## Results

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### #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?

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=10/3km.

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:

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:

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 takes 6 hours 15 minutes in walking a distance and riding back to starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride back both ways is:

Solution:

Time taken in walking both the ways = 7 hours 45 minutes ——– (i)
Time taken in walking one way and riding back = 6 hours 15 minutes ———– (ii)
By the equation (ii) × 2 – (i), we have,
Time taken by the man in riding both ways,
= 12 hours 30 minutes – 7 hours 45 minutes
= 4 hours 45 minutes.

### #5. 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:

Solution:
Let the total distance be 100 km.
Average speed=total distance covered time taken
=100/[(30/20) (60/40) (10/10)]
=100/[(3/2) (3/2) (1)]
=100/[(3 3 2)/2]
=(100×2)/8
=25kmph

### #6. 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:

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

### #7. 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:

Solution:

Time taken to complete the revolution:
A → 12/4 = 3 hours
B → 12/3 = 4 hours
C → 12 × 2/3 = 8 hours
Required time,
= LCM of 3, 4, 8.
= 24 hours.

### #8. 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:

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 – 2 = 48 km.

According to question,

72/x 4=48/x
or, 3/x 4=2/x
3x=2x 8
or,
x=8kmph

### #9. The speed of A and B are in the ratio 3 : 4. A takes 20 minutes more than B to reach a destination. Time in which A reach the destination?

Solution:

Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 (As Speed ∝ 1/Time, remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.

Then,
4x – 3x = 20/60

Or, x = 1/3

Hence, time taken by A = 4x

hours = 4 × 1/3
= 4/3 hours.