Compound Interest

Compound Interest: Interest on Interest

Under simple interest, the interest is worked out on the same original sum every year, so each year's interest is the same. Under compound interest (CI), the interest earned in a year is added to the sum at the end of that year, and the next year's interest is worked out on this bigger amount. The interest itself starts earning interest.

Take ₹47,000 at 10% per annum for 3 years:

The CI for 3 years is 4,700 + 5,170 + 5,687 = ₹15,557, against an SI of ₹14,100. In the first year CI and SI are equal. After that CI pulls ahead: the extra ₹470 in year 2 is 10% of the first year's ₹4,700 of interest, and the extra ₹987 in year 3 is 10% of the ₹9,870 earned in the first two years.

Each year the amount is multiplied by the same number, 1 + R/100. Three years at 10% means multiplying by 1.1 three times:

$$A = P\left(1 + \frac{R}{100}\right)^n, \qquad \text{CI} = A - P$$

Here P is the principal, R the rate per annum and n the number of years. For the table: 47,000 × 1.1 × 1.1 × 1.1 = 47,000 × 1.331 = ₹62,557, and CI = 62,557 − 47,000 = ₹15,557.

Words to watch.

• Amount is principal + interest. CI is the interest alone. The options nearly always carry both, so read the last line of the question before you pick.
• "Compounded annually" means the interest is added once a year. If a question says "compound interest" and names no period, take it as yearly.
• The rate is per annum (p.a.) unless the question says otherwise. "4% per half-year" is not 4% p.a.
• 1.331 means the amount is 133.1% of the sum, so the CI is 33.1% of it. Thinking of the CI as a percentage of the principal saves a step in many questions.

Most CI questions carry the same three wrong options: the SI (₹14,100 here), the amount in place of the CI (₹62,557), and the CI for one year less (₹9,870). Decide whether the question wants A or CI before you start.

The Multiplier Fractions

Working out 1.05 × 1.05 × 1.05 by hand is slow. Write the rate as a fraction and the multiplier becomes a neat fraction too: 10% is $\frac{1}{10}$, so every year the sum is multiplied by $\frac{11}{10}$.

The sum cancels the bottom. SSC picks the principal so that it divides by the denominator. Cancel first and multiply last:

• ₹12,800 at 12.5% for 2 years: A = 12,800 × $\frac{81}{64}$ = 200 × 81 = ₹16,200, so the CI is ₹3,400.
• ₹1,36,000 at 5% for 3 years: A = 1,36,000 × $\frac{9261}{8000}$ = 17 × 9,261 = ₹1,57,437, so the CI is ₹21,437.
• ₹21,600 at $16\frac{2}{3}\%$ for 2 years: A = 21,600 × $\frac{49}{36}$ = 600 × 49 = ₹29,400, so the CI is ₹7,800.

Parts instead of rupees. At 12.5% for 3 years, P : A = 512 : 729, so P : CI = 512 : (729 − 512) = 512 : 217. If the CI is ₹7,595, then 217 parts = ₹7,595, one part is ₹35, and P = 512 × 35 = ₹17,920. No dividing by 1.423828125.

Reading the rate off the numbers. The numerators are squares and cubes: 441 = $21^2$, 9,261 = $21^3$, 1,331 = $11^3$, 729 = $9^3$ = $27^2$ and 17,576 = $26^3$. When a sum grows to $\frac{9261}{8000}$ of itself in 3 years, the multiplier is $\frac{21}{20}$ each year, so the rate is 5%.

It helps to know the squares up to 30 and the cubes up to 25 by heart, since they are the numerators of these multipliers.

Year-by-Year Interest and the Two- and Three-Year Rates

Each year's interest is R% more than the year before, because the sum it is worked on has grown by R%. So the yearly interests themselves grow like a CI sum:

$$\text{Interest for year } k \text{ alone} = P \times \frac{R}{100} \times \left(1 + \frac{R}{100}\right)^{k - 1}$$

Per ₹1,000 at the two rates SSC uses most:

At 25% it is cleaner to start from ₹1,600: the years give ₹400, ₹500 and ₹625, so the CI is ₹900 for 2 years and ₹1,525 for 3 years.

One rate for the whole period. Since $(1 + r)^2 = 1 + 2r + r^2$, the CI for 2 years is a fixed percentage of P, and so is the CI for 3 years:

$$\text{2 years: } \left(2R + \frac{R^2}{100}\right)\% \qquad \text{3 years: } \left(3R + \frac{3R^2}{100} + \frac{R^3}{100^2}\right)\%$$

So the CI on ₹4,500 at 20% for 3 years is 72.8% of 4,500 = ₹3,276. Year by year it is 900 + 1,080 + 1,296, the same total.

Interest for one year alone. "Find the CI on ₹4,500 at 20% for the 3rd year." Only the 3rd year counts: 4,500 × 0.2 × $1.2^2$ = ₹1,296. The trap option is ₹3,276, which is the CI for all three years.

The year-to-year jump. The 2nd year's interest is more than the 1st year's by R% of the 1st year's interest: 1,080 − 900 = 180, which is 20% of 900. This is how you find the rate when the interests of two years are given (see Two Amounts or Two Interests Given below).

[SSC Tip] The 2-year rate is never just 2R. At 15% it is 32.25%, not 30%; at 12% it is 25.44%, not 24%. The 2R figure is the SI, and SSC nearly always puts it in the options.

Finding the Principal

What you do depends on what the question gives.

1. The amount is given. Divide by the multiplier, which is the same as multiplying by its upside-down fraction. "A sum amounts to ₹37,485 in 2 years at 5% CI." P = 37,485 × $\frac{400}{441}$ = 85 × 400 = ₹34,000.

2. The CI is given. The CI is a fixed share of P, so write P : CI in parts. At 20% for 3 years, P : A = 125 : 216, so P : CI = 125 : 91. "The CI on a sum for 3 years at 20% is ₹5,460." Then 91 parts = ₹5,460, one part is ₹60, and P = 125 × 60 = ₹7,500. In percentage form, 5,460 is 72.8% of P, and 5,460 ÷ 0.728 gives the same ₹7,500.

3. The interest of one year is given. "At 10% CI, the interest for the 2nd year alone is ₹4,070." The 2nd year's interest is 10% of the amount after one year, which is 1.1P. So 0.11P = 4,070 and P = ₹37,000. For the 3rd year alone it would be $0.1 \times 1.1^2 = 0.121$ times P.

Present value. "What sum must be invested now to get ₹x after n years?" is route 1 in other words: divide x by the multiplier for n years. Equal yearly instalments are built on the same idea, and they are in CI vs SI, Splitting Sums and Instalments.

Which route? Look at the word just before the rupee figure. "Amounts to" or "becomes" means route 1. "The compound interest is" means route 2. "The interest for the 2nd year" or "in the 3rd year" means route 3.

[SSC Tip] The usual mistake is to divide the CI by the SI percentage. For ₹5,460 at 20% for 3 years that gives 5,460 ÷ 0.6 = ₹9,100, and that figure will be an option. The other trap is treating an amount as if it were the CI.

Half-Yearly, Quarterly and Other Periods

When interest is added more often than once a year, work with the rate for one period and the number of periods. The rate in the question is still per annum.

$$\text{Half-yearly: } A = P\left(1 + \frac{R}{200}\right)^{2n} \qquad \text{Quarterly: } A = P\left(1 + \frac{R}{400}\right)^{4n}$$

Half-yearly. ₹3,04,000 at 10% p.a., compounded half-yearly, for $1\frac{1}{2}$ years. The rate is 5% per half-year and there are 3 half-years:

A = 3,04,000 × $\frac{9261}{8000}$ = 38 × 9,261 = ₹3,51,918, so the CI is ₹47,918.

Compounded yearly for the same $1\frac{1}{2}$ years it would be 3,04,000 × 1.1 × 1.05 = ₹3,51,120, a CI of ₹47,120. More frequent compounding always gives a little more.

Quarterly. ₹1,40,000 at 12% p.a., compounded quarterly, for 6 months. The rate is 3% per quarter and there are 2 quarters: A = 1,40,000 × $1.03^2$ = 1,40,000 × 1.0609 = ₹1,48,526, so the CI is ₹8,526.

Count periods from months. 9 months compounded quarterly is 3 periods. 18 months compounded half-yearly is 3 periods. 2 years compounded every 4 months is 6 periods at R/3. For any period of k months, the rate per period is R × $\frac{k}{12}$: 18% p.a. compounded every 8 months gives 12% per period, and 16 months is 2 such periods.

A rate already given per period. "4% per half-year" or "2% per quarter" is already the rate for one period, so don't divide it again. Only a per annum rate gets divided.

The two classic slips.

• Halving the rate but keeping the number of years as the number of periods. For 2 years half-yearly at 10% p.a. that uses 5% twice instead of 5% four times.
• Counting the periods correctly but keeping the full yearly rate. For the ₹3,04,000 above, 10% for 3 periods gives ₹4,04,624, a CI of ₹1,00,624, which is far too big.

Effective Annual Rate, and Half-Yearly vs Yearly

The effective annual rate is the yearly rate that gives the same interest in one year as the more frequent compounding does.

$$\text{Half-yearly: } \left(1 + \frac{R}{200}\right)^2 - 1 = \left(R + \frac{R^2}{400}\right)\% \qquad \text{Quarterly: } \left(1 + \frac{R}{400}\right)^4 - 1$$

So 20% p.a. compounded half-yearly is the same as 21% p.a. compounded yearly. The half-yearly figure for R is the 2-year figure for R/2 from the earlier table: 8% half-yearly and 4% for 2 years both give 8.16%.

Comparing two offers. "Which is better: 10% p.a. compounded half-yearly or 10.2% p.a. compounded yearly?" 10% half-yearly is 10 + $\frac{100}{400}$ = 10.25% effective, so it beats 10.2% by 0.05% a year. Quick check on ₹100: 105 after six months, then 110.25 after a year.

The gap after one year. In one year, half-yearly compounding beats yearly compounding by the interest earned on the first half-year's interest:

$$\text{Gap for 1 year} = P \times \left(\frac{R}{200}\right)^2$$

• ₹75,000 at 12% p.a.: the yearly CI is ₹9,000 and the half-yearly CI is 75,000 × 0.1236 = ₹9,270. The gap is 75,000 × $0.06^2$ = ₹270.
• Going back: "At 10% p.a., half-yearly compounding gives ₹225 more than yearly compounding in 1 year. Find the sum." 225 = P × $0.05^2$, so P = 225 × 400 = ₹90,000.

[Tier 2] Over 2 years. There is no neat one-line gap, so work both amounts. ₹80,000 at 20% p.a.: yearly gives 80,000 × 1.44 = ₹1,15,200; half-yearly is 10% for 4 periods, 80,000 × $1.1^4$ = 80,000 × 1.4641 = ₹1,17,128. The gap is ₹1,928.

Different Rates, Depreciation and Part of a Year

Different rates in different years. Multiply by each year's own multiplier:

$$A = P\left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right)\left(1 + \frac{R_3}{100}\right)$$

₹87,500 at 4% in the 1st year, 10% in the 2nd and 12% in the 3rd: A = 87,500 × $\frac{26}{25} \times \frac{11}{10} \times \frac{28}{25}$ = ₹1,12,112, so the CI is ₹24,612. Step by step it runs ₹91,000, ₹1,00,100, ₹1,12,112. The order of the rates does not change the answer. For two years the single equivalent rate is $R_1 + R_2 + \frac{R_1 R_2}{100}$, the successive-change rule (Chapter 3): 4% then 10% is 14.4%.

A missing rate comes out the same way. "₹38,000 earns 10% in the first year and amounts to ₹48,488 after 2 years. Find the second year's rate." After one year it is ₹41,800, and 48,488 ÷ 41,800 = 1.16, so the rate is 16%.

Depreciation. A machine or a car that loses R% of its value every year is CI running downwards:

$$\text{Value after } n \text{ years} = P\left(1 - \frac{R}{100}\right)^n$$

A machine bought for ₹5,00,000 loses 20% a year. After 2 years it is worth 5,00,000 × $\frac{16}{25}$ = ₹3,20,000. The loss in the 2nd year alone is 20% of ₹4,00,000 = ₹80,000. Simple depreciation (40% off, ₹3,00,000) is the trap. To go back in time, divide: a value of ₹3,20,000 after 2 years at 20% depreciation means a starting value of 3,20,000 ÷ 0.64 = ₹5,00,000. Growth works the same way with + in place of −.

Part of a year. With yearly compounding, interest is added only at the end of a full year. For the leftover part of a year, take SI on the amount reached so far:

$$A = P\left(1 + \frac{R}{100}\right)^n \times \left(1 + \frac{f \times R}{100}\right)$$

where f is the fraction of a year.

• ₹54,000 at 10% for $2\frac{1}{2}$ years: 54,000 × 1.21 = ₹65,340. Half a year at 10% p.a. is 5%, so A = 65,340 × 1.05 = ₹68,607 and the CI is ₹14,607.
• ₹1,80,000 at 15% for 2 years 4 months: 1,80,000 × $\frac{529}{400}$ = ₹2,38,050. Four months is $\frac{1}{3}$ of a year, so 5% more: A = ₹2,49,952.50 and the CI is ₹69,952.50.

The traps are SI for the half-year on the original ₹54,000 (CI ₹14,040) and a full third year (CI ₹17,874). If the question says half-yearly instead, $2\frac{1}{2}$ years is simply 5 periods at 5%, which gives about ₹68,919.20.

Finding the Rate

From the principal and the amount. Work out A ÷ P and spot the power:

$$\left(1 + \frac{R}{100}\right)^n = \frac{A}{P}$$

"₹4,800 amounts to ₹6,348 in 2 years at CI." $\frac{6348}{4800} = \frac{529}{400} = \left(\frac{23}{20}\right)^2$, so the yearly multiplier is $\frac{23}{20}$ and R = 15%.

A ratio can be given straight away: "A sum becomes $\frac{125}{64}$ of itself in 3 years." That is $\left(\frac{5}{4}\right)^3$, so the rate is 25%.

From the CI over 2 years. Add the CI to P first. "The CI on ₹58,000 for 2 years is ₹12,180." A = ₹70,180 and $\frac{70180}{58000} = \frac{121}{100}$, so R = 10%. In percentage form, 12,180 is 21% of 58,000, and $2R + \frac{R^2}{100} = 21$ gives R = 10. Halving 21% to 10.5% is SI thinking, and it is wrong.

From the CI over 3 years. "The CI on ₹15,625 for 3 years is ₹1,951." A = ₹17,576. Since 17,576 = $26^3$ and 15,625 = $25^3$, the multiplier is $\frac{26}{25}$ and R = 4%.

Half-yearly or quarterly. Find the rate per period, then turn it back into a yearly rate. "₹72,000 amounts to ₹79,380 in 1 year, compounded half-yearly." $\frac{79380}{72000} = \frac{441}{400} = \left(\frac{21}{20}\right)^2$ over 2 half-years, so the rate is 5% per half-year, which is 10% p.a. Answering 5% is the trap.

When the numbers are not neat. Use the options. Take each option's rate, look up its 2- or 3-year percentage in the earlier table, and see which one matches CI ÷ P. For a 2-year CI that is 25.44% of P, the rate is 12%, because 2 × 12 + $\frac{144}{100}$ = 25.44. It takes seconds and you never need a square root.

[SSC Tip] A rate that comes out of a CI question is always a little less than the SI-style answer (CI ÷ P ÷ n), because part of the CI is interest on interest. If your answer is bigger than the SI-style figure, something has gone wrong.

Two Amounts or Two Interests Given

Amounts after two consecutive years. The amount after year n + 1 is the amount after year n plus one year's interest on it. So the difference is one year's interest on the earlier amount:

$$R = \frac{A_{n+1} - A_n}{A_n} \times 100$$

"A sum amounts to ₹41,454 in 2 years and to ₹43,526.70 in 3 years at CI." The difference is ₹2,072.70, and 2,072.70 ÷ 41,454 = 0.05, so R = 5%. Then P = 41,454 × $\frac{400}{441}$ = ₹37,600.

The two traps are dividing ₹2,072.70 by the later amount (about 4.76%) and going back as if it were SI (41,454 − 2 × 2,072.70 = ₹37,308.60).

Amounts after n and m years. The ratio is the multiplier raised to the gap in years:

$$\frac{A_m}{A_n} = \left(1 + \frac{R}{100}\right)^{m - n}$$

"A sum becomes ₹28,800 in 2 years and ₹41,472 in 4 years." $\frac{41472}{28800}$ = 1.44 = $1.2^2$, so R = 20%, and P = 28,800 ÷ 1.44 = ₹20,000. With a 3-year gap, look for a cube.

Interests of two consecutive years. Each year's interest is R% more than the year before. "At CI, the interest for the 2nd year is ₹1,890 and for the 3rd year is ₹2,268." The rise is ₹378 on ₹1,890, which is 20%. The 1st year's interest was 1,890 ÷ 1.2 = ₹1,575, and that is 20% of P, so P = ₹7,875.

[Tier 2] CI for 1 year and for 2 years. "The CI on a sum for 1 year is ₹750 and for 2 years is ₹1,560." The 2nd year's interest alone is 1,560 − 750 = ₹810, which is ₹60 more than the 1st year's. So R = $\frac{60}{750} \times 100$ = 8%, and P = 750 ÷ 0.08 = ₹9,375.

In all of these, divide by the earlier figure: the earlier amount, or the earlier year's interest. Dividing by the later one gives a rate that is slightly too small.

Time: Doubling, Tripling and Powers

Time from P and A. Work out A ÷ P and count how many times the multiplier goes into it. "In how many years will ₹33,280 become ₹47,385 at 12.5% CI?" $\frac{47385}{33280} = \frac{729}{512} = \left(\frac{9}{8}\right)^3$, so the answer is 3 years.

With half-yearly or quarterly compounding the count comes out in periods, so turn it back into years: 3 half-years is $1\frac{1}{2}$ years, not 3 years.

When the ratio is not a neat power, go year by year. "In how many whole years will a sum first become more than double at 20% CI?" The multipliers run 1.2, 1.44, 1.728, 2.0736, so the answer is 4 years.

The powers rule. If a sum becomes x times itself in t years, it becomes $x^k$ times itself in kt years, because each block of t years multiplies it by x again.

• Doubles in 4 years: 4 times in 8 years, 8 times in 12 years, 16 times in 16 years.
• Becomes 3 times in 7 years: 9 times in 14 years, 27 times in 21 years.
• Going back: becomes 64 times in 12 years. Since 64 = $4^3$, it becomes 4 times in 4 years, and since 64 = $2^6$, it doubles in 2 years.

From one multiple to another. Write both as powers of the same base. "A sum becomes 25 times itself in 8 years. When will it become 125 times?" 25 = $5^2$, so it becomes 5 times every 4 years, and 125 = $5^3$ takes 12 years.

Compare with SI. Under SI the interest grows in a straight line: a sum that doubles in 4 years (100% interest) becomes 4 times in 12 years (300% interest). Under CI the multiples go up by powers, so 4 times takes only 8 years. SSC puts both answers in the options, so be sure which kind of interest the question is about.

A rough check for doubling. At CI a sum roughly doubles in 72 ÷ R years. At 12% that is about 6 years, and in fact $1.12^6$ is about 1.97. Use it only to throw out options that are far off, never to pick between close ones.

[SSC Tip] The time is proportional to the power, not to the multiple. Going from 2 times to 8 times is going from $2^1$ to $2^3$, so it takes three times as long, not four times.

Solved Examples

Question 1

What is the compound interest on ₹43,000 at 10% per annum for 3 years, compounded annually? (a) ₹12,900 (b) ₹14,233 (c) ₹57,233 (d) ₹9,030

Answer. I just multiply by 1.1 every year. After year 1 it's ₹47,300, after year 2 it's ₹52,030 and after year 3 it's ₹57,233. So the CI is 57,233 − 43,000 = (b) ₹14,233. Faster: 3 years at 10% is 33.1%, and 33.1% of 43,000 is 14,233.

Question 2

Find the amount on ₹44,800 at 12.5% per annum compound interest for 2 years. (a) ₹56,000 (b) ₹11,900 (c) ₹56,197.12 (d) ₹56,700

Answer. 12.5% is $\frac{1}{8}$, so every year the money becomes $\frac{9}{8}$ of itself. For 2 years: 44,800 × $\frac{81}{64}$. 44,800 ÷ 64 = 700, and 700 × 81 = 56,700. So the amount is (d) ₹56,700.

Question 3

What is the compound interest on ₹1,04,000 at 5% per annum for 3 years, interest compounded yearly? (a) ₹15,600 (b) ₹1,20,393 (c) ₹10,660 (d) ₹16,393

Answer. 5% means the multiplier is $\frac{21}{20}$, so 3 years is $\frac{9261}{8000}$. 1,04,000 ÷ 8,000 = 13, and 13 × 9,261 = ₹1,20,393. That's the amount, so the CI is 1,20,393 − 1,04,000 = (d) ₹16,393.

Question 4

What is the compound interest on ₹32,500 at 20% per annum for the 3rd year alone? (a) ₹6,500 (b) ₹7,800 (c) ₹9,360 (d) ₹23,660

Answer. Each year's interest is 20% more than the year before. Year 1: 20% of 32,500 = ₹6,500. Year 2: 6,500 × 1.2 = ₹7,800. Year 3: 7,800 × 1.2 = (c) ₹9,360. ₹23,660 is the CI for all 3 years together, not the 3rd year alone.

Question 5

The compound interest on a sum of money for 2 years at 8% per annum is ₹3,952. What is the sum? (a) ₹24,700 (b) ₹22,000 (c) ₹23,750 (d) ₹49,400

Answer. For 2 years at 8% the multiplier is $\frac{729}{625}$, so P : CI = 625 : 104. 104 parts = ₹3,952, so one part is ₹38, and P = 625 × 38 = (c) ₹23,750. Check: 2 years at 8% is 16.64%, and 16.64% of 23,750 is 3,952. ₹24,700 treats it as SI (3,952 ÷ 16%).

Question 6

A sum of money amounts to ₹21,870 in 3 years at 12.5% per annum compound interest. What is the sum? (a) ₹15,360 (b) ₹15,905.45 (c) ₹17,280 (d) ₹6,510

Answer. 3 years at 12.5% means multiplying by $\frac{729}{512}$. To go back I multiply by $\frac{512}{729}$: 21,870 ÷ 729 = 30, and 30 × 512 = (a) ₹15,360.

Question 7

Find the compound interest on ₹70,000 at 20% per annum for $1\frac{1}{2}$ years, if the interest is compounded half-yearly. (a) ₹21,000 (b) ₹22,400 (c) ₹23,170 (d) ₹50,960

Answer. Half-yearly means half the rate and double the periods: 10% per half-year for 3 half-years. 70,000 × $\left(\frac{11}{10}\right)^3$ = 70,000 × 1.331 = ₹93,170. So the CI is (c) ₹23,170. Not halving the rate gives ₹50,960.

Question 8

What is the compound interest on ₹93,750 for 9 months at 16% per annum, compounded quarterly? (a) ₹11,250 (b) ₹11,706 (c) ₹1,05,456 (d) ₹24,348

Answer. Quarterly means a quarter of the rate: 16 ÷ 4 = 4% per quarter. 9 months is 3 quarters. So A = 93,750 × $\left(\frac{26}{25}\right)^3$ = 93,750 × $\frac{17576}{15625}$ = 6 × 17,576 = ₹1,05,456, and the CI is (b) ₹11,706.

Question 9

Bank A pays 16% per annum compounded half-yearly, and Bank B pays 16.5% per annum compounded yearly. On a deposit of ₹50,000 for 1 year, which bank pays more interest, and by how much? (a) Bank A, by ₹70 (b) Bank B, by ₹250 (c) Bank B, by ₹70 (d) Both pay the same

Answer. Bank A gives 8% for each half-year: 50,000 × $1.08^2$ = 50,000 × 1.1664, so the interest is ₹8,320. Bank B gives 16.5% of 50,000 = ₹8,250. So the answer is (a) Bank A, by ₹70. In rates: 16% half-yearly is 16.64% effective, which beats 16.5%. Treating Bank A as a plain 16% (₹8,000) leads to 'Bank B, by ₹250'.

Question 10

[Tier 2] What is the difference between the compound interest on ₹1,12,000 at 10% per annum for 2 years when the interest is compounded half-yearly and when it is compounded yearly? (a) ₹280 (b) ₹560 (c) ₹616.70 (d) ₹24,136.70

Answer. Yearly: 2 years at 10% is 21%, so the CI is ₹23,520. Half-yearly: 5% for 4 half-years, and $1.05^4$ = 1.21550625, so the CI is 1,12,000 × 0.21550625 = ₹24,136.70. The difference is (c) ₹616.70.

Question 11

A sum of ₹18,000 is invested at compound interest at 5% for the first year, 10% for the second year and 20% for the third year. What is the compound interest at the end of 3 years? (a) ₹6,300 (b) ₹6,948 (c) ₹24,948 (d) ₹2,790

Answer. I multiply by each year's own multiplier. 18,000 × 1.05 = 18,900. 18,900 × 1.1 = 20,790. 20,790 × 1.2 = 24,948. So the CI is 24,948 − 18,000 = (b) ₹6,948.

Question 12

What is the compound interest on ₹1,44,000 at 15% per annum for $2\frac{1}{2}$ years, interest being compounded annually? (a) ₹54,000 (b) ₹57,240 (c) ₹60,723 (d) ₹75,006

Answer. First 2 full years at CI: 1,44,000 × $\frac{529}{400}$ = ₹1,90,440. For the last half-year the rate is 15 ÷ 2 = 7.5%, worked on ₹1,90,440: 1,90,440 × 1.075 = ₹2,04,723. So the CI is (c) ₹60,723. ₹57,240 takes the half-year's 7.5% on the original ₹1,44,000 instead of on ₹1,90,440.

Question 13

A machine is bought for ₹2,50,000. Its value goes down by 20% every year. What will its value be after 3 years? (a) ₹1,00,000 (b) ₹1,28,000 (c) ₹1,60,000 (d) ₹1,02,400

Answer. Every year the value becomes $\frac{4}{5}$ of what it was. After 1 year: ₹2,00,000. After 2 years: ₹1,60,000. After 3 years: ₹1,28,000. So the answer is (b) ₹1,28,000.

Question 14

A car loses 15% of its value every year. Its value now is ₹4,33,500. What was its value 2 years ago? (a) ₹5,73,303.75 (b) ₹6,19,285.71 (c) ₹5,10,000 (d) ₹6,00,000

Answer. Two years of 15% depreciation means the old value was multiplied by $0.85^2$ = 0.7225. So I divide: 4,33,500 ÷ 0.7225 = (d) ₹6,00,000. Check: 6,00,000 × 0.85 = 5,10,000, and 5,10,000 × 0.85 = 4,33,500. ₹5,73,303.75 adds 15% twice to today's value, which is wrong because the 15% was taken from the bigger, older values.

Question 15

A sum of ₹7,500 amounts to ₹10,800 in 2 years at compound interest, compounded annually. What is the rate of interest per annum? (a) 22% (b) 20% (c) 44% (d) 18%

Answer. A ÷ P = $\frac{10800}{7500} = \frac{36}{25} = \left(\frac{6}{5}\right)^2$. So each year the money is multiplied by $\frac{6}{5}$, which means (b) 20%.

Question 16

The compound interest on ₹27,000 for 3 years, compounded annually, is ₹8,937. What is the rate of interest per annum? (a) 10% (b) 11% (c) 33.1% (d) 12%

Answer. Amount = 27,000 + 8,937 = ₹35,937. A ÷ P = $\frac{35937}{27000}$, and 35,937 = $33^3$ while 27,000 = $30^3$. So the multiplier is $\frac{33}{30} = \frac{11}{10}$, and the rate is (a) 10%.

Question 17

A sum of ₹88,000 amounts to ₹1,01,871 in $1\frac{1}{2}$ years at compound interest, compounded half-yearly. What is the rate of interest per annum? (a) 5% (b) 15% (c) 7.5% (d) 10%

Answer. $\frac{101871}{88000} = \frac{9261}{8000} = \left(\frac{21}{20}\right)^3$ (both numbers divide by 11). The 3 is the number of half-years, so the rate is 5% per half-year. Per annum that is double: (d) 10%.

Question 18

A sum of money at compound interest amounts to ₹10,285 in 2 years and to ₹11,313.50 in 3 years. What is the sum? (a) ₹8,500 (b) ₹8,228 (c) ₹9,350 (d) ₹8,000

Answer. The two amounts are one year apart, so the extra ₹1,028.50 is one year's interest on ₹10,285. 1,028.50 ÷ 10,285 = 0.1, so the rate is 10%. Now go back 2 years: 10,285 ÷ 1.21 = (a) ₹8,500.

Question 19

A sum of money at compound interest amounts to ₹12,000 in 2 years and to ₹18,750 in 4 years. What is the sum? (a) ₹5,250 (b) ₹9,600 (c) ₹8,000 (d) ₹7,680

Answer. From year 2 to year 4 is 2 years, and $\frac{18750}{12000} = \frac{25}{16} = \left(\frac{5}{4}\right)^2$. So the yearly multiplier is $\frac{5}{4}$, which is 25%. Now go back 2 years from ₹12,000: 12,000 × $\frac{16}{25}$ = (d) ₹7,680.

Question 20

The compound interest on a sum for the 2nd year is ₹3,480 and for the 3rd year is ₹4,176. What is the sum? (a) ₹14,500 (b) ₹17,400 (c) ₹20,880 (d) ₹15,000

Answer. The 3rd year's interest is ₹696 more than the 2nd year's, and 696 ÷ 3,480 = 20%, so the rate is 20%. The 1st year's interest was 3,480 ÷ 1.2 = ₹2,900, and that is 20% of the sum. So the sum is 2,900 × 5 = (a) ₹14,500.

Question 21

[Tier 2] The compound interest on a sum of money for 1 year is ₹2,700, and for 2 years it is ₹5,724. What is the sum? (a) ₹23,850 (b) ₹25,200 (c) ₹47,700 (d) ₹22,500

Answer. The 2nd year's interest alone is 5,724 − 2,700 = ₹3,024. That is ₹324 more than the 1st year's, and the extra is interest on the 1st year's interest, so the rate is 324 ÷ 2,700 = 12%. The 1st year's ₹2,700 is 12% of the sum, so the sum is 2,700 ÷ 0.12 = (d) ₹22,500.

Question 22

In how much time will ₹1,52,000 amount to ₹1,75,959 at 10% per annum compound interest, compounded half-yearly? (a) 3 years (b) $1\frac{1}{2}$ years (c) 2 years (d) 1 year

Answer. Half-yearly at 10% p.a. is 5% per half-year, so the multiplier is $\frac{21}{20}$. $\frac{175959}{152000} = \frac{9261}{8000} = \left(\frac{21}{20}\right)^3$ (both numbers divide by 19), so it takes 3 half-years, which is (b) $1\frac{1}{2}$ years. '3 years' is the trap: the 3 counts half-years, not years.

Question 23

A sum of money at compound interest becomes 3 times itself in 9 years. In how many years will it become 81 times itself? (a) 36 years (b) 243 years (c) 360 years (d) 27 years

Answer. 81 = $3^4$, so the sum has to triple 4 times over. Each tripling takes 9 years, so it needs 4 × 9 = (a) 36 years. 360 years is the SI way: 2 times the sum as interest in 9 years, so 80 times takes 360 years.

Question 24

A sum of money at compound interest becomes 4 times itself in 14 years. In how many years will it become 32 times itself? (a) 112 years (b) $144\frac{2}{3}$ years (c) 35 years (d) 70 years

Answer. 4 = $2^2$, so the sum doubles twice in 14 years, once every 7 years. 32 = $2^5$ needs 5 doublings: 5 × 7 = (c) 35 years.