Answer:
He borrowed $14,769.23 at 6% and $5,230.77 at 7%
Step-by-step explanation:
Let
x ----> loan amount at 6%
20,000 -x ----> loan amount at 7%
we know that
The simple interest formula is equal to
[tex]I=P(rt)[/tex]
where
I is the Final Interest Value
P is the loan amount
r is the rate of interest
t is Number of Time Periods
in this problem we have
Loan at 6%
[tex]t=1\ years\\ P=x\\r=6\%=6/100=0.06[/tex]
[tex]I_1=0.06x[/tex] ----> equation A
Loan at 7%
[tex]t=1\ years\\ P=20,000-x\\r=7\%=7/100=0.07[/tex]
[tex]I_2=(20,000-x)0.07[/tex]
[tex]I_2=1,400-0.07x[/tex] ----> equation B
Remember that
the annual interest charge on the 6% loan is $520 more than on the 7% loan
so
[tex]I_1=I_2+520[/tex] ----> equation C
substitute equation A and equation B in equation C
[tex]0.06x=1,400-0.07x+520[/tex]
solve for x
[tex]0.06x+0.07x=1,400+520\\0.13x=1,920\\x=\$14,769.23\\[/tex]
Find the value of (20,000-x)
[tex]\$20,000-\$14,769.23=\$5,230.77[/tex]
therefore
He borrowed $14,769.23 at 6% and $5,230.77 at 7%
