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Anika [276]
4 years ago
13

Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest

and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000?
Mathematics
1 answer:
hjlf4 years ago
8 0

Answer:

In the 8% option , $5,000 should be invested

In the 12% option, $15,000 should be invested

Step-by-step explanation:

Let’s start with the total return.

Since the total return expected is 11%, the amount expected to be returned will be;

11/100 * 20,000 = $2,200

Now, let the amount invested at 8% interest be $x while the amount invested at 12% interest be $y

Mathematically;

x + y = 20,000 ••••••(i)

Now let’s work with the interest part;

On the 8% part, amount of interest expected is 8/100 * x

on the 12% part, amount of interest expected is 12/100 * x

8% of $x + 12% of $y = 2,200

Writing this fully mathematically, we have;

(8/100 * x) + (12/100 * y) = 2,200

8x/100 + 12y/100 = 2,200

Multiply through by 100

8x + 12y = 220,000 ••••••(ii)

Now we have two equations to solve simultaneously;

x + y = 20,000

8x + 12y = 220,000

From i, x = 20,000-y

Substitute this into ii

8(20,000-y) + 12y = 220,000

160,000-8y + 12y = 220,000

4y = 220,000 - 160,000

4y = 60,000

y = 60,000/4

y = $15,000

x = 20,000 -y

x = 20,000 - 15,000

x = $5,000

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Step-by-step explanation:

a) The compound interest formula can be used:

  A = P(1 +r/n)^(nt)

where P is the principal invested at annual rate r compounded n times per year for t years.

  A = $750,000(1 +.038/4)^(4·10) ≈ $1,094,748.09

Tamsyn's account will have a balance of $1,094,748.09 when she retires.

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b1) The amortization formula is good for this.

  A = P(r/n)/(1 -(1 +r/n)^(-nt))

where P is the amount earning interest at annual rate r compounded n times per year for t years.

  A = $1,094,748.09(0.049/12)/(1 - (1 +0.049/12)^(-12·15)) ≈ $8600.28

Tamsyn can withdraw $8600.28 per month for 15 years.

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b2) The account balance after n months will be ...

  B = P(1 +r/12)^n -A((1+r/12)^n -1)/(r/12)

Filling in the known values and solving for n, we have ...

  300,000 = 1,094,748.09(1.1.00408333^n) -8600.28(1.00408333^n -1)/.000408333

  300,000 = 1,094,748.09(1.1.00408333^n) -2,106,191.02(1.00408333^n -1)

  -1,806,191.02 = -1,011,442.93(1.00408333^n)

  1.785757 = 1.00408333^n

  n = log(1.785757)/log(1.00408333) = 142.3

After about 11 years 10 months, the account balance will be $300,000.

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