Arithmetic

[tehmatriks]

1. The problem statement, all variables and given/known data
The scale on a map is 1:25000. The lenght of a wall on the map is 3.2 mm. Find the actual length in metres.


2. Relevant equations



3. The attempt at a solution

[Mark44]

EDIT: Fixed momentary lapse of brain.
Each 1 mm distance on the map represents 25,000 mm (= 25 meters) on the ground. If the length of some feature on the map happened to be 10 mm, you would set up a proportion like this, where x represents the actual length of the feature, in mm.

\frac{1}{25000}=\frac{10}{x}

After solving for x in this equation, convert to meters.

[tehmatriks]

Each 1 mm distance on the map represents 25,000 mm (= 2.5 meters) on the ground. If the length of some feature on the map happened to be 10 mm, you would set up a proportion like this, where x represents the actual length of the feature, in mm.

\frac{1}{25000}=\frac{10}{x}

After solving for x in this equation, convert to meters.

25,000 x 3.2 = 80,000 x 10-3 = 80 cm

[Mentallic]

25,000 mm (= 2.5 meters)
Either a typo or a brain fart, 25m :tongue:

25,000 x 3.2 = 80,000 x 10-3 = 80 cm
It shouldn't be cm.
25,000 x 3.2 mm = 80,000 mm = 80,000 x 10-3 m = 80 m

[tehmatriks]

yea, that was a typo

thanks fellas, i've just gotten to this chapter and the examples always leave out the good stuff, so many thanks.

[Mark44]

Either a typo or a brain fart, 25m :tongue:
Unfortunately, the latter. It looked odd to me, but I still didn't catch it.

It shouldn't be cm.
25,000 x 3.2 mm = 80,000 mm = 80,000 x 10-3 m = 80 m