Question

How much more is nitrogen than oxygen?

A) $99\% $
B) $78\% $
C) $57\% $
D) None of these.

Answer 1

Hint: In this question, we are given a pie-chart which shows the division of gases in the atmosphere. We have been asked to find how much more nitrogen is from the oxygen. For this, write down their individual percentages and then simply subtract them from each other. You will get your answer.

Complete step-by-step solution:
We are given a pie-chart which shows the division of gases in the atmosphere. It can be seen that nitrogen is the highest among all the gases followed by oxygen and other gases. The content of the gases given in the pie chart, is written below:

Gases	Content (in percentage)
Nitrogen	$78\% $
Oxygen	$21\% $
Other gases	$1\% $

Now, we have been asked how much more nitrogen is than oxygen. For this, we simply need to subtract the content of oxygen from the content of nitrogen.
$ \to $ Percentage of nitrogen $ = 78\% $
$ \to $ Percentage of oxygen $ = 21\% $
Percentage by which nitrogen is more than oxygen= Percentage of nitrogen- Percentage of oxygen
Percentage by which nitrogen is more than oxygen $ = 78\% - 21\% $
Percentage by which nitrogen is more than oxygen $ = 57\% $

Therefore, nitrogen is more than oxygen by Option C) $57\% $.

Note: You can also find the degrees of these gases in the given pie chart using the given percentages. You have to simply divide each percentage by 100 and then multiply it with 360. After simplifying, you will get the degrees of area covered by each gas.
For example: take the gases given in this question and their percentage and use the steps mentioned above. You can see how it is done below:
$ \Rightarrow $ Nitrogen = $\dfrac{{78}}{{100}} \times 360 = 280.8^\circ $
$ \Rightarrow $ Oxygen = $\dfrac{{21}}{{100}} \times 360 = 75.60^\circ $
$ \Rightarrow $ Other gases = $\dfrac{1}{{100}} \times 360 = 3.6^\circ $
Now, we can add all of them to verify whether the answer is correct or not.
$ \Rightarrow 280.8^\circ + 75.6^\circ + 3.6^\circ = 360^\circ $
Therefore, our answer is correct.