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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?
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
Or,
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 = = 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.
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,
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 = = 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. Walking 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:
of usual time = Usual time + 16 minutes;
Hence, of usual time = 16 minutes;
Thus, Usual time = 16 × 3 = 48 minutes.
2nd method:
When speed goes down to
(i.e. 75%) time will go up to (or 133.33%) of the original time.
Since, the extra time required is 16 minutes; it should be equated to of the normal time.
Hence, the usual time required will be 48 minutes.
Solution:
1st method:of usual time = Usual time + 16 minutes;
Hence, of usual time = 16 minutes;
Thus, Usual time = 16 × 3 = 48 minutes.
2nd method:
When speed goes down to
(i.e. 75%) time will go up to (or 133.33%) of the original time.
Since, the extra time required is 16 minutes; it should be equated to 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:
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
=
Now,
100 × = 136t
Or, 25(4t + 3) = 136t
Or, 100t + 75 = 136t
Or, 36t = 75
Or, t = = 2.083 hours
Then, distance x km = 136 × 2.083 ≈ 283.33 km.
Solution:
Difference in time of departure between two trains = 45 min. = hour = 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
=
Now,
100 × = 136t
Or, 25(4t + 3) = 136t
Or, 100t + 75 = 136t
Or, 36t = 75
Or, t = = 2.083 hours
Then, distance x km = 136 × 2.083 ≈ 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:
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.
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:
Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
Solution:
Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
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 of his speed and B traveled with double of his speed, they would have met after 5 hours. The speed of A is:
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
Alternate
They meet after 6 hours if they walk towards each other i.e., their speed will be added.
So, their relative speed in opposite direction
Relative speed in opposite direction :
According to the question,
Solution:
A →_______60Km_________← BLet 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
Alternate
They meet after 6 hours if they walk towards each other i.e., their speed will be added.
So, their relative speed in opposite direction
Relative speed in opposite direction :
According to the question,
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 km/h. They will meet together at the starting place at the end of:
A → = 3 hours
B → = 4 hours
C → 12 × = 8 hours
Required time,
= LCM of 3, 4, 8.
= 24 hours.
Solution:
Time taken to complete the revolution:A → = 3 hours
B → = 4 hours
C → 12 × = 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:
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,
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,
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?
Ratio of time taken = 4 : 3 (As Speed ∝ When distance remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.
Then,
4x - 3x =
Or, x =
Hence, time taken by A = 4x
hours = 4 × = hours.
Solution:
Ratio of speed = 3 : 4Ratio of time taken = 4 : 3 (As Speed ∝ When distance remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.
Then,
4x - 3x =
Or, x =
Hence, time taken by A = 4x
hours = 4 × = hours.
10. A man covers half of his journey at 6 km/h and the remaining half at 3 km/h. His average speed is
Solution:
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