Question

A sweeper was cleaning a tunnel of length 120 kms. He was standing at a point which divides the length of the tunnel in the ratio of 5:7, from entry point to exit point. After seeing the sweeper the train whistles from a particular point outside the tunnel. After listening the whistle sweeper run towards the train and just came out safely. On the next day, again sweeper is cleaning at the same point and train again whistles from the same point but today sweeper come out from the exit point and just came out safely. Then find the ratio of speed of train to sweeper. Note: Train always approaches from the entry point of the tunnel? ​

Answer 1

Answer:

Correct option is B)

Let Distance between Train and Entrance of Tunnel when the Train Whistles be x and Length of Tunnel be AB

When The cat moves Towards the Entrance

⇒

T

x

=

8

3

×

C

AB

1.)

When the Cat Move Towards the Exit

⇒

T

x+AB

=

8

5

×

C

AB

2.)

On Dividing Equation(1) by Equation(2) we get

x+AB

x

=

5

3

x=

2

3AB

Putting value of x in Eq(1) we get

⇒

2T

3AB

=

8

3

×

C

AB

⇒T:C=4:1

Therefore Answer is (B)

Explanation:

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Answer 2

Answer:

The ratio of speed of train to sweeper is 6:5.

Explanation:

Given length of tunnel =  120 kms.

Sweeper was standing at a point which divides the length of the tunnel in the ratio of 5:7, from entry point to exit point.

Let the speed of the train be $v_{T}$ and of the sweeper be $v_{s}$.

let the train be at a distance of $x$ from the tunnel.

When the sweeper run towards the train, then the time can be related as

$\frac{x}{v_{T} } =\frac{5}{12} \times \frac{120}{v_{s}}\implies \frac{x}{v_{T} } = \frac{50}{v_{s}}$                        ...(1)

When the sweeper run towards the exit, then the time can be related as

$\frac{x+120}{v_{T} } =\frac{7}{12} \times \frac{120}{v_{s}}\implies \frac{x+120}{v_{T} } = \frac{70}{v_{s}}$               ...(2)

Dividing both the equations,                    

$\frac{\frac{x}{v_{T} }}{ \frac{x+120}{v_{T} } } = \frac{\frac{50}{v_{s}}}{\frac{70}{v_{s}}}$

$\implies{\frac{x}{{x+120}} ={\frac{50}{70}}$

$\implies{7x}=5(x+120)$

$\implies{2x}=120\implies x=60km\hr$

Using $x$ in equation (1),

$\frac{v_{T}}{v_{s}}= \frac{60}{50}= \frac{6}{5}$

Therefore, the ratio of speed of train to sweeper is 6:5.