$3,000 is invested in an account paying 4% interest and twice that amount is invested at 6%. How do you write the expression that represents the amount of interest earned on the money invested at 6%?

1 Answer
Aug 1, 2018

See explanation

Explanation:

Assumption: The interest given is the #ul("yearly interest rate")#

Twice the amount is #$6000#

Let the count of years be #n#

#color(brown)("At 6% "ul("simple")" interest calculated once per year is:")#

#$6000+[nxx(6/100xx$6000)]#

Factoring out the #$6000# giving:

#$6000(1+(6n)/100)#

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#color(brown)("At 6% annually "ul("compounded monthly")#

The 6% is 'spread' over each calculation cycle. There are 12 months in a year so instead of #6/100# we have #6/100xx1/12#

Note that the count in years is #n#. However ther are 12 calculations cycles in one year so the 'power' of #n# becomes #12n# instead.

#$6000(1+6/(12xx100))^(12xxn)#

#$6000(1+cancel(6)^1/cancel(1200)^600)^(12n)#

#$6000(1+1/600)^(12n)#