Mitchell took out a loan for $1100 at a 19.2% APR, compounded monthly, tobuy a scanner. If he will make monthly payments of $71.50 to pay off theloan, how many total payments will he have to make?ОА. 18Ов. 19ооо

Answer:The total number of the payments is 18 ⇒ answer AStep-by-step explanation:* Lets revise the rule of compounded monthly payment→ [tex]EMI=\frac{(A)(r)}{[1-\frac{1}{(1+r)^{n}}]}[/tex], where- A is the loan amount- r is monthly interest in decimal (R/12*100)) - n the total number of payments∵ A = $1100∵ EMI = $71.5- Interest rate is 19.2% APR∵ r = [tex]\frac{19.2}{12*100}=0.016[/tex]- Substitute these values in the rule to find n∴ 71.5 = [tex]\frac{1100(0.016)}{[1-\frac{1}{(1+0.016)^{n}}]}=\frac{17.6}{[1-\frac{1}{(1.016)^{n}}]}[/tex]- By using cross multiplication∴ 71.5[1 - [tex]\frac{1}{(1.016)^{n}}[/tex] ] = 17.6- Divide both sides by 71.5∴ 1 - [tex]\frac{1}{(1.016)^{n}}[/tex] = [tex]\frac{16}{65}[/tex]- Subtract 1 from both sides∴ - [tex]\frac{1}{(1.016)^{n}}[/tex] = - [tex]\frac{49}{65}[/tex]- Multiply both sides by -1∴ [tex]\frac{1}{(1.016)^{n}}[/tex] = [tex]\frac{49}{65}[/tex]- By using cross multiplication∴ 49[ [tex](1.016)^{n}[/tex] ] = 65- Divide both sides by 49∴ [tex](1.016)^{n}[/tex] = [tex]\frac{65}{49}[/tex]- Insert log for both sides∴ ㏒ [tex](1.016)^{n}[/tex] = log( [tex]\frac{65}{49}[/tex] )- Put n in-front of the ㏒∴ n㏒(1.016) = ㏒( [tex]\frac{65}{49}[/tex] )- Divide both sides by ㏒(1.016)∴ n = 17.8 ≅ 18 * The total number of the payments is 18