A dessert recipe calls for 50% melted chocolate and 50%

BTGmoderatorDC wrote:A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5

Chocolate percentage in the incorrect sauce: 40%.
Chocolate percentage in the pure chocolate: 100%.
Chocolate percentage in the mixture: 50%.

Let I = the incorrect sauce and C = the pure chocolate.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for I and C on the ends and the percentage for the mixture in the middle.
I 40%----------50%-----------100% C

Step 2: Calculate the distances between the percentages.
I 40%----10----50%----50-----100% C

Step 3: Determine the ratio in the mixture.
The ratio of I to C is equal to the RECIPROCAL of the distances in red.
I:C = 50:10 = 5:1.

Since I:C = (5 cups) : (1 cup), 1 of every 6 cups must be pure chocolate.
Thus:
Pure chocolate = (1/6)(15 cups) = 15/6 = 5/2 = 2.5 cups.

The correct answer is B.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html

BTGmoderatorDC wrote:A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5

Source: Veritas Prep

$$?\,\,\, = \,\,\,x = \# \,\,{\text{out}}\,\,\underline {{\text{sauce}}} \,\,{\text{cups}}\,\, = \,\,\# \,\,{\text{in}}\,\,\underline {{\text{100 % }}\,\,{\text{choco}}} \,\,{\text{cups}}$$
$$\matrix{
{{\rm{real}} \to {\rm{ideal}}} \cr
{15\,\,{\rm{cups}}} \cr

} \,\,\,\left\{ \matrix{
\,{\rm{choco}}\,:\,\,{2 \over 5}\left( {15} \right)\,\,{\rm{cups}} - x \cdot {2 \over 5}\,\,{\rm{cups}} + x\,\,{\rm{cups}}\,\,\,\,\, = \,\,\,\,{{2.5} \over 5}\left( {15} \right)\,\,{\rm{cups}} \hfill \cr
\,{\rm{rasp}}\,:\,\,{3 \over 5}\left( {15} \right)\,\,{\rm{cups}} - x \cdot {3 \over 5}\,\,{\rm{cups}} + 0\,\,{\rm{cups}}\,\,\,\,\, = \,\,\,\,{{2.5} \over 5}\left( {15} \right)\,\,{\rm{cups}} \hfill \cr} \right.$$
$${3 \over 5}\left( {15} \right)\,\,{\rm{cups}} - x \cdot {3 \over 5}\,\,{\rm{cups}}\,\,\,\,\, = \,\,\,\,{{2.5} \over 5}\left( {15} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{3 \over 5}x = {{0.5} \over 5}\left( {15} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = 2.5$$
$$\left[ {\,{2 \over 5}\left( {15} \right)\,\,{\rm{cups}} - x \cdot {2 \over 5}\,\,{\rm{cups}} + x\,\,{\rm{cups}}\,\,\,\,\, = \,\,\,\,{{2.5} \over 5}\left( {15} \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{3 \over 5}x = {{0.5} \over 5}\left( {15} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x = 2.5} \right]$$

The correct answer is (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.