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| Math Central | Quandaries & Queries |
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Question from Kenneth: Hello: An investor wants to divide $1000.00 into four amounts or accounts -- one account earning 2%, the second account earning 5%, the third earning 9%, and the fourth earning 12%. He wants the overall total return to equal 5%. Can the below calculation be reversed engineered into a more simple or basic arithmetic type calculation--something less algebraic? I thank you for your reply. ------------------------------- Use the variables w, x, y, z to represent the amount of money in each of the accounts (w is amount in 2%, x in 5%, etc.). (1) w + x + y + z = 1000 (2) 0.02w + .05x + .09y + .12z = .05(1000) There are only two equations and four unknowns, there will be lots (in fact, infinitely many) possible solutions. Lets see why. Using (1) solve for w: w = 1000 - x - y - z Plug this into 2.: .02(1000 - x - y - z) + .05x + .09y + .12z = 50 20 - .02x - .02y - .02z + .05x + .09y + .12 z = 50 Collect all like terms: .03x + .07y + .1z = 30 (3) Any combination of x, y, z that satisfy this equation (3) will give you the desired return of 5%. Remember, take the values that you obtain and use them to reconstruct An example of how to do this would be to try any old (reasonable) value for x and y, and solve for z: Say x = 100, y = 200: .03(100) + .07(200) + .1z = 30 3 + 14 + .1z = 30 w = 1000 - (100 + 200 + 170) |
Hi Kenneth,
My first reaction to this problem was to suggest investing the entire $1000 in the second account that earns 5% as required.
What if you want to diversify some and invest in two accounts. One must earn more than 5% and the other less than 5% (why?) and hence you have two choices, 9% and 2% or 12% and 2%. Set up the algebra in each of these two situations as you did above, and solve. In each case you get one solution. Before you solve you might want to make a guess as to how much you invest at each rate and see how close you come to making 5%.
What about three accounts. There are 4 possible choices: 2%, 5%, 9%; 2%, 5%, 12%; 5%, 9%, 12%; and 2%, 9%, 12%. You can immediately rule out 5%, 9% and 12% (why?) and hence set up the algebra in each of the three remaining situations, for example 2%, 5% and 9%. Now you can see why there might be many solutions. You can choose some amount to invest at 2%, say $100 and then try to solve for the other two amounts. What if you invest $200 at 5%? What about $900?
My reading of the problem suggests that this is the type of exploration expected.
Penny
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