Question

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

Answer 1

Answer:

The original speed of the train is 45km/hour.

Step-by-step explanation:

Sum of the two distances = 63+72= 135

Time = sum of speed / total time

Time = 135÷3= 45km/hour

Original Average speed = 45km/hour

Answer 2

$\bullet\:{\boxed{\tt{\red{{Given :}}}}}$

❍ A train travels at a certain average speed for a distance of 63 km.

❍ And travels a distance of 72 km at an average speed of 6 km/hr more than its original speed.

❍ Time taken by the train for whole journey = 3hours.

$\bullet\:{\boxed{\tt{\red{{To \: Find :}}}}}$

❍ The original average speed.

$\bullet\:{\boxed{\tt{\red{{Solution :}}}}}$

Let,

The original average speed of the train be x km/hr.

As we know,

$\star \: \sf Time ={{\boxed{{\purple{\sf \dfrac{Distance}{ Speed}}}}}} \: \star$

In first case, $\sf ( t_{1} )$

Time taken by the train to cover 63 km/hr = $\sf \dfrac{63}{x}$

In second case, $\sf ( t_{2} )$

Time taken to cover distance of 72 km at an average speed of 6 km/hr more than its original speed = $\sf \dfrac{72}{x + 6}$

According to the question,

Time taken by the train for whole journey = 3hours

$\longrightarrow \sf \: t_{1} + t_{2} = 3$

$\longrightarrow \sf \: \dfrac{63}{x} + \dfrac{72}{x + 6} = 3$

$\longrightarrow \sf \: \dfrac{ 63 (x + 6)+72x}{x(x + 6)} = 3$

$\longrightarrow \sf \: \dfrac{ 63 (x + 6)+72x}{ {x}^{2} + 6x} = 3$

$\longrightarrow \sf \: \dfrac{ 63x + 378+72x}{ {x}^{2} + 6x} = 3$

$\longrightarrow \sf \: 63x + 378+72x = 3 ({x}^{2} + 6x)$

$\longrightarrow \sf \: 63x + 378+72x = 3{x}^{2} + 18x$

$\longrightarrow \sf \: 135x + 378 = 3{x}^{2} + 18x$

$\longrightarrow \sf \: 0 = 3{x}^{2} + 18x - 135x - 378$

$\longrightarrow \sf \: 0 = 3{x}^{2} - 117x - 378$

$\longrightarrow \sf \: 3{x}^{2} - 117x - 378 = 0$

$\longrightarrow \sf \: 3({x}^{2} - 39x - 126) = 0$

$\longrightarrow \sf \: {x}^{2} - 39x - 126 = \dfrac{0}{3}$

$\longrightarrow \sf \: {x}^{2} - 39x - 126 = 0$

$\longrightarrow \sf \: {x}^{2} - 42x + 3x - 126 = 0$

$\longrightarrow \sf \: x(x - 42) + 3(x - 42) = 0$

$\longrightarrow \sf \: (x + 3)(x - 42) = 0$

$\longrightarrow \sf \: (x + 3) = 0 \: or \: (x - 42) = 0$

$\longrightarrow \sf \: x = - 3 \: (rejected) \: or \: x = 42$

But the speed cannot be negative.

$\therefore \sf {{\underline{\boxed{\purple{\sf{x = 42 }}}}}}$

So, The original average speed of the train be 42 km/hr.

$\bullet\:{\boxed{\tt{\red{{Verification :}}}}}$

We were given that,

Time taken by the train for whole journey = 3hours

$\longrightarrow \sf \: t_{1} + t_{2} = 3$

$\longrightarrow \sf \: \dfrac{63}{x} + \dfrac{72}{x + 6} = 3$

Putting x as 42

$\longrightarrow \sf \: \dfrac{63}{42} + \dfrac{72}{42 + 6} = 3$

$\longrightarrow \sf \: \dfrac{63}{42} + \dfrac{72}{48} = 3$

$\longrightarrow \sf \dfrac{504 + 504}{336} = 3$

$\longrightarrow \sf \dfrac{1008}{336} = 3$

$\longrightarrow \sf 1008 = 3(336)$

$\longrightarrow \sf 1008 = 1008$

Hence verified

${\rule{300pt}{3pt}}$

Hope it helps u sis !! ^w^