linear algebra

Question

ted finshes a 42km race in 2.9hrs. the first 2 hours are run at x km/h and the rest of the race is run 2km/hr faster. what does x =?

Answer 1

This problem doesn't say it is a D-I-R-T (Distance-Is-Rate-times-Time) problem;  it expects you to realize that.

 

Ted ran part of the race (the first distance) at rate x in time=2 hrs.

Then, Ted ran the last part of the race (the second distance) at rate (x+2) in time=0.9 hrs.

Ted ran the total distance in 2.9 hrs.

 

We don't need to know any distances; we are asked to find the value of x (the rate Ted ran for the first part of the race).

    The total distance = the first distance + the second distance

               42 km   =   [ (x km/hr) * (2 hr) ]   +   [ ((x+2) km/hr) * (0.9 hr) ]

                                  [note: hr cancels out, leaving only km]

                    42 = 2x + 0.9(x+2)

                    42 = 2x +0.9x + 1.8       (distribute the 0.9)

                    40.2 = 2.9x                 (subtract 1.8 from both sides)

                    13.9 km/hr = x            (divide by 2.9 hr; round)

 

Checking (very important):

   Is 45 = (13.9)(2) + (15.9)*(0.9)      ?

       42 = 42.11     ?     Close  (we rounded rate)

Answer 2

We know two pieces of information: For the first 2 hours, Ted ran at x km/h and the rest of it Ted ran x+2 km/h.  Using how long Ted ran at each speed, we can set up an equation based on this.

 

Ted ran a total of 2.9 hours, but we need to separate this into two parts, the initial 2 hours and the later 0.9 hours.  If we multiply the initial 2 hours by x and the later by x+2 (sine he ran 2 km/h faster for the last part of the race), we get the equation:

 

2x+0.9*(x+2)=42

 

So distributing and then simplifying we get:

 

2x+0.9x+1.8=42

 

2.9x=40.2

 

x=40.2/2.9

 

Therefore:

 

x= about 13.9 km/h

 

I hope this helps! :)