Time Speed and Distance Questions Answers

$begin{aligned} text{Total time taken for the journey }&= frac{120}{80} + frac{40}{40} &= 2.5 text{ hours} end{aligned}$
Average speed $= 160 times frac{2}{5} = 64$ km per hour

If he is away from $x$ days he spend $5x$ paise each day.
$therefore x times 5x=9243 text{ or } x =43$

Up rate = 13 + 5 = 2.6 km per hour

Down rate = 28 + 5 = 5.6 km per hour

$therefore$ Speed of the current $= frac{1}{2} (5.6 - 2.6)$
=1.5 km p.h.

Walking $frac{3}{4}$ of original speed means taking $frac{4}{3}$ of original time. Let the original time be t.

$therefore frac{4}{3}t = t + 1frac{1}{2} text{ hrs.}$

$therefore frac{1}{3}t = 1frac{1}{2} text{ hrs. or } t = 3 times 1frac{1}{2} text{ , } i.e., 4frac{1}{2} text{ hrs.}$

The train starting at 9 a.m. has travelled by 2 p.m. a distance of $5 times 25$ km i.e., 125 km

Relative speed of two trains = 35 - 25   i.e., 10 km per hour

The second train shall meet the first train in $frac{125}{10}$ i.e.,  12.5 hrs

Distance from Aligarh = 12.5 $times$ 35 =  437.5  km

Relative speed = 56 - 29 = 27 km p.h.

Length of the train $= 16 times Big(27 times frac{5}{18} Big) = 120 text{ m}$

Time gained $= 30 div 40 = 70 text{ min} = frac{70}{60} text{ hrs.}$

Let $x$ be the distance

$therefore dfrac{x}{3} - dfrac{x}{4} 70 times dfrac{1}{60} text{ or } x = 14 text { km }$

Distance travelled

$= 10 Big(frac{2 times 21 times 24} {21+24} Big) = 224 text{ km.}$

Total length to be crossed by the train
= 110 + 132 = 242 m

$text{Time} = frac{242}{Big(36 times frac{5}{18}Big)} text {seconds} = 24.2 text{ seconds}$

Time to cover (121 + 99), i.e., 220 meteres at the rate of (32 + 40), i.e., 72 km. p.h.

$= dfrac{220}{72 times frac{5}{18}} text{ seconds} = 11 text{ seconds}$