A receipe for soda requires w liters of water for every

Question

A receipe for soda requires w liters of water for every liter of syrup. If soda is made according to this recipe using m liters of syrup, and sold for j dollars a liter , what will be the gross profit if syrup costs k dollars a liter and water costs nothng.?

a. m(w+j-k)
b.jm(1/w +1)
c. m(jw -K)
d. (j-k)m
e. jm(1+w)-km

Pls explain..

ywilfred · Answer

1 litre of syrup requires w litres of water
So m litres of syrup will require mw litres of water

Cost of syrup is $k a litre, so cost of m litres of syrup is $km
Water costs nothing, so mw litres of water will cost $0
Total cost of making this mixture is $km

This mixture is (mw+w) litres
It is sold for $J a litre, so total income is $J(mw+w)

Therefore profit will be :

J(mw+w) - km
Jm(w+1)-km

rchadha · Answer

Does anyone have OA for this?

Here's my answer and expln:

H20:Syrup=w:1
Reqd ratio=wm:m

S.P=$j/litre
C.P=$k/litre

SP will include water weight also but CP will not:

profit=jwm-km=m(jw-k)

HIMALAYA · Answer

C. 
soda = s, syrup=sp, water=w. soda = sp+w
1 lt syrup needs w lt water.
total syrup used = m lt
total soda = mw
total revenue, tr, from the soda = jmw
total costs, tc, of syrup=mk
so, gp = tr-tc=jmw-mk=m(jw-k)

riteshgupta1 · Answer

it has to be E

m liter os syrup
mw liter of water.

J is selling price per liter

Total selling price = j (m + mw) => mj(1+w)

Total cost = mk

Profit is = selling price - cost

Profit = mj(1+w) - mk

Dilshod · Answer

It is E. 
(m+mw)j - mk = jm(1+w) - mk

A receipe for soda requires w liters of water for every

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