Question

David used part of $100,000 to purchase a house. Of the remaining portion, he invested 1 3 of it at 4 percent simple annual interest and 2 3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled $320, what was the purchase price of the house?

Answer 1

Answer:

The purchase price of the house is $94,000

Explanation:

Let the amount invested by David be b, then the amount used to purchase the house would be 100,000 - b

If he invested 1/3 of it at 4 percent simple annual interest and 2/3 of it at 6 percent simple annual interest. If after a year the income from the two investments totaled $320

Then,

$\frac{1}{3}b × 4% + \frac{2}{3}b × 6% = 320$

$\frac{b}{3} × \frac{4}{100} + \frac{2b}{3} × \frac{6}{100} = 320$

$\frac{4b}{300} + \frac{12b}{300} = 320$

$\frac{16b}{300} = 320$

16b = 320 × 300

b = 320 × 300/16

b = 6,000

Therefore, the cost of the house (100,000 - b)

= 100,000 - 6,000

= $94,000