# It takes Maliha two hours more to complete a 50 km journey than it takes ashin to complete a 40 km journey. If the average speed of Maliha for the...journey is 5 km/h less than Ashin, calculate the...

It takes Maliha two hours more to complete a 50 km journey than it takes ashin to complete a 40 km journey. If the average speed of Maliha for the...

journey is 5 km/h less than Ashin, calculate the average speed of each girl by using quadratic equation.

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We know that distance = speed*time. So we can set up equations.

Let Ashin's speed be x. That means that Maliha's speed is x-5.

So now we know that

40/x = 50/(x-5) - 2

because the time it took Maliha to go 50 km at her speed is 2 hours more than it took Ashin to go 40 km at her speed.

So then what you need to do is multiply both sides by x(x-5)

That will get you

40 (x-5) = 50x -2x(x-5)

Eventually, this will simplify down to

0 = x^2-10x-100

This factors out to (x=6.18)(x-16.18)

Since her speed must be positive, x = 16.18

So Ashin's speed is 16.18 kph and Maliha's speed is 11.18 kph

It took Ashin 2.472 hours to go 40 km and it took Maliha 4.472 hours to go 50 km

Here the speed of the Mahila and Ahin are to be determined. Let it be x km/h for Mahila.Then , Ashin's speed is x+5 km/h

The time taken by Mahila for 50 kms =50/x ...........(1)

The time taken by Ashin tofor 40 kms = 40(x+5)....(2)

By the condition that Mahila takes 2 hours extra for her 50kms than that of Ashin for 40kms, we get the relation between (1) and (2):

50/x hrs- 40/(x+5) hrs = 2 hrs.......(3). Now solve for x and you get Mahilas speed, and then x+5 km/h will be Ashin's speed. Multiplying by the x(x+5) both sides of (3):

50(x+5)-40x = 2x(x+5) or

50x+250-40x=2x^2+10x or

10x+250=2x^2+10x or

2x^2 = 125 or x = (125)^1/2 = 5(5)^(1/2) kms/h is Mahila's speed = 11.18033989 kms/h.

Ashin's speed = x+5 = 5+5(5)^(1/2) kms/h = 16.18033989 kms

Tally:

The time taken by Mahila ans AShin are 50/[5*5^(1/2) = 4.4721... hrs and 40/{5+5*5^(1/2) = 2.4721... hrs)