Interest calculations — SSC CGL Quant Notes

Concepts and rules

Simple interest SI = P × R × T/100, where P is the principal, R the rate per cent per annum and T the time in years. Amount A = P + SI = P(1 + RT/100).

Interest is charged only on the original principal, so the interest of every year is the same: P × R/100. This is why amounts at SI increase by a fixed step each year.

Convert time into years: months ÷ 12, days ÷ 365. When dates are given, count either the starting day or the ending day, not both.

If a sum becomes n times itself in T years at SI, the interest is (n − 1)P, so R = (n − 1) × 100/T. A sum that doubles in T years becomes 3 times in 2T years and 4 times in 3T years.

When the rate changes over periods, SI = P × (R1T1 + R2T2 + R3T3)/100. A change of ΔR in the rate changes the interest by P × ΔR × T/100.

Two parts lent at R1% and R2% give an average rate between R1 and R2; the parts are in the ratio (R2 − R) : (R − R1), the alligation rule.

Equal annual instalments at SI: if a debt of A is due at the end of n years and x is paid at the end of each year, each payment earns SI until the due date, so A = n × x + x × R × n(n − 1)/200.

Method and shortcuts

Step 1: list P, R, T, SI and A and mark the unknown. Step 2: put the time in years. Step 3: use SI = PRT/100 or A = P + SI and solve.

Two amounts after different times: the difference of the amounts is the SI for the difference of the times. Find the interest for 1 year, then subtract to reach P.

Percent view: in T years the interest is RT% of P. For 5 years at 8%, interest is 40% of P, so Amount is 140% of P and P = Amount × 100/140.

Splitting a sum between two rates: find the average rate = total yearly interest × 100/P, then use alligation on the two rates.

When it fails: the step-per-year idea is true only for SI. Under compound interest the yearly interest itself grows, so do not apply these shortcuts there.

Example 1: Direct SI

Question: Find the simple interest and the amount on ₹8,000 at 6% per annum for 3 years.

Solution: SI = 8,000 × 6 × 3/100 = ₹1,440. Amount = 8,000 + 1,440 = ₹9,440.

Answer: SI ₹1,440, amount ₹9,440

Example 2: Sum becoming n times

Question: A sum doubles itself in 8 years at simple interest. Find the rate, and the time in which it becomes 4 times itself.

Solution: Doubling means interest = P in 8 years, so R = 100/8 = 12.5%. To become 4 times, interest = 3P, which takes 3 × 8 = 24 years. Check: 3 × 100/12.5 = 24.

Answer: 12.5% per annum; 24 years

Example 3: Two amounts given

Question: A sum of money amounts to ₹9,800 in 3 years and to ₹11,000 in 5 years at simple interest. Find the sum and the rate.

Solution: SI for 2 years = 11,000 − 9,800 = ₹1,200, so SI for 1 year = ₹600. SI for 3 years = ₹1,800, so P = 9,800 − 1,800 = ₹8,000. R = 600 × 100/8,000 = 7.5%.

Answer: ₹8,000 at 7.5% per annum

Example 4: Effect of a higher rate

Question: Had a sum been lent at a rate 3% higher, it would have earned ₹360 more in 4 years. Find the sum. (a) ₹3,600 (b) ₹2,500 (c) ₹3,000 (d) ₹4,000

Solution: The extra interest comes only from the extra rate: P × 3 × 4/100 = 360, so P = 360 × 100/12 = ₹3,000. The original rate is not needed.

Answer: (c) ₹3,000

Example 5: Splitting a sum between two rates

Question: Meena lends ₹12,000 in two parts, one at 8% and the other at 12% simple interest. The total interest after 3 years is ₹3,420. How much did she lend at 12%? (a) ₹4,500 (b) ₹7,500 (c) ₹5,000 (d) ₹4,000

Solution: Yearly interest = 3,420 ÷ 3 = ₹1,140, so the average rate = 1,140 × 100/12,000 = 9.5%.

Alligation: (part at 8%) : (part at 12%) = (12 − 9.5) : (9.5 − 8) = 2.5 : 1.5 = 5 : 3. Part at 12% = 3/8 × 12,000 = ₹4,500. Check: 7,500 × 8% + 4,500 × 12% = 600 + 540 = 1,140 per year.

Answer: (a) ₹4,500

Example 6: Equal annual instalments at SI

Question: What equal annual instalment will discharge a debt of ₹6,450 due at the end of 4 years at 5% simple interest? (a) ₹1,600 (b) ₹1,550 (c) ₹1,450 (d) ₹1,500

Solution: Paid at the ends of years 1, 2, 3 and 4, the instalments earn interest for 3, 2, 1 and 0 years until the due date. Their values on the due date add to x × [4 + 5 × (3 + 2 + 1 + 0)/100] = x × (4 + 0.3) = 4.3x.

So 4.3x = 6,450 and x = ₹1,500. Check: 1,500 × (1.15 + 1.10 + 1.05 + 1) = 1,500 × 4.3 = 6,450.

Answer: (d) ₹1,500

Common mistakes

Leaving time in months. 9 months is 9/12 = 0.75 year, not 9.

Treating 'becomes 3 times' as interest = 3P. The interest is (3 − 1)P = 2P because the principal is part of the amount.

Dividing the instalment debt simply by n. Each instalment earns interest until the due date, so the divisor is n + R × n(n − 1)/200, not n.

Using the plain mean of two rates when the parts are unequal. Weight each rate by its part, or use alligation.

Counting both the start date and the end date when time is given as dates.

Practice questions

Q1. At what rate per cent per annum simple interest will ₹2,500 amount to ₹3,400 in 4 years? (a) 8% (b) 9% (c) 10% (d) 9.5%

Q2. At simple interest a sum becomes 7/4 of itself in 6 years. What is the rate of interest? (a) 12% (b) 15% (c) 12.5% (d) 10.5%

Q3. Ramesh borrowed ₹15,000 at simple interest: 8% for the first 2 years, 10% for the next 3 years and 12% after that. What total interest does he pay for 7 years? (a) ₹10,500 (b) ₹9,900 (c) ₹11,100 (d) ₹10,800

Answers

Q1. (b) SI = 3,400 − 2,500 = 900; R = 900 × 100/(2,500 × 4) = 9%.

Q2. (c) Interest = 7/4 − 1 = 3/4 of P in 6 years, so R = 75/6 = 12.5%.

Q3. (a) Rate-years = 8 × 2 + 10 × 3 + 12 × 2 = 16 + 30 + 24 = 70, so SI = 15,000 × 70/100 = ₹10,500.

Quick revision

SI = PRT/100; A = P + SI; yearly interest is constant.

n times in T years ⇒ R = (n − 1) × 100/T.

Difference of two amounts = SI for the difference of times.

Instalments at SI: A = n × x + x × R × n(n − 1)/200.

Concepts and rules

Compound interest adds each period's interest to the principal. Amount A = P(1 + R/100)ⁿ for n years at R% compounded annually, and CI = A − P.

Half-yearly compounding: use rate R/2 per period and 2n periods. Quarterly: rate R/4 and 4n periods.

Different rates in different years: A = P(1 + R1/100)(1 + R2/100)(1 + R3/100). The order of the rates does not change the final amount.

Part of a year (the usual SSC convention): for 2 years 4 months, A = P(1 + R/100)² × (1 + (R/3)/100), since 4 months is 1/3 of a year and that part earns R/3 %.

For 1 year compounded annually, CI = SI. For 2 years, CI − SI = P(R/100)². For 3 years, CI − SI = P(R/100)² × (3 + R/100).

Amounts at the end of successive years form a geometric progression with ratio (1 + R/100). So R% = (later amount − earlier amount) × 100/earlier amount for consecutive years.

If a sum becomes k times in n years at CI, it becomes k² times in 2n years and k³ times in 3n years.

The same formula gives population growth, P(1 + R/100)ⁿ, and depreciation of a machine, P(1 − R/100)ⁿ.

Method and shortcuts

Step 1: fix the rate per period and the number of periods. Step 2: write the multiplying factor as a fraction, for example 10% gives 11/10 and 5% gives 21/20. Step 3: multiply P by the factor once per period.

Effective rate for 2 years at R%: 2R + R²/100 per cent. At 10% this is 21%; at 20% it is 44%. Then CI = P × effective rate/100.

Year-wise view: the interest of year 2 exceeds that of year 1 by R% of year 1's interest. This explains CI − SI = P(R/100)² for 2 years.

Equal annual instalments at CI: the principal equals the sum of the present values, P = x/(1 + R/100) + x/(1 + R/100)² + … for each instalment x.

When it fails: the 2R + R²/100 shortcut is for exactly 2 periods at the same rate. For 3 periods or changing rates, multiply the factors.

Example 1: Basic CI

Question: Find the compound interest on ₹10,000 at 10% per annum for 2 years, compounded annually.

Solution: A = 10,000 × 11/10 × 11/10 = ₹12,100, so CI = ₹2,100. Shortcut: effective rate = 2 × 10 + 100/100 = 21%, and 21% of 10,000 = 2,100.

Answer: ₹2,100

Example 2: Half-yearly compounding

Question: Find the compound interest on ₹16,000 for 1 year at 10% per annum, compounded half-yearly.

Solution: Rate per half-year = 5% and there are 2 periods. A = 16,000 × 21/20 × 21/20 = ₹17,640, so CI = ₹1,640. Annual compounding would give only ₹1,600.

Answer: ₹1,640

Example 3: CI minus SI for 2 years

Question: The difference between compound and simple interest on a sum for 2 years at 5% per annum is ₹25. Find the sum.

Solution: CI − SI = P(R/100)² = P × 1/400 = 25, so P = 25 × 400 = ₹10,000. Check: CI = 10,000 × (1.1025 − 1) = 1,025 and SI = 1,000; difference 25.

Answer: ₹10,000

Example 4: Growth by powers

Question: A sum of money doubles itself in 5 years at compound interest. In how many years will it become 8 times itself? (a) 20 years (b) 10 years (c) 25 years (d) 15 years

Solution: Every 5 years the amount is multiplied by 2. Since 8 = 2 × 2 × 2, it needs 3 such blocks: 3 × 5 = 15 years. Simple interest reasoning (7 × 5 = 35 years) does not apply here.

Answer: (d) 15 years

Example 5: Two equal instalments at CI

Question: A loan of ₹4,200 is to be repaid in 2 equal annual instalments at 10% per annum compound interest. Find each instalment. (a) ₹2,310 (b) ₹2,420 (c) ₹2,520 (d) ₹2,400

Solution: Present values: x × 10/11 + x × 100/121 = x × (110 + 100)/121 = 210x/121. Setting 210x/121 = 4,200 gives x = 4,200 × 121/210 = ₹2,420.

Check year by year: after year 1 the debt is 4,620; paying 2,420 leaves 2,200; after year 2 it is 2,420, cleared by the second instalment.

Answer: (b) ₹2,420

Example 6: From two amounts to SI

Question: At compound interest a sum amounts to ₹6,050 in 2 years and ₹6,655 in 3 years. What is the simple interest on the same sum at the same rate for 4 years? (a) ₹2,200 (b) ₹1,800 (c) ₹2,000 (d) ₹2,400

Solution: Year 3 interest = 6,655 − 6,050 = 605, earned on 6,050, so R = 605 × 100/6,050 = 10%. Principal = 6,050 ÷ (11/10)² = 6,050 × 100/121 = ₹5,000.

SI for 4 years = 5,000 × 10 × 4/100 = ₹2,000.

Answer: (c) ₹2,000

Common mistakes

Forgetting to halve the rate and double the periods for half-yearly compounding.

Giving the amount when CI is asked. CI = A − P; subtract the principal at the end.

Using the 2-year difference formula P(R/100)² for 3 years. For 3 years multiply by (3 + R/100) as well.

Thinking a sum that doubles in 5 years becomes 4 times in 10 years at SI as well. That is true only for CI; at SI it becomes 3 times in 10 years.

Splitting a CI instalment loan as if each instalment were P/n plus interest. Use present values or a year-by-year table.

Practice questions

Q1. What is the compound interest on ₹12,500 at 8% per annum for 2 years, compounded annually? (a) ₹2,000 (b) ₹2,080 (c) ₹2,160 (d) ₹2,040

Q2. The difference between compound interest and simple interest on a sum for 3 years at 10% per annum is ₹93. What is the sum? (a) ₹3,000 (b) ₹3,100 (c) ₹2,700 (d) ₹3,300

Q3. The population of a town is 1,25,000. It rises by 8% in the first year and falls by 10% in the second year. What is the population after 2 years? (a) 1,22,500 (b) 1,23,000 (c) 1,21,000 (d) 1,21,500

Answers

Q1. (b) Effective rate = 2 × 8 + 64/100 = 16.64%; CI = 12,500 × 16.64/100 = ₹2,080.

Q2. (a) P × (10/100)² × (3 + 10/100) = P × 0.031 = 93, so P = ₹3,000.

Q3. (d) 1,25,000 × 108/100 × 90/100 = 1,35,000 × 0.9 = 1,21,500.

Quick revision

A = P(1 + R/100)ⁿ; CI = A − P; half-yearly R/2 and 2n periods.

2 years: effective rate 2R + R²/100; CI − SI = P(R/100)².

3 years: CI − SI = P(R/100)² × (3 + R/100).

Consecutive amounts: R% = difference × 100/earlier amount.

Instalments at CI: principal = sum of present values of the instalments.