# Bicyclist

• Jan 23rd 2009, 07:53 AM
metlx
Bicyclist
A bicylcist cycled from town A to town B and back. He returned into town A in 4 hours.
His speed on the flat part of the road was 16km/h. His speed on the slope (going up) was 12km/h, and his speed on the slope (going down) was 24km/h.

What is the distance between town A and B? (there is no info missing - thats all that it is).

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I think I manged to solve it with some logical thinking but I am still having problems doing it properly (with equations and all). I think the answer is 32km. (I won't explain how I got that because I don't want it to affect on your thinking..)
• Jan 23rd 2009, 08:34 AM
earboth
Quote:

Originally Posted by metlx
A bicylcist cycled from town A to town B and back. He returned into town A in 4 hours.
His speed on the flat part of the road was 16km/h. His speed on the slope (going up) was 12km/h, and his speed on the slope (going down) was 24km/h.

What is the distance between town A and B? (there is no info missing - thats all that it is).

----------------
I think I manged to solve it with some logical thinking but I am still having problems doing it properly (with equations and all). I think the answer is 32km. (I won't explain how I got that because I don't want it to affect on your thinking..)

I've got the same result and since we are the majority you are right! :D

What I've done:

1. Calculate the average speed of the trip from A to B:

$v_1 = \dfrac{2\cdot 16 \cdot 12}{16+12} = \dfrac{96}7\ \frac{km}{h}$

2. Calculate the average speed of the trip from B to A:

$v_2 = \dfrac{2\cdot 16 \cdot 24}{16+24} = 19.2\ \frac{km}{h}$

3. Calculate the average speed of the complete trip:

$v= \dfrac{2\cdot \frac{96}7 \cdot 19.2}{\frac{96}7+19.2} = 16\ \frac{km}{h}$

4. The total length of the trip is therefore $4\ h \cdot 16\ \frac{km}h = 64 \ km$. And that means the distance between A and B must be 32 km
• Jan 23rd 2009, 09:12 AM
metlx
Quote:

Originally Posted by earboth
1. Calculate the average speed of the trip from A to B:

$v_1 = \dfrac{2\cdot 16 \cdot 12}{16+12} = \dfrac{96}7\ \frac{km}{h}$

2. Calculate the average speed of the trip from B to A:

$v_2 = \dfrac{2\cdot 16 \cdot 24}{16+24} = 19.2\ \frac{km}{h}$

hmm. why 2 * 16 * 12 and 2 * 16* 24? why the 2 at the beginning?

btw, i got that he is cycling 2.5h (flat part + uphill) and 1.5 hour (downhill + flat part).
since the uphill speed : downhill speed = 1:2 -> that means the time he spends going up is twice as big as the time he spends going down.

so the hill part = 3x
2 hours up = 2* 12 = 24km
1 hour down = 1 *24 = 24km
then i guessed that 3x = 3 hours and that the remaining 1 hour is the time he spent on the flat part of the road (1 hour, both ways -> flat part = 8km in length)
flat part both ways = 2*8 km = 16km

24+24+16 = 64
AB = 32

but i did a logical jump (or whatever) there :P
the teacher wanted to know where i got that 3x = 3 hours.. i didnt have an explanation and i still dont have it..
• Jan 23rd 2009, 10:16 AM
Soroban
Herllo, metlx!

Quote:

A bicylcist cycled from town $A$ to town $B$ and back in 4 hours.
His speed on the flat part of the road was 16 km/h.
His speed going uphill was 12 km/h, and his speed going downhill was 24 km/h.

What is the distance between towns $A$ and $B$?

Code:

                      *                 x  *    *  y                   *          *                 *              *     A * - * - * - - - - - - - - - - * - * - * - * B       : - w - :                    : - - z - - :
Let $w,x,y,z$ be the distances for the four phases of the ride.

$\text{From }A\text{ to }B.$

. . $\begin{array}{cc}\text{He rode }w\text{ km at 16 kph.} & \text{This took: }\frac{w}{16}\text{ hours.} \\
\text{He rode }x\text{ km at 12 kph.} & \text{This took: }\frac{x}{12}\text{ hours.} \\
\text{He rode: }y\text{ km at 24 kph.} &\text{This took: }\frac{y}{24}\text{ hours.} \\
\text{He rode: }z\text{ km at 16 kph.} & \text{This took: }\frac{z}{16}\text{ hours.}\end{array}$

His ride from A to B took: . $\frac{w}{16} + \frac{x}{12} + \frac{y}{24} + \frac{z}{16}\:=\:\frac{3w+4x+2y+3z}{48}$ hours.

$\text{From }B\text{ to }A.$

. . $\begin{array}{cc}\text{He rode }z\text{ km at 16 kph.} & \text{This took: }\frac{z}{16}\text{ hours.} \\
\text{He rode }y\text{ km at 12 kph.} & \text{This took: }\frac{y}{12}\text{ hours.} \\
\text{He rode }x\text{ km at 24 kph.} & \text{This took: }\frac{x}{24}\text{ hours.} \\
\text{He rode }w\text{ km at 16 kph.} & \text{This took: }\frac{w}{16}\text{ hours.} \end{array}$

His ride from B to A took: . $\frac{z}{16} + \frac{y}{12} + \frac{x}{24} + \frac{w}{16} \:=\:\frac{3w + 2x + 4y + 3z}{48}$ hours.

The round trip took 4 hours.

. . $\frac{3w + 4x + 2y + 3z}{48} + \frac{3w + 2x + 4y + 3z}{48} \:=\:4 \quad\Rightarrow\quad 6w + 6x + 6y + 6z \:=\:192$

Therefore: . $w + x + y + z \:=\:32\text{ km}$

• Jan 23rd 2009, 11:09 AM
metlx
wow..just wow!

why are there 4 phases?
aren't there just 3 all together?
2 in one way: w (flat part) and x (uphill)
2 in the opposite direction: y (downhill) and w (flat part)
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