Math Problem Statement

Maricopa's Success scholarship fund receives a gift of $ 210000. The money is invested in stocks, bonds, and CDs. CDs pay 5.25 % interest, bonds pay 4.2 % interest, and stocks pay 7.2 % interest. Maricopa Success invests $ 10000 more in bonds than in CDs. The annual income from the investments is $ 11107.5.

How much did the scholarship invest in each venue

Solution

To solve this problem, we will let xx, yy, and zz represent the amounts invested in CDs, bonds, and stocks, respectively. We'll establish equations based on the problem's conditions and solve them step by step.


Step 1: Define the variables

  • xx: amount invested in CDs (earning 5.25% interest).
  • yy: amount invested in bonds (earning 4.2% interest).
  • zz: amount invested in stocks (earning 7.2% interest).

Step 2: Establish equations based on the problem

  1. Total investment: x+y+z=210000x + y + z = 210000

  2. Interest income: The annual income from the investments is $11,107.50: 0.0525x+0.042y+0.072z=11107.50.0525x + 0.042y + 0.072z = 11107.5

  3. Relationship between bonds and CDs: The investment in bonds is $10,000 more than in CDs: y=x+10000y = x + 10000


Step 3: Substitute y=x+10000y = x + 10000 into the equations

Substituting into the total investment equation:

x+(x+10000)+z=210000x + (x + 10000) + z = 210000 2x+z=200000(Equation 4)2x + z = 200000 \quad \text{(Equation 4)}

Substituting into the interest income equation:

0.0525x+0.042(x+10000)+0.072z=11107.50.0525x + 0.042(x + 10000) + 0.072z = 11107.5 Simplify: 0.0525x+0.042x+0.042(10000)+0.072z=11107.50.0525x + 0.042x + 0.042(10000) + 0.072z = 11107.5 0.0945x+420+0.072z=11107.50.0945x + 420 + 0.072z = 11107.5 0.0945x+0.072z=10687.5(Equation 5)0.0945x + 0.072z = 10687.5 \quad \text{(Equation 5)}


Step 4: Solve the system of equations

We now have two equations:

  1. 2x+z=2000002x + z = 200000 (Equation 4)
  2. 0.0945x+0.072z=10687.50.0945x + 0.072z = 10687.5 (Equation 5)

Solve for zz in terms of xx using Equation 4:

z=2000002xz = 200000 - 2x

Substitute z=2000002xz = 200000 - 2x into Equation 5:

0.0945x+0.072(2000002x)=10687.50.0945x + 0.072(200000 - 2x) = 10687.5 Expand: 0.0945x+0.072(200000)0.072(2x)=10687.50.0945x + 0.072(200000) - 0.072(2x) = 10687.5 0.0945x+144000.144x=10687.50.0945x + 14400 - 0.144x = 10687.5 Combine like terms: 0.0495x+14400=10687.5-0.0495x + 14400 = 10687.5 0.0495x=10687.514400-0.0495x = 10687.5 - 14400 0.0495x=3712.5-0.0495x = -3712.5 x=3712.50.0495=75000x = \frac{-3712.5}{-0.0495} = 75000


Step 5: Find yy and zz

Using x=75000x = 75000:

  • y=x+10000=75000+10000=85000y = x + 10000 = 75000 + 10000 = 85000
  • z=2000002x=2000002(75000)=50000z = 200000 - 2x = 200000 - 2(75000) = 50000

Final Answer:

  • Investment in CDs: x=75000x = 75000
  • Investment in bonds: y=85000y = 85000
  • Investment in stocks: z=50000z = 50000

Verification:

  1. Total investment: 75000+85000+50000=210000(Correct)75000 + 85000 + 50000 = 210000 \quad \text{(Correct)}

  2. Interest income: 0.0525(75000)+0.042(85000)+0.072(50000)=3937.5+3570+3600=11107.5(Correct)0.0525(75000) + 0.042(85000) + 0.072(50000) = 3937.5 + 3570 + 3600 = 11107.5 \quad \text{(Correct)}

Let me know if you'd like additional explanations!


Related Questions:

  1. How do we derive the equations for systems of linear equations?
  2. What strategies exist for solving systems of linear equations with substitution?
  3. Can we solve this problem using matrices or determinants?
  4. How would the solution change if the interest rates were different?
  5. What if the annual income constraint was different—how would that affect the investments?

Tip:

When solving word problems, break them into smaller parts to derive equations systematically. Always double-check your results for consistency!

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Equations
Linear Equations
Interest Calculations

Formulas

Total Investment: x + y + z = 210000
Interest Income: 0.0525x + 0.042y + 0.072z = 11107.5
Relationship: y = x + 10000

Theorems

Substitution Method
Solving Systems of Linear Equations

Suitable Grade Level

Grades 9-12