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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Drunk on May 16, 2011, 06:16:59 pm

Title: Rates Question
Post by: Drunk on May 16, 2011, 06:16:59 pm

Hey guys, help me out plz?

A bus is due to reach its destination 75 km away at a certain time. The bus usually travels
with an average speed of x km/h. Its start is delayed by 18 minutes but, by increasing its
average speed by 12.5 km/h, the driver arrives on time.

a Find x.
b How long did the journey actually take?

Title: Re: Rates Question
Post by: luken93 on May 16, 2011, 07:35:58 pm

Let T = Usual Time

So, usually 75 = xT
T = 75/x

However, 75 = (x + 12.5)(T - 3/10)

75 = (x + 12.5)(75/x - 3/10)

Solving for x gives x = -125/2 or x = 50
Since x must be >0, x = 50

75 = (50 + 12.5)(T - 3/10)

75 = 62.5(T - 3/10)

12/10 = T - 3/10

T = 1.5hrs

Title: Re: Rates Question
Post by: aznxD on May 16, 2011, 07:42:25 pm

x = Usual speed
x + 12.5 = Delayed speed
108 / x = Usual time taken
108 / x - 0.3 = Delayed time taken

D = st
D normal = x(75/x)
D delayed = (x + 12.5)((75/x)-0.3)

Equate distances to find x.
x(75/x) = (x + 12.5)((75/x)-0.3)
x = 50

Sub x = 50 into 108 / x - 0.3 (Delayed time taken)
Time taken = 1.2 hours = 75 mins