Mu Sigma Speed and Distance Quiz-1

Let the Distance Covered by Freight train = Passenger Train = D kms
Speed of Passenger Train = 60 km/hr
Time taken by Passenger Train to Travel distance D = 4 Hrs
Time taken by Freight Train to Travel Distance D = ( 4 + 2 ) = 6 Hrs
Distance = Speed × Time
Distance Freight = Speed Freight × Time Freight ........ (i)
Distance Passenger = Speed Passenger × Time Passenger ........ (ii)
Distance Freight = Distance Passenger
⇒ Speed Freight × Time Freight = Speed Passenger × Time Passenger
⇒ Speed Freight × 6 = 60 × 4
⇒ Speed Freight = 40 km/hr

Let 10\'splace digit is x and unit\'s place digit y

First milestone: 10x+y
Second milestone: 10y+x
Third milestone: 100x+y

Since the speed is uniform so
Distance covered in first Hr = Distance covered in Second Hr
(10y+x)-(10x+y) = (100x+y)-(10y+x)
By solving,
y=6x
but since x and y are digits so the only possible combination is x=1 and y=6,
So average speed = 45 Kmph

The distance between them is 110 km.
let the train A started at 4 am at 20 kmph
and train B started at 5 am with 25kmph speed.
=>distance travelled by train A in 1 hour(4am-5am).= 20×1= 20 km
remaining distance between them is now = 110-20 = 90 km.
now train A us moving with a velocity of 20 km/h and train B is moving with 25km/h
so the relative velocity between them = Velocity of A-(- velocity of B,) ,as they are moving in opposite direction.
so relative velocity = 20-(-25)
=> 20+25 =45 km/h
time taken to meet= (distance between them)/ (relative velocity)
=> 90/45= 2 hrs
so, they will meet after two hours of staring of train B
which is 7 am.

(2*30*60)/(30+60) = 40kmph

Let the speed of car be x kmph.

Therefore, the speed of train = 1.5x kmph

Distance travelled by train and car is same. The time difference is 12.5 minutes.

(75/x) - (75/1.5x) = (125/10x60)

x = 120

Going in opposite directions, the speeds of the trains are added.
200 + 1000 = 1200 mi/hr
d = r*t
t = d/r
t = 200/1200 = 1/6 hour until the trains meet
----
d = r*t = 1500 mi/hr * 1/6 hour = 250 miles traveled by the bee

Since train covers 60 miles in 80 minutes,

The distance covered by train in 20 minutes = (20/80) × 60

= 15 miles

Now we know that speed of aeroplane is 2 times the speed of the train

Hence, it will cover twice the distance as covered by the train

= 2 × 15

= 30 miles

Distance =120 km

Speed = 60kmph

Time Taken = (Distance Travelled) / (Speed) = 120/60 = 2 hrs

For Backward Journey

Distance =120 km

Speed = 40kmph

Time Taken = (Distance Travelled) / (Speed) = 120/40 = 3 hrs

Total Distance = 120 + 120 = 240

Total Time taken = 3 + 2 = 5

Avg. Speed = (Total Distance Travelled) / (Total Time)

Avg. Speed = 240/5 = 48kmph

Length of First train = 27 × A meters

Length of Second train = 17 × B meters

Speed of first train relative to second train = (A + B) m/s

First train has to cover (27A + 17B) meters in 23 seconds with speed (A+B) m/s

⇒ (27A + 17B) ÷ (A + B) = 23

⇒ 27A + 17B = 23 × (A + B)

⇒ 4A = 6B

⇒ A/B = 6/4 = 3/2

⇒ A : B = 3 : 2

Say L is the length of the train.

Speed of the train = 45 km/hr = 45 × (5/18) m/s = 12.5 m/s

Time taken by the train to cross the platform i.e. travel the distance (L + 100) m is 60 seconds.

⇒ (L + 100)/12.5 = 60 [Distance/speed = Time]

⇒ (L + 100) = 60 × 12.5 = 750

⇒ L = 650 m

∴ Time taken to cross an electric pole = (650/12.5) seconds = 52 seconds