Time and Distance Questions

We know that

Time = Distance/speed
Required time = (10 1/4)/2 + (10 1/4)/6
= 41/8 + 41/6
= 287/24 = 11.9 hours.

Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get:

x = 40 km.

Substitute sensible numbers and work out the problem. Then change the numbers back to letters. For example if the machine puts 6 caps on bottles in 2 minutes, it will put 6 /2 caps on per minute, or (6 /2) x 60 caps per hour. Putting letters back this is 60c/m.If you divide the required number of caps (b) by the caps per hour you will get time taken in hours. This gives bm/60c

Given the ratio of the circumference of front wheel  is 40 inches and back wheel is 70 inches. 

Distance covered = Circumference of the wheel × No. of Revolutions made by the wheel

If n is the number of revolutions made by back wheel, the number of revolutions made by front wheel is n+15

Distance covered by both the wheels is the same

i.e, 40*(n+15)=70n

 n=20 

                Front wheel      :      Back Wheel 

 Circumference       40      :      70 

 Revolutions           35      :      20

Distance covered  :  40×35 = 70×20 = 1400 inches.

Distance travelled by Ramu = 45 x 4 = 180 km

Somu travelled the same distance in 6 hours.

His speed = 180/6 = 30 km/hr

Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.

Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively.

Hence difference in the time taken = 0.5 hours = 30 minutes.

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.

Let the total distance be 'x' km.

Time taken to cover remaining 40% of x distance is   t1=40×x100×48

But given time taken to cover first 60% of x distance is   t2=t1+2060hrs

 $$\begin{gathered} t2=60×x100×48 \\ \end{gathered}$$

∴60×x100×48= 40×x100×48+2060

 x=80 km.

Let, Distance between A and B = d

Distance travelled by P while it meets Q = d + 11

Distance travelled by Q while it meets P = d – 11

Distance travelled by Q while it meets R = d + 9

Distance travelled by R while it meets Q = d – 9

Here the ratio of speeds of P & Q => SP : SQ = d + 11 : d – 11

The ratio of speeds of Q & R => SQ : SR = d + 9 : d – 9

But given Ratio of speeds of P & R => P : R = 3 : 2

 SPSR = SPSQxSQSR = d+11d+9d-11d-9

=> d+11d+9d-11d-9 = 3/2

=>  d = 1, 99

=> d = 99 satisfies.

Therefore, Distance between A and B = 99