# Time and Distance

A train with 120 wagons crosses Arun who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Arun ( he is riding on the horse) before it starts overtaking the Sriram( who is also riding on his horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Sriram do the Arun meets to Sriram?

Let the length of the train be L metres and speeds of the train Arun and Sriram be R, A and S respectively, then

$$\frac{L}{R - A} = 36$$  --------- (i)

and

$$\frac{L}{R + K} = 24$$ --------- (ii)

From eq.(i) and (ii)

3 (R − A) = 2 (R + K)

⇒ R = 3A + 2K

In 30 minutes (i.e 1800 seconds), the train covers 1800R (distance) but the Arun also covers 1800 A (distance) in the same time. Therefore distance between Arun and Sriram, when the train has just crossed Sriram

= 1800 (R - A) − 24 (A + K)

∴ Time required = $$\frac{1800 \left ( R - A \right ) - 24 \left ( A + K \right )}{\left ( A + K \right )}$$

= (3600 - 24) = 3576 s

Shahrukh starts from Barabanki to Fatehpur, 1 hour after Ajay starts. Shahrukh meets Kajol 1.5 hours after Shahrukh starts. If the speed of Shahrukh is at least 20km/h faster than the speed of Kajol.

What is the minimum speed of Shahrukh to overtake Ajay, before he meets Kajol?

Let t be the time after Kajol starts, when she meets Ajay, then

$$t = \frac{300}{x + y}$$

This should be less than 2.5 or (x + y) > 120

Since $$y = \frac{3x}{2}$$ ⇒ y > 72

This (y > 72) is greater than 67.5 km/h and hence Shahrukh will always overtake Ajay before he meets Kajol.

The ratio between the speeds of two trains is 7: 8. If the second train runs 400 kms in 4 hours, then the speed of the first train is ?

Let the speeds of two trains be 7x and 8x km/hr.

$$8x = \frac{400}{4}$$

⇒ x = 12.5 km/hr

So speed of first train is 12.5 × 7 = 87.5 km/hr

A boat sails 15 km of a river towards upstream in 5 hours. How long will it take to cover the same distance downstream, if the speed of current is one-fourth the speed of the boat in still water:

Upstream speed = B − S

Downstream speed = B + S

B − S = $$\frac{15}{5}$$ = 3 km/h

Again          B = 4S

Therefore    B − S = 3 = 3S

⇒ S = 1 and B = 4 km/h

Therefore    B + S = 5km/h

Therefore, Time during downstream = $$\frac{15}{5}$$ = 3h

A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is

Total time taken = $$\left ( \frac{160}{64} + \frac{160}{8} \right )$$ hrs

= $$\frac{9}{2}$$ hrs.

Average speed = $$\left ( 320 \times \frac{2}{9} \right )$$ km/hr

= 71.11 km/hr.

If Deepesh had walked 20 km/h faster he would have saved 1 hour in the distance of 60 km. what is the usual speed of Deepesh?

Let the original speed  be x km/h, then

$$\frac{600}{x} - \frac{600}{x + 20} = 1$$

⇒ $$600\left(\frac{x + 20 - x}{x\left(x + 20 \right )} \right ) = 1$$

⇒ x2 + 120x - 12000 = 0

⇒ (x - 120)(x - 100) = 0

⇒ x = 100 and x = 120

∴ original speed = 100 km/h

A person goes to his office at $$\frac{1}{3}$$rd of the speed at which he returns from his office. If the average speed during the whole trip is 12 m/h . what is the speed of the person while he was going to his office?

u = k , v= 3k

∴ $$\frac{2uv}{u + v}$$ ⇒ $$\frac{2 \times k \times 3k}{\left ( k + 3k \right )} = 12$$

⇒ 1.5 k = 12

⇒ k = 8 km/hr

The distance of the college and home of Rajeev is 80 km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4 km/h and thus he reached  to college  at the normal time. What is the changed (or increased) speed of Rajeev?

Let the normal speed be x km/h, then

$$\frac{80}{x} - \frac{80}{x + 4} = 1$$

⇒ x2 + 4x - 320 = 0
⇒ x(x + 20) - 16(x + 20) = 0
⇒ (x + 20)(x - 16) = 0

x = 16 km/h

∴ (x + 4) = 20 km/h

Therefore increased speed = 20 km/h

A train overtakes two girls who are walking in the  opposite direction in which the train  is going at the rate of 3 km/h and 6 km/h and passes them completely in 36 seconds and 30 seconds respectively. The length of the train is:

Let the length of the train be x meter, and let the speed of train be y km/h, then

$$x = \left ( y + 3 \right )\frac{5}{18} \times 36$$       ...........(1)

and $$x = \left ( y + 6 \right )\frac{5}{18} \times 30$$ ..........(2)

From eq. (1) and (2), we get

⇒ (y + 3) × 36 = (y + 6) × 30

⇒ y = 12 km/h

⇒ $$x = \left ( y + 3 \right ) \times \frac{5}{18} \times 36$$

⇒ x = 150 m

A person crosses a 600 m long street in 5 minutes, What is his speed in km per hour?

Speed = $$\left ( \frac{600}{5 \times 60} \right )$$ m/sec
= $$\left ( 2 \times \frac{18}{5} \right )$$ km/hr