A recipe for soda requires w litres of water for every litre of syrup. If soda is made according to this recipe using m litres of syrup, and sold for j dollars a litre, what will be the gross profit if syrup costs k dollars a litre and water costs nothing?

A. m(w+j-k)
B. jm[(1/w)+1]
C. m(jw-k)
D. (j-k)m
E. jm(1+w)-km
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Should be C.

lets say we have 5liters of water for 1 liter of syrup. That means when we have 5 liters of syrup we have 25 liters of water.

now just same idea. w*m ---> m(jw-k).

Remember Profit= Revenue-Cost.

for m litres of syrup water required is w*m litres. total litres of solution is m+ wm

Now this solution is sold for j$ /litre of the solution. So total overturn is j*(m+ wm) and total cost = m*k

Gross profit is jm(1+w)- mk

will any one provide solution of this question.
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Since every litre of syrup requires w litres of water and there are m litres of syrup, we need mw litres of water, and together with the m litres of syrup, we can make mw + m litres of soda.

Since we are selling the soda for j dollars a litre, we can make j(mw + m) dollars. However, since the syrup costs k dollars a litre (water costs nothing), we have a cost of km dollars. Thus, the total profit from selling the soda is:

j(mw + m) - km = jm(w + 1) - km

Answer: E
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