Quadratic Equation Solution21





In this page quadratic equation solution21 we are going to see solution of the word problems of the topic quadratic equation.

Question 28

The distance between two stations A and B is 192 km. Traveling by a fast train takes 48 minutes less than another train. Calculate the speed of the fast train if the speeds of two trains differ by 20 km/hr

Solution:

Distance between two stations A and B = 192 km

Fast train takes 48 minutes less then the time taken by the slow train

Let “x” be the speed taken by the fast train

speeds of two trains differ by 20 km/hr

So speed of slow train is “x – 20”

Time = Distance/speed

Let T1 be the time taken by the fast train

Let T2 be the time taken by the slow train

T1 = 192/x

T2 = 192/(x – 20)

We are going to change minutes as hours. For that we have to divide 48 by 60

48/60 = 4/5 hours

T1 – T2 = 4/5

[192/(x -20) – 192/x] = 4/5

192[ (x – x + 20)/x(x - 20)] = 4/5

192(20) /x² – 20 x = 4/5

3840 /x² – 20 x = 4/5

3840 (5) = 4 (x² – 20 x)

19200 = 4 x² – 80 x

Now we are going to divide the whole equation by 4,So we get

4800 = x² – 20 x

x² – 20 x – 4800 = 0

x² – 60 x + 40 x – 4800 = 0

x (x – 60) + 40 (x – 60) = 0

(x – 60) ( x + 40) = 0

x – 60 = 0             x + 40 = 0

 x = 60                       x = -40

Here x represents the speed of the slow train. So we should not take the negative value - 40 for x.

So speed of the faster train = 60 km/hr

Speed of slow train = x – 20

                                   = 60 - 20            

                                   = 40 km/hr

 Speed of the faster train = 60 km/hr

Speed of the slow train = 40 km/hr

Verification:

speeds of two trains differ by 20 km/hr

60 – 40 = 20

quadratic equation solution21