A bank lent $1.2 million for the development of three new products, with one loan each at 6%, 7%, and 8%. The amount lent at 8% was equal to the sum of the amounts lent at the other two rates, and the bank's annual income from the loans was $88,000. How much was lent to each rate?

Question

A bank lent $1.2 million for the development of three new products, with one loan each at 6%, 7%, and 8%. The amount lent at 8% was equal to the sum of the amounts lent at the other two rates, and the bank's annual income from the loans was $88,000. How much was lent to each rate?

Answer 1

To solve this problem, we can use a system of equations.

Let's denote the amount lent at 6% as x, the amount lent at 7% as y, and the amount lent at 8% as z.

We are given three pieces of information:

1. The amount lent at 8% was equal to the sum of the amounts lent at the other two rates: z = x + y.

2. The bank's annual income from the loans was $88,000. We can calculate the income from each loan as follows:
- Income from the loan at 6% is 0.06x.
- Income from the loan at 7% is 0.07y.
- Income from the loan at 8% is 0.08z.
So, we have the equation: 0.06x + 0.07y + 0.08z = $88,000.

Now, let's solve this system of equations.

From the first equation, we can rewrite z in terms of x and y: z = x + y.

Substituting this into the second equation, we get: 0.06x + 0.07y + 0.08(x + y) = $88,000.

Expanding and simplifying, we have: 0.06x + 0.07y + 0.08x + 0.08y = $88,000.

Combining like terms, we get: 0.14x + 0.15y = $88,000.

Dividing both sides by 0.01, we have: 14x + 15y = $8,800,000. (equation A)

Now, using the first equation (z = x + y), we can substitute this into equation A:

14x + 15y = $8,800,000 becomes 14x + 15(x + y) = $8,800,000.

Expanding and simplifying, we get: 14x + 15x + 15y = $8,800,000.

Combining like terms, we have: 29x + 15y = $8,800,000.

Now, we have a system of two linear equations:

Equation A: 14x + 15y = $8,800,000.

Equation B: 29x + 15y = $8,800,000.

We can solve this system of equations using substitution or elimination. I'll use elimination.

Subtracting equation A from equation B, we eliminate the y term:

(29x + 15y) - (14x + 15y) = $8,800,000 - $8,800,000.

This simplifies to: 15x = 0.

Dividing by 15, we find: x = 0.

Now, substituting this value of x back into equation A, we get:

14(0) + 15y = $8,800,000.

This simplifies to: 15y = $8,800,000.

Dividing by 15, we find: y = $586,666.67.

Now, we can substitute the values of x and y into the equation z = x + y:

z = 0 + $586,666.67.

Therefore, z = $586,666.67.

To summarize, the amount lent at 6% is $0, the amount lent at 7% is $586,666.67, and the amount lent at 8% is also $586,666.67.