Time and Distance

1.Two buses start from a bus terminal with a speed of 20 km/h at interval of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 8 minutes?

**Answer:**Option** 1**

Solution:

Let Speed of the man is x kmph.Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20+x) kmph.

Or, 20 *

Or, 200 = 160+8x

Or, 8x = 40

Hence, x = 5kmph.

A _____________M_______________B

A = Bus Terminal.

B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well.

Distance Covered by Bus in 10 min = AB = (

Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes.

Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each other. Let Speed of the man = x kmph.

Relative speed = 20 + x

To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 +x) kmph = distance covered by bus in 10 minuted with speed 20 kmph.

2.Waking ^{3}/_{4} of his normal speed, Rabi is 16 minutes late in reaching his office. The usual time taken by him to cover the distance between his home and office:

**Answer:**Option** 1**

Solution:

Hence,

Thus, Usual time = 16*3 = 48 minutes.

When speed goes down to

Since, the extra time required is 16 minutes; it should be equated to

Hence, the usual time required will be 48 minutes.

3.Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together:

**Answer:**Option** 1**

Solution:

Difference in time of departure between two trains = 45 min. = Let the distance be x km from Delhi where the two trains will be together.

Time taken to cover x km with speed 136 kmph be t hour

and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be

(t+

= (

Now,

100 *

Or, 25(4t+3) = 136t;

Or, 100t+75 = 136t;

Or, 36t = 75;

Or, t =

Then, distance x km = 136 * 2.084 = 283.33 km.

4.A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:

**Answer:**Option** 1**

Solution:

$$\begin{array}{rl}& \text{Let the total distance be 100 km}.\\ & \text{Average speed}\\ & =\frac{\text{total}\phantom{\rule{thinmathspace}{0ex}}\text{distance}\phantom{\rule{thinmathspace}{0ex}}\text{covered}}{\text{time}\phantom{\rule{thinmathspace}{0ex}}\text{taken}}\\ & =\frac{100}{\left[\left(\frac{30}{20}\right)+\left(\frac{60}{40}\right)+\left(\frac{10}{10}\right)\right]}\\ & =\frac{100}{\left[\left(\frac{3}{2}\right)+\left(\frac{3}{2}\right)+\left(1\right)\right]}\\ & =\frac{100}{\left[\frac{\left(3+3+2\right)}{2}\right]}\\ & =\frac{\left(100\times 2\right)}{8}\\ & =25\phantom{\rule{thinmathspace}{0ex}}kmph\end{array}$$

5.From two places, 60 km apart, A and B start towards each other at the same time and meet each other after 6 hour. If A traveled with ^{2}/_{3} of his speed and B traveled with double of his speed, they would have met after 5 hours. The speed of A is:

**Answer:**Option** 1**

Solution:

Let the speed of A= x kmph and that of B = y kmph;

According to the question;

X*6+y*6 = 60

Or, x+y = 10 --------- (i)

And,

{(

Or, 10x+30y = 180;

Or, x+3y = 18; ---------- (ii)

From equation (i)*3 - (ii)

3x+3y-x-3y = 30-18

Or, 2x = 12

Hence, x = 6 kmph.

6.A, B and C start together from the same place to walk round a circular path of length 12km. A walks at the rate of 4 km/h, B 3 km/h and C ^{3}/_{2} km/h. They will meet together at the starting place at the end of:

**Answer:**Option** 1**

Solution:

Time taken to complete the revolution:A→

B→

C→12*

Required time,

= LCM of 3, 4, 8.

= 24 hours.

7.Ravi and Ajay start simultaneously from a place A towards B 60 km apart. Ravi's speed is 4km/h less than that of Ajay. Ajay, after reaching B, turns back and meets Ravi at a places 12 km away from B. Ravi's speed is:

**Answer:**Option** 1**

Solution:

Ajay →

A ________ 60 km _________ B

Ravi →

Let the speed of Ravi be x kmph;

Hence, Ajay's speed = (x+4) kmph;

Distance covered by Ajay = 60 + 12 = 72 km;

Distance covered by Ravi = 60 - 12 = 48 km.

According to question,

Or,

3x = 2x+8

Or, x = 8 kmph.

8.A man covers half of his journey at 6km/h and the remaining half at 3 km/h. His average speed is

**Answer:**Option** 1**

Solution:

$$\begin{array}{rl}& \text{Average}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\\ & =\frac{2xy}{\left(x+y\right)}\\ & =\frac{2\times 6\times 3}{\left(6+3\right)}\\ & =\frac{36}{9}\\ & =4\phantom{\rule{thinmathspace}{0ex}}\text{kmph}\end{array}$$

9.Two guns are fired from the same place at an interval of 6 minutes. A person approaching the place observes that 5 minutes 52 seconds have elapsed between the hearings of the sound of the two guns. If the velocity of the sound is 330 m/sec, the man was approaching that place at what speed (in km/h)?

**Answer:**Option** 1**

Solution:

Difference of time = 6 min-5 mins. 52 secs = 8 secs.Distance covered by man in 5 mins. 52 secs = distance covered by sound in 8 secs

= 330*8 = 2640 m.

Hence, speed of man

=

=

= 2640 *

= 27 kmph.

10.Running at ^{5}/_{4} of his usual speed, an athlete improves his timing by 5 minutes. The time he usually takes to run the same distance is:

**Answer:**Option** 1**

Solution:

When the athlete walks at → Usual time - (

→(

Usual time = 25 minutes.