Zoho Time speed & Distances Quiz 1

Speed = \left ( \frac{540}{8*60} \right ) m/sec

= 1.125 m/sec

Converting m/sec to km/hr 

= \left ( 1.125*\frac{18}{5} \right ) km/hr 

= 4.050 km/hr

Explanation - Distance = (300 x 7) = 2100 km.

 

Speed = Distance/Time

 

Speed =  \frac{2100}{\frac{11}{3}} km/hr.        [We can write 2\tfrac{5}{3} hours as 11/3 hours]

 

Required speed = 2100 \times (\frac{3}{11}) km/hr  = 572.72 km/hr

Explanation - Let the actual distance traveled be x km.

 

\frac{x}{15}=\frac{x+30}{20}

 

20x = 15x + 450

5x = 450

x = 90 km.

Explanation - Let the speed of the bus be X kmph.

Then, speed of the car = 130/100* X = 13/10*X kmph

\frac{55}{x}- \frac{55}{\frac{13}{10}x} = \frac{95}{10} \times 60

\frac{55}{x}- \frac{550}{13x} = 0.158

0.158 X = 715 - 550/13

0.158 X = 12.69

X = 12.69/0.158

X = 80.31 kmph

Explanation - Due to stoppages, it covers 15 km less.

Time taken to cover 15 km = (15/40) * 60 min = 10min

Explanation - Let the duration of the flight be x hours.

Then, \frac{400}{x}-\frac{400}{x+\frac{1}{2}} = 150

\frac{400}{x} - \frac{800}{2x+1} = 150

400 (\frac{1}{x} - \frac{2}{(2x+1)}) = 150

\frac{1}{x} - \frac{2}{(2x+1)} = \frac{150}{400}

X = \frac{5}{6}

\frac{\frac{1}{2}X}{18} + \frac{\frac{1}{2}X}{24} = 6

x/18 + x/24 = 12

 X = 123.42

Explanation - Let the speed of two cars be 4x and 5x km/hr.

Then, 5x = (250/3) = 83.33

X = (83.33/3) = 27.77

Speed of first cars = (4 x 27.77) km/hr = 111.080 km/hr

Total time taken = ( \frac{175}{75}+\frac{175}{95}) hrs = 4.175 hrs.

Average speed = \left ( \frac{350}{4.175} \right) = 83.83km/hr

Time taken = 1 hr 40 min 48 sec 

= 1hr 40 \tfrac{4}{5} min  

= 1 \tfrac{51}{75} hrs 

= \frac{126}{75}

Let the actual speed be x km/hr.

Then 5/7x * 126/75 = 42

X = \left ( \frac{42*7*75}{5*126} \right ) = 35kmph