Quadratic Equation Solution22





In this page quadratic equation solution22 we are going to see solution of the word problems of the topic quadratic equation.

Question 29

A train covers a distance of 300 km at a certain average speed. If its speed was decreased by 10km/hr, the journey would take 1 hour longer. What is the average speed.

Solution:

Let “x” be the average speed of the train

If its speed was decreased by 10km/hr, the journey would take 1 hour longer

So “x – 10” be the decreased speed

Time = Distance/Speed

Let T1 be the time taken by the train to cover the distance in the speed of x km/hr

 Let T2 be the time taken by the train to cover the distance in the speed of (x - 10) km/hr

T1 = 300/x

T2 = 300/(x – 10)

T1 – T2 = 1

[300/x] -  [300/(x-10)] = 1

300 [(1/(x - 10) – 1/(x)] = 1

300 [ x –  x + 10]/[x (x -10)] =  1

3000/(x² – 10 x) = 1

3000 = x² – 10 x

x² – 10 x = 3000

x² – 10 x – 3000 = 0

x² – 60 x + 50 x – 3000 = 0

x (x – 60) + 50 (x – 60) = 0

(x + 50)(x – 60) = 0

x + 50 = 0            x – 60 = 0

 x = - 50                     x = 60

Here x represents the speed of the train. So we should not take the negative value - 50 for x.

So speed of the 60 km/hr

Verification:

Time taken by the train with speed 60 km/hr

Time = Distance/speed

          = 300/60

          = 5 hours

Time taken by the train with speed 50 km/hr

Time = Distance/speed

          = 300/50

          = 6  hours

difference between time taken = 6 - 5 

                                           = 1 hour

quadratic equation solution22