Lesson Loan problems
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<H2>Loan problems</H2> <H3>Problem 1</H3>You want to buy a car. The loan amount is $25,000. The company is offering a 5% interest rate for 48 months (4 years). What will your monthly payments be? <B>Solution</B> <pre> Use the formula for the monthly payment for a loan M = {{{P*(r/(1-(1+r)^(-n)))}}} where P is the loan amount; r = {{{0.05/12}}} is the effective interest rate per month; n is the number of payments (same as the number of months); M is the monthly payment. In this problem P = $25000; r = {{{0.05/12}}}. Substitute these values into the formula and get for monthly payment M = {{{25000*(((0.05/12))/(1-(1+0.05/12)^(-48)))}}} = $575.73. <U>ANSWER</U>. The monthly payment is $575.73. In total, you will pay 4*12*575.73 = 27,635.04 dollars in 4 years. The difference $27,635.04 - $25,000 = $2,635.04 is the interest you pay to financial company. </pre> <H3>Problem 2</H3>You purchased a refrigerator at Sears for $2,500 (no sales tax added), and you decide to pay for it in 4 years. Sears' interest rate is 10%. What is the monthly payment that you need to pay? <B>Solution</B> <pre> Use the formula for the monthly payment for a loan M = {{{P*(r/(1-(1+r)^(-n)))}}} where P is the loan amount; r = {{{0.1/12}}} is the effective interest rate per month; n is the number of payments (same as the number of months in 4 years); M is the monthly payment. In this problem P = $2500; r = {{{0.1/12}}}; n = 4*12 = 48. Substitute these values into the formula and get for monthly payment M = {{{2500*(((0.1/12))/(1-(1+0.1/12)^(-48)))}}} = $63.41 (rounded). Thus, the monthly payment is $61.41. <U>ANSWER</U> In total, you will have to repay 4*12*61.41 = 2,947.68 dollars in 4 years. The difference $2,947.68 - $2,500 = $447.68 = $370,681.60 is the interest you pay for this loan. </pre> <H3>Problem 3</H3>Jessica borrowed $5000 from the bank in order to buy a new piano. She will pay it off by equal payments at the end of 2 years. The interest rate is 8% compounded weekly. Determine the size of the payments and the total interest paid. <B>Solution</B> <pre> Use the formula for the periodical payment for a loan M = {{{P*(r/(1-(1+r)^(-n)))}}} where P is the loan amount; r = {{{0.08/52}}} is the effective interest rate per period (per week in this problem); n is the number of payments (total number of weeks in this problem); M is the payment for the period (weekly payment in this problem). In this problem P = $5000; r = {{{0.08/52}}}; n = 2*52 = 104. Substitute these values into the formula and get for monthly payment M = {{{5000*(((0.08/52))/(1-(1+0.08/52)^(-104)))}}} = $52.07. Thus, the weekly payment is $52.07. In total, Jane will have to repay 2*52*52.07 = 5415.28 dollars in 2 years. The difference $5,415.28 - $5,000 = $415.28 is the interest she will pay for this loan. </pre> <H3>Problem 4</H3>You can afford a $300 per month car payment. You've found a 5 year loan at 8% interest. What the maximum loan could be? <B>Solution</B> <pre> Use the formula for the monthly payment for a loan M = {{{P*(r/(1-(1+r)^(-n)))}}} where P is the loan amount; r = {{{0.03/12}}} is the effective interest rate per month; n is the number of payments (same as the number of months); M is the monthly payment. From this formula, the expression for the maximum loan is P = {{{M/((r/(1-(1+r)^(-n))))}}}. In this problem M = $300; r = {{{0.08/12}}}, n = 5*12 = 60 monthly payments. Substitute these values into the formula and get for the maximum loan amount P = {{{300/(((0.08/12)/(1-(1+0.08/12)^(-60))))}}} = $14,795.53. <U>ANSWER</U>. The maximum loan amount is $14,795.53. </pre> <H3>Problem 5</H3>A bank loaned $15,000 total, part of it at 8% per year, the rest at 18% per year. If the interest the bank received in one year totaled $2000, how much was loaned at 8%? <B>Solution</B> <pre> Let x be the amount loaned at 8%; then the rest, (15000-x), is the amount loaned at 18%. Write equation for the total annual interest 0.08x + 0.18*(15000-x) = 2000. Simplify and solve for x 0.08x + 0.18*15000 - 0.18x = 2000 0.08x - 0.18x = 2000 - 0.18*15000 -0.1x = -700 x = -700/(-0.1) = 70000. <U>ANSWER</U>. $7000 was loaned at 8%. </pre> <H3>Problem 6</H3>Yankee Construction agreed to lease payments of $762.79 on construction equipment to be made at the end of each month for six years. Financing is at 15% compounded monthly. What is the value of the original lease contract? <B>Solution</B> <pre> In this problem, the value of the original lease contract is the amount of a loan, which requires monthly payments of $762.79 at the end of each month for six years at 15% compounded monthly. Use the standard formula for monthly payments of a loan M = {{{P*(r/(1-(1+r)^(-n)))}}}, (1) where P is the loan amount; r = {{{0.15/12}}} is the effective interest rate per month; n is the number of payments (same as the number of months, n = 6*12 = 72); M is the monthly payment. In this problem M = $762.79; r = {{{0.15/12}}}. Substitute these values into the formula and get this equation 762.79 = {{{P*(((0.15/12))/(1-(1+0.15/12)^(-72)))}}}. (2) In equation (2), calculate separately the coefficient (the multiplier, or the factor) {{{((0.15/12))/(1-(1+0.15/12)^(-72))}}} = 0.021145013 (rounded). Now from equation (2), find the <U>ANSWER</U> P = {{{762.79/0.021145013}}} = 36074.23. <U>ANSWER</U>. The value of the original lease contract is $36074.23. </pre> <H3>Problem 7</H3>If a loan of $300,000 is to be settled by $3,000 monthly payments for 15 years, what interest rate compounded monthly is charged on the loan? <B>Solution</B> <pre> Write a loan equation M = {{{(L*r)/(1-(1+r)^(-n))}}}, where M is the monthly payment; L is the loaned amount; r is the monthly effective rate as a decimal; n is the number of payments (= the number of months). In our case this equation takes the form 3000 = {{{(300000*r)/(1-(1+r)^(-15*12))}}}, or {{{3000/300000}}} = {{{r/(1-(1+r)^(-180))}}}. 0.01 = {{{r/(1-(1+r)^(-180))}}}. In this equation, r is the unknown. This equation is unsolvable algebraically, so use the numerical methods. You may use any of numerous online calculators. I used online calculator DESMOS of common use at web-site www.desmos.com/calculator It gave me r = 0.00729945, approximately. Here is the link to the DESMOS solution. https://www.desmos.com/calculator/ey97hhvirs Click on the intersection point to get its coordinates. This value r = 0.00729945 is the monthly effective rate - so, the annual nominal rate is 12 times this value {{{r[annual]}}} = 12*0.00729945 = 0.0875934, or about 0.0876, which corresponds to 8.76%. <U>ANSWER</U>. In this problem, the monthly compounded interest rate is about 8.76 %. To check my solution, I substituted this value r= 0.00729945 into the loan function f(r) = {{{r/(1-(1+r)^(-180))}}} = {{{0.00729945/(1-1.00729945^(-180))}}}. I got the value 0.0100000030, which is quite close to 0.01, so the solution is confirmed. </pre> <H3>Problem 8</H3>A $38,000 loan bears interest at 10% compounded semi-annually and is to be repaid in semi-annual payments of $2,000 each. How many semi-annual payments must be the debtor make? <B>Solution</B> <pre> Use the standard formula for the semi-annual payment for a loan P = {{{L*(r/(1-(1+r)^(-n))))}}}, where L is the loan amount; r = {{{0.1/2}}} is the effective semi-annual compounding interest rate; n is the number of payments; P is the semi-annual payment. In this problem P = $2000; r = {{{0.1/2}}} = 0.05. Substitute these values into the formula and get for monthly payment 2000 = {{{38000*(0.05/(1-1.05^(-n))))}}}. In this equation, n is the unknown: we should find n from this equation. Simplify step by step {{{2000/38000}}} = {{{(0.05/(1-1.05^(-n)))}}}, 0.052631579 = {{{(0.05/(1-1.05^(-n)))}}}, {{{0.052631579/0.05}}} = {{{(1/(1-1.05^(-n)))}}}, 1.05263158 = {{{(1/(1-1.05^(-n)))}}}, {{{1/1.05263158}}} = {{{1-1.05^(-n)}}}, 0.95 = {{{1-1.05^(-n)}}}, {{{1.05^(-n)}}} = 1 - 0.95, {{{1.05^(-n)}}} = 0.05, {{{1/1.05^n}}} = 0.05, 1.05^n = 1/0.05, 1.05^n = 20, n*log(1.05) = log(20), n = {{{log((20))/log((1.05))}}} = 61.4. So, 61 full semi-annual payments should be made of 2,000 each, and then the last,62-th payment, should be made of the lesser amount. <U>ANSWER</U>. 61 full semi-annual payments should be made of 2,000 each, and then the last, 62-th payment, should be made of the lesser amount. The total number of semi-annual payments is 62. </pre> <H3>Problem 9</H3>How many quarterly payments of $15,000 will be necessary to pay off a debt of $175,000 if the interest rate charged is 8% compounded quarterly? <B>Solution</B> <pre> Use the standard formula for the quarterly payment for a loan P = {{{L*(r/(1-(1+r)^(-n))))}}}, where L is the loan amount; r = {{{0.08/4}}} = 0.02 is the effective quarterly compounding interest rate; n is the number of payments; P is the quarterly payment. In this problem P = $15000; r = {{{0.08/4}}} = 0.02. Substitute these values into the formula and get for quarterly payment 15000 = {{{175000*(0.02/(1-1.02^(-n))))}}}. In this equation, n is the unknown: we should find n from this equation. Simplify step by step {{{15000/175000}}} = {{{(0.02/(1-1.02^(-n)))}}}, 0.085714286 = {{{(0.02/(1-1.02^(-n)))}}}, {{{0.085714286/0.02}}} = {{{(1/(1-1.02^(-n)))}}}, 4.2857143 = {{{(1/(1-1.02^(-n)))}}}, {{{1/4.2857143}}} = {{{1-1.02^(-n)}}}, 0.233333333 = {{{1-1.02^(-n)}}}, {{{1.02^(-n)}}} = 1 - 0.233333333, {{{1.02^(-n)}}} = 0.766666667, {{{1/1.02^n}}} = 0.766666667, 1.02^n = 1/0.766666667, 1.02^n = 1.304347826 , n*log(1.02) = log(1.304347826), n = {{{log((1.304347826))/log((1.02))}}} = 13.4. So, 13 full semi-annual payments should be made of 2,000 each, and then the last,14-th payment, should be made of the lesser amount. <U>ANSWER</U>. 13 full semi-annual payments should be made of 2,000 each, and then the last, 14-th payment, should be made of the lesser amount. The total number of semi-annual payments is 14. </pre> <H3>Problem 10</H3>A company "Warehouse, Inc" is going to loan $16,000 to construct a warehouse. The warehouse will provide the storage space-value at $3,600 per year. It will have the annual maintenance and operating cost of $360 per year. The loan interest is 10% compounded yearly. How long the warehouse must function to recoup the cost of the loan and the operating cost? (a) 8 years. (b) 9 years. (c) 7 years. (d) 6 years. <B>Solution</B> <pre> Let PMT be the annual payment for the loan (which is some constant value over the years). Then an inequality for the project to be profitable is PMT + 360 <= 3600, (1) which implies PMT <= 3600 - 360 = 3240. (2) Now, let's calculate PMT for 6, 7, 8 and 9 years with the interest rate r = 10% = 0.1. Use the standard formula for n = 6 years PMT = {{{L*(r/(1-(1+r)^(-n)))}}} = {{{16000*(0.1/(1-1.1^(-6)))}}} = 3673.72; for n = 7 years PMT = {{{L*(r/(1-(1+r)^(-n)))}}} = {{{16000*(0.1/(1-1.1^(-7)))}}} = 3286.49; for n = 8 years PMT = {{{L*(r/(1-(1+r)^(-n)))}}} = {{{16000*(0.1/(1-1.1^(-8)))}}} = 2999.10; for n = 9 years PMT = {{{L*(r/(1-(1+r)^(-n)))}}} = {{{16000*(0.1/(1-1.1^(-9)))}}} = 2778.25. Comparing these numbers with 3240 in the right side of (2), you see that to be profitable, the warehouse should function at least 8 years. <<<---=== <U>ANSWER</U> </pre> My other lessons on Finance problems in this site are - <A HREF=https://www.algebra.com/algebra/homework/Finance/Problems-on-simple-interest-accounts.lesson>Problems on simple interest accounts</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Problems-on-discretely-compounded-accounts.lesson>Problems on discretely compounded accounts</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Problems-on-continuously-compounded-accounts-.lesson>Problems on continuously compounded accounts</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Find-future-value-of-an-Ordinary-Annuity.lesson>Find future value of an Ordinary Annuity</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Find-a-value-of-a-regular-deposits-for-an-Ordinary-Annuity.lesson>Find regular deposits for an Ordinary Annuity</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/How-long-will-it-takes-for-an-ordinary-annuity-to-get-an-assigned-value.lesson>How long will it take for an ordinary annuity to get an assigned value?</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Find-future-value-of-an-Annuity-Due-saving-plan.lesson>Find future value for an Annuity Due saving plan</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Regular-withdrawals-from-an-annuity-account.lesson>Regular withdrawals from an annuity account</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Ordinary-annuity-account-with-non-zero-initial-deposit-as-a-combined-total-of-two-accounts.lesson>Ordinary annuity account with non-zero initial deposit as a combined total of two accounts</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Annual-depositing-and-semi-annual-compounding-in-ordinary-annuity-saving-plan.lesson>Annual depositing and semi-annual compounding in ordinary annuity saving plan</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/21-add-Regular-withdrawals-from-a-compounded-account.lesson>Variable withdrawals from a compounded account (sinking fund)</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Present-value-of-an-ordinary-annuity-saving-plan.lesson>Present value of an ordinary annuity cumulative saving plan</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Starting-value-of-a-sinking-fund.lesson>Problems on sinking funds</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Find-the-compounding-rate-of-an-ordinary-annuity.lesson>Find the compounding rate of an ordinary annuity</A> - <A HREF=https://www.algebra.com/algebra/homework/Finance/Accumulate-money-using-ordinary-annuity-then-spend-money-via-sinking-fund.lesson>Accumulate money using ordinary annuity; 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Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
