Question

6. A man invested $16,000. He invested a portion at 6% and the balance at 8%. If he receives $260 every 3

months, how much did he invest at each rate?
1 =
Let p = money invested at 6%. _512000
16,000 - p = money invested at 8% $4,000.
3 1
yr (We need to express time in terms of years. 3 months is of a year.)
12
I = $260
Check
260 = p(0.06 yrs £ ¥r) + (16,000 – p(0.08/yr) ( 4 yr)
260 (2000)(0.0 Gdyncdys)*
(1500072,000.00.0Vgn Gray)
260=180+80
260:260
2602020157 0.02(163000 €)
200=0.05p+320 -0.010
260=390-0.005p
200-320 -0.005
-Go
0.005
0=12000

So I have already solved this problem but I don’t really understand how I solved it if anyone could help me I would really appreciate it!

Answer 1

If the interest for 3 months is $260, then the total interest for the year is 4 times $260 or $1040.

Let x = amount invested at one of the rates, say 6%. This leaves the rest of the $16,000 or (16000-x) to be invested at the other rate, at 8%.

The equation is basically that the interest earned for 1 year at 6% plus the interest earned for the year at 8% is equal to the total interest for the year, or $1040.

.06x + .08(16000-x) = 1040

.06x + .08(16000) - .08x = 1040

-.02x + 1280 = 1040

Subtract 1280 from each side:

-.02x = 1040 - 1280

-.02x = -240

To solve for x, divide both sides by -.02, remembering that a negative divided by a negative is a negative.

x = 240/.02

Either divide with a calculator, or multiply numerator and denominator by 100 to eliminate the decimal.

x = 24000/2 = $12000 at 6%

16000-x = 16000-12000 = $4000 at 8%