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# A recipe for soda requires w litres of water for every litre

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A recipe for soda requires w litres of water for every litre [#permalink]  29 Apr 2008, 05:01
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Jan 2014, 08:26, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: recipe [#permalink]  29 Apr 2008, 05:33
If soda requires w liters of water for 1 liter of syrup, there will be w*m liters of soda for for m liters of syrup.

Soda costs j dollars a liter so wm*j will be the selling cost of soda. Syrup costs k dollars, so it will cost m*k to produce soda.

gross profit: wmj-mj = m(jw-k)
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Re: recipe [#permalink]  29 Apr 2008, 15:45
A recipe for soda requires w litres of water for every liter of syrup. If the soda is made according to this recipe using m litres of syrup, and sold for j dollars a liter, what will be the gross profit if the syrup costs k dollars a liter and water costs nothing ?

m(w+j-k)
jm(1/w + 1)
m(jw -k)
(j-k) m
jm(1+w) - km

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.
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Re: recipe [#permalink]  29 Apr 2008, 16:02
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A recipe for soda requires w litres of water for every liter of syrup. If the soda is made according to this recipe using m litres of syrup, and sold for j dollars a liter, what will be the gross profit if the syrup costs k dollars a liter and water costs nothing ?

m(w+j-k)
jm(1/w + 1)
m(jw -k)
(j-k) m
jm(1+w) - km

Water for m liters of Syrup = WM
So total soda is wm + m liters

Selling price (wm + m)j
Cost Price = mk

Profit = mj(w+1) - mk
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word problem [#permalink]  07 Mar 2010, 09:31
word problem
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Re: word problem [#permalink]  07 Mar 2010, 09:49
raghavs wrote:
word problem

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
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Re: word problem [#permalink]  07 Mar 2010, 10:00
raghavs wrote:
word problem

Given Soda has w:1 ratio of Water to Syrup

Soda is made using m litres of Syrup. THis means m*w litres of water. Therefore total Soda made in litres = w*m+m = m(1+w)

Selling price of soda per litre = j dollars
Selling price of m(1+w) litres of soda = j*m(1+w) dollars

Cost of syrup per litre = k dollars
Cost of m litres of syrup used = km dollars

Gross Profit = Selling price of Soda - Cost price of Syrup = jm(1+w) - km... (Ans E or 5)
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Re: A recipe for soda requires w litres of water for every liter [#permalink]  07 Jan 2014, 08:17
will any one provide solution of this question.
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Re: A recipe for soda requires w litres of water for every liter [#permalink]  07 Jan 2014, 08:48
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Expert's post
sanjoo wrote:
will any one provide solution of this question.

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

Since soda requires $$w$$ liters of water for every liter of syrup, then $$m$$ liters of syrup gives $$water+syrup=mw+m$$ liters of soda.

Sales = $$j(mw+m)$$ dollars.
Cost = $$km$$ dollars.

Profit = $$j(mw+m)-km$$.

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Re: A recipe for soda requires w litres of water for every litre [#permalink]  11 Mar 2015, 06:21
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Re: A recipe for soda requires w litres of water for every litre   [#permalink] 11 Mar 2015, 06:21
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