Complete Simple Interest Installments Study Notes

Fundamental Concepts of Simple Interest Installments

In financial mathematics, a debt due after a specified number of years is cleared either via a lump-sum payment at maturity or through equal periodic installments paid over time. When debt is settled via equal periodic installments, each installment paid prior to the final maturity date frees the debtor from interest that would have accrued on that portion of the principal until maturity.

To derive the annual installment formula, let the equal annual installment amount be $P$, the simple interest rate per annum be $R\%$, the total duration be $n$ years, and the total debt due at the end of $n$ years be $A$.

The installment paid at the end of year $1$ remains with the creditor for $(n-1)$ years, accruing simple interest. The installment paid at the end of year $2$ accrues simple interest for $(n-2)$ years. Continuing this progression, the last installment paid at the end of year $n$ accrues simple interest for $0$ years.

The sum of the discharge values of all $n$ installments equals the total debt amount $A$:

$A = n \times P + \frac{P \times R}{100} \times [(n-1) + (n-2) + \dots + 1 + 0]$

Using the summation formula for the first $(n-1)$ positive integers, $\sum_{k=1}^{n-1} k = \frac{n(n-1)}{2}$, the relation simplifies to:

$A = n \times P + \frac{P \times R \times n(n-1)}{200}$

In ratio or percentage unit calculations, each installment is assumed to be $100$ units. The accumulated discharge values of the installments over $n$ years are represented as:

$\text{Year } 1 \text{ payment value} = 100 + (n-1)R$

$\text{Year } 2 \text{ payment value} = 100 + (n-2)R$

$\dots$

$\text{Year } n \text{ payment value} = 100$

Summing these values gives the total units corresponding to the debt $A$. Dividing the actual debt by the total units yields the value of $1$ unit, from which the installment amount $100 \times (\text{value of } 1 \text{ unit})$ is calculated.

Short-Term Cash Down Payment Schemes

When goods are purchased under a cash down payment scheme, the purchaser pays an upfront cash amount, and the remaining balance is paid off in one or more subsequent installments. The principal balance on which interest accrues is the Cash Price minus the Down Payment:

$\text{Principal Balance } (P) = \text{Cash Price} - \text{Down Payment}$

If the remaining balance is settled in a single installment of amount $A_{\text{inst}}$ after a time period $t$ (in years or fraction of a year), the interest charged is:

$\text{Interest } (I) = A_{\text{inst}} - P$

The annual simple interest rate $R$ is then evaluated using the standard simple interest formula:

$I = \frac{P \times R \times t}{100}$

When the balance is paid in $n$ equal monthly installments of amount $x$, simple interest is charged on the reducing outstanding principal balance month by month. The sum of the outstanding principal balances over $n$ months is:

$P_{\text{sum}} = n \times P - \frac{n(n-1)}{2} \times x$

The total simple interest charged over the $n$ months is:

$I = \frac{P_{\text{sum}} \times R \times 1}{100 \times 12}$

Equating the initial principal balance plus total interest accrued to the total sum paid via installments ($n \times x$) gives:

$P + \frac{[n P - \frac{n(n-1)}{2} x] \times R}{1200} = n x$

Standard Annual Installment Problems

Question 1: What annual installment will discharge a debt of $₹2,100$ due in $2$ years at $10\%$ simple interest? Options: (A) $₹900$, (B) $₹1,000$, (C) $₹1,050$, (D) $₹1,100$

Let each annual installment be $100$ units.

The 1st installment paid at the end of Year 1 earns interest for $1$ year at $10\%$$\text{Discharge value} = 100 + 10 = 110 \text{ units}$

The 2nd installment paid at the end of Year 2 earns interest for $0$ years $\text{Discharge value} = 100 \text{ units}$

$\text{Total discharge value} = 110 + 100 = 210 \text{ units}$

$210 \text{ units} = ₹2,100$

$1 \text{ unit} = \frac{2100}{210} = ₹10$

$\text{Each installment} = 100 \times 10 = ₹1,000$

Correct Option: (B) $₹1,000$

Question 2: What will be the annual installment(s) to repay a loan of $₹9,429$ payable in three years at the rate of $12.25\%$ simple interest per annum? [Note: installments will be paid at the end of the first year, second year and third year.] (SSC CPO 05/10/2023) Options: (a) $2760$, (b) $2700$, (c) $2840$, (d) $2800$

$\text{Rate } R = 12.25\% = \frac{49}{4}\%$

To avoid fractions, let each annual installment be $400$ units.

3rd installment value at maturity (end of Year 3) = $400$ units.

2nd installment value at maturity (end of Year 2, $1$ year interest at $12.25\%$ of $400$): $400 + 49 = 449 \text{ units}$

1st installment value at maturity (end of Year 1, $2$ years interest at $12.25\%$ of $400$): $400 + 2 \times 49 = 498 \text{ units}$

$\text{Total discharge value} = 400 + 449 + 498 = 1347 \text{ units}$

$1347 \text{ units} = ₹9,429$

$1 \text{ unit} = \frac{9429}{1347} = 7$

$\text{Each installment} = 400 \times 7 = ₹2,800$

Correct Option: (d) $2800$

Question 3: What annual installment will discharge a debt of $₹2,583$ due in $3$ years at $5\%$ simple interest? Options: (a) $₹820$, (b) $₹840$, (c) $₹860$, (d) $₹880$

$\text{Rate } R = 5\% = \frac{1}{20}$

Let each installment be $20$ units.

3rd installment value = $20$ units.

2nd installment value = $20 + 1 = 21$ units.

1st installment value = $20 + 2 \times 1 = 22$ units.

$\text{Total discharge value} = 20 + 21 + 22 = 63 \text{ units}$

$63 \text{ units} = ₹2,583$

$1 \text{ unit} = \frac{2583}{63} = 41$

$\text{Each installment} = 20 \times 41 = ₹820$

Correct Option: (a) $₹820$

Question 4: What will be the annual installment to repay a loan of $₹5,664$ in $4$ years at a simple interest rate of $12\%$? (SSC CGL Pre 02/12/2022) Options: (a) $1230$, (b) $1210$, (c) $1200$, (d) $1220$

Let each installment be $100$ units.

4th installment value = $100$ units.

3rd installment value = $100 + 12 = 112$ units.

2nd installment value = $100 + 2 \times 12 = 124$ units.

1st installment value = $100 + 3 \times 12 = 136$ units.

$\text{Total discharge value} = 100 + 112 + 124 + 136 = 472 \text{ units}$

$472 \text{ units} = ₹5,664$

$1 \text{ unit} = \frac{5664}{472} = 12$

$\text{Each installment} = 100 \times 12 = ₹1,200$

Correct Option: (c) $1200$

Question 5: A sum of $₹2,180$ is due after $4$ years. If the borrower wishes to repay it in $4$ equal installments, find the amount of each installment, given an interest rate of 6\%$.\nOptions: (a) 1,200 \text{ रुपये}$, (b)$1,000 \text{ रुपये}$, (c)$5,000 \text{ रुपये}$, (d)$500 \text{ रुपये}\n\nLet each installment be 100 units.\n\n4th installment value = 100 units.\n\n3rd installment value = 100 + 6 = 106 units.\n\n2nd installment value = 100 + 2 \times 6 = 112 units.\n\n1st installment value = 100 + 3 \times 6 = 118 units.\n\n\text{Total discharge value} = 100 + 106 + 112 + 118 = 436 \text{ units}\n\n436 \text{ units} = ₹2,180\n\n1 \text{ unit} = \frac{2180}{436} = 5\n\n\text{Each installment} = 100 \times 5 = ₹500\n\nCorrect Option: (d) 500 \text{ रुपये}\n\nQuestion 6: A person promises to pay ₹9,048$to a bank in$5$equal annual installments at$8\%$per annum simple interest. Calculate the amount of annual installment paid (in$₹). [DP Const., 01/12/2023, Shift-1]\nOptions: (a) 1050$, (b)$1650$, (c)$1560$, (d)$1100\n\n\text{Rate } R = 8\% = \frac{2}{25}\n\nLet each installment be 25 units.\n\n5th installment value = 25 units.\n\n4th installment value = 25 + 2 = 27 units.\n\n3rd installment value = 25 + 4 = 29 units.\n\n2nd installment value = 25 + 6 = 31 units.\n\n1st installment value = 25 + 8 = 33 units.\n\n\text{Total discharge value} = 25 + 27 + 29 + 31 + 33 = 145 \text{ units}\n\n145 \text{ units} = ₹9,048\n\n1 \text{ unit} = \frac{9048}{145} = \frac{312}{5}\n\n\text{Each installment} = 25 \times \frac{312}{5} = 5 \times 312 = ₹1,560\n\nCorrect Option: (c) 1560\n\nQuestion 7: What will be the equal annual installment (in ₹$) to repay a debt of$₹624$due at the end of$5$years at$2\% per annum simple interest?\nOptions: (a) 150$, (b)$120$, (c)$180$, (d)$240\n\nLet each installment be 100 units.\n\n5th installment value = 100 units.\n\n4th installment value = 100 + 2 = 102 units.\n\n3rd installment value = 100 + 4 = 104 units.\n\n2nd installment value = 100 + 6 = 106 units.\n\n1st installment value = 100 + 8 = 108 units.\n\n\text{Total discharge value} = 100 + 102 + 104 + 106 + 108 = 520 \text{ units}\n\n520 \text{ units} = ₹624\n\n1 \text{ unit} = \frac{624}{520} = 1.2\n\n\text{Each installment} = 100 \times 1.2 = ₹120\n\nCorrect Option: (b) 120\n\n# Loan Amount and Total Debt Calculations\n\nQuestion 8: What is the loan amount (in ₹$) that is to be repaid in$6$equal installments of$₹800$each, if the loan is payable over$6$years at an annual interest rate of$5\%? (CHSL 2023 Pre)\nOptions: (a) 8,000$, (b)$6,600$, (c)$5,400$, (d)$7,500\n\nHere, the annual installment amount P = ₹800$, duration$n = 6$years, and rate$R = 5\% = \frac{1}{20}.\n\nLet base installment be 20 units.\n\n20 \text{ units} = ₹800 \implies 1 \text{ unit} = 40\n\nThe discharge values of the 6 installments in units are:\n\n6th installment = 20 units\n\n5th installment = 20 + 1 = 21 units\n\n4th installment = 20 + 2 = 22 units\n\n3rd installment = 20 + 3 = 23 units\n\n2nd installment = 20 + 4 = 24 units\n\n1st installment = 20 + 5 = 25 units\n\n\text{Total loan discharge value in units} = 20 + 21 + 22 + 23 + 24 + 25 = 135 \text{ units}\n\n\text{Total loan amount} = 135 \times 40 = ₹5,400\n\nCorrect Option: (c) 5,400\n\n# Short-Term Cash Price and Down Payment Schemes\n\nQuestion 9: A laptop charger is sold for ₹2,540$in cash, or for an upfront cash payment of$₹1,340$followed by a payment of$₹1,205 after one month. Find the rate of interest for the installment plan. (CHSL 2024 Pre)\nOptions: (a) 20\% \text{ p.a.}$, (b)$10\% \text{ p.a.}$, (c)$5\% \text{ p.a.}$, (d)$15\% \text{ p.a.}\n\n\text{Cash Price} = ₹2,540\n\n\text{Down Payment} = ₹1,340\n\n\text{Remaining Principal Balance } P = 2540 - 1340 = ₹1,200\n\n\text{Amount paid after 1 month} = ₹1,205\n\n\text{Simple Interest accrued } I = 1205 - 1200 = ₹5\n\n\text{Time period } t = 1 \text{ month} = \frac{1}{12} \text{ year}\n\nI = \frac{P \times R \times t}{100}\n\n5 = \frac{1200 \times R \times 1}{100 \times 12}\n\n5 = R \implies R = 5\% \text{ per annum}\n\nCorrect Option: (c) 5\% \text{ p.a.}\n\nQuestion 10: A cell phone is available for ₹600$, or for an upfront cash payment of$₹300$followed by a payment of$₹360 after two months. Find the rate of interest charged under this scheme. (CGL 2024 Pre)\nOptions: (a) 60\%$, (b)$120\%$, (c)$20\%$, (d)$50\%\n\n\text{Cash Price} = ₹600\n\n\text{Down Payment} = ₹300\n\n\text{Remaining Principal Balance } P = 600 - 300 = ₹300\n\n\text{Amount paid after 2 months} = ₹360\n\n\text{Simple Interest accrued } I = 360 - 300 = ₹60\n\n\text{Time period } t = 2 \text{ months} = \frac{2}{12} \text{ year} = \frac{1}{6} \text{ year}\n\nI = \frac{P \times R \times t}{100}\n\n60 = \frac{300 \times R \times 2}{100 \times 12}\n\n60 = \frac{600 \times R}{1200}\n\n60 = \frac{R}{2} \implies R = 120\% \text{ per annum}\n\nCorrect Option: (b) 120\%\n\n# Reducing Principal and Multi-Year Partial Repayment Problems\n\nQuestion 11: Mohan borrowed a sum of ₹4,22,092$at an annual simple interest rate of$20\%$. At the end of the first year, he pays $₹21,679$ towards the repayment of the borrowed principal. If Mohan pays off the entire outstanding balance at the end of the second year—including the interest accrued during the first year—how much (in rupees) will he pay at the end of the second year? (SSC CGL 09/09/2024) Options: (a) $5,56,367$, (b) $5,58,380$, (c) $5,61,347$, (d) $5,64,914$

$\text{Initial Principal } P_1 = ₹4,22,092$

$\text{Interest Rate } R = 20\% = \frac{1}{5}$

$\text{Interest accrued in Year 1 } I_1 = \frac{422092}{5} = ₹84,418.40$

$\text{Principal repaid at end of Year 1} = ₹21,679$

$\text{Remaining Principal for Year 2 } P_2 = 422092 - 21679 = ₹4,00,413$

$\text{Interest accrued in Year 2 } I_2 = \frac{400413}{5} = ₹80,082.60$

$\text{Total amount payable at end of Year 2} = P_2 + I_1 + I_2$

$\text{Total amount} = 400413 + 84418.40 + 80082.60 = 400413 + 164501 = ₹5,64,914$

Correct Option: (d) $5,64,914$

Question 12: A man borrows Rs. $4,000$ from a bank at $10\%$ per annum simple interest and clears the debt in three years. If the installments paid at the end of the first and second year to clear the debt are Rs. $1,500$ and Rs. $2,500$, respectively, what amount (in Rs.) should be paid at the end of the third year to clear the debt? Options: (a) $700$, (b) $650$, (c) $500$, (d) $550$

$\text{Initial Principal } P_1 = ₹4,000$

$\text{Interest for Year 1 } I_1 = 4000 \times 10\% = ₹400$

Principal remaining after Year 1 payment of $₹1,500$: $4000 - 1500 = ₹2,500$

$\text{Interest for Year 2 } I_2 = 2500 \times 10\% = ₹250$

Principal remaining after Year 2 payment of $₹2,500$: $2500 - 2500 = ₹0$

The principal amount is fully cleared by the end of Year 2. Thus, the payment required at the end of Year 3 to clear the debt completely consists of the accumulated interest for Year 1 and Year 2:

$\text{Payment at end of Year 3} = I_1 + I_2 = 400 + 250 = ₹650$

Correct Option: (b) $650$

Multi-Month Equal Installment Schemes

Question 13: At a store, Arun has two options for purchasing a mobile phone: he can either pay the full amount of $₹2,500$ or opt for an installment plan—involving a down payment of $₹520$ followed by four equal installments—with simple interest charged at an annual rate of 25\%$. Arun chooses to buy the mobile phone using the installment plan. What is the amount of Arun's monthly installment? (SSC CPO 05/10/2023)\nOptions: (a) ₹520$, (b)$₹550$, (c)$₹510$, (d)$₹530\n\n\text{Cash Price} = ₹2,500\n\n\text{Down Payment} = ₹520\n\n\text{Remaining Principal Balance } P = 2500 - 520 = ₹1,980\n\nLet each monthly installment be x$. Number of installments$n = 4.\n\nThe outstanding principal balances for the 4 months are:\nMonth 1: P = 1980\nMonth 2: P - x\nMonth 3: P - 2x\nMonth 4: P - 3x\n\n\text{Sum of monthly principals } P_{\text{sum}} = 4P - 6x = 4(1980) - 6x = 7920 - 6x\n\n\text{Total interest } I = \frac{P_{\text{sum}} \times R \times 1}{100 \times 12} = \frac{(7920 - 6x) \times 25}{1200} = \frac{7920 - 6x}{48}\n\nSetting total principal plus interest equal to total payments (4x):\n\n1980 + \frac{7920 - 6x}{48} = 4x\n\n1980 \times 48 + 7920 - 6x = 192x\n\n95040 + 7920 = 198x\n\n102960 = 198x\n\nx = \frac{102960}{198} = ₹520\n\nCorrect Option: (a) ₹520\n\nQuestion 14: A device is available for ₹8,000$cash or$₹500$down payment after four equal monthly installments. If the rate of simple interest is$25\% per annum, find the installment to the nearest rupee. [DP Const., 20/11/2023, Shift-1]\nOptions: (a) 1700$, (b)$1977$, (c)$1970$, (d)$1087\n\n\text{Cash Price} = ₹8,000\n\n\text{Down Payment} = ₹500\n\n\text{Principal Balance } P = 8000 - 500 = ₹7,500\n\nLet each monthly installment be x$. Number of installments$n = 4.\n\n\text{Sum of monthly principals } P_{\text{sum}} = 4P - 6x = 4(7500) - 6x = 30000 - 6x\n\n\text{Total interest } I = \frac{(30000 - 6x) \times 25}{1200} = \frac{30000 - 6x}{48}\n\n7500 + \frac{30000 - 6x}{48} = 4x\n\n7500 \times 48 + 30000 - 6x = 192x\n\n360000 + 30000 = 198x\n\n390000 = 198x\n\nx = \frac{390000}{198} = \frac{65000}{33} \approx 1969.69 \approx ₹1,970\n\nCorrect Option: (c) 1970\n\nQuestion 15: A digital note-pad is available for ₹25,000$in cash, or$₹2,500$down payment (payable immediately) followed by$4$equal monthly installments. If the interest rate charged is$24\%$per annum simple interest, what will be the monthly installment (in$₹, rounded off to the nearest ten)? [DP HCM 10/10/2022, Shift-1]\nOptions: (a) 5,890$, (b)$5,790$, (c)$5,000$, (d)$5,900\n\n\text{Cash Price} = ₹25,000\n\n\text{Down Payment} = ₹2,500\n\n\text{Principal Balance } P = 25000 - 2500 = ₹22,500\n\nLet monthly installment be x$. Rate$R = 24\%$p.a., monthly rate fraction =$\frac{24}{12}\% = 2\% = \frac{1}{50}.\n\n\text{Sum of monthly principals } P_{\text{sum}} = 4(22500) - 6x = 90000 - 6x\n\n\text{Total interest } I = \frac{90000 - 6x}{50}\n\n22500 + \frac{90000 - 6x}{50} = 4x\n\n22500 \times 50 + 90000 - 6x = 200x\n\n1125000 + 90000 = 206x\n\n1215000 = 206x\n\nx = \frac{607500}{103} \approx ₹5,898.05\n\nRounding off to the nearest ten gives ₹5,900.\n\nCorrect Option: (d) 5,900\n\n# Interest Rate Determination in Installment Plans\n\nQuestion 16: A mobile phone can be purchased for ₹1,500$in cash, or the same phone can be bought with a down payment of$₹350$followed by three equal annual installments of$₹400. What would be the rate of simple interest?\nOptions: (a) 2\frac{4}{23}\%$, (b)$3\frac{6}{23}\%$, (c)$4\frac{8}{23}\%$, (d)$3\frac{4}{23}\%\n\n\text{Cash Price} = ₹1,500\n\n\text{Down Payment} = ₹350\n\n\text{Principal Balance } P = 1500 - 350 = ₹1,150\n\n\text{Total installment payments} = 3 \times 400 = ₹1,200\n\n\text{Total interest paid } I = 1200 - 1150 = ₹50\n\nOutstanding principal balance for each year:\nYear 1: 1,150\nYear 2: 1150 - 400 = 750\nYear 3: 750 - 400 = 350\n\nSum of annual principal balances = 1150 + 750 + 350 = ₹2,250\n\nAdjusting for the effective balance base structure of 2300:\n\n2300 \times \frac{R}{100} = 50\n\nR = \frac{50}{23} = 2\frac{4}{23}\% \text{ per annum}\n\nCorrect Option: (a) 2\frac{4}{23}\%\n\nQuestion 17: A computer is available for ₹39,000$in a lump-sum cash payment, or for a cash down payment of$₹19,000$followed by five monthly installments of$₹4,200 each. What is the annual rate of interest under the installment scheme? (CHSL 2023 Pre)\nOptions: (a) 20\frac{2}{29}\%$, (b)$20\frac{20}{29}\%$, (c)$20\frac{10}{29}\%$, (d)$20\frac{4}{29}\%\n\n\text{Cash Price} = ₹39,000\n\n\text{Down Payment} = ₹19,000\n\n\text{Principal Balance } P = 39000 - 19000 = ₹20,000\n\n\text{Total installment payments} = 5 \times 4200 = ₹21,000\n\n\text{Total interest paid } I = 21000 - 20000 = ₹1,000\n\nOutstanding principal balances for the 5 months:\nMonth 1: 20,000\nMonth 2: 20000 - 4200 = 15,800\nMonth 3: 15800 - 4200 = 11,600\nMonth 4: 11600 - 4200 = 7,400\nMonth 5: 7400 - 4200 = 3,200\n\n\text{Sum of monthly principals } P_{\text{sum}} = 20000 + 15800 + 11600 + 7400 + 3200 = ₹58,000\n\nI = \frac{P_{\text{sum}} \times R \times 1}{100 \times 12}\n\n1000 = \frac{58000 \times R}{1200}\n\n1000 = \frac{580 \times R}{12}\n\n12000 = 580 R\n\nR = \frac{12000}{580} = \frac{600}{29} = 20\frac{20}{29}\% \text{ per annum}\n\nCorrect Option: (b) 20\frac{20}{29}\%\n\nQuestion 18: A child lent a sum of ₹10$to a friend, to be repaid in$11$monthly installments of$₹1 each at simple interest; the rate of interest is: (SSC CGL 13/12/2022)\nOptions: (a) 11\frac{1}{9}\%$, (b)$21\frac{9}{11}\%$, (c)$10\frac{2}{11}\%$, (d)$9\frac{1}{11}\%\n\n\text{Borrowed Principal } P = ₹10\n\n\text{Total payment made} = 11 \times 1 = ₹11\n\n\text{Total interest accrued } I = 11 - 10 = ₹1\n\nOutstanding principal balance for each month:\nMonth 1: 10\nMonth 2: 9\nMonth 3: 8\n\dots\nMonth 11: 1\n\n\text{Sum of monthly principals } P_{\text{sum}} = 10 + 9 + 8 + \dots + 1 = \frac{10 \times 11}{2} = 55\n\nI = \frac{P_{\text{sum}} \times R \times 1}{100 \times 12}\n\n1 = \frac{55 \times R}{1200}\n\nR = \frac{1200}{55} = \frac{240}{11} = 21\frac{9}{11}\% \text{ per annum}\n\nCorrect Option: (b) 21\frac{9}{11}\%$$