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Difficulty: MediumAlgebraic Word Problems and Modeling

An investor allocated a total principal of $10,000\$10,000 between two accounts, Account X and Account Y. Account X pays simple annual interest at a rate of 6%6\%, while Account Y pays simple annual interest at a rate of 8%8\%. Let xx represent the amount of money, in dollars, invested in Account X, and let dd represent the total annual interest earned, in dollars, from both accounts after one year. Which of the following statements must be true? Select all that apply.

  1. The total annual interest dd satisfies 600≤d≤800600 \le d \le 800.Answer
  2. The amount invested in Account X can be represented as x=800−d0.02x = \frac{800 - d}{0.02}.Answer
  3. C
    If d=750d = 750, the amount invested in Account X is greater than the amount invested in Account Y.
  4. If equal amounts were invested in both accounts, the total annual interest earned is $700\$700.Answer
  5. E
    If d=680d = 680, the ratio of the amount invested in Account X to the amount invested in Account Y is 22 to 33.

Answer

The true statements are that the total annual interest dd satisfies 600≤d≤800600 \le d \le 800, the amount in Account X is given by x=800−d0.02x = \frac{800 - d}{0.02}, and equal investment in both accounts yields $700\$700 in total interest.
The total interest equation d=800−0.02xd = 800 - 0.02x dictates all valid relationships. Because 0≤x≤10,0000 \le x \le 10,000, the bounds for dd are strictly between 600600 and 800800. Rearranging the equation yields x=800−d0.02x = \frac{800 - d}{0.02}, which correctly models the Account X investment. Substituting x=5,000x = 5,000 yields d=700d = 700, confirming the equal-allocation scenario.

Step-by-Step Solution

1
Formulate the total interest model as a linear equation in terms of xx.
d=0.06x+0.08(10,000−x)=800−0.02xd = 0.06x + 0.08(10,000 - x) = 800 - 0.02x
The interest earned from Account X is 0.06x0.06x and the interest from Account Y is 0.08(10,000−x)0.08(10,000 - x).
2
Determine the range of possible interest values for 0≤x≤10,0000 \le x \le 10,000.
When x=10,000x = 10,000, d=600d = 600. When x=0x = 0, d=800d = 800. Thus 600≤d≤800600 \le d \le 800.
The minimum and maximum interest values occur at the extreme allocation bounds.
3
Rearrange the interest equation to express xx in terms of dd.
0.02x=800−d  ⟹  x=800−d0.020.02x = 800 - d \implies x = \frac{800 - d}{0.02}
Isolating xx provides a formula for calculating the Account X principal directly from the total interest.
4
Evaluate the specific numerical cases given in the options.
Equal investment (x=5,000x = 5,000) gives d=800−100=700d = 800 - 100 = 700. For d=750d = 750, x=2,500x = 2,500 (Account Y = 7,5007,500). For d=680d = 680, x=6,000x = 6,000 (Account Y = 4,0004,000, ratio 3:23:2).
Substituting specific values tests the validity of each conditional statement.

Key Concept

Linear Modeling and Algebraic Rate Allocation
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Topics

Algebraic Word Problems and Modeling

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