Question

To save on helium costs, a balloon is inflated with both helium and nitrogen gas. Between the two gases, the balloon can be inflated up to 8 litres in volume. The density of helium is 0.2 grams per litre and the density of nitrogen is 1.30 grams per litre. The balloon must be filled so that the volumetric average density of the balloon is lower than that of air, which has a density of 1.20 grams per litre. Which of the following systems of inequalities best describes how the balloon will be filled, if $x$ represents the number of litres of helium and $y$ represents the number of litres of nitrogen?
A. $\left\{ \begin{align}
  & x+y>8 \\
 & 20x+130y>120 \\
\end{align} \right.$
B. $\left\{ \begin{align}
  & x+y=8 \\
 & \dfrac{0.2x+1.30y}{2}<1.20 \\
\end{align} \right.$
C. $\left\{ \begin{align}
  & x+y\le 8 \\
 & 0.20\left( \dfrac{x}{x++y} \right)+1.30\left( \dfrac{y}{x++y} \right)<1.20 \\
\end{align} \right.$
D. $\left\{ \begin{align}
  & x+y\le 8 \\
 & 0.20x+1.30y<1.20 \\
\end{align} \right.$

Answer 1

Hint: To form the first linear inequality, take the sum of the volumes of both the gases and substitute it with the conditions that this sum cannot be greater than 8. To form the second linear inequality, take the sum of the masses of both the gases and divide it by the total volume of both the gases and substitute it with the condition that this ratio cannot be greater than the density of air, that is, 1.20 grams per litre. Now, to calculate the mass, use the formula : mass = volume $\times $ density.

Complete step by step answer:
Here, we have been provided with conditions in the above question and we have to form two linear inequalities using those conditions.
Now, the first condition is that the balloon can be inflated up to 8 litres in volume. So, the sum of volume of helium and nitrogen in the balloon cannot exceed 8 litres. Since, we have been given,
Volume of helium = $x$
Volume of nitrogen = $y$
Therefore, total volume of both the gases = $x+y$
Now, according to the question, $x+y$ cannot exceed 8.
$\Rightarrow x+y\le 8\ldots \ldots \ldots \left( i \right)$
Let us move to the second condition now. It is given that the balloon must be filled so that the volumetric average density of the balloon is lower than that of air. Here, we know that,
Volumetric average density = $\dfrac{\text{Total mass of all the gases}}{\text{Total volume of all the gases}}$
So, applying the above formula, we get,
Volumetric average density = $\dfrac{\text{Mass of helium + Mass of nitrogen}}{\text{Volume of helium + Volume of nitrogen}}$
Let us calculate the mass of helium and nitrogen.
1. For helium gas, we have been given :
Density = 0.20 grams per litre
Volume = $x$ litres
We know that mass = volume $\times $ density. Therefore,
Mass of helium = $0.20\times x=0.20x$ grams.
2. For nitrogen gas, we have been given :
Density = 1.30 grams per litre
Volume = $y$ litres
We know that mass = volume $\times $ density. Therefore,
Mass of nitrogen = $1.30\times y=1.30y$ grams.
Substituting all the values in the formula for volumetric average density, we get,
Volumetric average density = $\dfrac{\text{0}\text{.20}x+1.30y}{x+y}=0.20\left( \dfrac{x}{x+y} \right)+1.30\left( \dfrac{y}{x+y} \right)$
Now, it is given that this volumetric average density must be lower than that of air, which has a density of 1.20 grams per litre. Therefore, we have,
$0.20\left( \dfrac{x}{x+y} \right)+1.30\left( \dfrac{y}{x+y} \right)<1.20\ldots \ldots \ldots \left( ii \right)$
So, from equation (i) and (ii), the two inequalities are :
$\left\{ \begin{align}
  & x+y\le 8 \\
 & 0.20\left( \dfrac{x}{x++y} \right)+1.30\left( \dfrac{y}{x++y} \right)<1.20 \\
\end{align} \right.$
Hence, option C is the correct answer.

Note:
One must remember the formula of calculation of mass and volumetric average density to solve the above question. You may note that we can easily eliminate the other options A, B and D. Clearly, we can see that in these inequalities, the unit of L.H.S is gram and the unit of R.H.S is grams per litre, that means L.H.S contains unit of mass and R.H.S contains unit of density which cannot be possible, hence options A, B and D can be easily eliminated this way. But this process can only be applied if options are given.