A bank loans $12,000 , part at a rate of 8% the other is 18%. if the interest received is at $1000, how much was loaned at 8%?

Question

in question 

Answer 1

Let's define some terms, so the process and answer can be used for similar problems:

 

total loan principle: P = $12,000

interest rate 1: R = 8%

interest rate 2: S = 18%

interest earned: I = $1000

 

We want to find the principle for each portion of the loan, using 'x' and 'y' to emphasize these are the unknowns we are solving for

principle 1: x = ?

principle 2: y = ?

 

The important relationships are

P = x + y

I = R x + S y

 

Solve the system of equations by elminating x₂:

y = P - x

I = R x + S (P - x)

  = R x + S P - S x

  = S P + x (R - S)

 

x (R - S) = I - S P

x = (I - S P)/(R - S)

 

Plug in numbers:

x = (1000 - 0.18 * 12,000) / (0.08 - 0.18)

   = $11,600

y = 12,000 - 11,600

   = $400

 

 

Answer 2

Assuming annual interest and 1 year:

A = p(1+r/n)^nt

becomes

A = p(1+r)

 

If we invest $x at 8%

We invest $(1200-x) at 18%

The final amount is A=13000

 

13000 = x(1.08) + (12000-x)(1.18)

 

13000 = 1.08x - 1.18x + 14160

 

-1160 = -.1x

 

x = -1160/-.1

 

x = $11,600 loaned at 8%

 

This leaves $400 loaned at 18%

 

Check:  11,600(1.08) + 400(1.18) = $13,000 final amount

 

 

Answer 3

Assuming a one-year term:

 

X + Y = $12,000

.08X + .18Y = $1,000

 

Express X in terms of Y

X = 12,000 - Y

 

Substitute in second equation:

 

.08(12,000 - Y) + .18Y = $1,000

$960 - .08Y + .18Y = $1,000

.10Y = $40

Y = $400

 

Proof:

 

X + 400 = $12,000

Therefore X = $11,600

 

$11,600*.08 + $400*.18 =  $1,000

$928 + $72 = $1,000