Question

A merchant has two loans totaling $20000. The simple interest rates are 6% and 7%. If the annual interest charge on the 6% loan is $520 more than on the 7% loan, how much did he borrow at each rate?

Answer 1

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

$I=P(rt)$

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%

$t=1\ years\\ P=x\\r=6\%=6/100=0.06$

$I_1=0.06x$  ----> equation A

Loan at 7%

$t=1\ years\\ P=20,000-x\\r=7\%=7/100=0.07$

$I_2=(20,000-x)0.07$

$I_2=1,400-0.07x$  ----> equation B

Remember that

the annual interest charge on the 6% loan is $520 more than on the 7% loan

so

$I_1=I_2+520$ ----> equation C

substitute equation A and equation B in equation C

$0.06x=1,400-0.07x+520$

solve for x

$0.06x+0.07x=1,400+520\\0.13x=1,920\\x=\ 14,769.23$

Find the value of (20,000-x)

$\ 20,000-\ 14,769.23=\ 5,230.77$

therefore

He borrowed $14,769.23 at 6% and $5,230.77 at 7%