Unitary Method Test - 2

Distance=Speed × Time

When speed is 60 km/h

Distance travelled by car = 60 × 10

= 600 km

When speed is 80 km/h

Time taken to complete the journey = 600/80

 = 7.5 hours

When speed is 20 km/hr he reaches 15 minutes late when Speed is 50 km/hr he reaches 10 minutes earlier.

Let Distance be D and T = D/S

D/20 − D/50 = 25min

50D−20D / 100 = 25 / 60 hr

D = 25/9 km = 2.7 km

Let length of the train is X and length of the bridge is Y.

Speed of the First train = 50 km = (50 × 5)/18 = 13.88 m/sec

Trains are moving in opposite direction, so their relative speed would be,

S = 50 + 42 = 92 km/h = (92 × 5)/18 = 25.55 m/sec

train crosses another train having half of its length in 10 sec.

So, Train covered (X + X/2) = 3X/2 m in 10 sec.

Relative Speed = 3X/ (2 × 10)

3X/20 = 20/3 m/sec

X = 6.66 m/sec = 66.6 m

Length of the train = 66.6 m

Train crosses bridge in 41 sec. That is

Train needs to cover (66.6 + Y) m in 41 sec.

Speed of the train = (66.6 + Y)/41 m/sec

13.88 = (66.6 + Y)/41

569.08 = 66.6 + Y

Y = 569.08 – 66.6 = 502.48 m

Length of the bridge = 502.48 m.

Since the trains travel in the same direction relative speed of the trains

= (75 - 43) km/hr = 27 km/hr

= (27×5/18) m/sec = 15/2 m/sec

= 7.5 m/sec

Therefore, length of the faster train == Speed × Time = (7.5 × 10) m =75 m

Given swimmer speed in still water = x = 11 km/hr.

Rate of stream = y = 5 km/hr.

Therefore, speed downstream = x + y = 11 + 5 = 16 km/hr.

Distance covered = average speed × time travelled

So, according to given question.

Distance covered = 60× 3 = 180 km

Their relative speed = (30 + 6) km/h = 36 km/h = (36×5/18) m/sec

= 10m/sec

Required time = 250/10 sec = 25 sec