angus invested $18,000 part at 3% and part at 5%.?

Question

if the total interest at the end of the year is $600 how much did he invest at each rate

Answer 1

Let us assume he had invested x $ at 3%.
Then he would have invested (18000-x) $ at 5%
Total interest rate = $600
Therefore
3*x/100+5*(18000-x)/100 = 600
which on solving should give x = 15000 and 18000-x = 3000
Therefore he invested $15000 at 3% and $3000 at 5%

Answer 2

If x is investd at 3% and y at 5%, we have x + y = 18,000 and 3x + 5y = 60000 or 3x = 60000 - 5y and substituting in the other equation, we have

60000 - 5y + 3 y = 54000 (we multiplied the first equation by 3)

or - 2y = 54000 - 60000 = - 6000 or y = 3000

So, x = 18000 - y = 15000

Let us check.

15000 at 3% gives 450 dollars and 3000 at 5% gives 150 dollars totalling 600 dollars as interest. So, our calculation is OK.

Answer 3

Sum invested=x - Sum invested=18000-x
R=5% - R=3%
T=1 yr. - T=1 yr.

SI1=PRT/100 - SI2=PRT/100
=(x*5*1)/100 - =(18000-x)*3*1/100
=5x/100 - =3(18000-x)/100





Total SI=SI1+SI2
600=(5x+54000-3x)/100
600*100=2x+54000
60000=2x+54000
60000-54000=2x
6000=2x
3000=x

Sum invested at 5%=x=$3000
Sum invested at 3%=18000-x=18000-3000=$15,000