Answer: Option C. -> 120 km/hr
Suppose the three trains meet x hours after 9 pm
Let the speed of train C be y km/hr
Distance travelled by train A in (x + 3) hrs = Distance travelled by train B in x hrs
⇒ 60 (x + 3) = 90x
⇒ 30x = 180
⇒ x = 6
⇒ Also, distance travelled by train B in x hrs + distance travelled by train C in x hrs = 1260 km
⇒ 90x + yx = 1260
⇒ 540 + 6y = 1260
⇒ 6y = 720
⇒ y = 120
Hence, speed of train C = 120 km/hr

Answer: Option A. -> 0.03 sec
Error = Time taken to cover 10 m at 300 m/sec
    = $$\frac{10}{300}$$ sec
    = $$\frac{1}{30}$$ sec
    = 0.03 sec

Answer: Option B. -> 100 km/hr, 80 km/hr
Let the speeds of the train and the car be x km/hr and y km/hr respectively.
Then,
$$\eqalign{
& \Rightarrow \frac{{160}}{x} + \frac{{600}}{y} = 8 \cr
& \Rightarrow \frac{{20}}{x} + \frac{{75}}{y} = 1.....(i) \cr} $$
And,
$$\eqalign{
& \Rightarrow \frac{{240}}{x} + \frac{{520}}{y} = 8\frac{1}{5} \cr
& \Rightarrow \frac{{240}}{x} + \frac{{520}}{y} = \frac{{41}}{5}.....(ii) \cr} $$
Multiplying (i) by 12 and subtracting (ii) from it, we get :
$$\eqalign{
& \Rightarrow \frac{{380}}{y} = 12 - \frac{{41}}{5} \cr
& \Rightarrow \frac{{380}}{y} = \frac{{19}}{5} \cr
& \Rightarrow y = \left( {380 \times \frac{5}{{19}}} \right) \cr
& \Rightarrow y = 100 \cr} $$
Putting y = 100 in equation (i), we get :
$$\eqalign{
& \Rightarrow \frac{{20}}{x} + \frac{3}{4} = 1 \cr
& \Rightarrow \frac{{20}}{x} = \frac{1}{4} \cr
& \Rightarrow x = 80 \cr} $$
∴ 100 km/hr, 80 km/hr

Answer: Option B. -> 6 hours
Speed of Karan = 60 kmph
Time = 9 hrs
Distance = Speed × Time
Distance = 60 × 9
Distance = 540 km
∴ Time taken to cover 540 km at 90 km/ph
= $$\frac{540}{90}$$ hours
= 6 hours

Answer: Option D. -> 24 km
Let the distance covered by man be x km
Journey covered by rail = $$\frac{2x}{15}$$
Journey covered by bus = $$\frac{9x}{20}$$
Remaining covered by cycle = 10 km
$$\eqalign{
& \therefore x\left( {1 - \frac{2}{{15}} - \frac{9}{{20}}} \right) = 10 \cr
& \Rightarrow x\left( {\frac{{60 - 8 - 27}}{{60}}} \right) = 10 \cr
& \Rightarrow x\left( {\frac{{25}}{{60}}} \right) = 10 \cr
& \Rightarrow x = \frac{{10 \times 60}}{{25}} \cr
& \Rightarrow x = 24{\text{ km}} \cr} $$

Answer: Option B. -> 132 km
Suppose the trains meet after x hrs
Then,
Distance covered by 1st train in x hrs + Distance covered by 2nd train in $$\left( {x - \frac{3}{4}} \right)$$  hrs = 232 km
$$\eqalign{
& \Rightarrow 48x + 50\left( {x - \frac{3}{4}} \right) = 232 \cr
& \Rightarrow 98x = 232 + \frac{{75}}{2} \cr
& \Rightarrow 98x = \frac{{539}}{2} \cr
& \Rightarrow x = \frac{{539}}{{196}}{\text{ hrs}} \cr} $$
Required distance = Distance travelled by 1st train in $$\left( {\frac{{539}}{{196}}} \right)$$  hrs
$$\eqalign{
& = \left( {48 \times \frac{{539}}{{196}}} \right){\text{km}} \cr
& = 132{\text{ km}} \cr} $$

Answer: Option B. -> 50 km/hr
Actual time taken for the journey :
= 4 hrs 15 min - (10 + 2 × 5 + 3) min
= 4 hrs 15 min - 23 min
= 3 hrs 52 min
= $$3\frac{26}{30}$$ hrs
= $$\frac{116}{30}$$ hrs
∴ Average speed :
$$\eqalign{
& = \left( {\frac{{580}}{3} \times \frac{{30}}{{116}}} \right){\text{ km/hr}} \cr
& = 50{\text{ km/hr}} \cr} $$

Answer: Option B. -> 7t
Distance covered by first person in time t = $$\frac{2}{8}$$ round = $$\frac{1}{4}$$ round
Distance covered by second person in time t = $$\frac{3}{8}$$ round
Speed of the first person = $$\frac{1}{4t}$$
Speed of the second person = $$\frac{3}{8t}$$
Since the two persons start from A and B respectively, so they shall meet each other when there is a difference of $$\frac{7}{8}$$ round between the two.
Relative speed of A and B :
$$\eqalign{
& = \left( {\frac{3}{{8t}} - \frac{1}{{4t}}} \right) \cr
& = \frac{1}{{8t}} \cr} $$
Time taken to cover $$\frac{7}{8}$$ round at this speed :
$$\eqalign{
& = \left( {\frac{7}{8} \times 8t} \right) \cr
& = 7t \cr} $$

Answer: Option D. -> $$\frac{9}{7}x$$
Let the speed of train Z be z mph
Distance travelled by train X in 1 hr = x miles
Relative speed of train Z w.r.t. train X = (z - x) mph
To catch train X at 5.30 am, train Z will have to cover x miles at relative speed in 3 hr 30 min, i.e., $$\frac{7}{2}$$ hrs
$$\eqalign{
& \therefore \left( {z - x} \right) \times \frac{7}{2} = x \cr
& \Rightarrow \frac{7}{2}z = \frac{9}{2}x \cr
& \Rightarrow z = \left( {\frac{9}{2} \times \frac{2}{7}} \right)x \cr
& \Rightarrow z = \frac{9}{7}x \cr} $$

Answer: Option B. -> 72 km/hr
1 m/sec = $$\frac{18}{5}$$ km/hr
Car cover 20 metres in a second
∴ 20 m/sec = $$\frac{20 ×18}{5}$$  = 72 km/hr