Question

Solve the problem. A $98,000 trust is to be invested in bonds paying 7%, CDs paying 5%, and mortgages paying 10%. The bond and CD investment together must equal the mortgage investment. To earn a $7550 annual income from the investments, how much should the bank invest in bonds?

Answer 1

The Bond Investment is $1000.

Explanation:

Let the B denotes the Bonds Investment

CD's denoted by C

Mortgage denoted By M

and we have given that

B+C+M = 98,000 ------------------(I)

Also

B+C=M or B+C-M=0 ------------------(II)

Subtracting Eq(I) from Eq(II) we get

2M=98,000

M = 49000

Now Eq(II) Becomes

B+C = 49000

And

B= 49000-C -----------------(III)

also we have given that

Interest on Bonds = B*(7/100) = 0.07B

Interest on CD's = 0.05C

Interest on Mortgage = 0.1M

Total Interest = 0.07B + 0.05C +0.1M = 7550

(0.07)(49000-C) + (0.05)C + (0.1)(49000) =7550

3430 - 0.07C +0.05C +4900 = 7550

-0.02C + 8330 = 7550

Subtracting 8330 from both sides we get

-0.02C = - 780

Dividing -0.02 Both Sides We get

C = 39000

So substituting C =39000 in Eq(III) We get

B = 49000 -39000

B = $1000

So the Bond Investment to earn $7550 annual income from the investment is $1000.

* All Rupees are in $.

Answer 2

Answer: bonds = $10000

CD = $39000

Mortgage = $49000

Explanation:

Let x represent the amount that should be invested in the bond.

Let y represent the amount that should be invested in the CD.

Let z represent the amount that should be invested in the mortgage.

A $98,000 trust is to be invested in bonds paying 7%, CDs paying 5%, and mortgages paying 10%. It means that

x + y + z = 98000 - - - - - - -1

The bond and CD investment together must equal the mortgage investment. It means that

z = x + y

To earn a $7550 annual income from the investments, it means that

0.07x + 0.05y + 0.1z = 7550- - - - -- 2

Substituting z = x + y into equation 1 and equation 2, it becomes

x + y + x + y = 98000

2x + 2y = 98000

Dividing through by 2,

x + y = 49000

x = 49000 - y - - - - - - -3

0.07x + 0.05y + 0.1(x + y) = 7550

0.07x + 0.05y + 0.1x + 0.1y = 7550

0.17x + 0.15y = 7550- - - - - - - - -4

Substituting equation 3 into equation 4, it becomes

0.17(49000 - y) + 0.15y = 7550

8330 - 0.17y + 0.15y = 7550

- 0.17y + 0.15y = 7550 - 8330

- 0.02y = - 780

y = - 780/-0.02

y = 39000

x = 49000 - y = 49000 - 39000

x = 10000

Substituting x = 10000 and y = 39000 into z = x + y 1, it becomes

z = 10000 + 39000 = 49000