Quadratic Equation Solution23





In this page quadratic equation solution23 we are going to see solution of the word problems of the topic quadratic equation.

Question 30

The time taken by a train to travel a distance of 250 km was reduced by 5/4 hours when average speed was increased by 10km/hr. Calculate the average speed.

Solution:

Distance to be covered = 250 km

Let “x” be the required average speed

If the average speed was increased by 10 km/hr

“x + 10” be the increased speed

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

Let T2 be the time taken to cover the distance in the average speed of (x + 10) km/hr

Time = Distance/Speed

 T 1 = 250/x

 T 2  = 250/(x + 10)

T1 – T2 = 5/4

250/x – 250/(x + 10) = 5/4

250 [ 1/x – 1/(x + 10) ] = 5/4

250 [ (x + 10 – x)/x(x + 10)] = 5/4

2500/(x2 + 10 x) = 5/4

2500 (4) = 5 (x2 + 10 x)

10000 = 5 x2 + 10 x

Now we are going to divide the whole equation by 5, so we get

  x2 + 10 x = 2000

  x2 + 10 x – 2000 = 0

  x2 + 50 x  - 40 x – 2000 = 0

x (x + 50) – 40 (x + 50) = 0

(x – 40) (x + 50) = 0

x – 40 = 0                 x + 50 = 0

 x = 40                            x = -50

Therefore the required average speed = 40 km/hr

Increased speed = (40 + 10) = 50 km/hr

Verification:

Time taken to cover the distance of 40 km/hr

 Time = distance/speed

           = 250/40

           = 25/4

Time taken to cover the distance of 50 km/hr

 Time = distance/speed

           = 250/50

           = 5

Difference between time taken = (25/4) – 5

                                           = (25- 20)/4

                                           = 5/4 hours   

quadratic equation solution23