Unitary Method Test - 6

Let the speed of man is u and thatof current is v

Downstream speed will be u + v

Upstream speed will be u − v

He travels 20 km downstreamin 4 hrs

So, downstream speed is 20/4 = 5 km/hr

Similarly, upstream speed will be 12/4 =3 km/hr

So, u + v = 5

⇒ u – v = 3
Subtracting both the equations;

⇒ 2v = 2

⇒v = 1 km/hr

Therefore,option a is correct.

$\frac{14}{x-y}-\frac{14}{x+y}=7$
$\Rightarrow \frac{14x+14y-14x+14y}{x^2-y^2}=7$
$\Rightarrow 28y=7(x^2-y^2)$
$\Rightarrow x^2=4y+y^2\dots ..\left(1\right)$
$\text{And,}\frac{14}{(2x-y)}-\frac{14}{(2x+y)}=1$
$\Rightarrow 28y=(4x^2-y^2)$
$\Rightarrow x^2=\frac{28y+y^2}{4}\dots ..\left(2\right)$

Now, comparing (1) and (2) we get,
$\Rightarrow 4y+y^2=\frac{28y+y^2}{4}$
$\Rightarrow 16y+4y^2=28y+y^2$
$\Rightarrow 3y^2=12y$
$\therefore y=4$
$"The speed of the current is"=4 mph$

Therefore,option c is correct.

Let x= speed of the boat in stillwater
and y= speed of the current
Down speed =(x + y)
Up speed =(x−y)
According to the problem, we have

$\frac{12}{x+y}$=2 and $\frac{12}{x-y}$=4

2x +2y = 12

x + y = 6 … (1)

4x – 4y = 12

x – y = 3 … (2)

Adding equation (1) and (2)

2x = 9

= x = 4.5

Therefore, speed of boat in still water is 4.5 km/h.

Therefore,option c is correct.

Train was halted for half an hour.

So, total time taken in Journey = 25 - (1/2) =24.5 hours.

Average speed in Whole Journey = 1000/24.5 =48.9 km/hr.

The average speed on return journey is 30%less than onward journey.

Therefore, ratio of average speed of onwardand return journey,

= 100/70 = 10/7.

Let average speed of onward journey = 10x.

Average speed on return journey = 7x.

Average speed on whole journey = (10x +7x)/2

48.9 = (10x + 7x)/2

97.8 = 17x

x = 5.75.

Average speed on return = 7x = 7 × 5.75 = 40.25km/h

= 40 km/h(approx.)

Therefore,option a is correct.

Downstream speed = 30/2 = 10 km/h

Upstream speed  = 30/6 = 5 km/h
Speed of boat in still water =Upstream + Downstream / 2
= $\frac{10+5}{2}$ = 7.5km/h

Therefore, option d is correct.

6 stations = 20 minutes
1 hour = 60 minutes.
Therefore for 1 hour Audi car crosses = 

$\frac{6\times 60}{20}$

= 6 × 3 = 18 stations.

Therefore, option c is correct.

Distance travelled by car in 3hrs=81km

So, speed of the car = 84/2 = 42 km

Distance travelled by car in 4 hrs = 42 ×4 = 168 km

Therefore, option c is correct.

Given a person travels 75 kmin 5 hrs in a car.

Then speed of the car is distance/time = 75/5 =15 km/ hr

Therefore, option d is correct.

Let distance ne x

Speed in 1st trip = 60km/hr

∴ Time taken x/60hrs.

speed in return trip = 30 km/hr

∴ Time taken = x/30hrs.

Avg. speed =Total Distance /Total time

=$\frac{x+x}{x/60+x/30}$

=$\frac{2x\times 60}{x+2x}$
=40

Therefore, option a is correct.

Let the speed of Joseph in his first trip be x km per hour.

∴ Timetaken in his first trip

= 30x hrs.

and the speed of Joseph in his second trip = 2x km perhour

∴ Timetaken in his second trip

= 600/2x = 300/x hrs.

$\frac{time taken in second trip}{time taken in first trip}=\frac{\frac{800}{x}}{\frac{80}{x}}=10$

Therefore,option c is correct.

Let thespeed of the bike be x kmph.
$Clearly,\frac{220}{x}-\frac{220}{x+2}=1$
=$\frac{220(x+2)-220x}{x(x+2)}=1$

= 220x + 440 – 220x = 1 (x(x+2))

= 440 = x² + 2x

= x² + 2x – 440

X = 20

Therefore, option a is correct.

Let the time taken by trainbe t hrs.

Then, 

$\Rightarrow 30t=25(t+\frac{10}{60})$

$\Rightarrow 30t=25t+\frac{25}{6}$

$\Rightarrow 5t=\frac{25}{6}$

$\Rightarrow t=\frac{5}{6}hrs$

Required length of the total journey (d) =s×t

= 30×$\frac{5}{6}$=25 km

∴ The distance travelled by thetrain is 25 km.

Therefore, option b is correct.

Speed of Bus= 48 km/hr

= $\frac{1000\times 48}{60\times 60}$m/sec

= 13.3 m/sec

Distance travelled in 1 sec= 10 m

Distance travelled in 10 sec = 10 × 13.3 = 266 m

Speed = distance / time

∴time = Distance/Speed

Given, S = 63 km/hr 

= 63000/3600 m/sec = 17.5 m/sec

∴ time = 700/17.5 = 40 sec

Therefore, option a is correct.

Speed \(=\left(60 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{sec}=\frac{50}{3} \mathrm{m} / \mathrm{sec}\)

Length of the train \(=(\text { Speed } \times \text { Time })\)

\(\therefore\) Length of the train \(=\left(\frac{50}{3} \times 9\right) m=150 m\)