Prots

[Johnrohit]

Food C containd 20% protein by weight and 80% carbs. Food D contains 75% protein and 25% carbs by weight.In what ratio should Food C & D be mixed such that the resulting mix contains 45% of protein by weight? (ratios for C: D presented below)

A) 3:2
B)5:6
C)6:5
D)4:3

[mevicks]

Food C containd 20% protein by weight and 80% carbs. Food D contains 75% protein and 25% carbs by weight.In what ratio should Food C & D be mixed such that the resulting mix contains 45% of protein by weight? (ratios for C: D presented below)

A) 3:2
B)5:6
C)6:5
D)4:3

Food C:
P________________________C
(1/5)----------------------------(4/5)

Food D:
P________________________C
(3/4)----------------------------(1/4)

Final Mix protein is 45/100 --> 9/20
Make the denominators of protein C&D 20

Final Mix:
P___________F_____________C
(4/20)--------(9/20)---------(15/20)

C/D = (15/20 - 9/20)/(9/20 - 4/20) = 6/5

Answer C

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Hi Johnrohit,

You can solve this question using the mixture formula:

Part/Whole = (total protein)/(total of all) = (.2C + .75D)/(C + D) = .45

Now we do algebra:

.2C + .75D = .45C + .45D

.30D = .25C

.30/.25 = C/D

30/25 = 6/5

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

[theCodeToGMAT]

Shorter Approach

Directly apply Weighted Average formula: [N1/N2 = (M2-M)/(M-M1)]

N1/N2 = 75-45/45-20 = 30/25 = 6/5

[spoiler]{C}[/spoiler]

[GMATGuruNY]

Food C containd 20% protein by weight and 80% carbs. Food D contains 75% protein and 25% carbs by weight.In what ratio should Food C & D be mixed such that the resulting mix contains 45% of protein by weight? (ratios for C: D presented below)

A) 3:2
B)5:6
C)6:5
D)4:3

Protein in C = 20%.
Protein in D = 75%,
Protein in the MIXTURE of C and D = 45%.

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 C and D on the ends and the percentage for the mixture in the middle.
C 20%-----------45%-----------75% D

Step 2: Calculate the distances between the percentages.
C 20%----25-----45%----30-----75% D

Step 3: Determine the ratio in the mixture.
The required ratio of C to D is equal to the RECIPROCAL of the distances in red.
C : D = 30:25 = 6:5.

The correct answer is C.

For two similar problems, check here:

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

An alternate approach is to PLUG IN THE ANSWERS.

Answer choice C: C : D = 6:5
6 parts 20% protein = 6*20 = 120.
5 parts 75% protein = 5*75 = 375.
Average per 11 parts = (120 + 375)/11 = 495/11 = 45%.
Success!