# It takes Meihua 2 hours more to complete a 50 km journey than it takes Ailin to complete a 40 km journey. If the average speed of Meihua for the...journey is 5 km/h less than Ailin, calculate the...

It takes Meihua 2 hours more to complete a 50 km journey than it takes Ailin to complete a 40 km journey. If the average speed of Meihua for the...

journey is 5 km/h less than Ailin, calculate the average speed of each girl.

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Ailin runs 40 km in T hours at Va speed: 40 = Va*T

Meihua runs 5 km/h slower than Ailin: Va = Vm - 5

Meihua runs 40 km in 2 hours longer than T.

So,

Vm = Va - 5

40 = Va*T

50 = Vm(T + 2)

50 = (Va - 5)*(T + 2) = Va*T + 2Va - 5T - 10

50 = 40 + 2Va - 5*40/Va - 10

0 = -20 + 2Va - 5*40/Va

0 = -20Va + 2Va^2 - 5*40

Using the quadratic formula, Va = -6.18034 or 16.1803

Taking the positive value, Ailin traveled 16.2 km/h. Thus Meihua traveled Va - 5 = 11.2 km/h.

Let A and M be the average speed of Allin and Mehua.

Then by the condition s of the problem:

2+40/A = 50/M (1)

A-M=5 (2)

From (1) M =50A/(2A+40)=25A/(A+20). Substitute this value of M in (2) to get:

A-25A/(A+20)=5 .

Multiply by A+20 to get:

A(A+20)-25A=5(A+20)

A^2-5A=5A+100

Add -5A+25 to both sides:

A^2-10A+25 =100+25 =125

(A-5)^2 = 125

A-5 =sqrt(125)=5sqrt5

A=** 5+5sqrt5 = 16.18033989 **kmph

M = A-5 =** 5sqrt5** = **11.18033989** kmph.

Tally:

Time taken for M to complete 50 km = 50 /(5sqrt5) =2qsrt5.Time take. for A to mplete 40 km = 40/{5+5sqrt5} = 8/{ 1+sqrt5} =(8/4){sqrt5-1}=2{sqrt5-1}, which is less than 2hrs of M's time to go 50 kms. So it is talleyed.