Respuesta :
If Mike is willing to pay no more than an effective rate of 8.000% annually, the loans that meet his criteria are loan X and loan Z. Of those two, the lowest would be loan X. I hope the answer will help you :)
Answer:
b. X and Z
Step-by-step explanation:
Since, the effective rate is,
[tex]r=(1+\frac{i}{n})^i-1[/tex]
Where, i is the nominal rate,
n is the number of compounding periods,
For loan X,
i = 7.815 % = 0.07815,
n = 2, ( 1 year = 2 semiannual )
Thus, the effective rate would be,
[tex]r=(1+\frac{0.07815}{2})^2-1[/tex]
[tex]=0.079676855625[/tex]
[tex]=7.9676855625\%\approx 7.968\%[/tex]
Since, 7.968 % < 8.000 %,
⇒ Loan X meets Mike's criteria,
For loan Y,
i = 7.724 % = 0.07724,
n = 12 ( 1 year = 12 months ),
Thus, the effective rate would be,
[tex]r = ( 1+\frac{0.07724}{12})^{12}-1[/tex]
[tex]=0.0800339518197[/tex]
[tex]=8.00339518197\% \approx 8.003\%[/tex]
Since, 8.003 % > 8.000 %,
⇒ Loan Y does not meet his criteria,
For loan Z,
i = 7.698 % = 0.07698,
n = 52 ( 1 year = 52 weeks ),
Thus, the effective rate would be,
[tex]r = (1+\frac{0.07698}{52})^{52}-1[/tex]
[tex]=0.0799589986135[/tex]
[tex]=7.99589986135\%[/tex]
[tex]\approx 7.996\%[/tex]
Since, 7.996 % < 8.000 %,
⇒ Loan Z meets his criteria.
Therefore, Option 'b' is correct.