Time and Distance Questions and answers

Average speed = \( \Large \frac{Total \  distance }{Total \  time} \)

S = \( \Large \frac{450 m}{1\frac{1}{2}s}\)

=300 m/s

\( \Large \frac{d}{4} - \frac{d}{5} \) = \(  \Large \frac{15}{60} \)

= \( \Large \frac{5d - 4d}{20} \) = \(  \Large \frac{15}{60} \)

= \( \Large \frac{d}{20}\) = \(  \Large \frac{1}{4} \)

=> d = 5 km

Let the total number of passenger be 'x' that had boarded at Delhi.
Now, at Ghaziabad
110 get down
So the remaining passenger = x - 110
Now again 110 got in
So Net passenger at Ghaziabad = (x - 110) + 100
=x - 10

Now at Aligarh
50% passenger got down
Remaining passenger = \( \Large \frac{x - 10}{2} \)
Now again 25 got in
Net passenger at Aligarh = \( \Large \frac{x - 10}{2} + 25\)

At Kanpur,
5 got down and 50 get in
\( \large \left[ \frac{x - 10}{2} +25\right] - 5 +50 = 200\)
\( \large \left[ \frac{x - 10}{2}\right] = 130\)
X - 10 = 260
X = 270

So the number of passenger boarded at Delhi is 270

r = 35 cm = \( \Large \frac{35}{100000}km \)

Now the distance covered in 1 revolution = \( \Large 2 \times \pi \times r \)

= \( \Large 2 \times \frac{22}{7} \times \frac{35}{100000} = \frac{220}{10000}\) = \( \large \frac{22}{100000}km \)

In 500 revolutions,

Distance covered = \( \Large 500 \times \frac{22}{10000} km \)

= \( \Large \frac{22}{20}km \) = \( \frac{11}{10}km \)

So \( \Large \frac{11}{10}\) km is covered in 1 minute.

60 min = 1 hr, distance covered

=\( \Large \frac{60 \times 11}{10}km \)

= 66 km

Hence, speed = distance/time

= 66km / 1 hr = 66 Km/hr

\( \Large 8\frac{1}{2}\) = \( \Large \frac{x}{12} + \frac{72 - x}{4\frac{1}{2}}\)
X = 54 km

\( \Large \left(\frac{d}{80}+ \frac{d}{40} \right) - \left(\frac{d}{60}+ \frac{d}{60} \right) = 2 \)

\( \Large \frac{3d + 6d - 4d - 4d}{240} = 2 \)

D = 480 km

t = \( \Large \frac{50}{4 + 6} = 5 hrs\)

\( \Large \frac{d}{4} - \frac{d}{5} = \frac{7\frac{1}{2}}{60} \)

\( \Large \frac{5d - 4d}{20} = \frac{15}{120} = \frac{1}8{}\)

\( \Large \frac{d}{20} = \frac{1}{8}\)

D = 2.5 km.