## anonymous one year ago On a certain map, a distance of 40 miles is represented by 1cm. How many miles are represented by 2.2cm on the map ?

1. anonymous

okay so we know that 2 cm is 80 miles, now we just need to find what the .2 on the map is

2. anonymous

Should I multiply ?

3. anonymous

okay, so if I recall, you should find 20% of 40 and add 80 to that, since the .2 in 2.2 is really 0.20 which is equal to 20%

4. anonymous

5. anonymous

Which is 160

6. anonymous

no

7. anonymous

80+8=88

8. anonymous

I mean 88

9. anonymous

right

10. anonymous

Sorry I'm not so focused

11. anonymous

I am not really sure if the .2 in 2.2 stands for a .02 or a .20

12. anonymous

but I think this is the answer: 88

13. anonymous

@OregonDuck

14. mathstudent55

Since a map scale is a ratio, map scale problems can usually be solved by a proportion.

15. mathstudent55

You are given a real distance and a map distance: Real distance Map distance 40 miles 1 cm This allows you to establish a ratio, which is the scale of the map: 40 miles : 1 cm

16. mathstudent55

Now you are given a map distance and an unknown real distance We can let x represent the unknown real distance. Real distance Map distance x 2.2 cm This can be written as a ratio: x : 2.2 cm

17. mathstudent55

The map scale ratio equals the ratio with the unknown real distance, so we write an equation showing the ratios are equal. An equation that is two equal ratios is called a proportion. Also, there are several ways of writing a ration. The ratio of x to y can be written in these ways: x to y x : y x/y Let's write an equation with our ratios, but at the same time let's change the ratios to the fraction form: $$\large \dfrac{40~miles}{1~cm} = \dfrac{x}{2.2~cm}$$

18. anonymous

Should I multiply 40 by 2.2 ?

19. mathstudent55

Now we need to solve the proportion for x. We cross multiply: $$1 \times x = 40 \times 2.2$$ Now we simplify: $$x = 88$$ Answer: the distance is 88 miles

20. anonymous

I understand it so much better now thank you so much I really appreciated that @mathstudent55 and @heretohelpalways :)

21. anonymous