Hi nikitathegreat,D1/D2 = S1/S2 only holds when each whole trip runs at one constant speed. Your second trip does not, and that is where it breaks.
You wrote that trip's speed as
13x, from x + 12x. Speeds of back-to-back segments never add. Walking one stretch at x and driving another at 12x does not happen at 13x, and 13x is not the average speed either.
Build on time instead, segment by segment. Let a be the distance to the house, b the distance to the hall, and v the walking speed:
- Walk straight to the hall: time =
b/v- Walk home, then drive: time =
a/v + (a+b)/(12v)Set them equal and multiply through by 12v:
12b = 12a + (a + b)12b = 13a + b11b = 13aSo
a/b = 11/13, choice
C. Note the driving distance is
a + b: back home first, then all the way past the starting point to the hall. Folding that into a single 13x speed hides it.
Why speeds cannot add. Walk 2 miles at 2 mph, then drive 12 miles at 12 mph.
- Real time:
2/2 + 12/12 = 2 hours
- Adding the speeds, 14 miles at 14 mph:
1 hour
Multi-speed trips have to be handled one segment at a time, through time.
Answer: Cnikitathegreat
House (GC) -------------------George---------Concert Hall (GH)
Since we are saying time is constant here, can we say that D1/D2 = S1/S2?
We need to find GH/ GC = GH = George's distance to his house and GC is George distance to his concert hall.
So, when D1/D2 = S1/S2
GC/GH+GH+GC= 1x/13x since George to concert hall speed is x and 13x (x+12x) for D2
So, we get 6GC = GH
6GC/ GC = 6:1
So, I am getting the answer 6:1
Can you tell me what is wrong here?