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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?

Explanation:

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 *10/60 = 8/60(20+x)
Or, 200 = 160+8x
Or, 8x = 40
Hence, x = 5kmph.

Detailed Explanation:
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 = (20/60) * 10 = 10/3 km.

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:

Explanation:

Solution:
1st method:

4/3 of usual time = Usual time + 16 minutes;
Hence, 1/3rd of usual time = 16 minutes;
Thus, Usual time = 16*3 = 48 minutes.

2nd method:

When speed goes down to
3/4th (i.e. 75%) time will go up to 4/3rd (or 133.33%) of the original time.
Since, the extra time required is 16 minutes; it should be equated to 1/3rd of the normal time.
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:

Explanation:

Solution:
Difference in time of departure between two trains = 45 min. = 45/60 hour = 3/4 hour.
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+3/4)
= ((4t+3)/4);

Now,
100 * (4t+3)/4 = 136t;
Or, 25(4t+3) = 136t;
Or, 100t+75 = 136t;
Or, 36t = 75;
Or, t = 75/36 = 2.08 hours;
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:

Explanation:

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

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:

Explanation:

Solution:
A →_______60Km_________← B
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,

{(2x/3)*5}+(2y*5) = 60
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:

Explanation:

Solution:
Time taken to complete the revolution:
A→12/4 = 3 hours;
B→ 12/3 = 4hours;
C→12*2/3 = 8 hours;
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:

Explanation:

Solution:

Ajay → (x+4) kmph.
A ________ 60 km _________ B
Ravi → x kmph.

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,

72/(x+4) = 48/x
Or, 3/(x+4) = 2/x
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

Explanation:

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

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)?

Explanation:

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
= 2640/5 min. 52 secs;
= 2640m/352 secs
= 2640 * 18/352 * 5 kmph
= 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: