Question

How do you find the cost per ounce of a mixture of \[200\] oz of a cologne that costs \[\$ 5.50\] per ounce and \[500\] oz of a cologne that costs \[\$ 2.00\] per ounce?

Answer 1

Hint: According to the question, we have to find out the cost per ounce of a mixture. So, we have to take two cases here. In the first case we have to find out the total cost for \[200\] oz when the cost of cologne is \[\$ 5.50\]. In the second case, we have to find out the total cost for \[500\] oz when the cost of cologne is \[\$ 2.00\]. We have to divide both the results to find out the cost per ounce of mixture.

Complete step by step solution:
Case-1
We know that in case-1, the total cost of a cologne per ounce \[ = \$ 5.50\]
So, total cost for \[200\]oz if the cost of a cologne is \[\$ 5.50 = 200 \times \$ 5.50 = \$ 1100\]
Case-2
We know that in case-2, the total cost of a cologne per ounce \[ = \$ 2.00\]
So, total cost for \[500\] oz if the cost of a cologne is \[\$ 2.00 = 500 \times \$ 2.00 = \$ 1000\]
Now, we will try to calculate the cost per oz of the mixer. We will calculate it by dividing the total cost of cologne in both the cases to the total number of oz, and we get:
Cost per oz of the mixer \[ = \dfrac{{1100 + 1000}}{{200 + 500}}\]
When we add \[1100\] with \[1000\], then we get \[2100\]. When we add \[200\] with \[500\], then we get \[700\]. When we put these two results in the numerator and the denominator respectively, then we get:
\[ = \dfrac{{2100}}{{700}}\]
Now, we need to divide \[2100\] by \[700\]. We know that both the zeros in the numerator and in the denominator will get cancel out, and we get:
\[ = \dfrac{{21}}{7}\]
When we divide \[21\] by \[7\], then we know that we will get \[3\] as result, so:
\[\dfrac{{21}}{7} = 3\]
Therefore, the result is \[\$ 3\]. So, the cost per oz of the mixer is \[\$ 3\].

Note: The above question is an application using linear models which comes under the algebra section in Mathematics. For example, in cricket they use linear models to model the run rate of any given team.