Question

Jovita divides $12,000 into three investments: a savings account paying 4% annual interest, a bond paying 6%, and a money market fund paying 9%. The annual interest from the three accounts is $860, and she has 2 times as much invested in the bond as in the savings account. What amount does she have invested in each account?

Answer 1

Answer:

She invested $2000 in the savings account,

$4000 in the bond account AND

$6000 in the Money market fund

Step-by-step explanation:

Please refer to the attached image for explanations

Answer 2

Answer:

$2080 was invested in the savings account, $4241 was invested in the bond account, & $6539 was invested in the money fund.

Step-by-step explanation:

Let a, b, and c be the fractions of the $12,000 that is invested in the savings, bond, & money fund respectively.

Let S1, S2, S3 be the total monies from the savings, bond, & money fund after the year with interest respectively.

We know that a+b+c=1

and

S1+S2+S3 = 12000 + 860

S1+S2+S3 = 12860

From, the question, we know that b = 2a, and c = 1 - 3a based on the condition of 2 times more in the bond account that the savings account.

Thus;

S1 = a(12000)(1.04) = 12480a

S2 = 2a(12000)(1.06) = 25,440a

S3 = (1 - 3a)(12000)(1.09) = 13080 - 39240a

We already know that;

S1+S2+S3 = 12860

Thus, Adding the 3 equations gives;

12480a + 25,440a + 13080 - 39240a = 12860

Thus, simplifying gives;

13080 - 12860 = 39240a - 12480a - 25,440a

220 = 1320a

a = 220/1320

a = 0.1667

So,b = 2 x 0.1667 = 0.3334

c = 1 - (3 x 0.1667)

c = 1 - 0.5

c = 0.5

Thus,

S1 = 12480a = 12480 x 0.1667 ≈ $2080

S2 = 25,440a = 25440 x 0.1667 ≈ $4241

S3 = 13080 - 39240(0.1667) ≈ 6539

So, $2080 was invested in the savings account, $4241 was invested in the bond account, & $6539 was invested in the money fund.