Question

The Kwons invested $6000, part at 7 3/4% and part at 10%, and earned $567 in interest. how much did they invest at each rate?

Answer 1

a = amount invested at 7¾% or 7.75%.

how much is 7.75% of "a"? (7.75/100) * "a", namely 0.0775a.

b = amount invested at 10%

how much is 10% of "b"? (10/100) * "b", namely 0.10b.

we know the total amount invested is $6000, so whatever "a" and "b" might be, we know that a + b = 6000.

we also know that the yielded amount in interest is $567, so if we simply add their interest, that'd be 0.0775a + 0.10b = 567.

`\stackrel{\textit{using the 1st equation}}{a+b=6000}\implies b=6000-a \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{0.0775a~~ + ~~0.10(6000-a)~~ = ~~567}\implies 0.0775a+600-0.10a=567 \\\\\\ 600-0.0225a=567\implies -0.0225a=-33 \implies a=\cfrac{-33}{-0.0225} \\\\\\ a=1466(2)/(3)\implies \boxed{a=1466.\overline{66}}\hspace{5em}b=4533(1)/(3)\implies \stackrel{ 6000~~ - ~~a }{\boxed{b=4533.\overline{33}}}`