This is the fourth part of the Solving Investment Problems Series. In this part, we discuss a problem which is very similar to the third part. We discuss an investment at two different interest rates.

**Problem**

Mr. Garett invested a part of $20 000 at a bank at 4% yearly interest. How much does he have to invest at another bank at a 8% yearly interest so that the total interest of the money is 7%.

*Solution and Explanation*

Let x be the money invested at 8%

(1) We know that the interest of 20,000 invested at 4% yearly interest is

20,000(0.04)

(2) We also know that the interest of the money invested at 8% is

(0.08)(x)

(3) The interest of total amount of money invested is 7%. So,

(20,000 + x)(0.07)

Now, the interest in (1) added to the interest in (2) is equal to the interest in (3). Therefore,

20,000(0.04) + (0.08)(x) = (20,000 + x)(0.07)

Simplifying, we have

800 + 0.08x = 1400 + 0.07x

To eliminate the decimal point, we multiply both sides by 100. That is

80000 + 8x = 140000 + 7x

8x – 7x = 140000 – 80000

x = 60000

This means that he has to invest $60,000 at 8% interest in order for the total to be 7% of the entire investment.

Check:

$20,000 x 0.04 = $800

$60,000 x 0.08 = 4800

Adding the two interest, we have $5600. We check if this is really 7% of the total investment.

Our total investment is $80,000.

Now, $80,000 x 0.07 = $5600.