Simple interest for AFCAT: the formula, reverse questions and shortcuts

Simple interest is one of the most predictable topics in AFCAT Numerical Ability. There is one formula, the numbers are usually round, and every question is a variation on "three of these four values are given; find the fourth". The marks are there for anyone who has practised the reverse questions: finding the principal, the rate or the time rather than the interest.

This guide covers the formula and why it behaves the way it does, then the question types that appear again and again, each with a fast method.

The formula and what it means

SI = P × R × T ÷ 100

• P is the principal (the sum lent or borrowed).
• R is the rate per cent per year.
• T is the time in years.
• Amount = P + SI.

Simple interest is charged on the original principal only. So the interest is the same every year: ₹5,000 at 8% earns ₹400 in year one, ₹400 in year two and ₹400 in year three. That steady, straight-line growth is what makes simple interest questions easy to reverse.

Rearranged forms, which you should know without deriving:

Units: convert time to years

The rate is per year, so time must be in years.

• 6 months = 1/2 year; 8 months = 2/3 year; 2 years 6 months = 2.5 years.
• 73 days = 73/365 = 1/5 year.

If a question gives a rate "per half-year" or "per month", convert either the rate or the time so that both use the same period.

Doubling, tripling and "n times"

A sum becomes n times itself when the interest equals (n − 1) × P. Put that into the formula and P cancels:

R × T = 100 × (n − 1)

• Doubling: R × T = 100. A sum that doubles in 8 years has R = 12.5%.
• Tripling: R × T = 200. At 10%, a sum triples in 20 years.
• Because simple interest grows in a straight line, if a sum doubles in 5 years (interest = P), it triples in 10 years (interest = 2P) and becomes four times in 15 years.

Two amounts after two different periods

If a sum amounts to A₁ after T₁ years and A₂ after T₂ years at simple interest:

• Interest for one year = (A₂ − A₁) ÷ (T₂ − T₁).
• Principal = A₁ − T₁ × (one year's interest).

This works because the only difference between the two amounts is the extra years of identical interest.

Worked examples

Example 1 (principal from amount). A sum amounts to ₹6,500 in 3 years at 10% a year. Find the principal.

• Interest for 3 years is 30% of P, so the amount is 1.3 × P.
• P = 6,500 ÷ 1.3 = ₹5,000.

Example 2 (two amounts). A sum amounts to ₹7,800 in 3 years and ₹9,000 in 5 years at simple interest. Find the principal and the rate.

• Two extra years earned 9,000 − 7,800 = ₹1,200, so one year's interest = ₹600.
• Principal = 7,800 − 3 × 600 = ₹6,000.
• Rate = 600 ÷ 6,000 × 100 = 10%.

Example 3 (n times). A sum doubles in 8 years at simple interest. In how many years will it become four times itself?

• Doubling means interest = P in 8 years.
• Four times means interest = 3P, which takes 3 × 8 = 24 years.

Example 4 (rate change). Had the rate been 2% higher, ₹5,000 would have earned ₹300 more interest. For how many years was it lent?

• The extra interest comes only from the extra 2%: 5,000 × 2 × T ÷ 100 = 300.
• 100T = 300, so T = 3 years.

Example 5 (a sum split at two rates). ₹10,000 is lent in two parts, one at 8% and the other at 12% a year. The total yearly interest is ₹1,040. Find the two parts.

• Let the part at 8% be x. Then 0.08x + 0.12(10,000 − x) = 1,040.
• 1,200 − 0.04x = 1,040, so x = 4,000.
• The parts are ₹4,000 at 8% and ₹6,000 at 12%.
• Faster by alligation: the average rate is 1,040 ÷ 10,000 = 10.4%. The ratio is (12 − 10.4) : (10.4 − 8) = 1.6 : 2.4 = 2 : 3, giving ₹4,000 and ₹6,000. See mixtures and alligation for why this works.

Example 6 (time in months). Find the simple interest on ₹4,800 at 7.5% a year for 8 months.

• T = 8/12 = 2/3 year.
• SI = 4,800 × 7.5 × 2/3 ÷ 100 = ₹240.

Simple and compound interest side by side

Because the first year is identical, a question can give you a one-year compound figure and expect you to treat it as simple interest. The two only differ from year two onwards. The next guide, compound interest, builds on this.

Common mistakes

• Time not in years. Convert months and days first.
• Mixing up SI and amount. "Amounts to ₹6,500" is P + SI, not SI.
• Treating doubling as proportional to n. Doubling takes T years, but four times takes 3T years, not 2T or 4T.
• Forgetting the principal is fixed. In simple interest, the second year's interest is not calculated on the first year's amount.

Practice set

1. Find the SI on ₹4,000 at 6% a year for 5 years.
2. A sum triples in 10 years at simple interest. Find the rate.
3. At what rate does ₹3,000 earn ₹900 in 3 years?
4. Find the SI on ₹12,000 at 9% a year for 2 years 6 months.
5. What principal earns ₹1,120 in 4 years at 7% a year?
6. A sum amounts to ₹6,200 in 3 years and ₹7,000 in 5 years at simple interest. Find the principal and rate.
7. At what rate will a sum become 5/4 of itself in 2.5 years?
8. In how many years will ₹2,500 amount to ₹3,250 at 6% a year?

Answers:

1. ₹1,200. 4,000 × 6 × 5 ÷ 100.
2. 20%. Interest = 2P in 10 years, so R × 10 = 200.
3. 10%. 900 × 100 ÷ (3,000 × 3) = 90,000 ÷ 9,000.
4. ₹2,700. 12,000 × 9 × 2.5 ÷ 100.
5. ₹4,000. 1,120 × 100 ÷ (7 × 4) = 1,12,000 ÷ 28.
6. ₹5,000 at 8%. One year's interest = 800 ÷ 2 = ₹400. P = 6,200 − 1,200 = ₹5,000. R = 400 ÷ 5,000 × 100.
7. 10%. Interest = P/4 = 25% of P over 2.5 years, so 25 ÷ 2.5 = 10% a year.
8. 5 years. Interest ₹750. One year's interest = 2,500 × 6 ÷ 100 = ₹150, and 750 ÷ 150 = 5.

What to do next

• Learn the rearranged formulas in the table until you can write any of them in five seconds.
• Practise ten "two amounts" and ten "n times" questions, the two types that reward a shortcut most.
• Move on to compound interest, and revise percentage if multipliers still feel slow.