EDIT: lol I just realized that you've already figured it out. Just giving my explanation anyway so that you don't get discouraged from coming to the forums if you need help.
Here it is:
Remember the equation r=d/t
So we need the rate, distance, and time
We have our given--
Distance:
9km for the first half
6km for the next third
3km for the last stretch
Time:
Remember that what we are looking for is kilometers per *hour* so you have to bear in mind that the unit you use in your equation is in the hour form. 79 minutes, converted to hours, is equal to 79/60. So we have 79/60 hours in total.
Rate:
Let the normal speed = x
For the 9km part, the rate is normal, so it is x
For the 6km part, the rate, increased by 25%, is 1.25x
For the 3km part, the rate of 1.25x is increased further by 20%, so that gives us 1.20(1.25x) or simply 1.50x
What we are looking for is the rate. However, each segment of the race was completed at different times. If you think about it, the time it took to complete each segment of the race should equal to 79/60 hours when added together. I mean:
(79/60 hours) = (hours it took for the 9k segment) + (hours it took for the 6k segment) + (hours it took for the 3k segment)
So we can have a different value of time for each of the 3 segments, which when added up equals 79/60 hr.
Let:
z_1 = time in hours it took for the 9k segment
z_2 = time in hours it took for the 9k segment
z_3 = time in hours it took for the 9k segment
Using the formula r=d/t, we can derive that:
x=(9km) / (z_1)
1.25x=(6km) / (z_2)
1.50x=(3km) / (z_3)
Remember that for this problem, (79/60) = (z_1 + z_2 + z_3). So if we isolate the z variables from the last 3 equations like so...
z_1 = (9km) / (x)
z_2 = (6km) / (1.25x)
z_3 = (3km) / (1.50x)
We can substitute the (79/60 hr) = (z_1 + z_2 + z_3) equation with:
(79/60 hr) = ( (9km)/(x) ) + ( (6km)/(1.25x) ) + ( (3km) / (1.50x) )
We want to get x. How do we do that? Somehow we have to isolate x on one side of the equation. Notice that on the right side of the equation, all the x's are on the denominators. Since they have different coefficients, we cannot simply add up those 3 fractions. We can, however, use the distributive property to "bring out" the x and use it later for substitution. Take for example:
A = (B/D) + (C/2D)
is equal to:
A = (1/D) * (B + (C/2))
or:
A = (B + (C/2)) / D
Get it? so we apply the same property to the equation:
(79/60 hr) = ( (9km)/(x) ) + ( (6km)/(1.25x) ) + ( (3km) / (1.50x) )
is equal to:
(79/60 hr) = ((9km)+(6km/1.25)+(3km/1.50)) / x
We can begin isolating x by moving it to the left side, which would allow x to become part of the numerator. We can do this using simple algebra - multiply both sides by x to cancel out x on the right side and to move x on the left:
(x) * (79/60 hr) = ((9km)+(6km/1.25)+(3km/1.50))
Then divide the entire right side by (79/60 hr), once again using algebra:
(x) = ((9km)+(6km/1.25)+(3km/1.50)) / (79/60 hr)
Plug it into the calculator and what have you got?
x = (9 + 4.8 + 2)km / (1.3167)hr
x = 12 km/hr
l33tness.
