Quiz-1

Solution:
Speed of yacht downstream

= Distance covered/Time taken=\frac{45}{3.5} = 12.85 km/h

Now, the speed of boat upstream

= Distance covered /Time taken=\frac{45}{7.1} =6.33 km/h

Now, the speed of the boat

= (Speed downstream+Speed upstream)/2

= \frac{12.85+6.33}{2} = 9.59 kmph

Solution:
Let the speed of goods train be x km/hr.
Then, (67+x)\times (\frac{5}{18})=\frac{293.5}{11}

⇒ x= 29.05 km/hr

Let the speed of the stream be x km/hr. Then,

Speed downstream = (15 + x) km/hr,

Speed upstream = (15 - x) km/hr.

Distance traveled downstream= 30 Km

Distance traveled upstream= 30 Km

Time = distance /speed

Total time taken = 4.5 hrs

Time downstream + Time upstream = 4.5

(\frac{30}{15+x})+(\frac{30}{15-x})=4.5

\frac{900}{(225-x^{2})}=\frac{9}{2}

225-x^{2}=200

x = 5 km/hr.

Let the speed of the cyclist be x km/h and wind be y km/h
\frac{1}{(x+y)}=\frac{3}{60}
x+y=20..........(1)

And
\frac{1}{(x-y)}=\frac{4}{60}
x-y=15.........(2)
On solving both the equations we get:
x=\frac{35}{2}
y=\frac{5}{2}
Now, time taken by cyclist without wind: (\frac{2}{35})\times 60=\frac{24}{7} min

Let the bike travel a distance of 2d km, total.

The first half of the distance takes d km/40 kmph = \frac{d}{40} hr.

The second half of the distance takes d km/60 kmph = \frac{d}{60} hr.

So the total time taken to cover 2d km = \frac{d}{40} +\frac{d}{60} = \frac{100d}{2400} hr.

The average speed of the car = Total distance/ total time = 2d/[100d/2400]

= 2d\times \frac{2400}{100d}

= 48 kmph.

Let the speed of the current be x km/hr
Thus upward speed =(6-x) km/hr
and downward speed =(6+x) km/hr

Time= Distance/ speed

Time taken to row up = 3 times time taken to row down

Let distance traveled be y, then
\frac{y}{6-x}= \frac{3y}{6+x}

⇒ x= 3 km/hr

Solution:
Speed=(45\times \frac{5}{18})m/sec=\frac{25}{2}m/sec
Time=30 sec
Let the length of the bridge be x metres.
Then,
\frac{130+x}{30}=\frac{25}{2}
2(130+x)=750
x=245 m

let x be the length of the train.

\frac{x+142}{38} = \frac{x}{23}

= 23\times (x+42) = 38x

= 23x + 3266 = 38x

= 3266 = 15x

= x = 217.73 m.

Let the distance of train be any value ( if you don't want to solve by taking x. )

Let the length of each train = L.C.M. of the time given 29 and 19 (for simplicity) = 551 meters.

Passing a pole means, crossing its own length.

So, for train 1.

Distance covered in passing a pole = length of train 1 =551 meter.

Time taken = 29 sec.

So, Speed of train 1 = 551/29 = 19 m/s

For train 2:

Speed of train 2 = \frac{551}{19} = 29 m/s

( These data can be assumed in solving relatively. If x would have been taken in place of 551, the process would have been the same. If confused, think x=551. )

Opposite direction movement, time taken to cross:

In opposite direction, the relative speed is added up.

So, relative speed = 19 + 29 = 48 m/s

Relative Distance covered for both trains to cross each other = sum of the length of both trains

= 551 + 551 = 1102 m

So, time taken for both trains to cross each other= distance/ relative speed

= \frac{1102}{48} sec

=22.9 sec.

Since the train and man are moving in opposite directions,

The relative speed = 75-(-8.8) kph = 75 + 8.8 = 83.8 kph

83.8 km/h in m/s = 83.5\times \frac{5}{18}= 23.27m/s

Total distance to be covered = 135 m (length of the train)

So, time = distance/speed = 135 divided by 23.27 which equals 5.80 s.

So, the required time is 5.80 seconds.