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Home » 2. A train, traveling at a uniform speed for km, would have taken minutes less to travel the same distance if its speed were km/hr more. Find the original speed of the train.
2. A train, traveling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.
Class 10 | ICSE | Maths | Quadratic Equations | RD Sharma
2. A train, traveling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.

Solution:

Let the original speed of train be x km/hr

When increased by 5, speed of the train =\left( x+5 \right) km/hr

Using, speed = distance/ time

Time taken by the train for original uniform speed to cover 360 km =360/x hr.

And, time taken by the train for increased speed to cover 360 km =360/\left( x+5 \right) hr.

Given, that the difference in the times is 48 mins. ⇒ 48/60 hour

This can be expressed as below:

\frac{360}{x}-\frac{360}{\left( X+5 \right)}=\frac{48}{60}

\frac{360\left( x+5 \right)-360x}{x\left( x+5 \right)}=\frac{4}{5}

\frac{360+1800-360x}{{{x}^{2}}+5x}=\frac{4}{5}

1800\left( 5 \right)=4\left( {{x}^{2}}+5x \right)

9000=4{{x}^{2}}+20x

4{{x}^{2}}+20x-9000=0

{{x}^{2}}+5x-2250=0

{{x}^{2}}+50x-45x-2250=- [by factorisation method]

x\left( x+50 \right)-45\left( x+50 \right)=0

\left( x+50 \right)\left( x-45 \right)=0

\therefore x=-50 or x=45

Since, the speed of the train can never be negative x=-50 is not considered.

Therefore, the original speed of train is 45 km/hr.

 

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