Question

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


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Answer 1

Step-by-step explanation:

average speed = distance /time

speed = 73km + 63km / 3

speed =136/3

speed= 45.333333333km/hr

Answer 2

$\star$ SOLUTION:

Given;

• A train travels at a certain average speed for a distance of 63km.
• And then travels a distance of 72km at an average speed of 6km/hr more than its original speed.

To be found:

The original average speed if it takes 3 hours to complete the total journey.

Let the original average speed of the train be y km/h

Then,

The speed for the journey of 72 km will be (y+6) km/h

Now.

According to the question,

$\frac{63}{y}+\frac{72}{y+6}=3$

[As, Time = Distance/ Speed ]

On dividing both the sides by 3, we get,

$= \frac{21}{y} +\frac{24}{y+6}=1$

$\implies 21(y+6)+24y = y(y+6)$

$\implies 21y+126+24y = y^{2}+6y$

$\implies y^{2} - 39y - 126=0$

$\implies y^{2} - 42y +3y-126=0$

[By factorization]

$\implies y(y-42)+3(y-42) = 0$

$\implies (y+3)(y-42)=0$

$\implies \boxed{y=(-3)}, \boxed{y=(42)}$

Although, y is the average speed of the train, so y cannot be negative.

Hence,

The original average speed of the train is 42km/h.