Question

You invested $26,000 into accounts paying 6% and 9% annual interest, respectively if the total interest and for the year was 2190, how much was invested at each rate?

Answer 1

Answer:

$5000 is invested in 6% interest account and $21000 is invested in 9% interest account.

Step-by-step explanation:

Let the amount of investment in the account paying 6% interest is $x and that in the account paying 9% interest is $y.

So, x + y = 26000 .......... (1)

And, the interest of $x after 1 year will be $\frac{x \times 6}{100} = 0.06x$ and the interest of $y after 1 year will be $\frac{y \times 9}{100} = 0.09y$.

From the given condition, we can write

0.06x + 0.09y = 2190

⇒ 6x + 9y = 219000 ......... (2)

Now, solving equations (1) and (2) we get,

9y - 6y = 219000 - 26000 × 6

⇒ 3y = 63000

⇒ y = $21000

So, from equation (1) we get, x = 26000 - y = $5000.

Therefore, $5000 is invested in the 6% interest account and $21000 is invested in the 9% interest account. (Answer)