Questions: In planning her retirement, Liza deposits some money at 1.5% interest, with twice as much deposited at 2%. Find the amount deposited at each rate if the total annual interest income is 825. She deposited at 1.5% and at 2%

In planning her retirement, Liza deposits some money at 1.5% interest, with twice as much deposited at 2%. Find the amount deposited at each rate if the total annual interest income is 825.

She deposited  at 1.5% and  at 2%
Transcript text: In planning her retirement, Liza deposits some money at $1.5 \%$ interest, with twice as much deposited at $2 \%$. Find the amount deposited at each rate if the total annual interest income is $\$ 825$. She deposited $\$$ $\square$ at $1.5 \%$ and $\$$ $\square$ at $2 \%$
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Solution

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Solution Steps

To solve this problem, we need to set up a system of linear equations based on the given information. Let \( x \) be the amount deposited at 1.5% interest and \( y \) be the amount deposited at 2% interest. We know that the total interest income is $825. We also know that twice as much is deposited at 2% as at 1.5%. This gives us two equations:

  1. \( 0.015x + 0.02y = 825 \)
  2. \( y = 2x \)

We can solve these equations simultaneously to find the values of \( x \) and \( y \).

Step 1: Define Variables and Equations

Let \( x \) be the amount deposited at \( 1.5\% \) interest and \( y \) be the amount deposited at \( 2\% \) interest. We are given the following information:

  1. The total annual interest income is \( \$825 \).
  2. Twice as much is deposited at \( 2\% \) as at \( 1.5\% \).

This gives us the following system of equations: \[ 0.015x + 0.02y = 825 \] \[ y = 2x \]

Step 2: Substitute and Solve

Substitute \( y = 2x \) into the first equation: \[ 0.015x + 0.02(2x) = 825 \] \[ 0.015x + 0.04x = 825 \] \[ 0.055x = 825 \] \[ x = \frac{825}{0.055} = 15000 \]

Step 3: Calculate \( y \)

Using \( y = 2x \): \[ y = 2 \times 15000 = 30000 \]

Final Answer

Liza deposited \( \$15000 \) at \( 1.5\% \) and \( \$30000 \) at \( 2\% \).

\[ \boxed{x = 15000} \] \[ \boxed{y = 30000} \]

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