Solve using systems

Question

Use the information in the circle graph to determine what percent of Earth's atmosphere is nitrogen, is oxygen, and is other gases.

 

?Nitrogen: This is 12% more than three times the sum of the recent oxygen and the percent other gases.

 

?Other gases: This is 20% less than the percent oxygen.

Answer 1

nitrogen = 3(oxygen + other gases) + .12(oxygen + other gases)

 

other gases = oxygen - .20oxygen

 

nitrogen + oxygen + other gases = 100% or 1.00

 

 [3(oxygen + other gases) + .12(oxygen + other gases)]  +  [oxygen]  +  [oxygen - .20oxygen]  = 100% or 1.00

 

[3(oxygen + oxygen - .20oxygen) + .12(oxygen + oxygen - .20oxygen)]  + [ oxygen]  +  oxygen - .20oxygen = 1.00

 

6 oxygen - .60 oxygen + .12 oxygen + .12 oxygen - .024 oxygen + oxygen + oxygen - .20 oxygen = 1.00

 

8.24 oxygen - .824 oxygen = 1.00      7.416 oxygen = 1.00    Divide both sides by 7.416.  oxygen = .1348 or 13.48%

 

other gases = oxygen - .20 oxygen      other gases = .1348 - .20(.1348)    other gases = .10784 or 10.78%

 

nitrogen = 3(oxygen + other gases) + .12(oxygen + other gases)   nitrogen = 3(.1348 + .1078) + .12(.1348 + .1078)

 

nitrogen = .756912 or 75.69%

 

 

 

 

Answer 2

N = percent of Nitrogen       x = %oxygen        g = % other gases

 

N = 3(x+g) + .12*3(x+g) = 3.36x + 3.36g

g = x - .2x = .8x

 

 

 

N + x + g = 100

(3.36x + 3.36g) + x + (.8x) = 100

5.16 x + 3.36(.8x) = 100

7.848x = 100

x = 12.7%

g = .8(.127) = 10.2%

N =100 - 12.7-10.2 = 77.1%

 

I just looked up the actual statistics....N = 78%, X = 21%, g = 1%

 

So since I don't have a picture of your circle graph, I think at least my nitrogen level is close. It is just a math problem and not reality I guess.

 

Hope this helped