WeCP Speed, Time and Distance Quiz 1

Given that time taken for riding both ways will be 2 hrs lesser than the time needed for waking one way and riding back.

 

Therefore,

time needed for riding one way = time needed for waking one way -2 hours

Given that time taken in walking one way and riding back = 5 hrs 45 mins

Hence, the time he would take to walk both ways

= 5 hrs 45 mins + 2 hrs

= 7 hrs 45 mins

Distance = 600 meters = 0.6 Kms

Time = 5 mins = \frac{1}{12} Hour

Speed = \frac{Distance}{Time} = \frac{0.6}{\frac{1}{12}} = 7.2 \text{ km/hr}

Speed of the bus excluding stoppages = 54 kmph

Speed of the bus including stoppages = 45 kmph

 

Loss in speed when including stoppages = 54 - 45 = 9kmph

In 1 hour, the bus covers 9 km less due to stoppages.

 

Hence, time in which the bus stops per hour

= Time taken to cover 9 km

= \frac{distance}{speed} = \frac{9}{54} hour = \frac{1}{6} hour = \frac{60}{6} min = 10 min

Speed downstream = (12 + 4)km/hr = 16 km/hr;

Speed upstream = (12 - 4) km/hr = 8 km/hr.

Let the distance between A and B be x km, Then,

\frac{x}{16} + \frac{(x / 2)}{8} = 19 \Leftrightarrow \frac{x}{16} + \frac{x}{16} = 19

\Leftrightarrow \frac{2x}{16} = 19 \Leftrightarrow \frac{x}{16} = 152\ km

Hence Option B is correct

Rate upstream = 8 - 1.5 = 6.5 kmph

Rate downstream = 8 + 1.5 = 9.5 kmph

Time taken to go upstream =\frac{61.75}{6.5}c= 9.5 hr.

Time taken to go downstream =\frac{61.75}{9.5}= 6.5 hr.

Total time = 9.5 + 6.5 = 16 hrs.

Hence, option C is correct.

\text{ Average Speed } = \frac{2 \times 21 \times 24}{21 + 24} = 22.4 \text{ kmph}

\text{ Total distance }= 22.4 \times 10 = 224 kms

Time = 1 hr 40 min 48 sec

1 \text{ hrs } + \frac{40}{60} \text{ hr } + \frac{48}{3600} \text{ hr }

 

1 \text{ hrs } + \frac{2}{3} \text{ hr } + \frac{1}{75} \text{ hr } = \frac{126}{75} \text{ hr }

 

Distance = 42 kms

Speed = \frac{distance}{time} = \frac{42}{\frac{126}{75}} = \frac{42 \times 75}{126} = 25 \text{ kmph }

 

\frac{5}{7} of the actual speed = 25

 

Actual speed = 25 \times \frac{7}{5} = 35 \text{ kmph }

Let the speed of the motorboat in still water be x kmph. Then,

Speed downstream = (x + 2) kmph; Speed upstream = (x - 2) kmph.

\therefore \frac{6}{x+2} + \frac{6}{x-2} = \frac{33}{60} \Leftrightarrow 11x^2 - 240x - 44 = 0

\Rightarrow 11x^2 - 242x + 2x - 44 = 0  \Leftrightarrow (x - 22)(11x + 2) = 0 \Leftrightarrow x = 22

Hence, the speed of a motorboat in still water = 22 kmph.

Hence, option D is correct.

Given,

Upstream rate = 3.5 kmph, Speed of a Khushpreet in still water  = 5 kmph

Let the downstream rate be x kmph, then

By the short trick approach, we get

Speed of the man (boat) in still water =\frac{1}{2}

[DOWNSTREAM rate + UPSTREAM rate]

\Rightarrow 5 = \frac{1}{2} (x + 3.5)  or   x = 6.5 kmph

Hence, option C is correct.

To solve this question, we can apply a short trick approach;

x \left ( \frac{n-1}{n+1} \right ) km/hr

Given,

x = Person speed = 18 kmph

n = the no. of times of hours taken by the boat in the 2nd scenario (upstream) = 3

By the short trick approach, we get

\Rightarrow 18\left( \frac{3 - 1}{3 + 1} \right) = \frac{18\times 2}{4} = 9 kmph

Hence, option C is correct.