Time and Distance Questions

Let the actual distance be t kms

Now, According to the data,

$$\begin{gathered} 5t + 860 = 6t \\ => 560t + 860 = 6t \\ => 60t + 8 = 72t \\ => 8 = 12t \\ => t = 8/12 = 2/3 hrs \end{gathered}$$

Now, the total distance = 6t = 6 x 2/3 = 2 x 2 = 4 kms.

Length of train A = 48 x 9 x 5/18 = 120 mts

Length of train B = 48 x 24 x 5/18 - 120

=> 320 - 120 = 200 mts.

As we know that Speed = Distance / Time
for average speed = Total Distance / Total Time Taken
Thus, Total Distance = 325 + 470 = 795 km
Thus, Total Speed = 7.5 hrs
Average Speed = 795/7.5 => 106 kmph.

Let the distance between home and school is 'x'.
Let actual time to reach be 't'.
Thus, x/4 = t + 4 ---- (1)
and x/6 = t - 30 -----(2)
Solving equation (1) and (2)
=> x = 98 mts.

Let the initial speed of Akhil be '4p' kmph

Then speed after decrease in speed = '3p' kmph

We know that, 

Change in speed == Change in time

According to the given data,

$$\begin{gathered} 243p - 244p = 35 - 2560 \\ => p = 12 \end{gathered}$$

Hence, the initial speed of Akhil = 4p = 4 x 12 = 48 kmph.

Let the original Speed be "s" kmph
And the usual time be "t" hrs
Given that if the bus is running at 9s/10 kmph the time is 22 hrs
=> [9s/10] x 22 = t x s
=> 99/5 = t
=> t = 19.8 hrs

Hence, if bus runs at its own speed, the time saved = 22 - 19.8 = 2.2 hrs.

Given Karthik can cover the distance of 200 kms resting 2 hrs per day in 22 days.

Let the initial speed be '1'

Hence, Time = 22(24hrs - 2)

Noe new speed = 2/3

Let the number of days he take be 'D'

Therefore, 2224-2x3200=D24-2x2250

D = 11 x 3 x 5/4 = 41.25 days. = 41 1/4 days.

Here distance d = 800 mts

speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s

time t = 800x1860x5= 48 sec.