Engineering Fundamentals: An Introduction to Engineering
Engineering Fundamentals: An Introduction to Engineering
6th Edition
ISBN: 9780357112311
Author: Saeed Moaveni
Publisher: Cengage Learning US
Question
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Chapter 20, Problem 20P

(a)

To determine

Find the monthly payment amount.

(a)

Expert Solution
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Answer to Problem 20P

The monthly payment amount is $304.15.

Explanation of Solution

Given data:

The normal interest rate (i) is 8%,

The present value (P) is $15000,

The number of years (n) is 60 months that is 5 years,

The number of interest compounding periods per year (m) is given monthly therefore 12.

Formula used:

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

A=P[(im)(1+im)nm(1+im)nm1] (1)

Here,

P is the Present cost,

i is the normal interest rate,

n is the number of years,

m is the number of interest compounding periods per year.

Calculation:

Substitute $15000 for P, 8% for i, 5 for n, and 12 for m in equation (1) to find A.

A=($15000)[((8%)12)(1+(8%)12)(5)(12)(1+(8%)12)(5)(12)1]=($15000)[(0.0812)(1+0.0812)60(1+0.0812)601]=($15000)[(0.0812)(1+0.0812)60(1+0.0812)601]=$304.15

Therefore, the monthly payment amount is $304.15.

Conclusion:

Thus, the monthly payment amount is $304.15.

(b)

To determine

Find the effective interest rate.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

The effective interest rate is 10%.

Explanation of Solution

Given data:

The present value (P) is $15000,

The number of years (n) is 60 months that is 5 years,

The number of interest compounding periods per year (m) is given monthly therefore 12.

Formula used:

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

A=P[(im)(1+im)nm(1+im)nm1] (2)

Here,

P is the Present cost,

i is the normal interest rate,

n is the number of years,

m is the number of interest compounding periods per year.

Formula to calculate the effective interest rate is,

ieff=(1+im)m1 (3)

Calculation:

It is given that the bank charges 4.5% of the loan amount at the time the bank gives loan.  Therefore, the total loan amount given by the bank is,

Loanamount=P(4.5%)P (4)

Substitute $15000 for P in equation (4) to find the Loanamount.

Loanamount=($15000)(4.5%)($15000)=($15000)(0.045)($15000)=($15000)($675)=$14325

Hence, the present value (P) is $14325.

Substitute $304.15 for A, $14325 for P, 5 for n, and 12 for m in equation (2) to find i.

($304.15)=($14325)[(i12)(1+i12)(5)(12)(1+i12)(5)(12)1]

Reduce the equation as,

[(i12)(1+i12)60(1+i12)601]=0.0212 (5)

By using trial and error method, calculate the value of i in equation (5).

(i) Interest rate of 5%:

Substitute 5% for i in equation (5).

[((5%)12)(1+(5%)12)60(1+(5%)12)601]=0.0212[(0.0512)(1+0.0512)60(1+0.0512)601]=0.02120.0188=0.0212 (6)

From the equation (6), it is clear that L.H.SR.H.S. Therefore, the value of i should not be 5%.

(ii) Interest rate of 10%:

Substitute 10% for i in equation (5).

[((10%)12)(1+(10%)12)60(1+(10%)12)601]=0.0212[(0.112)(1+0.112)60(1+0.112)601]=0.02120.0212=0.0212 (7)

From the equation (7), it is clear that L.H.S=R.H.S. Therefore, the value of i is 10%.

Substitute 10% for i, and 12 for m in equation (3) to find ieff.

ieff=(1+(10%)12)121=(1+0.112)121=0.1(or)10%

Therefore, the effective interest rate is 10%.

Conclusion:

Thus, the effective interest rate is 10%.

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