Question

Jolene invests her savings in two bank accounts, one paying 6 percent and the other paying 10 percent simple interest per year. She puts twice as much in the lower-yielding account because it is less risky. Her annual interest is 10428 dollars. How much did she invest at each rate? Amount invested at 6 percent interest is $ Amount invested at 10 percent interest is $

Answer 1

Answer:

Amount invested at 10% = $47,400

Amount invested at 6% = $94,800

Step-by-step explanation:

Let

Amount invested at 10% = X

Amount invested at 6% = 2X

According to given condition

Interest received at 10% + Interest received at 6% = 10428

X10% + 2X6% = 10428

0.1X + 0.12X = 10428

0.22X = 10428

X = 10428/0.22

X = 47400

So

Amount invested at 10% = X = 47,400

Amount invested at 6% = 2X = 2*47400 = 94,800

Check

47400*10% + 94800*6% = 10428

4740 + 5688 = 10428

10428 = 10428

Answer 2

Answer:

The amount invested at 6% is $94,800

&

The amount invested at 10% is $47,400.

Step-by-step explanation:

Total yearly interest for the two accounts is: $10,428

Let x be the amount invested at 6%

& y be the amount invested at 10%

From the question we can get 2 equations as;

x = 2y --------------------------Equation 1

0.06x + 0.10y = 10428 ----------Equation 2

Substitute for x in Equation 2 we get;

0.06 (2y) + 0.10y = 10428

0.12y + 0.10y = 10428

0.22y = 10428

Divide the above equation by 0.22, we get;

y =  $\frac{10428}{0.22}$

y = $47,400

Let us substitute the value of y in Equation 1 we get;

x = 2(47400)

x = $94,800

Now to check our answer let us put in the simple interest formula. If we get the sum of the two interests equal to 10428 then our answers are correct:

0.06 x 94800 + 0.10 x 47400

= 5688 + 4740

= $10,428

Hence the amount invested at 6% is $94,800 and the amount invested at 10% is $47,400.