In this guide
Simple interest is the easiest interest topic, and it is where you should build your speed before moving to compound interest. There is one formula. Every question gives you three of its four quantities, or something that lets you find them, and asks for the fourth.
The questions that slow candidates down are the ones that hide the formula: "a sum amounts to ₹6,720 in 3 years and ₹7,200 in 5 years", or "₹10,000 is lent partly at 8% and partly at 10%". Each has a quick method, set out below.
The formula, and why it is a straight line
- SI = P × R × T ÷ 100
- Amount (A) = P + SI
Here P is the principal, R the rate per cent per year, and T the time in years.
Why it looks like this: in simple interest, the interest is always calculated on the original principal. Each year earns the same R% of P, so after T years the interest is T lots of (R/100 × P). Nothing is added back to the principal, which is the difference from compound interest.
Because the interest is the same every year, SI grows in a straight line. That single fact is behind most shortcuts in this chapter.
The four uses
| To find | Use |
|---|---|
| Interest | SI = P × R × T ÷ 100 |
| Principal | P = SI × 100 ÷ (R × T) |
| Rate | R = SI × 100 ÷ (P × T) |
| Time | T = SI × 100 ÷ (P × R) |
Example: the SI on ₹5,000 at 8% a year for 3 years is 5,000 × 8 × 3 ÷ 100 = ₹1,200.
Amount as a multiplier
In T years at R%, the interest is RT% of P, so A = P × (1 + RT/100).
At 8% for 3 years, RT = 24, so A = 1.24 × P. If you know the amount, divide by 1.24 to get the principal. This avoids setting up an equation.
Doubling and tripling
A sum doubles when the interest equals the principal, that is, when RT% = 100%. So:
- Doubles: R × T = 100.
- Triples: R × T = 200 (interest = 2P).
- Becomes n times: R × T = 100(n − 1).
Why: "becomes n times" means the interest is (n − 1) times the principal, and interest is RT% of P.
The two-amounts method
If a sum amounts to A₁ after T₁ years and A₂ after T₂ years, the difference A₂ − A₁ is the interest for (T₂ − T₁) years.
Why it works: the principal is inside both amounts and cancels when you subtract. What remains is pure interest, which is the same every year. Find one year's interest, then work back to the principal.
Time in months or days
Always convert time into years. 8 months = 8/12 = 2/3 year. 73 days = 73/365 = 1/5 year. If the rate is per year and the time is in months, forgetting to convert is the most common slip in this chapter.
Worked questions at NTPC level
Q1. A sum amounts to ₹7,440 in 3 years at 8% simple interest. Find the principal.
RT = 24, so A = 1.24 × P. P = 7,440 ÷ 1.24 = ₹6,000.
Check: interest = 6,000 × 8 × 3 ÷ 100 = 1,440, and 6,000 + 1,440 = 7,440.
Q2. A sum amounts to ₹6,720 in 3 years and to ₹7,200 in 5 years at simple interest. Find the principal and the rate.
Interest for 2 years = 7,200 − 6,720 = 480, so one year's interest = 240.
Interest for 3 years = 720, so P = 6,720 − 720 = ₹6,000.
Rate = 240 ÷ 6,000 × 100 = 4%.
Q3. A sum doubles in 8 years at simple interest. Find the rate, and the time in which it will triple.
Doubling: R × 8 = 100, so R = 12.5%.
Tripling: R × T = 200, so T = 200 ÷ 12.5 = 16 years.
Q4. ₹10,000 is lent in two parts, one at 8% and the other at 10% simple interest. The total interest for one year is ₹920. Find the two parts.
Let the part at 8% be x. Then 0.08x + 0.10(10,000 − x) = 920.
So 1,000 − 0.02x = 920, giving 0.02x = 80 and x = 4,000.
Parts: ₹4,000 at 8% and ₹6,000 at 10%. Check: 320 + 600 = 920.
Q5. Had the rate been 2% higher, the simple interest on a sum for 3 years would have been ₹360 more. Find the sum.
The extra interest comes only from the extra 2%: P × 2 × 3 ÷ 100 = 360.
So 6P ÷ 100 = 360, and P = ₹6,000.
Q6. 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.
Common mistakes
- Not converting months or days into years.
- Giving the amount when the question asks for interest, or the other way round.
- Using R × T = 100 for tripling. Tripling needs 200.
- Treating two amounts as two principals. In the two-amounts method, subtract to get interest, not a new principal.
- Applying SI shortcuts to compound interest. The straight-line rules only hold for simple interest.
Practice set
- Find the simple interest on ₹6,000 at 5% for 4 years.
- At what rate does a sum double in 10 years?
- Find the amount on ₹4,500 at 10% simple interest for 2 years.
- Find the principal if the simple interest is ₹900 at 6% for 3 years.
- In how many years will ₹2,000 earn ₹480 at 8% simple interest?
- A sum amounts to ₹5,600 in 2 years and ₹6,200 in 4 years at simple interest. Find the principal and rate.
- At what rate will a sum become five times itself in 16 years?
- The simple interest on a sum for 5 years is 2/5 of the sum. Find the rate.
Answers:
- ₹1,200. 6,000 × 5 × 4 ÷ 100.
- 10%. R × 10 = 100.
- ₹5,400. RT = 20, so 4,500 × 1.2.
- ₹5,000. 900 × 100 ÷ (6 × 3) = 90,000 ÷ 18.
- 3 years. 480 × 100 ÷ (2,000 × 8) = 48,000 ÷ 16,000.
- ₹5,000 at 6%. Two years' interest = 600, so one year = 300; P = 5,600 − 600 = 5,000; R = 300 ÷ 5,000 × 100 = 6.
- 25%. Five times means interest = 4P, so RT = 400 and R = 400 ÷ 16.
- 8%. Interest is 40% of P over 5 years, so 40 ÷ 5 = 8% a year.
What to do next
- Write the four-use table and the "becomes n times" rule on a revision card.
- Solve 15 simple interest questions with a 40-second target each, converting months to years every time.
- Move on to compound interest, then practise mixed SI–CI sets.
- If multipliers feel slow, revise percentage.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Railway Recruitment Boards website .
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