A loan is to be amortized by a group of five end of year payments forming anascending arithmetic progression. The initial payment was to be P 5,000 and thedifference between successive payments was to be P500. But the loan wasrenegotiated to provide for the payment of equal rather than uniformly varyingsums. If the interest rate of the loan was 15%, what was the annual payment?P 5861P 5530P 5420P 5713

A loan is a sum of money borrowed from a third party, such as a bank, and returned over time in…

A loan is to be amortized by a group of five end of year payments forming an ascending arithmetic progression. The initial payment was to be P 5,000 and the difference between successive payments was to be P50O. But the loan was renegotiated to provide for the payment of equal rather than uniformly varying sums. If the interest rate of the loan was 15%, what was the annual payment? O P 5861 P 5530 P 5420 O P 5713

A loan is to be amortized by a group of five end of year payments forming an ascending arithmetic progression. The initial payment was to be P 5,000 and the difference between successive payments was to be P50O. But the loan was renegotiated to provide for the payment of equal rather than uniformly varying sums. If the interest rate of the loan was 15%, what was the annual payment? O P 5861 P 5530 P 5420 O P 5713

Corporate Fin Focused Approach

5th Edition

ISBN:9781285660516

Author:EHRHARDT

Publisher:EHRHARDT

Chapter4: Time Value Of Money

Section4.17: Amortized Loans

Problem 1ST

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A loan is to be amortized by n level annual payments of X where n > 5. You are given (1) The amount of interest in the first payment is 604.00 (2) The amount of interest in the third payment is 593.75 (3) The amount of interest in the fifth payment is 582.45 Calculate X.
1.A LOAN WAS TO BE AMORTTIZED BY A GROUP OF SIX END-OF YEAR PAYMENTS FORMING AM ASCENDING ARITHMETIC PROGRESSION. THE INITIAL PAYMENT WAS TO BE 15,000 Php AND THE DIFFERENCE BETWEEN SUCCESSIVE PAYMENTS WAS TO BE 1000 Php. BUT THE LOAN WAS REGOTIANATED TO PROVIDE FOR THE PAYMENT OF EQUAL RATHER THAN UNIFROMLY VARYING SUMS. IF THE INTEREST RATE OF THE LOAN WAS 16%, WHAT WAS THE ANNUAL PAYMENT? FROM PROBLEM 1., FIND THE EQUIVALENT ANNUAL PAYMENT OF THE FOLLOWING OBLIGATIONS AT 21% INTEREST?
2.A certain loan was to be amortized based on the schedule below at 15% interest. What is the equivalent annual uniform payment of the loan? End of Year Payment 1 P 7,000 2 P 6,500 3 P 6.000 4 P 5,500 5 P 5.000 6 P 4,500 7 P 4,000 a. P 5365.80 b. P 5333.00 c. P 6475.12 d. P 5775.08 PLEASE SHOW YOUR SOLUTIONS. THANK YOU!
Using the relation R= 1 - ( 1 + 1)-1), 1-(1+i)-n find the amortization schedule for a loan of GH¢4000 to be paid off in 2 years with quarterly payments. The interest on the loan is 7.5% annual interest compounded quarterly.
A loan is to be repaid after a term of 5 years. If the principal amount is at P215,000 and the total amount to be repaid is at P275,000, what was the corresponding interest rate if it compounded quarterly? * 4.35% 4.53% 4.76% 4.95%
A LOAN OF P80,000 IS MADE FOR A PERIOD OF 18 MONTHS AT A SIMPLE INTEREST RATE OF 10%. WHAT FUTURE AMOUNT IS DUE AT THE END OF THE LOAN PERIOD? O a P92,100.00 O b. P92,020.00 O c. P92,000.00 O d. P93,000.00
The initial payment to a loan was to be P571 and the difference between successive payments was to be P6,195 forming an increasing arithmetic progression. But the loan was renegotiated to provide for the payment of equal rather than uniformity varying sums. If the interest rate of the loan was 10.90%, what was the equivalent annual payments if the payment is to be made for 6 years? Write your final answer in 4 decimal places.
A loan was to be amortized by a 5 end-of-year payments forming an ascending arithmetic progression. The initial payment was to be P10,000 and the difference between successive payment was to be P1,000. If the loan is to be paid in a single payment after 3 years what should the single payment be? The interest of the loan is 10% pa. Show your cash flow diagram. O P 69,150 O P 64,405 O P 49,756 O P59,588
This questi Consider a loan of $7700 at 6.8% compounded semiannually, with 18 semiannual payments. Find the following. (a) the payment necessary to amortize the loan (b) the total payments and the total amount of interest paid based on the calculated semiannual payments (c) the total payments and total amount of interest paid based upon an amortization table. (a) The semiannual payment needed to amortize this loan is $ 545.94. (Round to the nearest cent as needed.) (b) The total amount of the payments is $ 9827.07 (Round to the nearest cent as needed.) The total amount of interest paid is $ 2127.07 (Round to the nearest cent as needed.) (c) The total payment for this loan from the amortization table is $ (Round to the nearest cent as needed.) The total interest from the amortization table is $ (Round to the nearest cent as needed.)
3. A debt of P3,500 is to be amortized by 6 equal semiannual payments with interest at 6%compounded semiannually. What is the total interest to be paid during the term of the loan?
A LOAN OF P80,000 IS MADE FOR A PERIOD OF 18 MONTHS AT A SIMPLE INTEREST RATE OF 10%. WHAT DUE AT THE END OF THE LOAN PERIOD? O a. P92,100.00 O b. P93,000.00 O c. P92,020.00 O d. P92,000.00
Consider a loan of $7700 at 6.8% compounded semiannually, with 18 semiannual payments. Find the following. (a) the payment necessary to amortize the loan (b) the total payments and the total amount of interest paid based on the calculated semiannual payments (c) the total payments and total amount of interest paid based upon an amortization table. (a) The semiannual payment needed to amortize this loan is $ (Round to the nearest cent as needed.) (b) The total amount of the payments is $ (Round to the nearest cent as needed.) The total amount of interest paid is $. (Round to the nearest cent as needed.) (c) The total payment for this loan from the amortization table is $ (Round to the nearest cent as needed.) The total interest from the amortization table is $ (Round to the nearest cent as needed.)