I need help with my maths with percentages?

Question

Someone borrows $10 000 from a finance company, some at 8% per annum interest and the rest at 12% per annum interest, at the end of the first year the person still owed $10 000 and had paid a total of $944 in interest. How much of the $10 000 had they borrowed at 8% and how much at 12%?

Answer 1

Let

A = amount borrowed at 8%

B = amount borrowed at 12%

Total amount borrowed = 10,000

Interest rates

A ==>  8% or .08

B ==> 12% or .12

Total interest ==> 944

===============

A + B = 10000  ==> Eq. 1

.08A + .12B = 944 ==> Eq. 3

A + B = 10000

A = 10000 - B

Substitute the above value into Eq. 2

.08(10000 - B) + .12B = 944

800 - .08B + .12B = 944

.04B = 944 - 800

.04B = 144

B = 144/.04

B = $ 3600  ==> amount borrowed at 8%

A + B = 10000

A + 3600 = 10000

A = 10000 - 3600

A = $ 6,400  ==> amount borrowed at 12%

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To check

A = 6400

B = 3600

A + B = 10000 ==> Eq. 1

6400 + 3600 = 10000

10000 = 10000

.08A + .12B = 944 ==> Eq. 2

.08(6400) + .12(3600) = 944

512 + 432 = 944

944 = 944

Answer 2

suppose money borrowed at 8% was x $, and so $(10000 - x) at 12%

Total interest = ((x *1*8) + ((10000-x)*1*12))/100

944 = (8x +120000 - 12x)/100

94400 = -4x + 120000

4x = 120000 - 94400

4x =25600

x = $6400

so $ 6400 at 8% and $3600 at 12%

Answer 3

let x=the amount borrowed at 12%

10000-x=amount borrowed at 8%



.12x+.08(10000-x)=944

.12x+800-.08x=944

.04x=144

x=3600 at 12%

10000-3600=6400 at 8%

check

.12*3600=432

6400*.08=512

432+512=944

Answer 4

Amount borrowed at 8% (x):

0.08x + 0.12($10,000 - x) = $944

2x + 3($10,000 - x) = $23,600

2x + $30,000 - 3x = $23,600

x = $6,400

Amount borrowed at 12%:

= $10,000 - $6,400

= $3,600

Answer: at 8%, $6,400; at 12%, $3,600

Proof (interest paid was $944):

= 0.08($6,400) + 0.12($3,600)

= $512 + $432

= $944