A bank loaned out 17,000, part of it af the rate of 8% per year and the rest at 16% per year. If the interest received in one year totaled $2000, how much was loaned at 8%?

Answer 1

Let's denote the amount loaned at 8% as \( x \). Consequently, the amount loaned at 16% will be \( 17000 - x \).

The interest earned from the amount loaned at 8% in one year can be expressed as: \[ 0.08x \]

The interest earned from the amount loaned at 16% in one year can be expressed as: \[ 0.16(17000 - x) \]

According to the problem, the total interest received in one year amounts to $2000. Thus, we can set up the following equation: \[ 0.08x + 0.16(17000 - x) = 2000 \]

Now, let's simplify this equation:

1. Distribute \( 0.16 \): \[ 0.08x + 0.16 \times 17000 - 0.16x = 2000 \] \[ 0.08x + 2720 - 0.16x = 2000 \]

2. Combine like terms: \[ -0.08x + 2720 = 2000 \]

3. Subtract 2720 from both sides: \[ -0.08x = 2000 - 2720 \] \[ -0.08x = -720 \]

4. Divide by -0.08: \[ x = \frac{-720}{-0.08} \] \[ x = 9000 \]

Thus, the amount loaned at 8% is \(\boxed{9000}\).

To double-check this result, we can confirm the amounts:

• Amount at 8%: \( 9000 \)
• Amount at 16%: \( 17000 - 9000 = 8000 \)

Now, calculating the interest:

• Interest from 8%: \[ 0.08 \times 9000 = 720 \]

• Interest from 16%: \[ 0.16 \times 8000 = 1280 \]

Adding both interests gives: \[ 720 + 1280 = 2000 \]

Indeed, the total interest is correct, confirming the amount loaned at 8% is \(\boxed{9000}\).