Question

If a train runs at 40 kmph it reaches its destination late by 11 minutes.But if it runs at 50kmph it is late by 5 minutes only. Find the distance to be covered by the train. ​

Answer 1

Answer:

Therefore the distance covered by train = 20 Km

Step-by-step explanation:

mark me as brainliest

follow me

Answer 2

$\overline{\underline{\boxed{\sf GIVEN \darr}}}$

If a train runs at 40 kmph it reaches its destination late by 11 minutes.But if it runs at 50kmph it is late by 5 minutes only. Find the actual time of the

$\overline{\underline{\boxed{\sf SOLUTION \darr}}}$

Let us assume :

The time taken be t hrs

The distance covered be S km

Given that :

1st Case

When a train runs at 40 km/h and reaches 11 mins late

2nd Case

When the train runs at 50 km/h and reaches 5 mins late

Henceforth,

The difference in time = 11 mins - 5 mins → 6 mins

Formula for Time :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{ \small \underline{ \blue{\boxed{ \purple{\sf Time = \frac{Distance}{Speed}}}}}}$

Now, using this formula we can get time

If the train runs at 40 km/h

• Time = S/40 hr

If the train runs at 50 km/h

• Time = S/50 hr

Now, the equation is :

$\sf \mapsto \frac{S}{40} - \frac{S}{50} = 6 \: mins$

But The both the times are in hour whereas 6 is in minutes

Therefore, we need to convert 6 into hr and we know that :

60 mins = 1 hr

1 min = 1/60 hr

So, 6 min equals

6/60 hr = 1/10 hr

Now,

$\sf \leadsto \frac{50S}{2000} - \frac{40S}{2000} = \frac{1}{10}$

$\sf \leadsto \frac{10S}{2000} = \frac{1}{10}$

By cross multiplying we get

$\sf \leadsto10 \times 10S = 2000 \times 1$

$\sf \leadsto100S = 2000$

Transposing 100 to the other side we get 1/100

$\sf \leadsto S = \frac{2000}{100}$

$\sf \leadsto S = 20$

Henceforth, the distance covered is 20 km