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AFCAT Numerical Ability · Simple and Compound Interest

Paying Back a Loan in Instalments

A loan at compound interest is cleared when its payments, each divided by its own compound factor to bring it back to today, add up to the loan; or, year by year, when the balance carried forward reaches zero.

All Simple and Compound Interest notes

Why this matters

No AFCAT past-year question is on this page, but over a quarter of it is HARD and few rows are EASY. Every row uses one of two methods: bring each payment back to today, or carry the balance forward a year at a time. Carrying the balance forward when the payments are equal turns a one-line sum into a long one.

Concept 1 of 3: Equal instalments

A payment due in a year is worth less today, because money held today would grow. At 10%, ₹1,100 due in a year is worth ₹1,000 now. A loan equals the sum of what its instalments are worth today.

Definition

  • Equal yearly instalments xx at R%R\%, with r=R100r = \tfrac{R}{100}: x1+r+x(1+r)2+⋯=loan\dfrac{x}{1 + r} + \dfrac{x}{(1 + r)^2} + \dots = \text{loan}.
  • Two instalments: x=loan×(1+r)22+rx = \text{loan} \times \dfrac{(1 + r)^2}{2 + r}.
  • Use fractions where they are neat: at 25% the divisors give 45\tfrac45 and 1625\tfrac{16}{25}; at 12.5% they give 89\tfrac89 and 6481\tfrac{64}{81}.
  • Total interest paid = (number of instalments × instalment) minus the loan.
  • Half-yearly instalments: use the half-yearly rate and count half-years.

Two equal instalments

x=loan×(1+r)22+rx = \text{loan} \times \dfrac{(1 + r)^2}{2 + r}

Worked example

₹4,200 is borrowed at 10% compound interest and repaid in two equal yearly instalments. Find each instalment.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 1 · Simple and Compound Interest · Instalments

A sum of ₹11000 was taken as a loan. This is to be repaid in two equal annual instalments. If the rate of interest is 20% compounded annually, then the value of each instalment is

An instalment is not loan ÷ number

₹4,200 in two equal instalments at 10% is ₹2,420 each, not ₹2,100 plus some simple interest. Each payment pays off a different amount of today's money.
Drill 15 more on equal instalments

Concept 2 of 3: The loan from its payments

The same idea run the other way: given the payments, the loan is the sum of their present values, each payment divided by its own factor.

Definition

  • Loan =x11+r+x2(1+r)2+x3(1+r)3= \dfrac{x_1}{1 + r} + \dfrac{x_2}{(1 + r)^2} + \dfrac{x_3}{(1 + r)^3}; the payments may be different.
  • A down payment is paid today, so it counts at its full value.
  • Cash price against an instalment plan: bring the whole plan to today and compare it with the cash price.

Loan from its payments

loan=∑kxk(1+r)k\text{loan} = \sum_k \dfrac{x_k}{(1 + r)^k}

Worked example

A loan at 10% compound interest is cleared by ₹2,200 after one year and ₹3,630 after two years. Find the loan.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Simple and Compound Interest · Instalments

A loan is to be paid back in 22 equal annual installments of Rs. 2,523\text{Rs. } 2,523 each at 16%16\% compound interest per annum. Find the principal amount.

Each payment has its own factor

₹3,630 due in 2 years at 10% is worth 3630÷1.21=30003630 \div 1.21 = 3000 today. Dividing every payment by 1.1 overstates the loan.
Drill 7 more on the loan from its payments

Concept 3 of 3: Carrying the balance forward

When the earlier payments are given and the last one is asked, follow the debt a year at a time: grow it by the rate, take off the payment, and repeat. What is owed at the end is the final payment.

Definition

  • Each year: new balance = old balance × (1+r)(1 + r) minus the payment.
  • The final payment is the last balance after it has grown for its year.
  • The same answer comes from growing the loan to the end, growing each earlier payment to the end, and subtracting.
  • A last payment at a part of a year needs a logarithm at compound interest; such rows are rare.

Balance after a year

Bk=Bk−1(1+r)−xkB_k = B_{k-1}(1 + r) - x_k

Worked example

₹10,000 is borrowed at 10% compound interest. ₹5,000 is repaid after one year and ₹2,000 after two years. What clears the debt at the end of the third year?
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Simple and Compound Interest · Instalments

A person borrows Rs. 8,000\text{Rs. } 8,000 at 15%15\% compound interest per annum. He repays Rs. 3,000\text{Rs. } 3,000 at the end of first year and Rs. 4,000\text{Rs. } 4,000 at the end of second year. What amount should he pay at the end of third year to clear all dues?

Grow before you subtract

A payment at the end of a year comes after that year's interest. Taking it off first and then adding interest charges less interest than the loan really owes.
Drill 8 more on carrying the balance forward

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Equal instalments

    Two equal instalments

    x=loan×(1+r)22+rx = \text{loan} \times \dfrac{(1 + r)^2}{2 + r}
  • The loan from its payments

    Loan from its payments

    loan=∑kxk(1+r)k\text{loan} = \sum_k \dfrac{x_k}{(1 + r)^k}
  • Carrying the balance forward

    Balance after a year

    Bk=Bk−1(1+r)−xkB_k = B_{k-1}(1 + r) - x_k

Watch out for (3)

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