Question

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. if it takes 3 hrs to complete total journey,what is the original average speed

Answer 1

Hey

Here is your answer,

Given that  distance = 63 km.

Let original speed of train = x km/hr.

time = distance / time =  63/x hrs.


And it travels a distance of 72 kkm at a average speed of 6 km/hr more than the original speed.


distance = 72 km ; speed = (x + 6) km/hr .

time = 72/(x+6) hrs.

If it takes 3 hours to complete the whole journey

63/x + 72/(x + 6) = 3 hrs


⇒ 63(x + 6) + 72x = 3x(x + 6)
⇒ 21(x + 6) + 24x = x(x+6)
⇒ 45x + 21×6 = x2 + 6x
⇒ x2 -  39x - 126 = 0
⇒ x2 - 39x - 126 = 0
⇒ (x - 42)(x + 3) = 0


∴ x = 42 km/hr

∴ the original average speed = 42 km/hr

Hope it helps you!

Answer 2

Heya !!

Here's your answer.. ⬇⬇

_______________________________

➡ Given that a train travels 63km at certain average speed.

so, Distance = 63km

let that original speed = x km/hr

As we know, Speed = Distance/Time

Time = Distance/speed
Time = ( 63/x ) hrs.

➡ Now, given that train travels 72km at average speed of 6 km/hr more that original speed.

Distance = 72km
Average speed = (x + 6) km/hr

Time = Distance/speed
Time = 72/( x + 6 ) hrs.

➡ It completes total journey in 3hrs.

Time of Total journey = 3hrs.

63/x + 72/( x + 6 ) = 3

( 63x + 378 + 72x )/( x^2 + 6x ) = 3

63x + 72x + 378 = 3x^2 + 18x

135x + 378 = 3x^2 + 18x

3x^2 + 18x - 135x - 378 = 0

3x^2 - 117x - 378 = 0

3 ( x^2 - 39x - 126 ) = 0

x^2 - 39 - 126 = 0

x^2 - 42x + 3x - 126 = 0

x ( x - 42 ) + 3 ( x - 42 ) = 0

( x + 3 ) ( x - 42 ) = 0

( x + 3 ) = 0 and ( x - 42 ) = 0

x = -3 and x = 42

Speed is never in negative.

Hence, x = 42 km/hr

Original speed of train is 42km/hr.
________________________________

Hope it helps..

Thanks :))