Two boys begin together to write out a booklet containing 817 lines. The first boy start with first line, writing at rate of 200 lines an hour and the second boys start with a last line. He writes line 817 and so on backwards proceeding at the rate of 150 lines an hour. An what line will they meet?

A has already gone 3.75 km when B starts of the remaining 48 km.
A walks 3.75 km and B walks 4.25 km in one hour in opposite direction i.e., they together pass over (3.75 + 4.25) = 8 km. in hour.

Therefore , 48 km. are passed over in 48/8= 6 hours.

? A meets B in 6 hours after B started and, therefore, they meet at a distance of (4.25 X 6 ) = 25.5km. from Q.

Let the distance between Delhi and Kanpur be y km .
Suppose the train leaving from Delhi is A and the train leaving from Kanpur B

A's speed = y/ (10 a.m - 5 a.m) = y/5 km/hr.
B's Speed = y/ (2 p.m - 7 a.m) = y/7 km/hr.

Since B starts two hours later than A, the distance already covered by A at the start of B = 2y/5 km.

Remaining distance = y- 2y/5 = 3y/5 km.

Relative speed of approach of two trains = (y/5 + y/7)
= 12y/35 km/hr.

Time taken to cover the remaining distance by both trains = (3y / 5) / (12y / 35)
= (3/5) x (35/12) = 7/4 hrs.
= 1 hr. 45 min.

? The two train will meet at (7 a.m + 1hr.45 min)
= 8.45 a.m

Let the distance between Meerut and Delhi be y km.
Average speed of the train leaving Meerut = y/4 km/hr.
Average speed of the train leaving Delhi = 2y/7 km/hr.

suppose they meet x hrs. after 6 a.m
then, xy/4 + 2y (x-2)/7 = y
? x/4 + 2x-4/7 = 1
? 15x = 44

? x = 44/15 = 2 hrs. 56 min
So, the train meet at 8:56 a.m

Let the required time = T min.
Then, distance covered in T + 11 min at 40 km/hr = distance covered in T + 5 min at 50 km/hr.
? 40 ( T + 11 ) / 60 = 50 (T + 5 ) / 60
? T = 19 min.

Suppose the man covers first distance in x hrs and second distance in y hrs.
Then, 4x + 5y = 35 and 5x + 4y = 37
Solving these equations, we get
x = 5 and y = 3

? Total time taken = 5 + 3 hrs.
= 8 hrs.

Speed of the car A = (5 / 6) x 90
= 75 km/hr.

? Required time = [ 88 / (90 +75)] x 60
= 32 minutes

The ratio of speeds = The ratio of distances, when time is constant
? The ratio of distance covered by leopard to the tiger = 12 : 25
Again, ratio of rounds made by leopard to the tiger = 12 : 25
hence, leopard makes 48 rounds, when tiger makes 100 rounds

Initial speed of police = 10 m/s
Increased speed of police = 20 m/s
Speed of thief = 15 m/s
Initial difference between thief and police = 250 m
After 5 seconds difference between thief and police = 250 - (5 x 10) = 200 m
After 10 seconds more, the difference between thief and police = 200 + (5 x 10) = 250 m.
Now, the time required by police to catch the thief = 250/5 = 50 s
Distance travelled = 50 x 20 = 1000 m
Total time = 50 + 15 = 65 sec
Total distance = 1000 + (15 x 10) = 1150 m

Let pele covers D km in 1 hours. So Maradona takes (2 h - 40 min) = 1 h 20 min to cover D km.

Let speed of Maradona and pele be M and P respectively than.
D = M x 4/3 and D = P x 1
M/P = 3/4
Again, 300/M - 300/P = 1
300/3k - 300/4k = 1
k = 25
M = 3k = 75 km/h and
P = 4k = 100 km/h

The sum of their speeds = 615/15 = 43 km/h
Notice that they are actually exchanging their speeds. Only then they can arrive at the same time at their respective destinations.It means the difference in speeds is 3 km/h.
Thus, x + ( x + 3) = 43
x = 20 and x + 3 = 23
The concept is very similar to the case when after meeting each other they returned to their own places of departure. It can be solved through option also.