Question

After covering a distance of 30 km,there is some defect in a train engine and there after,its speed is reduced to 4/5 of its original speed.Consequently, the train reaches its destination late by 45 minutes.Had it happened after covering 18 km more,the train would have reached 9 minutes earlier.Find the speed of the train and the distance of journey.

Answer 1

Answer:

..........,.....,..,........,........

Step-by-step explanation:

After covering a distance of 30 km,there is some defect in a train engine and there after,its speed is reduced to 4/5 of its original speed.Consequently, the train reaches its destination late by 45 minutes.Had it happened after covering 18 km more,the train would have reached 9 minutes earlier.Find the speed of the train and the distance of journey.

Answer 2

${\pmb{\underline{\sf{Required \ Solution ... }}}}$

• Covering a distance of 30 km
• Speed is reduced to 4/5 of its original speed

Let the Original speed of the train be x km/hr and the Distance of journey be y km.

» Time taken = (y/x) hr

✎ First Condition:

When defect in the engine occurs after covering a distance of 30 kilometre.

We've,

• speed for the first 30 km = x km/hr
• speed for the remaining (y - 30)km = 4x/5 km/hr
• Time taken to cover 30 kilometre = 30/x hrs

» Time taken to cover (y - 30)km = ${\sf{ \dfrac{y-30}{(4x/5)} hr }}$

$\colon\implies{\sf{ \dfrac{5}{4x} (y - 30) hr }}$

According to the Question, we have;

$\colon\implies{\sf{ \dfrac{30}{x} + \dfrac{5}{4x} (y-30) = \dfrac{y}{x} + \dfrac{45}{60} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{30}{x} + \dfrac{5y-150}{4x} = \dfrac{y}{x} + \dfrac{3}{4} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{120+5y-150}{ \cancel{4x} } = \dfrac{4y+3x}{ \cancel{4x} } }} \\ \\ \\ \colon\implies{\sf{ 120 + 5y - 150 = 4y + 3x }} \\ \\ \\ \colon\implies{\sf{ 5y - 4y = 30+3x }} \\ \\ \colon\implies{\sf{ 3x-y+30 = 0 \ \ \ \ \ \ \ \cdots[1] }}$

✎ Second condition:

When defect in the engine occurs after covering a distance of 48 kilometres.

• speed for first 48 km = x km/hr
• speed for the remaining (y - 48)km = 4x/5 km/hr
• Time taken to cover 48 km = 48/x hr

» Time taken to cover (y - 48)km = ${\sf{ \dfrac{y-48}{(4x/5)} hr }}$

$\colon\implies{\pmb{\sf{ \left( \dfrac{5(y-48)}{4x} hr \right) }}}$

According to the Question, The train now reaches 9 minutes Earlier.

As We know that, Firstly Train was reached 45 late but now this time it reaches 9 minutes Earlier so Train now is 36 minutes Later.

$\colon\implies{\sf{ \dfrac{48}{x} + \dfrac{5(y-48)}{4x} = \dfrac{y}{x} + \dfrac{36}{60} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{48}{x} + \dfrac{5y-240)}{4x} = \dfrac{y}{x} + \dfrac{3}{5} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{192+5y-240}{4x} = \dfrac{5y+3x}{5x} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{5y-48}{4} = \dfrac{5y+3x}{5} }} \\ \\ \\ \colon\implies{\sf{ 5(5y-48) = 4(5y+3x) }} \\ \\ \\ \colon\implies{\sf{ 25y - 240 = 20y + 12x }} \\ \\ \\ \colon\implies{\sf{ 12x-5y+240 = 0 \ \ \ \ \ \ \ \ \cdots[2] }}$

${\pmb{\underline{\sf{Simultaneous \ Equation ... }}}}$

We have to use the system of Simultaneous as:-

$\circ \ {\sf\green{ 3x - y + 30 = 0 }} \\ \circ \ {\sf\green{ 12x - 5y + 240 = 0 }}$

By Using Multiplication Method, we've that:

$\colon\implies{\sf{ \dfrac{x}{-240+150} = \dfrac{-y}{720-360} = \dfrac{1}{-15+12} }} \\ \\ \\ \colon\implies{\sf{ \dfrac{x}{-90} = \dfrac{-y}{360} = \dfrac{1}{-3} }} \\ \\ \\ \colon\implies{\sf{ -3x = -90 , -3y = -360 }} \\ \\ \\ \colon\implies{\sf{ \left( x = \dfrac{90}{3} \right) , \left( y = \dfrac{360}{3} \right) }} \\ \\ \\ \colon\implies{\sf{ x = 30 , y = 120 }}$

Hence,

The Original speed of the train is 30 km/hr and the Distance of the journey is 120 km.