Answer: Option C. -> 25 m/s
1 km/hr = $$\frac{5}{18}$$
∴ 90 km/hr
= 90 × $$\frac{5}{18}$$
= 25 m/s

Answer: Option C. -> 3.750 km
Relative speed :
$$\eqalign{
& = \left( {40 - 45} \right) \times \frac{5}{{18}} \cr
& = \frac{{25}}{{18}}{\text{ m/s}} \cr} $$
∴ Required distance :
$$\eqalign{
& {\text{ = }}\frac{{25}}{{18}} \times 45 \times 60 \cr
& = 3750{\text{ metres or 3}}{\text{.75 km}} \cr} $$

Answer: Option C. -> 78 km/hr
Given, Ratio of speed of trains = 6 : 7
Second train covers 364 kms in 4 hours
Then, its speed = $$\frac{364}{4}$$ = 91 km/hr
In the question it is given that speed of the second train = 7 units
But actual speed = 91 km/hr
i.e., 7 units → 91 km
1 unit → 13 km
Therefore,
Speed of the first train is :
= 6R
= 6 × 13
= 78 km/hr

Answer: Option C. -> 3.3 km
Distance between the fort and the man :
= 330 × 10
= 3300 m
= 3.3 km

Answer: Option B. -> 2.005 km
We know, 1 km = 1000 metres
⇒ 2 km 5 metres = 2 km + $$\frac{5}{1000}$$  km
= 2 km + 0.005 km
= 2.005 km

Answer: Option D. -> 4 km/hr
$$\eqalign{
& {\text{Average speed :}} \cr
& {\text{ = }}\frac{{2xy}}{{x + y}} \cr
& = \frac{{2 \times 16 \times y}}{{16 + y}} \cr
& \Rightarrow 6.4 = \frac{{32y}}{{16 + y}} \cr
& \Rightarrow 102.4 + 6.4 = 32y \cr
& \Rightarrow 102.4 = 25.6y \cr
& \Rightarrow y = \frac{{102.4}}{{25.6}} \cr
& \Rightarrow y = 4{\text{ km/hr}} \cr} $$

Answer: Option B. -> 20 min
Their relative speed in same direction
= (90 - 75) kmph
= 15 kmph
Time taken by jeep to catch car :
$$\eqalign{
& = \frac{5}{15} {\text{ hour}} \cr
& = \frac{1}{3} {\text{ hour}} \cr
& = \frac{1}{3} \times 60 {\text{ min}} \cr
& = 20 {\text{ min}} \cr}$$

Answer: Option C. -> 1 : 4
Let the distance covered by the cyclist be x and the time taken be y
Then, required ratio :
$$\eqalign{
& = \frac{{\frac{1}{2}x}}{{2y}}:\frac{x}{y} \cr
& = \frac{1}{4}:1 \cr
& = 1:4 \cr} $$

Answer: Option C. -> 5 pm
Suppose the thief is overtaken x hrs after 2.30 pm
The, distance covered by the thief in x hrs = Distance covered by owner in $$\left( {x - \frac{1}{2}} \right)$$  hrs
$$\eqalign{
& \therefore 60x = 75\left( {x - \frac{1}{2}} \right) \cr
& \Leftrightarrow 15x = \frac{{75}}{2} \cr
& \Leftrightarrow x = \frac{5}{2}{\text{ hrs}} \cr} $$
So, the theif is overtaken at 5 pm

Answer: Option A. -> $$\frac{{r + t}}{{r - t}}$$
Let the speed of the faster and slower cyclists be x km/hr and y km/hr respectively
Then,
$$\eqalign{
& \frac{k}{{x - y}} = r \cr
& \left( {x - y} \right)r = k.....(i) \cr
& {\text{And,}} \cr
& \frac{k}{{x + y}} = t \cr
& \left( {x + y} \right)t = k.....(ii) \cr} $$
From (i) and (ii), we have :
$$\eqalign{
& \Rightarrow \left( {x - y} \right)r = \left( {x + y} \right)t \cr
& \Rightarrow xr - yr = xt + yt \cr
& \Rightarrow xr - xt = yr + yt \cr
& \Rightarrow x\left( {r - t} \right) = y\left( {r + t} \right) \cr
& \Rightarrow \frac{x}{y} = \frac{{r + t}}{{r - t}} \cr} $$