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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?

A.

5 km/h

B.

C.

4 km/h

D.

3 km/h

Answer with explanation

Answer: Option AExplanation

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,

20×1060=860×(20+x)20×1060=860×(20+x)

Or, 200 = 160 + 8x

Or, 8x = 40

Hence, x = 5 kmph.

**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 = 2060×102060×10

= 103103 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

Workspace

A train Express A leaves Delhi at 5 a.m and reaches Mumbai at 9 a.m.Another train Express B leaves Mumbai at 7 a.m and reaches Delhi at 10.30 a.m.At what time do they cross each other after 7 a.m ?

A.

50m

B.

52m

C.

56m

D.

54m

Answer with explanation

Answer: Option CExplanation

Speed of Express A = x/4kmph

Speed of Express B = 2x/7kmph =>(t=3.30 = 3(1/2) = 7/2)

Trains will meet T hours after 7 a.m

(x/4)(y+2) + (2x/7)y = x

(y+2)/4 + (2y/7) = 1

(7y+14+8y)/28 = 1

15y = 28-14

Y =(14/15)×60 = 56 min

Workspace

A.

9

B.

6

C.

7

D.

8

Answer with explanation

Answer: Option AExplanation

The second car overtake the first car in x hours

Distance covered by the first car in x hours = Distance covered by the second car in x hours

10x = x/2[2a + (x-1)d]

10x = x/2[2*8 + (x-1)1/2]

x = 40 -31 = 9

Workspace

A.

1

B.

2

C.

3

D.

4

Answer with explanation

Answer: Option BExplanation

First person speed = 25 m/s * 18/5 = 90 kmph

Second person speed = 35 m/s * 18/5 = 126 kmph

First person covers 90 * 10 = 900km

900/450 = 2

Workspace

A thief is spotted by a policeman from a distance of 200 metre. When the policeman starts chasing , the thief also starts running. If the speed of the thief be 16kmph and that of policeman be 20kmph, how far the thief will have run before he is overtaken?

A.

750 m

B.

650 m

C.

700 m

D.

800 m

Answer with explanation

Answer: Option DExplanation

d = 200 m, a = 16kmph = 40/9 m/s, b = 20kmph = 50/9 m/s

Required Distance D = d*(a/b-a)b= 200*(40/9/10/9) = 800m

Workspace

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