Unitary Method Test - 7

Using formula speed =distance / time

Speedof Bunty = $\frac{5000}{20\times 60}$

Distance travelled in 10 seconds = 4.1 × 10 = 41 ft.

Therefore, option a is correct.

Let the length of the other train be x m.

Sum of their lengths =(100+x) m

Relativespeed = $(35+37)kmph=72kmph=(72\times 5/18)=20m/sec$

$\frac{100+x}{\frac{37}{2}}=10$

2x = 170
x = 85

Therefore, the length of the other train is 85 m.

Therefore, option a is correct.

Suppose he move 3 km downstreamin x hours. Then,
Speed downstream = $\left(\frac{3}{x}\right)$ km/hr

Speed upstream =  $\left(\frac{2}{x}\right)$ km/hr
$\therefore \frac{30}{\left(\frac{3}{x}\right)}+\frac{30}{\left(\frac{2}{x}\right)}=10\text{or}x=\frac{2}{5}$

So, Speed downstream = 7.5 km/hr, Speed upstream= 5 km/hr.
Rate of the stream = $\frac{2}{5}$   km/hr = 1 km/hr.

Therefore, option c is correct.

Given Speed oftrain is 70.4 km/hr

Distance travelled in 40 min is $70.4\times \frac{40}{60}=46.9km=46900m$

Therefore, option a is correct.

the total distance be x

Timetaken to travel first half distance is $\frac{\frac{x}{2}}{30}=\frac{x}{60}$

Time taken to travel second half distance is $\frac{\frac{x}{2}}{50}=\frac{x}{100}$

Averagespeed is $\frac{x}{\frac{x}{50}+\frac{x}{100}}=\frac{300}{8}=37.5$

Therefore, option b is correct.

Speed of train \(=120 \mathrm{km} / \mathrm{hr}=120 \times \frac{5}{18} \mathrm{m} / \mathrm{s}=\frac{100}{3} \mathrm{m} / \mathrm{s}\)

Time taken to cross the electric pole = 9 sec

\(\therefore\) Length of the train \(=\left(\frac{100}{3} \times 9\right) m=300 m\)

Let the length of the platform be x m.

Then a distance of \((x+300) \mathrm{m}\) is to be covered in 24 seconds at speed of \(\frac{100}{3} \mathrm{m} / \mathrm{s}\)

\(\therefore(x+300)=\left(\frac{100}{3} \times 24\right) m\)

\(\Rightarrow x+300=800\)

\(\Rightarrow x=500 m\)

Distance covered by train A before train Bleaves Ludhiana = 50×3=150 km

Relative speed = (65−50) km/hr = 15 km/hr

∴ Time taken to cross each other$\frac{150}{15}hrs$=  hrs = 10 hrs

∴ Required time = 2 pm+ 10 hrs = 12 am

Therefore, option a is correct.

Speed of Car = 40 km/hr

Time = 4hr

∴ distance travelin 4 hours = 40×4=160 km

If the speed is 80 km/hr then timetaken to cover 160 km 

= distance / time = 160/80 = 2 hr

∴ Difference in bothtime = 4 – 2 = 2hr = 120 min

Therefore,option a is correct.

⇒ Speed of man = 20 km/hr

⇒ Time needed = 10 min = 10/60hr

⇒ Distance= Speed × Time
$=20\times \frac{10}{60}$

= 3.33 km

⇒ Distance = 3.33 km = (3.33×1000)m = 3330 m

∴ the length of the bridge is 3330 m

Therefore, option b is correct.

Speed downstream = Speed of Boat in stillwater + Speed of the stream  $=(15+2) k m / h r=17 k m / h r$

Time taken to travel 68 km downstream =$\frac{Dis\tan ce}{Speed}=\left(\frac{85}{17}\right)hrs = 5hrs.$

Therefore, option c is correct.

Weight =mg

Where, m= mass of object

g= gravity and mg= forceexerted on an object due to the acceleration of gravity.

Therefore, option a is correct.

Speed =4 m/s, time = 40 minutes = 40×60 seconds = 2400 s

Distance= speed × time

= 4 m/s× 2400 s = 9600 m = 9 km 600 m

Therefore,option c is correct.

Length of 1^st train =2x × 2 = 4x
Length of 2^nd train =6x × 2 =12x
Total distance = 16x
Speed = 2x+ 6x = 8x

∴ time = $\frac{16x}{8x}$ = 2 sec

Therefore, option a is correct.

$54\frac{km}{hr}$

$\frac{54\times 1000m}{60min}$

$\frac{54\times 1000}{60}m/min$

$900m/min$

Therefore, option b is correct.

The length of his journey is givenas: 54+48+50 km + 300+46+6 m 

= 152 km 352 m

Thusthe correct answer is 152 km 352 m

Therefore,option a is correct.