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Difficulty: Very hardSystems of Linear Equations

An investor allocates a sum of money across three accounts—AA, BB, and CC—which earn simple annual interest rates of 5%5\%, 4%4\%, and 6%6\%, respectively. The total interest earned from all three accounts in one year is $1,060\$1,060. The amount invested in Account CC equals the total amount invested in Accounts AA and BB combined. If the amount in Account AA were increased by 50%50\% and the amount in Account BB were decreased by 25%25\% while Account CC remained unchanged, the total annual interest earned would increase by $110\$110. What is the total amount, in dollars, invested across all three accounts?

  1. A
    $10,000\$10,000
  2. B
    $15,000\$15,000
  3. C
    $18,000\$18,000
  4. $20,000\$20,000Answer
  5. E
    $24,000\$24,000

Answer

The total amount invested across all three accounts is $20,000\$20,000.
The system of linear equations derived from the problem constraints uniquely solves to a=$6,000a = \$6,000, b=$4,000b = \$4,000, and c=$10,000c = \$10,000. Summing these three amounts gives a total investment of $20,000\$20,000.

Step-by-Step Solution

1
Define variables and translate the problem statements into algebraic equations.
Let aa, bb, and cc be the amounts invested in Accounts AA, BB, and CC.
Equation 1 (Total Interest): 0.05a+0.04b+0.06c=1,060  ⟹  5a+4b+6c=106,0000.05a + 0.04b + 0.06c = 1,060 \implies 5a + 4b + 6c = 106,000.
Equation 2 (Relationship among principal amounts): c=a+b  ⟹  a+b−c=0c = a + b \implies a + b - c = 0.
Equation 3 (Change in interest): Increasing aa by 50%50\% adds 0.05(0.50a)=0.025a0.05(0.50a) = 0.025a interest. Decreasing bb by 25%25\% reduces interest by 0.04(0.25b)=0.01b0.04(0.25b) = 0.01b. The net change is 0.025a−0.01b=110  ⟹  25a−10b=110,000  ⟹  5a−2b=22,0000.025a - 0.01b = 110 \implies 25a - 10b = 110,000 \implies 5a - 2b = 22,000.
Establishing a complete 3-variable system of linear equations is necessary to determine the unknowns.
2
Substitute c=a+bc = a + b into Equation 1 to reduce the system to two variables.
5a+4b+6(a+b)=106,000  ⟹  11a+10b=106,0005a + 4b + 6(a + b) = 106,000 \implies 11a + 10b = 106,000.
Using substitution eliminates variable cc, leaving a linear system in aa and bb.
3
Solve the two-variable system using elimination.
We have:
(1) 11a+10b=106,00011a + 10b = 106,000
(2) 5a−2b=22,0005a - 2b = 22,000
Multiply (2) by 5: 25a−10b=110,00025a - 10b = 110,000.
Add this to (1): (11a+10b)+(25a−10b)=106,000+110,000  ⟹  36a=216,000  ⟹  a=6,000(11a + 10b) + (25a - 10b) = 106,000 + 110,000 \implies 36a = 216,000 \implies a = 6,000.
Substitute a=6,000a = 6,000 back into (2): 5(6,000)−2b=22,000  ⟹  30,000−2b=22,000  ⟹  2b=8,000  ⟹  b=4,0005(6,000) - 2b = 22,000 \implies 30,000 - 2b = 22,000 \implies 2b = 8,000 \implies b = 4,000.
Elimination allows straightforward calculation of individual values for aa and bb.
4
Calculate cc and find the total sum invested across all three accounts.
c=a+b=6,000+4,000=10,000c = a + b = 6,000 + 4,000 = 10,000.
Total investment =a+b+c=6,000+4,000+10,000=20,000= a + b + c = 6,000 + 4,000 + 10,000 = 20,000.
The question asks for the total amount invested in all three accounts combined.

Key Concept

Systems of Linear Equations in Three Variables
Estimated Time:2m 0s
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Topics

Systems of Linear Equations

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