Unitary Method Test - 5

Since the trains travel in the same directionrelative speed of the trains

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

=$\left(27\times \frac{5}{18}\right)m/sec=\frac{15}{2}m/sec$

= 7.5 m/sec

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

Therefore, option d is correct.

Speed of train = 150km/hr = 150×$\frac{5}{18}=\frac{125}{3}$m/sec

Time = distance / speed

30=$\frac{350+x}{\frac{125}{3}}$

x= 900m

Speed of train = 900/ 250 = 2.5 m/s

Therefore, option b is correct.

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.

Therefore, option b is correct.

Let the speed of train Y be x km/hr

Speed of X relative to Y = (100 – x) km/hr

=[(100-x)×$\frac{5}{18}$]m/sec

= 500 - $\frac{5x}{18}=100$

$=500-\frac{5x}{18}=100$$=\frac{500}{\frac{500-5x}{18}}=100$

= 9000 = 100(500 – 5x)

= 9000 = 50000 – 500x

= 50000 – 9000 = 500x

= 500x = 41000

X = 82km/hr

Therefore, option d is correct.

Distance covered = average speed × timetravelled

So, according to given question.

Distance covered = 60  3 = 180 km

Therefore, option b is correct.

Their relative speed = (30 + 6) km/h = 36 km/h= (36×$\frac{5}{18}$) m/sec

= 10m/sec

Required time = $\frac{250}{10}sec= 25 sec$ 

Therefore, option b is correct.

Let S₁ be speedof A and S₂ be speed of B also S₁> S₂ 

$\therefore 1 \times S_1 + 1 \times S_2 = 35 km ... \left(1\right) (When travels in opposite direction)$

S₁ + S₂ = 35 … (1)

When travels in same direction the distancetravelled has difference of 35km.

i.e., 5 × S₁ - 5 × S₂= 35 km ... (2)

S₁ - S₂ = 7 … (2)

Adding (1) and (2) we get

2S₁ = 42

S₁= 21 km/hr

A scooter travel in 50 km = 4 hr.

The scooter travels in 1 km= 4/ 50 hr

Therefore,

The scooter travels in 500 km

=$\frac{4}{50}\times 500hr$

=40 hrs

Therefore, option b is correct.

1 leap of Dog = 3 leap of Cat
Hence, 1 leap of Cat =$\frac{1}{3}$ leap of Dog

Speed of Dog = 4 dog leaps/minute
Speed of Cat = 9 cat leaps/minute or 3 dog leaps/minute
Time after which Dog catches the Cat =$\frac{20}{4-3}$=20 min

Therefore, option b is correct.

Let the total distance travelled be x km.

 Then,total time taken =

$\frac{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{10}+\frac{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{20}+\frac{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{20}=\frac{x}{40}+\frac{x}{80}+\frac{x}{80}$

$\frac{2x+x+x}{80}=\frac{4x}{80}=\frac{x}{20}hrs$

Then average speed = total distance travelled/x total time taken

$\frac{x}{x/20}\frac{km}{hr}$ = 20 km/hr

Therefore, option b is correct.

let speed of B = x km/hr

Let speed of A = 2x km/hr

Given, 3x = 2×120 km/hr

⇒ x = 80 km/hr

Distance covered by thenafter 10 min. = 3×20 = 60 km

So, remaining distance = (80−60) km = 20 km

Timetaken by B to cover 20 km = $\frac{20}{\frac{80}{60}}$=$\frac{20\times 60}{80}$

= 15 min

Total time = 20+15 = 35 min

Therefore, option d is correct.

$dis\tan ce=speed \times time time=5 min$
$= 5 min = \frac{1\times 5}{60}=\frac{1}{12}hr$

Speed = 12 m/sec

12 m/sec = $\frac{12\times 18}{5}$=43.2km/hr

= distance = 43.2×$\frac{1}{12}$

= distance= 3.6 km

Therefore, option c is correct.

Let the man's normal rate of walking be xKm/hr and time taken initially be t hr.

As per question (x+2) (t-3) = 36 and also xt =36,

xt + 2t – 3x – 6 = 36

36 + 2t – 3x – 6 = 36

2t -3x = 6

2×$\frac{36}{x}-3x=6$

$\frac{72-3x^2}{x}=6$

== 3x² + 6x – 72

= x = 4

Solving both we get x = 4 Km/hr and t = 9 hr. 

Therefore, option c iscorrect.

Speed of Car = 120 km/hr

Time = 5hr

Therefore, distance travel in 5 hours =120×5=600 km

If the speed is 150 km/hr then time taken tocover 600 km

= distance / time = 600/150 = 4 hr

Therefore, Difference in both time = 5 - 4 =1hr = 60 min

Therefore, option a is correct.

Distance = 14 km

Speed of man when he goes from A to B =14 km/h

Time taken = 14/14 = 1 hour

Speed of man when he goes from B to A =21 km/h

Time taken = 14/21 = 2/3 hours

TotalDistance = 14 × 2 = 28 km

Total time = 1+ $\frac{2}{3}=\frac{5}{3}$

Average speed =

=$\frac{28}{5/3}$

=$\frac{28\times 3}{5}$

=$\frac{84}{5}$

= 16.8 kmph

Therefore, option d is correct.