CI vs SI, Splitting Sums and Instalments

How CI and SI Drift Apart

Here compound interest is set next to simple interest, using the CI formula and multipliers from the Compound Interest section.

Take ₹20,000 at 10% for 3 years, both ways.

Where does the extra come from?

• Year 1: both earn 10% of ₹20,000, so there is no difference. For 1 year compounded yearly, CI = SI.
• Year 2: CI also earns 10% on the first year's interest of ₹2,000. That is the extra ₹200.
• Year 3: CI earns 10% on the ₹4,200 of interest from the first two years, which is ₹420. That ₹420 is 2 × 200 (interest on two years of plain interest) plus ₹20 (interest on the earlier ₹200 extra).

So the gap is "interest on interest". With $r = \frac{R}{100}$:

$$\text{CI} - \text{SI (2 years)} = P r^2 = P\left(\frac{R}{100}\right)^2$$

$$\text{CI} - \text{SI (3 years)} = P r^2 (3 + r) = \frac{P R^2 (300 + R)}{100^3}$$

Check with the table: 20,000 × 0.01 = 200, and 200 × 3.1 = 620.

Half-yearly compounding for 1 year. Now the interest is added after 6 months, so the second half-year earns interest on the first half's interest. With the half-yearly rate $\frac{R}{2}$,

$$\text{CI} - \text{SI (1 year, half-yearly)} = P\left(\frac{R}{200}\right)^2$$

₹40,000 at 10% a year, compounded half-yearly, is 5% per half-year, so the difference is 40,000 × $\left(\frac{5}{100}\right)^2$ = ₹100. Check: 40,000 × 1.05 × 1.05 = 44,100, so the CI is ₹4,100 against an SI of ₹4,000.

[SSC Tip] "Find the difference between CI and SI for 1 year" with yearly compounding has the answer 0, and SSC sometimes puts it in the options to catch students who use the 2-year formula. With half-yearly or quarterly compounding the difference is not 0, so read how the interest is compounded before you pick a formula.

The 2-Year Difference: Finding D, P or R

Call the 2-year difference D. Everything comes from one formula, turned around as needed:

$$D = P\left(\frac{R}{100}\right)^2 \qquad P = D\left(\frac{100}{R}\right)^2 \qquad R = 100\sqrt{\frac{D}{P}}$$

Finding D. ₹42,500 at 8% for 2 years: D = 42,500 × $\frac{64}{10000}$ = ₹272. Check: SI = ₹6,800, and CI = 42,500 × 1.08 × 1.08 − 42,500 = 49,572 − 42,500 = ₹7,072.

Finding P. "The difference between CI and SI on a sum for 2 years at 12% is ₹252." P = 252 × $\frac{10000}{144}$ = ₹17,500.

Finding R. "On ₹1,12,500 the difference for 2 years is ₹180." $\frac{D}{P} = \frac{180}{112500} = \frac{1}{625}$, so $\frac{R}{100} = \frac{1}{25}$ and R = 4%.

Rates as fractions are quickest. D is just P divided by a square.

So at $12\frac{1}{2}\%$ a sum of ₹9,600 has D = 9,600 ÷ 64 = ₹150, and a difference of ₹75 at 5% means P = 75 × 400 = ₹30,000.

Different rates in the two years. If the rate is $R_1$% in the first year and $R_2$% in the second, the second year earns $R_2$% on the first year's interest:

$$D = P \times \frac{R_1}{100} \times \frac{R_2}{100}$$

₹25,000 at 8% then 10%: CI = 25,000 × 1.08 × 1.1 − 25,000 = ₹4,700, SI = 18% of 25,000 = ₹4,500, and D = 25,000 × 0.08 × 0.1 = ₹200. Here SI means simple interest at the same two rates, 8% for the first year and 10% for the second.

[SSC Tip] The usual slips are using R instead of R² (for ₹42,500 at 8% that gives ₹3,400, which is just one year's interest) and doubling the right answer. The 2-year difference is always small next to the interest, because it is only interest on one year's interest.

The 3-Year Difference

For 3 years the gap grows faster, because the interest on interest itself starts earning:

$$D_3 = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right) = \frac{P R^2 (300 + R)}{100^3}$$

Finding $D_3$. ₹12,000 at 10% for 3 years: the 2-year difference would be 12,000 × 0.01 = ₹120, and $D_3$ = 120 × 3.1 = ₹372. Check: CI = 12,000 × 1.331 − 12,000 = ₹3,972 and SI = ₹3,600.

So $D_3$ is a little more than 3 times $D_2$: 3.05 times at 5%, 3.1 times at 10%, 3.2 times at 20%.

The difference year by year. If you prefer to build it up, the extra under CI in each year is

• Year 1: 0
• Year 2: $P r^2$ (interest on year 1's interest)
• Year 3: $P r^2 (2 + r)$ (interest on the first two years' interest, including the earlier extra)

For ₹12,000 at 10% that is 0, ₹120 and ₹252, which add to ₹372.

Finding P. "The difference between CI and SI on a sum for 3 years at 5% is ₹183." At 5%, $D_3 = P \times \frac{1}{400} \times \frac{61}{20}$, so P = 183 × $\frac{400 \times 20}{61}$ = ₹24,000. Check: CI = 24,000 × 1.157625 − 24,000 = ₹3,783 and SI = ₹3,600.

[Tier 2] Finding R from the two differences. Since $D_3 = D_2 \times \left(3 + \frac{R}{100}\right)$,

$$\frac{D_3}{D_2} = \frac{300 + R}{100}$$

"For a sum, the difference between CI and SI is ₹400 for 2 years and ₹1,232 for 3 years." $\frac{1232}{400}$ = 3.08, so R = 8%, and then P = 400 × $\left(\frac{100}{8}\right)^2$ = 400 × 156.25 = ₹62,500.

[SSC Tip] Three times the 2-year difference (₹360 for ₹12,000 at 10%) is the standard wrong option. The extra $\frac{R}{100}$ in the bracket is the interest on the interest on the interest, and at 10% it adds ₹12 here.

CI and SI on the Same Sum: Finding R and P

Very often SSC gives both interests on the same sum, at the same rate and for the same 2 years, and asks for the rate or the sum.

The SI for 2 years is two years of plain interest. The CI is the same plus interest on the first year's interest. So the extra, CI − SI, is R% of one year's SI, which is half the 2-year SI:

$$\text{CI} - \text{SI} = \frac{\text{SI}}{2} \times \frac{R}{100} \Rightarrow R = \frac{2(\text{CI} - \text{SI})}{\text{SI}} \times 100$$

"On a certain sum, the SI for 2 years is ₹4,200 and the CI for 2 years at the same rate is ₹4,347. Find the rate and the sum."

• One year's SI = 4,200 ÷ 2 = ₹2,100.
• The extra ₹147 is interest on ₹2,100, so R = $\frac{147}{2100} \times 100$ = 7%.
• P = 2,100 × $\frac{100}{7}$ = ₹30,000.
• Check: 30,000 × 1.07 × 1.07 = 34,347, so the CI is ₹4,347.

Dividing ₹147 by the whole ₹4,200 gives 3.5%, the usual trap. The extra interest grows on one year's interest, not on two.

SI for 1 year and CI for 2 years. "The SI on a sum for 1 year is ₹1,750 and the CI on it for 2 years at the same rate is ₹3,745." Two years of plain interest is ₹3,500, so the extra is ₹245, which is 14% of ₹1,750. So R = 14% and P = 1,750 × $\frac{100}{14}$ = ₹12,500.

SI for 3 years and CI for 2 years. "The SI on a sum for 3 years is ₹4,500 and the CI on the same sum for 2 years is ₹3,090, at the same rate." One year's SI = ₹1,500. The CI for 2 years = 3,000 + extra, so the extra is ₹90, which is 6% of 1,500. So R = 6% and P = 1,500 × $\frac{100}{6}$ = ₹25,000. Check: 25,000 × 1.1236 − 25,000 = 3,090.

For 3 years the link is not linear, so in an MCQ the quickest way is to test the options: a rate that gives the right CI on the principal found from the SI is the answer.

Whatever is given, get one year's SI first. The rate is then the extra interest as a percentage of one year's SI, and the principal is one year's SI × $\frac{100}{R}$.

From SI to CI and Back

SI given, CI wanted (2 years, same rate). The CI is the SI plus interest on one year's SI:

$$\text{CI}_2 = \text{SI}_2 \times \left(1 + \frac{R}{200}\right)$$

"The SI on a sum for 2 years at 6% is ₹2,880. Find the CI on the same sum at the same rate for 2 years." CI = 2,880 × 1.03 = ₹2,966.40. Check: P = 2,880 ÷ 0.12 = ₹24,000, and 24,000 × 1.06 × 1.06 = 26,966.40.

CI given, SI wanted. Divide by the same factor:

$$\text{SI}_2 = \text{CI}_2 \times \frac{200}{200 + R}$$

"The CI on a sum for 2 years at 10% is ₹5,250." SI = 5,250 × $\frac{200}{210}$ = ₹5,000, and the sum is ₹25,000.

Different rate or time. If the rate or the time changes between the SI and the CI, find the principal first.

"The SI on a sum for 3 years at 8% is ₹4,320. Find the CI on the same sum for 2 years at 10%."

• P = 4,320 × $\frac{100}{24}$ = ₹18,000.
• CI = 21% of 18,000 = ₹3,780.

3 years. From one year's SI, $\text{CI}_3 = \text{SI}_1 \times (3 + 3r + r^2)$, where $r = \frac{R}{100}$. "The SI for 1 year at 10% is ₹1,500. Find the CI for 3 years." CI = 1,500 × 3.31 = ₹4,965. Check: P = ₹15,000, and 15,000 × 1.331 − 15,000 = 4,965.

CI as a multiple of one year's SI

So at 20%, if one year's SI is ₹1,350, the CI for 3 years is 1,350 × 3.64 = ₹4,914.

Watch out for "CI = SI × $\left(1 + \frac{R}{100}\right)$", a common wrong shortcut. It puts a full year's interest on both years of SI. Only the first year's interest has had time to earn more, so the correct factor for 2 years is $1 + \frac{R}{200}$.

SI Scheme versus CI Scheme

At the same rate, CI and SI are equal for the first year and CI is ahead after that. So a person who borrows at SI and lends at CI (same rate, same sum) gains exactly the CI − SI difference.

"Raju borrows ₹80,000 at 5% simple interest and lends it at 5% compounded yearly."

• Gain in 2 years = 80,000 × 0.0025 = ₹200.
• Gain in 3 years = 200 × 3.05 = ₹610.

Doing it the other way round (borrowing at CI and lending at SI) gives a loss of the same size.

At different rates, compare the total percentage each scheme gives over the full time. CI for 2 years at R% is $\left(2R + \frac{R^2}{100}\right)\%$ (from the Compound Interest section), and for 3 years it is the 3-year multiplier minus 1.

"₹40,000 is to be invested. Scheme A pays 11% simple interest; Scheme B pays 10% compounded yearly."

So the winner can change with the time, and you must work out the time the question gives.

Break-even rate. 10% CI for 2 years gives 21%, which is the same as 10.5% SI for 2 years. Over 3 years it gives 33.1%, which matches SI at about 11.03% a year. An SI rate above these wins, and one below loses.

"CI is always better" is true only at the same rate. When the SI rate is higher, work out both totals. Over 2 years, R% CI and $\left(R + \frac{R^2}{200}\right)\%$ SI give exactly the same interest, so a question built that way has "both are equal" as its answer.

Splitting a Sum so the CI Amounts Are Equal

A sum is split into parts that are lent for different times at the same compound rate, and the parts must grow to the same amount. The part lent for longer must start smaller.

If part 1 is lent for $t_1$ years and part 2 for $t_2$ years, with $t_2 > t_1$,

$$P_1\left(1 + \frac{R}{100}\right)^{t_1} = P_2\left(1 + \frac{R}{100}\right)^{t_2} \Rightarrow P_1 : P_2 = \left(1 + \frac{R}{100}\right)^{t_2 - t_1} : 1$$

Only the gap in time matters, not the times themselves.

"A father divides ₹44,200 between Rohan, aged 13, and Sonia, aged 15, so that each gets the same amount at 18. The money earns 10% compounded yearly." Rohan's money stays 5 years and Sonia's stays 3 years, a gap of 2 years. So Sonia : Rohan = 1.21 : 1 = 121 : 100.

• One part = 44,200 ÷ 221 = ₹200.
• Sonia gets ₹24,200 and Rohan gets ₹20,000.
• Check: 20,000 × $1.1^5$ = 32,210.20 and 24,200 × $1.1^3$ = 32,210.20.

Three parts. "₹49,650 is split into three parts lent at 10% CI for 1, 2 and 3 years so that the amounts are equal." The 1-year part must be $1.1^2$ times the 3-year part, and the 2-year part 1.1 times it. So the ratio is 121 : 110 : 100, which adds to 331. One part is ₹150, so the parts are ₹18,150, ₹16,500 and ₹15,000, and each grows to ₹19,965.

Fraction rates keep it whole. At 20% the multiplier is $\frac{6}{5}$, so a 2-year gap gives 36 : 25. At 5% it is $\frac{21}{20}$, so a 2-year gap gives 441 : 400, and a 1-year gap at 25% gives 5 : 4.

The SI version (in the Simple Interest section) uses 100 + RT for each part, so there the actual times matter and not just the gap. Mixing up the two methods is a common slip.

The younger child waits longer, so gets the smaller share now; if your ratio gives the younger child more, turn it round. The three-part split above works exactly like equal instalments, which come next.

Equal Annual Instalments under CI

When a loan at compound interest is repaid in equal yearly instalments, each instalment pays off its own share of the loan. An instalment of x paid after k years is worth only $\frac{x}{(1 + r)^k}$ today; this is its present value. The loan equals the sum of these:

$$P = \frac{x}{1 + r} + \frac{x}{(1 + r)^2} + \cdots + \frac{x}{(1 + r)^n}, \qquad r = \frac{R}{100}$$

Two instalments at 10%. "A loan of ₹39,900 is repaid in 2 equal annual instalments at 10% CI." $P = x\left(\frac{10}{11} + \frac{100}{121}\right) = x \times \frac{210}{121}$, so x = 39,900 × $\frac{121}{210}$ = ₹22,990.

Run it forward to check.

• End of year 1: 39,900 × 1.1 = 43,890; pay 22,990; ₹20,900 is still owed.
• End of year 2: 20,900 × 1.1 = 22,990; pay 22,990; nothing is owed.

The ratio method. Write the multiplier as a fraction, $1 + r = \frac{b}{a}$, and take the instalment as $b^n$ units. Then every present value is a whole number of units.

"A loan of ₹27,300 is repaid in 3 equal annual instalments at 20% CI." The loan is 455 units, so one unit is 27,300 ÷ 455 = ₹60, and the instalment is 216 × 60 = ₹12,960. Run forward: 27,300 × 1.2 = 32,760 − 12,960 = 19,800; then 19,800 × 1.2 = 23,760 − 12,960 = 10,800; then 10,800 × 1.2 = 12,960, which the last instalment clears.

Total interest paid = all the instalments − the loan = 3 × 12,960 − 27,300 = ₹11,580.

[SSC Tip] Two wrong options appear in most of these questions: the CI amount divided by n (27,300 × 1.728 ÷ 3 = ₹15,724.80), and the loan divided by n with one year's interest added (₹10,920). Present values are the safe method, and a forward run settles any doubt.

Loans Paid in Parts, What Is Still Owed, and Reinvested Returns

Unequal payments. Run the loan forward one year at a time: add the year's interest, take off the payment, carry on.

"Kiran borrows ₹50,000 at 10% CI. She pays ₹20,000 at the end of the first year and ₹20,000 at the end of the second. What must she pay at the end of the third year to clear the loan?"

• Year 1: 50,000 × 1.1 = 55,000; pay 20,000; ₹35,000 owed.
• Year 2: 35,000 × 1.1 = 38,500; pay 20,000; ₹18,500 owed.
• Year 3: 18,500 × 1.1 = ₹20,350.

Taking both payments straight off the 3-year amount (50,000 × 1.331 − 40,000 = ₹26,550) is wrong, because each payment stops interest from building on that money. That route works only if the payments also carry interest to the end: 66,550 − 20,000 × 1.21 − 20,000 × 1.1 = 66,550 − 24,200 − 22,000 = 20,350, which matches.

What is still owed. After some instalments are paid, the balance is wherever the forward run stands. In the 3-instalment loan of ₹27,300 at 20% worked above, ₹19,800 is owed right after the first instalment and ₹10,800 after the second. Another way to see it: the amount still owed is the present value of the instalments not yet paid. After the first instalment, the two that are left are worth $12960 \times \frac{5}{6} + 12960 \times \frac{25}{36}$ = 10,800 + 9,000 = 19,800.

Reinvested returns. When a business adds its profit back to its capital every year, the capital grows exactly like CI.

"A shop starts with ₹1,60,000 and earns a profit of 25% a year on its capital, all of which is put back into the shop." After 3 years the capital is 1,60,000 × $\left(\frac{5}{4}\right)^3$ = ₹3,12,500. If the owner took the profit out each year instead, the capital would stay at ₹1,60,000 and he would have drawn 3 × 40,000 = ₹1,20,000, so ₹2,80,000 in all, which is the SI picture.

If only part of the profit goes back in, the capital grows at that part of the rate. With a 25% profit and 60% of it reinvested, the capital grows at 15% a year.

[Tier 2] Half-yearly instalments. Use the half-yearly rate and count half-years. A loan of ₹10,500 repaid in 2 equal half-yearly instalments at 20% a year compounded half-yearly is the 10% two-instalment case: x = 10,500 × $\frac{121}{210}$ = ₹6,050.

Solved Examples

Question 1

What is the difference between the compound interest (compounded annually) and the simple interest on ₹32,000 for 2 years at 7.5% per annum? (a) ₹360 (b) ₹4,980 (c) ₹180 (d) ₹4,800

Answer. D = P × $\left(\frac{R}{100}\right)^2$ = 32,000 × $\frac{7.5 \times 7.5}{10000}$ = 32,000 × 0.005625 = (c) ₹180. Check: SI = ₹4,800 and CI = 32,000 × 1.075 × 1.075 − 32,000 = 36,980 − 32,000 = ₹4,980.

Question 2

The difference between the compound interest and the simple interest on a sum for 2 years at 15% per annum is ₹450. What is the sum? (a) ₹20,000 (b) ₹3,000 (c) ₹10,000 (d) ₹30,000

Answer. P = D × $\left(\frac{100}{R}\right)^2$ = 450 × $\frac{10000}{225}$ = (a) ₹20,000. Check: SI = ₹6,000 and CI = 20,000 × 1.3225 − 20,000 = ₹6,450, a gap of ₹450. Dividing by 15% only once (450 ÷ 0.15) gives ₹3,000.

Question 3

On ₹48,000, the compound interest for 2 years exceeds the simple interest for 2 years by ₹120. What is the rate of interest per annum? (a) 0.25% (b) 5% (c) 2.5% (d) 10%

Answer. $\left(\frac{R}{100}\right)^2 = \frac{120}{48000} = \frac{1}{400}$, so $\frac{R}{100} = \frac{1}{20}$ and R = (b) 5%. Check: 48,000 ÷ 400 = 120.

Question 4

What is the difference between the compound interest and the simple interest on ₹16,000 for 3 years at 5% per annum? (a) ₹40 (b) ₹120 (c) ₹2,522 (d) ₹122

Answer. $D_3 = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right)$ = 16,000 × $\frac{1}{400}$ × 3.05 = 40 × 3.05 = (d) ₹122. Check: CI = 16,000 × 1.157625 − 16,000 = ₹2,522 and SI = ₹2,400. ₹120 is 3 × 40, which misses the interest on the extra interest.

Question 5

[Tier 2] At the same rate of interest, the difference between the compound interest and the simple interest on a sum is ₹200 for 2 years and ₹608 for 3 years. What is the sum? (a) ₹1,25,000 (b) ₹31,250 (c) ₹50,000 (d) ₹2,00,000

Answer. $\frac{D_3}{D_2} = 3 + \frac{R}{100}$, and $\frac{608}{200}$ = 3.04, so R = 4%. Then P = 200 × $\left(\frac{100}{4}\right)^2$ = 200 × 625 = (a) ₹1,25,000. Check: 1,25,000 × 0.0016 = 200 and 200 × 3.04 = 608.

Question 6

What is the difference between the compound interest, compounded half-yearly, and the simple interest on ₹64,000 for 1 year at 12.5% per annum? (a) ₹1,000 (b) ₹0 (c) ₹250 (d) ₹8,250

Answer. Half-yearly, the rate is 6.25% = $\frac{1}{16}$ per half-year for 2 half-years, so D = 64,000 × $\left(\frac{1}{16}\right)^2$ = 64,000 ÷ 256 = (c) ₹250. Check: 64,000 × $\frac{17}{16} \times \frac{17}{16}$ = 72,250, so the CI is ₹8,250 against an SI of ₹8,000.

Question 7

The simple interest on a sum for 2 years is ₹5,600, and the compound interest on the same sum at the same rate for 2 years is ₹5,824. What are the rate and the sum? (a) 4%, ₹70,000 (b) 8%, ₹35,000 (c) 16%, ₹17,500 (d) 8%, ₹70,000

Answer. One year's SI = ₹2,800. The extra ₹224 is interest on ₹2,800, so R = $\frac{224}{2800} \times 100$ = 8%, and P = 2,800 × $\frac{100}{8}$ = ₹35,000. So the answer is (b) 8%, ₹35,000. Check: 35,000 × 1.08 × 1.08 = 40,824, so the CI is ₹5,824. Dividing ₹224 by the 2-year SI of ₹5,600 gives the trap 4%.

Question 8

The simple interest on a sum at 12% per annum for 2 years is ₹4,320. What is the compound interest on the same sum at the same rate for 2 years? (a) ₹4,838.40 (b) ₹4,665.60 (c) ₹4,492.80 (d) ₹4,579.20

Answer. CI for 2 years = SI × $\left(1 + \frac{R}{200}\right)$ = 4,320 × 1.06 = (d) ₹4,579.20. The extra ₹259.20 is 12% of one year's SI of ₹2,160. Check: P = 4,320 ÷ 0.24 = ₹18,000, and 18,000 × 1.2544 − 18,000 = ₹4,579.20.

Question 9

The compound interest on a sum for 2 years at 5% per annum is ₹1,230. What is the simple interest on the same sum at the same rate for 2 years? (a) ₹1,200 (b) ₹1,171.43 (c) ₹1,168.50 (d) ₹1,215

Answer. SI = CI × $\frac{200}{200 + R}$ = 1,230 × $\frac{200}{205}$ = (a) ₹1,200. Check: the sum is 1,200 ÷ 0.1 = ₹12,000, and 12,000 × 1.1025 − 12,000 = ₹1,230.

Question 10

The simple interest on a sum for 3 years at 9% per annum is ₹6,480. What is the compound interest on the same sum for 2 years at 10% per annum? (a) ₹4,800 (b) ₹4,514.40 (c) ₹5,040 (d) ₹7,128

Answer. The rate and the time change, so find the sum first: P = 6,480 × $\frac{100}{27}$ = ₹24,000. CI at 10% for 2 years = 21% of 24,000 = (c) ₹5,040.

Question 11

Sanjay borrowed ₹75,000 at 8% per annum simple interest and lent the whole sum at 8% per annum compound interest, compounded annually. What is his gain after 3 years? (a) ₹1,440 (b) ₹1,478.40 (c) ₹480 (d) ₹960

Answer. His gain is the 3-year CI − SI on ₹75,000 at 8%. The 2-year difference is 75,000 × 0.0064 = ₹480, and $D_3$ = 480 × 3.08 = (b) ₹1,478.40. Check: CI = 75,000 × 1.259712 − 75,000 = ₹19,478.40 and SI = ₹18,000.

Question 12

Priya has ₹60,000 to invest for 3 years. Scheme X pays 10.5% per annum simple interest, and Scheme Y pays 10% per annum compounded annually. Which scheme gives more interest, and by how much? (a) Scheme X, by ₹960 (b) Both give the same (c) Scheme Y, by ₹300 (d) Scheme Y, by ₹960

Answer. X gives 10.5 × 3 = 31.5% = ₹18,900. Y gives 33.1% ($1.1^3$ = 1.331) = ₹19,860. So the answer is (d) Scheme Y, by ₹960. Over 2 years the two would be equal (21% each), which is why "Both give the same" is a trap; over 3 years CI pulls ahead.

Question 13

₹50,460 is divided into two parts. One part is lent for 2 years and the other for 4 years, both at 5% per annum compound interest, and the two amounts turn out equal. What is the smaller part? (a) ₹24,000 (b) ₹26,460 (c) ₹25,230 (d) ₹23,600

Answer. The gap is 2 years, so 2-year part : 4-year part = $1.05^2$ : 1 = 441 : 400. One part = 50,460 ÷ 841 = ₹60, so the parts are ₹26,460 and (a) ₹24,000. The smaller part is the one lent for 4 years. Check: 26,460 × 1.1025 = 29,172.15 and 24,000 × $1.05^4$ = 29,172.15.

Question 14

A man divides ₹24,400 between his sons, aged 15 and 17, so that each gets the same amount at the age of 21. The money earns 20% per annum compounded annually. What is the elder son's share? (a) ₹10,000 (b) ₹12,200 (c) ₹14,400 (d) ₹13,000

Answer. The younger son's money stays 6 years and the elder's 4 years, a gap of 2 years. So elder : younger = $\left(\frac{6}{5}\right)^2$ : 1 = 36 : 25. One part = 24,400 ÷ 61 = ₹400, so the elder son gets 36 × 400 = (c) ₹14,400 and the younger ₹10,000. Check: 14,400 × $1.2^4$ = 29,859.84 and 10,000 × $1.2^6$ = 29,859.84.

Question 15

A loan of ₹33,600 is to be repaid in 2 equal annual instalments at 10% per annum compound interest. What is each instalment? (a) ₹20,328 (b) ₹19,360 (c) ₹20,160 (d) ₹18,480

Answer. At 10%, the loan is 210 units when the instalment is 121 units. One unit = 33,600 ÷ 210 = ₹160, so the instalment = 121 × 160 = (b) ₹19,360. Run forward: 33,600 × 1.1 = 36,960; pay 19,360, leaving 17,600; 17,600 × 1.1 = 19,360, which the second instalment clears. Halving the CI amount (33,600 × 1.21 ÷ 2) gives ₹20,328, which is too much.

Question 16

A loan is cleared by three equal annual instalments of ₹55,566 each at 5% per annum compound interest. How much was the loan? (a) ₹1,66,698 (b) ₹1,44,000 (c) ₹1,58,760 (d) ₹1,51,320

Answer. Present values: 55,566 × $\frac{20}{21}$ = 52,920, 55,566 × $\left(\frac{20}{21}\right)^2$ = 50,400, and 55,566 × $\left(\frac{20}{21}\right)^3$ = 48,000. The loan is their total: 52,920 + 50,400 + 48,000 = (d) ₹1,51,320.

Question 17

A sum borrowed at 20% per annum compound interest is repaid in 2 equal annual instalments of ₹14,400 each. How much interest is paid in all? (a) ₹6,800 (b) ₹9,680 (c) ₹8,800 (d) ₹4,400

Answer. At 20%, the instalment is 36 units and the loan 55 units. One unit = 14,400 ÷ 36 = ₹400, so the loan = 55 × 400 = ₹22,000. Total paid = 2 × 14,400 = ₹28,800, so the interest is 28,800 − 22,000 = (a) ₹6,800. Run forward: 22,000 × 1.2 = 26,400 − 14,400 = 12,000; then 12,000 × 1.2 = 14,400.

Question 18

A loan of ₹66,200 at 10% per annum compound interest is to be repaid in 3 equal annual instalments. How much is still owed right after the second instalment is paid? (a) ₹26,620 (b) ₹46,200 (c) ₹24,200 (d) ₹22,000

Answer. At 10%, the loan is 3,310 units for an instalment of 1,331 units, so one unit = 66,200 ÷ 3,310 = ₹20 and the instalment is 1,331 × 20 = ₹26,620. Run forward: 66,200 × 1.1 = 72,820 − 26,620 = 46,200; then 46,200 × 1.1 = 50,820 − 26,620 = (c) ₹24,200. That balance grows to ₹26,620 in the third year, which the last instalment clears. ₹46,200 is what is owed after the first instalment, not the second.

Question 19

Leela borrowed ₹40,000 at 15% per annum compound interest. She paid ₹16,000 at the end of the first year and ₹18,000 at the end of the second year. How much must she pay at the end of the third year to clear the loan? (a) ₹18,975 (b) ₹26,835 (c) ₹16,500 (d) ₹6,900

Answer. Run it forward. Year 1: 40,000 × 1.15 = 46,000 − 16,000 = ₹30,000. Year 2: 30,000 × 1.15 = 34,500 − 18,000 = ₹16,500. Year 3: 16,500 × 1.15 = (a) ₹18,975. Taking both payments off the full 3-year amount (60,835 − 34,000) gives ₹26,835, which keeps charging interest on money already paid back.

Question 20

Ankit starts a business with ₹2,00,000. Every year it earns a profit of 15% on its capital, and he puts all the profit back into the business. What is the capital after 3 years? (a) ₹2,90,000 (b) ₹3,04,175 (c) ₹2,64,500 (d) ₹3,00,000

Answer. Putting the profit back every year is compounding: 2,00,000 × $1.15^3$ = 2,00,000 × 1.520875 = (b) ₹3,04,175. Year by year: ₹2,30,000, then ₹2,64,500, then ₹3,04,175.

Question 21

[Tier 2] On a certain sum, the compound interest for 2 years at 8% for the first year and 12% for the second year is ₹288 more than the simple interest at the same rates. What is the sum? (a) ₹45,000 (b) ₹20,000 (c) ₹30,000 (d) ₹28,800

Answer. In the second year, the first year's interest earns 12% more interest, so D = P × $\frac{8}{100} \times \frac{12}{100}$ = $\frac{96P}{10000}$. So P = 288 × $\frac{10000}{96}$ = (c) ₹30,000. Check: CI = 30,000 × 1.08 × 1.12 − 30,000 = ₹6,288, and SI = 20% of 30,000 = ₹6,000.