Question

Last year, Yoko had $10,000 to invest. She invested some of it into an account that paid 10% simple interest per year, and she invested the rest in an account that paid 6% simple interest per year. After one year she received a total of $640 in interest. How much did she invest in each account?

Answer 1

Final answer:

At 10% interest, $1 000 was invested.

At 6% interest, $9 000 was invested.

Let "x" be the amount invested at 10%. The annual interest for this investment would be $ 0.1x

Now, the investment at 6% is (10 000 - x), and it would have an annual interest of 0.06(10 000 - x)

We will use the following equation to solve for this

$interest_1+interest_2=totalinterest$

interest 1 = 0.1x

interest 2 = 0.06(10 000 - x)

total interest = 640

Substitute these values

$interest_1+interest_2=totalinterest$$0.1x+0.06(10000-x)=640$

Solve for x

$0.1x+0.06(10000-x)=640$$0.1x-0.06x=640-\lbrack(0.06)(10000)\rbrack$$0.04x=640-600$$0.04x=40$

Divide both sides by 0.04

$\frac{0.04x}{0.04}=\frac{40}{0.04}$$x=1000$

Since the investment at 6% is equal to 10000-x,

$interest_2=10000-x$$=10000-1000$$=9000$

Final answer:

At 10% interest, $1 000 was invested.

At 6% interest, $9 000 was invested.

To check,

$0.1x+0.06(10000-x)=640$$0.1(1000)+0.06(10000-(1000))=640$$640=640$