MTH 120 - Solving Mixture Problems

MTH 120 - Solving Mixture Problems - Solving Mixture...

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Unformatted text preview: Solving Mixture Problems If soybean meal is 16% protein and corn meal is 9% protein, how many pounds of each should be mixed to get a 350‐lb. mixture that is 12% protein? Note: This problem could have read, “If soybean oil is 16% protein and corn oil is 9% protein, how many gallons of each should be mixed to get a 350‐gal. mixture that is 12% protein?” and the following solution would still apply with the only change being the “amounts” would be in gal. instead of lbs. Step 1. Familiarize yourself with the problem. You should plan to read the word problem (quite) a few times, and draw/ construct a diagram that describes the situation pictorially and contains all relevant information.. The first few times you read the word problem you are trying to obtain an overall understanding of the situation and the question or questions you are being asked to answer… Once you can envision/imagine the situation being described, then you should read the word problem at least one or two more times to identify what information is relevant and irrelevant, and capture the relevant information on your diagram… Soybean Meal Corm Meal Mixture 16% Protein 6 9% Protein 12% Protein ? lbs. + = ? lbs. 350 lbs. Before we can begin Step 2. Translate the problem into 1 or more equations, we must convert any unknowns into variables… Soybean Meal Corm Meal Mixture 16% Protein 9% Protein 12% Protein S lbs. Continued on next page… + = C lbs. 350 lbs. Soybean Meal Corm Meal Mixture 16% Protein 9% Protein 12% Protein S lbs. Step 2. + = C lbs. 350 lbs. Translate the problem into 2 equations. Note: There is an alternate method for solving mixture problems that requires only one equation; however, we will focus only on the more common 2‐equation method here. The two equations required to solve mixture problems can be generalized as follows… Amt. of A + Amt. of B = Amt. of Mixture [Equation 1] (% in A) x (Amt. of A) + (% in B) x (Amt. of B) = (% in Mixture) x (Amt. of Mixture) [Equation 2] Using our diagram, let’s identify the critical information; then, create our two equations… Amt. of A = S lbs. = S Amt. of B = C lbs. = C Amt. of Mixture = 350 lbs. = 350 Therefore, Equation 1 is… S + C = 350 % in A = 16% = 0.16 % in B = 9% = 0.09 % in Mixture = 12% = 0.12 Therefore, Equation 2 is… 0.16S + 0.09C = 0.12(350) [Equation 2] Before we can begin Step 3. Solve the 2 equations, we should simplify the right hand side of Equation 2… 0.16S + 0.09C = 0.12(350) ⇒ 0.16S + 0.09C = 42 [New Equation 2] Now that we have our 2 equations (see below), we can proceed to Step 3… S + C = 350 [Equation 1] 0.16S + 0.09C = 42 [New Equation 2] Continued on next page… S + C = 350 0.16S + 0.09C = 42 Step 3. [Equation 1] [New Equation 2] Solve the 2 equations. To avoid having to deal with negative numbers when using the Elimination Method to solve these 2 equations, we will choose to elimination the variable in New Equation 2 with the smaller coefficient. Since 0.09 is smaller than 0.16, we will choose to eliminate the variable by multiplying Equation 1 by –0.09, then by adding the New Equation 1 and the New Equation 2… –0.09(S + C = 350) –0.09S + –0.09C = –0.09(350) –0.09S + –0.09C = –31.5 [New Equation 1] 0.16S + 0.09C = 42 [New Equation 2] ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 0.07S = 10.5 0.07S / 0.07 = 10.5 / 0.07 S = 150 Use the original Equation 1 and the value found for S to find C… S + C = 350 [Equation 1] 150 + C = 350 –150 –150 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ C = 200 Step 4. Check your answers. Use the original Equation 2 to check the values you found for C and S… 0.16S + 0.09C = 0.12(350) [Equation 2] Does S = 150 and C = 200 satisfy 0.16S + 0.09C = 0.12(350)? 0.16(150) + 0.09(200) = 0.12(350) 24 + 18 = 42 42 = 42 Yes, our values for S and C do check! Continued on next page… S = 150 C = 200 Step 5. State the answer clearly. To state the correct answer or answers clearly, you should reread the problem one more time to be certain you have answered the actual question or questions being asked… “If soybean meal is 16% protein and corn meal is 9% protein, how many pounds of each should be mixed to get a 350‐lb. mixture that is 12% protein?” Then, state your answer or answers clearly… 150 lbs of Soybean Meal and 200 lbs. of Corn Meal DONE!!! ...
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This note was uploaded on 02/29/2012 for the course MATH 120 taught by Professor Adjunct during the Fall '11 term at Northern Virginia Community College.

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