Answer: 5:3

Explanation:
X ------------ Y ------------ Z
If ‘d’ is the distance between X and Y, then ‘d’ is the distance between Y and Z.
Now the total time for the batsman to row from X to Z is 4 hours. Therefore, time to row from X to Y is 2 hours.
Also the time for the boats man to row from X to Y and back is 10 hours. Hence, time required to row from Y to X is 8 hours.
If, a: speed of boats man in still water
b: speed of the river
d/(a + b) = 2; d/(a – b) = 8
2*(a + b) = 8*(a – b)
a + b = 4a – 4b
3a = 5b
a:b = 5:3

Answer: 7.2 sec

Explanation:
Speed = (Distance travelled by train) / Time
180/Time = 90 × 5/18
Time = 36 / 5 = 7 × 1/5 seconds

Answer: 45 min

Explanation:
Speed is inversely proportional to time, i.e. Speed α 1 / time
If the speed increases to 3/2 times the normal speed, then the time decreases to 2/3 times the original time
1/3 of original time = 15 min
Original time = 3 × 15 = 45 min

Answer: 4.8 km/hr

Explanation:
Upstream speed = 3/5 / 10/60 = 3/5 × 60/10 = 3.6 km/hr
Downstream speed = 3/5 × 60/6 = 6 km / hr
Speed in still water = 6 + 3.6 / 2 = 9.6 / 2 = 4.8 km/hr

Answer: 5 hrs 24 mins

Explanation:
Speed of still water = 5 km/hr
Speed with stream = 5/(45/60) = 20/3 km/hr
Speed of water current = 20/3 – 5 = 5/3 km/hr
Speed against current = 5 – 5/3 = 10/3 km/hr
Time taken = 12/(20/3) + 12/(10+3)
18+36 / 10 = 54/10 = 5.4 hrs
5 hrs and 24 mins

Answer: 3*11/12 kmph and 7/12 kmph

Explanation:
Upstream speed = d/t = 20/6 = 10/3 km/hr.
Downstream speed = 18/4 = 9/2 km/hr
Speed of boat in still water = (10/3 + 9/2) / 2
(20 + 27) / 6 × 2 = 47/12 = 3 11/12 km/hr
Speed of water current = (9/2 – 10/3)/2 = 27 – 20 / 12
7/12 km/hr

Answer: 50m and 175m

Explanation:
Let the length of the two trains be X & y m.
When the trains are travelling in opposite direction, Relative speed = 36 + 45 = 81 km/hr
81× 5/18 = 45/2 m/s
Distance covered = Sum of lengths of trains = (X + Y) m
X + Y / 10 = 45/2
X + Y = 225 => (1)
When the trains run in opposite directions,
The relative speed = 45 – 36 = 9 km/hr, ie. 9 × 5/18 = 5/2 m/s
Distance covered = X m
X/20 = 5/2
X = 50 m
Y = 225 – 50 = 175 m

Answer: 10 am

Explanation:
Let the two riders meet after x hours from the time the first rider starts
20x + 25(x-1)=110
20x + 25x – 25 = 110
4x = 135
x = 135/45 = 3 hrs
they meet at 10 am.

Answer: 75 m

Explanation:
 9 km/hr = 9 × 5/18 = 2.5 m/s and 18km/hr = 2 × 2.5 = 5 m/s
Time taken for the train to cross the first cyclist = Length/Relative speed = L/s – 2.5) = 10
L = 10S – 25 =>(1)
Time taken for the train to cross the second cyclist= Length/Relative speed
L / s – 5 = 15
L= 15S – 75 => (2)
(2) – (1)
5S -50= 0
S= 10m/s. substituting the value of S in equation (1), we have
L = 10 × 10 – 25 = 75 m

Answer: 75 min

Explanation:
On any normal day, Betty has a fixed speed , ‘s’ and fixed amount of time ‘t’ to reach her school. So the distance ‘d’ is,
D = S × T
In the given situation, betty travelled an extra 15 minutes, but at a reduced speed. So Betty covered the distance, ‘d’ at speed
S × 80/100 and in t + 15 minutes
d( s × 80/100) (t + 15)
As the distance in both the cases is the same, we can say that ,
S × t = ( s × 80/100) (t + 15)
Hence, t = 4/5t + 12
1/5t = 12 or t = 60 minutes
Hence, today she has taken t + 15 = 75 minutes