do question 3 and 4 plz What annual expenditure for 10 years is equivalent to sp
ID: 1101555 • Letter: D
Question
do question 3 and 4 plz
What annual expenditure for 10 years is equivalent to spending $1000 at the end of the first year. $2000 at the end of the fourth year, $3000 at the end of the eight year, if interest is at 8 percent per year? An inventor has been offered $12,000 per year for the next five years and $6000 annually for the following seven years. At what price could he afford to sell his invention to earn 10 percent? A loan of $12,000 is to be financed to assist in buying an automobile. On the basis of compounding for 42 months, the end of month equal payment is quoted as $445. What nominal interest rate is being charged? Bob Pearson borrowed $ 25,000 from a bank at an interest rate of 10% compounded monthly. The loan will be repaid in 36 equal installements over three years. Immediately after his 20 th payment, Bob desired to pay the remainder of the loan in a single payment. Compute the total amount he must pay. Use semiannual compounding. A standby generator was purchased 6 years ago for $4200. Similar equipment had shown an economic life of 15 years with a salvage of 15% of the first cost The generator is no longer needed and is to be sold for $1800. The interest rate is 8 %. What is the difference between the actual and anticipated equivalent annual capital recovery cost? The lining of a chemical tank must be replaced every 3 years at a cost of $1800. A new type of lining is available that is more resistant to corrosion. The new lining costs $3100. If the MARR is 12% and taxes and insurance are 4% of the first cost annually, how long must the improved lining last to be more economical than the present lining?Explanation / Answer
3) Let nominal interest charged be r%
Hence Present value of all EMIs = Principal amount
=> $ 12,000 = 445 (1/(1+r/12)^1+1/(1+r/12)^2+...+1/(1+r/12)^42)
=> r/12 = (monthly rate) = IRR of -12000 followed by 445 (42 times) = 2.544825%
=> Answer = r = 27.0538% compounded monthly
4) Bob pays EMI on $ 25000 at 10% compounded monthly
Let EMI = X
Hence Present value of all EMIs = Principle
=> $ 25,000 = X*(1/(1+10%/12)^1+1/(1+10%/12)^2+...+1/(1+10%/12)^36)
=> X = $ 806.6797
Hence, Loan amount remaining after 20th payment = $ 12,036.60 = ANswer
LOAN AMOUNT 25,000.00 RATE OF INTEREST PER ANNUM 10.00% RATE OF INTEREST PER MONTH 0.8333% NO OF INSTALLMENT 36 EMI $ 806.6797Related Questions
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