- 29 Oct '04 08:32Consider two cities (A and B) that are between mountain ranges (the trip has several uphill and downhill parts).

Mary's car does 72 kph downhill, 63 kph when it's flat and 56 kph uphill.

Mary takes 4 hours to go from city A to town B and 4h40min to return.

How far are the two cities?

(disregard traffic or acceleration) - 29 Oct '04 09:36

Let's devide the route in three parts; X km uphill on the way to B, Y km flat and Z km downhill on the way to B.*Originally posted by Palynka***Consider two cities (A and B) that are between mountain ranges (the trip has several uphill and downhill parts).**

Mary's car does 72 kph downhill, 63 kph when it's flat and 56 kph uphill.

Mary takes 4 hours to go from city A to town B and 4h40min to return.

How far are the two cities?

(disregard traffic or acceleration)

So X/56 + Y/63 + Z/72 = 4 hours

On the way back she has X km downhill, Y km flat and Z km uphill.

So X/72 + Y/63 + Z/56 = 4 2/3 hour

Multiplying the equations with 504 gives:

9X+8Y+7Z=2016

7X+8Y+9Z=2352

Substracting the second from the first;

-2X + 2Z = 336

So Z = 168 + X

We take this result and place it in one of the equations we had:

9X + 8Y + 7(168 + X) = 2016

Solving for Y gives;

Y = 105 - 2X

So all possible solutions for (X, Y, Z) are of the form (X, 105-2X, 168+X)

The distance between A and B is X+Y+Z = 273 km - 29 Oct '04 22:42

I'm sure the proper method is much more sophisticated. But if I got stranded on a desert island without my trusty calculator and needed to do this I'd just pick a number and square it seeing if it was higher or lower then readjust and continue.*Originally posted by piderman***Care to explain how it works (and the mechanics behind it)? Say I want to take the square root of 5764801**

For 5764801 my guesses were:

2000 too low

3000 too high(increasing by a large amount here means less guesses as now you have a range.)

2500 too high (halving my range)

2250 too low (again halving my range. Keep doing this.)

2375 too low

2438 too high (sticking to whole numbers as they are easier to square in my head.)

2407 too high

2391 too low

2399 too low

2403 too high

2401 exactly right

Takes some time but it can be done this way. Of course if the number doesn't have an exact square root it takes longer depending on how accurate you want your answer. - 31 Oct '04 12:50

The mechanics on how it works is not too hard to see once you see the method. It's the method that is hard to type down :p*Originally posted by piderman***Care to explain how it works (and the mechanics behind it)? Say I want to take the square root of 5764801**

OK, the process is supposed to look like a division process.

devide the number in parts of two, starting on the right;

5 76 48 01

We start with the 5. The biggest square smaller then 5 is 2*2, so the first digit of the root is a TWO

5-4 = 1 and we take the next two digits of the square making 176

Now we had 2*2, instead of multiplying we add the two two's now, giving a 4. Now we need to find x such that (40+x)*x is as close under 176 as possible.

In this situation x = 4 since 44*4=176. So the second digit of our root is FOUR

176-176=0 and we add the next two digits of our square to make 48

We had 44*4, and now instead of multiplying we add the two numbers to make 44+4=48. Now we need to find x so that (480+x)*x is as close under 48 as possible.

480*0=0 so the third digit of our root is ZERO.

48-0 = 48 and adding the last two digits of our sqaure to it gives 4801.

480+0=480, so now we need to find x so that (4800+x)*x is as close under 4801 as possible. This time x=1 and this leaves remainder 0, meaning we're done. The last digit of our root is 1

So sqrt(5764801)=2401.

The idea behind this method is that you start with a square of size 5, and take a square out of it (here size 4). What remains is an edge consisting of a square and two rectangles. Now you increase the size of the sides by a factor 10 (this is adding the next two digits of our square, as the square we had increases by a factor 100).

Now we try to find x such that 20x + 20x + x^2 approximates the edge as close as possible. above can be rewritten as (40+x)*x wich is the formula we solved.

You keep getting edges as remainder, and you keep approximating it as close as possible, increasing the edges by a factor 10 each time, until you're done.

I hope this made sense, it's much easier to explain on paper