Given data:

Total amount borrowed (T) from bank is $ 12000,

In First bank, uniform series payment at the end of each year (A1) is $ 2595.78, and the number of years (n1) is 6,

In Second bank, uniform series payment at the end of each month (A2) is $ 198.87, and the number of periods (n2) is 6 years that is 72 months.

Formula used:

Consider the following expression obtained from the payment details of first bank,

T=A1(PA, i , n1) (1)

Here,

T is the total borrowed amount,

A1 is the uniform series payment in the first bank,

P is the present cost,

A is the uniform series payment,

i is the normal interest rate,

n1 is the number of years.

The equation (1) is rewritten with respect to the given values is,

$ 12000=($ 2595.78)(PA, i , 6) (2)

Consider the following expression obtained from the payment details of second bank,

T=A2(PA, i, n2) (3)

Here,

A1 is the uniform series payment in the second bank,

n2 is the number of periods in months.

The equation (3) is rewritten with respect to the given values is,

$ 12000=($ 198.87)(PA, i , 72) (4)

Formula to calculate the present cost for the given uniform series payment is,

(PA, i, n)=[(1+i)n−1i(1+i)n] (5)

Formula to calculate the present cost for the given uniform series payment when the interest compounds m times per year is,

(PA, i, n)=[(1+im)nm−1(im)(1+im)nm] (6)

Calculation:

Case 1: Interest rate of first bank

Substitute the equation (5) in equation (2) and substitute 6 for n to find i.

$ 12000=($ 2595.78)[(1+i)6−1i(1+i)6][(1+i)6−1i(1+i)6]=$ 12000$ 2595.78[(1+i)6−1i(1+i)6]=4.623 (7)

By using trial and error method, calculate the value of i in equation (7).

(i) Interest rate of 5 %:

Substitute 5 % for i in equation (7).

[(1+(5 %))6−1(5 %)(1+(5 %))6]=4.623[(1+0.05)6−1(0.05)(1+0.05)6]=4.623

Reduce the equation as,

5.076=4.623 (8)

From the equation (8), it is clear that L.H.S≠R.H.S. Therefore, the value of i should not be 5 %.

(ii) Interest rate of 8 %:

Substitute 8 % for i in equation (7).

[(1+(8 %))6−1(8 %)(1+(8 %))6]=4.623[(1+0.08)6−1(0.08)(1+0.08)6]=4.623

Reduce the equation as,

4.623=4.623 (9)

From the equation (9), it is clear that L.H.S=R.H.S. Therefore, the value of i is 8 %.

Case 2: Interest rate of second bank:

Substitute the equation (6) in equation (4), and substitute 6 for n, and 12 for m to find i.

$ 12000=($ 198.87)[(1+i12)(6)(12)−1(i12)(1+i12)(6)(12)]

Rewrite the equation as follows,

[(1+i12)72−1(i12)(1+i12)72]=$ 12000$ 198.87[(1+i12)72−1(i12)(1+i12)72]=60.34 (10)

By using trial and error method, calculate the value of i in equation (10).

(i) Interest rate of 5 %:

Substitute 5 % for i in equation (10).

[(1+(5 %)12)72−1((5 %)12)(1+(5 %)12)72]=60.34[(1+0.0512)72−1(0.0512)(1+0.0512)72]=60.34

Reduce the equation as,

62.1=60.34 (11)

From the equation (11), it is clear that L.H.S≠R.H.S. Therefore, the value of i should not be 5 %.

(ii) Interest rate of 6 %:

Substitute 6 % for i in equation (10).

[(1+(6 %)12)72−1((6 %)12)(1+(6 %)12)72]=60.34[(1+0.0612)72−1(0.0612)(1+0.0612)72]=60.34

Reduce the equation as,

60.34=60.34 (12)

From the equation (12), it is clear that L.H.S=R.H.S. Therefore, the value of i is 6 %.

Therefore, from the analysis the interest rate of the first bank and second bank is 8 % and 6 %.

The second bank is highly preferable because it provides less interest rate than the first bank.

Conclusion:

Thus, the interest rate of the first bank and second bank is 8 % and 6 %.