sailing that she may just reach the shore as she begins to sink is
- (a)
9.17 km/h
- (b)
0.97 km/h
- (c)
55 km/h
- (d)
5.5 km/h
In 1 h water entered into ship = (20-12) = 8 ton
Now, it will take 10 h to allow to enter 50 tonnes of water into ship and in this time ship has to cover 55 km of distance.
Hence, required speed = 5.5 km/h
In case of 200m, A beats S by 20 m and N by 40 m. If S and N are running a race of 100m with exactly same speed as befoe, then by how many metres will S beat N?
- (a)
11.11 m
- (b)
10 m
- (c)
12 m
- (d)
25 m
Let the speeds of S and N be x and y respectively.
\(\therefore \\ \) \(\frac { 180 }{ x } =\frac { 160 }{ y } \Rightarrow \frac { x }{ y } =\frac { 180 }{ 160 } =\frac { 9 }{ 8 } \)
Let S beats N by d metre.
\(\therefore \\ \) \(\frac { 100 }{ x } =\frac { 100-d }{ y } \Rightarrow \frac { x }{ y } =\frac { 100 }{ 100-d } \)
\(\frac { 100 }{ 100-d } =\frac { 9 }{ 8 } \)
\(900-9d=800\Rightarrow d=11.11m\)
A man can walk up moving "up" escalator in 30 s. The same man can walk down this moving "up" escalator in 90 s. Assume that his walking speed is same upwards and downwards. How much time will he take to walk up the escalator, when it is not moving?
- (a)
30 s
- (b)
45 s
- (c)
60 s
- (d)
90 s
Let his walking speed be S km/h and speed of escalator be e km/h.
The distance between the top and bottom be d.
According to the question,
\(\frac { d }{ S+e } =30s\quad and\frac { d }{ S-e } =90s\)
\(\Rightarrow \) 30S + 30e = 90S - 90e
\(\Rightarrow \) -60S = -120 e, S=2e
Time taken to travel up the escalator alone is 45 s.
A man travelled 5 miles in the second hour of his trip. This was 1/4th more than he travelled in the first hour. In the third hour, he travelled 1/5 th more than he did in the second. How far he travels in the three hours?
- (a)
6 miles
- (b)
20 miles
- (c)
13-4/5 miles
- (d)
15 miles
Distance is 2 nd hour = 5 miles.
Distance in 1st hour let be n miles.
\(\therefore \quad n+\frac { 1 }{ 4 } n=5\quad \Rightarrow n=4miles\)
Then, in 3rd hour = 5 miles+ \(\frac { 1 }{ 5 } (5)=6miles\)
Total = 5 + 4 + 6 = 15 miles.