A man decides to travel 80 km in 8 h party by foot and party on a bicycle. If his speed on foot is 8 km/h and on bicycle 16 km/h, what distance would he travel on foot?

x/20 + y/70 = (x + y)/50
? (70x + 20y)/1400 = (x + y)/50
? 42x = 8y
? x/y = 4/21

Let the time taken in first third part of the journey be x minutes, then the time required in second third part of the journey is 3x/2 and in the last third part of the journey time required is 15x/8.
Therefore, x + (3x/2) + (15x/8) = 350min
? x = 80 min

Speed of wind (sound) / Relative speed of soldier and terrorist = Time utilised / Difference in time
1188/x = 330/5
x = 18 km/h and solder is faster.

In case of increasing gap between two objects.
Speed of sound / speed of tiger = Time utilised / Difference in time
? 1195.2 / x = 83 / 7
x = 100.8 km/h

In 20 minutes the difference between man and his son = 20 x 20 = 400 m
Distance travelled by dog when he goes towards son = (400/40) x 60 = 600 m and time required is 10 minutes
In 10 minutes the remaining difference between man and son = 400 - (20 x 10) = 200 m

Note: Relative speed of dog with child is 40 km/h and the same with man is 100 km/h.
Time taken by dog to meet the man = 200/100 = 2 min
In 2 minute the remaining distance between child and man = 200 - ( 2 x 20) = 160 m
Now, the time taken by dog to meet the child again = 160/40 = 4 min
In 4 minutes he covers 4 x 60 = 240 m distance while going towards the son.
In 4 minute the remaining distance between man and child = 160 - (4 x 20) = 80 m
Time required by dog to meet man once again = 80/100 = 0.8 min
In 0.8 min remaining distance between man and child = 80 - (0.8 x 20) = 64 min
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

Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20)
= (4 x 120)/12 = 40 km/h

Let two trains meet at a km from 'X'
[Time taken by M to cover (450 - a)km] - [Time taken by the L to cover a km ] = 40/60
? (450 - a)/80 - a/60 = 40/60
? (450 - a)/80 = 40/60 + a/60
? (450 - a)/8 - (a + 40) / 6 = 0
? 3(450 - a) - 4(a + 40) = 0
? 7a = 1190
? a = 1190 / 7 = 170
Time taken by L to cover 170 km = 170/60 h = 2 h 50 min
So, the two trains will meet 2 h 50 min after 6:00 pm. it means that the two trains will meet at 8 : 50 pm .

Let speed of A = 5x
Then, spend of B = 6x
Given that,speed of B = 6x = 90
? x = 90/6 = 15
? Speed of A = 5x = 5x = 5 x 15 = 75 km/h
Let A and B meet after T h. Then,
75 x T + 90 x T =88
? T = 88/165 h = (88 x 60)/165 = 32 min

Required number of hours is the number of terms of the series 40 + 45 + 50 +... as speed increases every hour.
Given, sum of the series is 385
? a = 40, d=5, s = 385 and n= ?
Using S = n/2[2a + (n - 1)d]
? 385 = n/2[80 + 5n - 5]
? 770 = 5n² + 75n
? n² + 15n - 154 = 0
? n² + 22n - 7n - 154 = 0
? n(n + 22) - 7(n + 22) = 0
? (n + 22) (n - 7) = 0
? n = 7
Remaining part = 1/12 - 1/20 = (5 - 3)/60 = 2/60 = 1/30 i.e., 1/30 th part is filled by B in 1 min
Hence, required time to fill the whole tank = (165 + 1 + 1) min = 167 min

Time taken by express train to cover 75 km including stoppage
= (60 / 100 x 75) min + = 48 min
Time taken by express train to cover 600 km.
= Time taken by it to cover 75 km
= (48 / 75 x 525 ) min + (60 / 100 x 75 ) min
= (336 + 45) min = 381
Time taken by local train to9 cover 25 km including stoppage
= (60 / 50 x 25) min + 1 min = 31 min
In 31 min, distance covered = 25 km
In(31 x 12) min. distance covered
= (25 / 31 x 31 x 12 ) = 300 km
In last 9 min, distance covered
= (25 / 31 x 9 ) = 7.25 km
? Required total distance
= 300 + 7.25 = 307.25 km