Shortcuts & Speed Methods

SI as R% of P per Year

Interest questions are percentage questions with time added, and in Tier 1 you have about 36 seconds for each. Every method below says when it applies, works one short example and ends with the cases where it goes wrong. Several of those cases are an SI shortcut used on a CI question, or the reverse, so first settle which kind of interest the question is about.

One year's interest first

When to use it: any simple interest question where you would normally write $\text{SI} = \frac{P \times R \times T}{100}$.

Fast way: under SI the interest is the same every year, and it is just R% of P. Find one year's interest, then multiply by the number of years.

For example: ₹7,400 at 9% for 5 years. One year's interest is 9% of 7,400 = ₹666, so five years give 666 × 5 = ₹3,330.

The total % trick

Over T years, the SI is (R × T)% of P. So 6% for 7 years is 42% of the sum. If that interest is ₹2,268, then 42% of P = 2,268 and P = 2,268 × $\frac{100}{42}$ = ₹5,400. One line, no formula.

The same idea handles 'the interest is a fraction of the sum'. If the SI is $\frac{2}{5}$ of the sum in 8 years, then R × 8 = 40 and R = 5%. In general, SI = $\frac{1}{n}$ of P means $R \times T = \frac{100}{n}$.

Months and days

• A rate per month is 12 times bigger per year: 1.5% a month is 18% a year. ₹6,200 for 10 months at 1.5% a month is 15% of 6,200 = ₹930.
• For days, use a 365-day year and look for a neat fraction: 73 days is $\frac{1}{5}$ of a year, 146 days is $\frac{2}{5}$ and 219 days is $\frac{3}{5}$. For ₹12,500 at 9% for 219 days, one year's interest is ₹1,125, so the answer is 1,125 × $\frac{3}{5}$ = ₹675.

Where it fails:

• Under compound interest the yearly interest grows, so one year's interest × T works only for SI.
• Reading 1.5% a month as 1.5% a year. Check the unit of the rate before anything else.
• 'The interest is $\frac{2}{5}$ of the amount' is a different question. Then SI : P = 2 : 3, so R × T = $66\frac{2}{3}$, not 40.
• When you count days between two dates, count the day the money is lent or the day it comes back, not both.

[SSC Tip] If R × T lands on a fraction from the percentage table (12.5% = $\frac{1}{8}$, 20% = $\frac{1}{5}$, 25% = $\frac{1}{4}$), use the fraction. 5% for 4 years is $\frac{1}{5}$ of P, so on ₹13,750 the interest is ₹2,750 at a glance.

Amounts After Two Different Times

Gap ÷ years between

When to use it: a sum at SI amounts to one figure after $t_1$ years and another after $t_2$ years, and the sum or the rate is asked.

Fast way: the gap between the two amounts is pure interest for the years in between. So

$$\text{one year's SI} = \frac{A_2 - A_1}{t_2 - t_1}$$

Take $t_1$ years of that off $A_1$ to get P, and then R = one year's SI ÷ P × 100.

For example: a sum amounts to ₹8,432 in 3 years and ₹9,520 in 5 years. One year's SI = 1,088 ÷ 2 = ₹544. P = 8,432 − 3 × 544 = 8,432 − 1,632 = ₹6,800, and R = $\frac{544}{6800} \times 100$ = 8%.

If only P is asked, one line does it:

$$P = \frac{t_2 A_1 - t_1 A_2}{t_2 - t_1}$$

Here (5 × 8,432 − 3 × 9,520) ÷ 2 = 13,600 ÷ 2 = 6,800.

The rate goes up

'Had the rate been 3% higher, the sum would have earned ₹486 more in 4 years.' The extra 3% for 4 years is 12% of P, so P = 486 × $\frac{100}{12}$ = ₹4,050. You never need the original rate.

The rate changes on the way

Add the percentages period by period. At 6% for the first 2 years, 8% for the next 3 and 10% after that, 7 years give 12 + 24 + 20 = 56% of P. If the interest is ₹4,592, P = 4,592 × $\frac{100}{56}$ = ₹8,200.

Where it fails:

• Dividing the gap by $t_2$ instead of $t_2 - t_1$. In the example, 1,088 ÷ 5 = ₹217.60 is not the yearly interest.
• Under CI the yearly steps grow, so the gap is not a flat yearly interest. CI amounts use the ratio method, which comes later in this section.
• If the question gives two interests instead of two amounts, each one ÷ its own years is already the yearly SI. Don't subtract anything.

[SSC Tip] Under SI the amounts after 1, 2, 3 and more years go up by equal steps. If you know the amounts after 2 and 6 years, the amount after 4 years is just their average.

A Sum Becoming n Times Itself

When to use it: a sum doubles, triples or becomes $\frac{8}{5}$ of itself in T years at SI, and the rate, the time, or the time for another multiple is asked.

Fast way: becoming n times means the interest is (n − 1) times the sum. So

$$R \times T = 100(n - 1) \qquad R = \frac{100(n - 1)}{T}$$

For example: a sum triples in 16 years. The interest is 2P, so R = $\frac{200}{16}$ = 12.5%.

Fractions work the same way. A sum that becomes $\frac{8}{5}$ of itself in 7.5 years has earned $\frac{3}{5}$ of itself, so R × 7.5 = 60 and R = 8%.

Some papers write 'increases by 200%' instead of 'becomes 3 times'. They mean the same thing: the interest is 2P.

Scaling to another multiple

At the same rate, the time grows with n − 1, not with n:

$$\frac{T_2}{T_1} = \frac{n_2 - 1}{n_1 - 1}$$

For example: the sum that triples in 16 years becomes 5 times itself in 16 × $\frac{4}{2}$ = 32 years. A sum that doubles in 9 years becomes 7 times itself in 9 × 6 = 54 years.

Where it fails:

• Using n instead of n − 1. For 'triples in 16 years', 300 ÷ 16 gives 18.75%, a common wrong option.
• Scaling by n. If a sum triples in 16 years, 6 times needs an interest of 5P, so it takes 16 × $\frac{5}{2}$ = 40 years, not 32.
• Compound interest scales by powers, not by n − 1. A sum that doubles in 5 years becomes 4 times itself in 15 years at SI but in 10 years at CI.

For doubling, keep R × T = 100 in your head: doubling in 8 years means 12.5%, at 6.25% it takes 16 years, and at 8% it takes 12.5 years.

Splitting a Sum Under SI

Two rates and a total interest

When to use it: a sum is lent in two parts at two rates, the total interest is given, and one part is asked.

Fast way: turn the total interest into one average yearly rate, then use the alligation cross (Chapter 6) on the rates:

lower-rate part : higher-rate part = (higher rate − average) : (average − lower rate)

For example: ₹32,000 is lent partly at 5% and the rest at 9%, and the total interest in 4 years is ₹8,320. The yearly interest is ₹2,080, so the average rate is $\frac{2080}{32000} \times 100$ = 6.5%. Parts = (9 − 6.5) : (6.5 − 5) = 2.5 : 1.5 = 5 : 3, so ₹20,000 at 5% and ₹12,000 at 9%. Check: 1,000 + 1,080 = 2,080.

Parts that earn equal interest

If $P_1 R_1 T_1 = P_2 R_2 T_2$, each part goes as $\frac{1}{R \times T}$.

For example: split ₹11,700 so that one part at 5% for 4 years earns the same as the other at 6% for 5 years. $P_1 : P_2 = \frac{1}{20} : \frac{1}{30}$ = 3 : 2, so the parts are ₹7,020 and ₹4,680. Each earns ₹1,404.

Parts that reach equal amounts

Now each part grows to $\frac{P(100 + RT)}{100}$, so the parts go as $\frac{1}{100 + RT}$.

One more: ₹18,600 is split between A, aged 10, and B, aged 14, so that each gets the same amount at 18, at 4% SI. A's money runs 8 years (100 + 32 = 132) and B's runs 4 years (100 + 16 = 116). A : B = $\frac{1}{132} : \frac{1}{116}$ = 116 : 132 = 29 : 33, so A gets ₹8,700 and B gets ₹9,900. Both grow to ₹11,484.

Where it fails:

• Flipping the alligation. The part at the lower rate gets (higher rate − average). The flip gives 3 : 5 here.
• Using $\frac{1}{RT}$ when the amounts (not the interests) are equal. For A and B that gives $\frac{1}{32} : \frac{1}{16}$ = 1 : 2, which is wrong.
• Forgetting to make the average yearly. ₹8,320 is 26% of ₹32,000, but that is 4 years' worth.

As a check, the average rate always sits between the two rates, closer to the rate on the bigger part. In an equal-amount split, the younger child (whose money runs longer) gets the smaller share.

CI as Successive Percentage Increases

When to use it: CI for 2 or 3 years at one rate, or at different rates in different years.

Fast way: compound interest is one percentage increase after another. Two increases of R% make one increase of

$$2R + \frac{R^2}{100} \text{ percent}$$

Three years at R% make $3R + \frac{3R^2}{100} + \frac{R^3}{100^2}$ percent. With different rates, $R_1\%$ and then $R_2\%$ make $R_1 + R_2 + \frac{R_1 R_2}{100}$ percent.

For example: CI on ₹7,500 at 8% for 2 years. 16 + 0.64 = 16.64%, and 16.64% of 7,500 = ₹1,248.

Worth knowing by heart:

Another one: CI on ₹12,000 at 5% for 3 years is 15.7625% of 12,000 = ₹1,891.50.

Different rates: 10% in the first year and 20% in the second give 10 + 20 + 2 = 32%. On ₹8,000 that is a CI of ₹2,560. The order of the rates doesn't change the answer.

Where it fails:

• Stopping at 2R. That is the SI. The $\frac{R^2}{100}$ part is the interest on the first year's interest, which is what makes it CI.
• Writing $3R + \frac{R^2}{100}$ for three years. If you don't remember the 3-year form, combine in two steps: at 10%, two years give 21%, and one more 10% gives 21 + 10 + 2.1 = 33.1%.
• Using the combined % on anything other than the starting sum.

For 2 years, then, CI is the SI plus $\frac{R^2}{100}$% of P. Once you know the SI, the CI is a small step away, and options equal to the SI or the amount are traps.

Fraction Multipliers and the 10% and 20% Tables

When to use it: CI or an amount over 2 to 4 years when the rate has a neat fraction.

Fast way: each year multiplies the amount by $1 + \frac{R}{100}$. Write it as a fraction and pick P so the denominators cancel.

For example: ₹7,680 at 12.5% for 3 years. The amount is 7,680 × $\frac{729}{512}$ = 15 × 729 = ₹10,935, so CI = ₹3,255. 7,680 is 15 × 512, which is why it cancels so neatly.

The 10% and 20% tables (on ₹1,000)

Each year's interest is the previous year's interest × 1.1 (or × 1.2).

For example: CI on ₹3,750 at 20% for 3 years is 72.8% of 3,750 = ₹2,730. The interest for the 3rd year alone on ₹5,000 at 20% is 28.8% of 5,000 = ₹1,440.

Where it fails:

• The fraction helps only when P cancels. ₹10,000 at 12.5% for 3 years gives about ₹14,238.28, so use decimals there.
• Mixing up 'interest for the 3rd year' with 'CI for 3 years'. At 20% they are 28.8% and 72.8% of P, and both sit in the options.
• The table is per ₹1,000. Scale it: on ₹6,000 at 10%, the 2nd year's interest is 6 × 110 = ₹660.

Under CI each year's interest is $1 + \frac{R}{100}$ times the last one: 200, 240, 288 at 20%. If two back-to-back years' interests are given, their ratio gives the rate at once.

Half-Yearly and Quarterly Compounding

When to use it: the interest is compounded half-yearly or quarterly.

Fast way: cut the rate and multiply the periods.

• Half-yearly: rate ÷ 2, and periods = years × 2.
• Quarterly: rate ÷ 4, and periods = years × 4.

Then use the same multipliers and tables as before.

For example: ₹28,000 at 10% a year, compounded half-yearly, for 1.5 years. That is 5% for 3 half-years: 28,000 × $\left(\frac{21}{20}\right)^3$ = 28,000 × $\frac{9261}{8000}$ = ₹32,413.50, so CI = ₹4,413.50.

A quarterly one: ₹40,000 at 8% a year, compounded quarterly, for 6 months. That is 2% for 2 quarters: 2 + 2 + 0.04 = 4.04% of 40,000 = ₹1,616.

Effective yearly rate

10% compounded half-yearly is 5% twice, which is 10.25% for the year. 8% compounded quarterly is 2% four times, about 8.24% for the year. For half-yearly compounding at 2r% a year, the effective rate is $2r + \frac{r^2}{100}$ percent.

Not the same: compounded yearly for 1.5 years

Here the whole year is compounded and the half-year is simple interest on the new amount. ₹28,000 at 10%: 28,000 × 1.1 = 30,800, then 5% more for the half-year gives 30,800 × 1.05 = ₹32,340. The CI is ₹4,340, which is ₹73.50 less than with half-yearly compounding.

Where it fails:

• Halving the rate but not doubling the periods, or the other way round. 10% for 3 periods on ₹28,000 gives ₹37,268, the trap.
• If the question already says '4% per half-year', don't halve it again.
• For one year, CI equals SI only with yearly compounding. Half-yearly compounding beats SI even in the first year.

[SSC Tip] The same rate compounded more often gives more interest, but only a little more. If a quarterly answer is far above the yearly one, it is a trap.

Rate from Two Amounts, and Powers of the Growth Factor

Back-to-back amounts

When to use it: a sum at CI amounts to one figure after n years and another after n + 1 years.

Fast way: the gap is one year's interest on the earlier amount. So

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

and P is the earlier amount divided by the multiplier n times.

For example: ₹19,600 after 2 years and ₹21,952 after 3 years. The gap is ₹2,352, and $\frac{2352}{19600}$ = 0.12, so R = 12%. 12% is $\frac{28}{25}$, so P = 19,600 × $\frac{625}{784}$ = ₹15,625.

The same works for back-to-back years' interests. If the 2nd year's interest is ₹1,650 and the 3rd year's is ₹1,782, R = $\frac{132}{1650} \times 100$ = 8%.

Two years apart: ₹6,050 after 2 years and ₹7,320.50 after 4 years. The ratio is 1.21 = $1.1^2$, so R = 10% and P = 6,050 ÷ 1.21 = ₹5,000.

Doubling, tripling and powers

Fast way: if a sum becomes n times itself in T years at CI, it becomes $n^2$ times in 2T years, $n^3$ times in 3T years, and so on.

For example: a sum doubles in 6 years, so it becomes 8 times ($2^3$) in 18 years and 16 times in 24 years. A sum that becomes 9 times in 10 years triples in 5 years and becomes 27 times in 15 years. A sum that becomes 8 times in 15 years doubles in 5 and becomes 4 times in 10.

Where it fails:

• Dividing the gap by the later amount. $\frac{2352}{21952}$ is about 10.7%, a near-miss option.
• Two years apart means the ratio is $\left(1 + \frac{R}{100}\right)^2$, not $1 + \frac{2R}{100}$.
• Mixing up SI and CI. At SI, doubling in 6 years means 8 times in 42 years (7 lots of the interest), not 18.

Spot the power first. 4, 8, 16, 32 and 64 are powers of 2, and 9, 27 and 81 are powers of 3. Write the new multiple as a power of the given one and multiply the time by that power.

CI Minus SI

When to use it: the difference between CI and SI on the same sum at the same rate for the same time, or the sum or rate from that difference.

Fast way (2 years): in year 1, CI and SI are equal. In year 2, CI also earns interest on year 1's interest. So

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

For example: ₹22,500 at 8% for 2 years. The difference is 22,500 × $\frac{64}{10000}$ = ₹144.

Finding P: P = difference × $\left(\frac{100}{R}\right)^2$. A difference of ₹45 at 5% gives P = 45 × 400 = ₹18,000. Handy multipliers: 4% → × 625, 5% → × 400, 10% → × 100, 20% → × 25.

Finding R: a difference of ₹63 on ₹17,500 gives $R^2 = \frac{63 \times 10000}{17500}$ = 36, so R = 6%.

Three years

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

Quick values: the 3-year difference is 0.7625% of P at 5%, 3.1% at 10% and 12.8% at 20%. On ₹36,000 at 5% that is ₹274.50.

Two differences give R: the 3-year difference ÷ the 2-year difference = $3 + \frac{R}{100}$. If they are ₹72 and ₹223.20, the ratio is 3.1, so R = 10% and P = 72 × 100 = ₹7,200.

Where it fails:

• Tripling the 2-year difference for 3 years. At 10% the 3-year difference is 3.1 times the 2-year one, not 3 times.
• Writing $\frac{P R}{100^2}$ with R instead of $R^2$. Square the rate.
• With half-yearly compounding, use the half-year rate and count half-years. For 1 year at 10% compounded half-yearly, the difference is P × $\left(\frac{5}{100}\right)^2$.

[SSC Tip] The 2-year difference is tiny next to P: 0.25% of P at 5% and 1% at 10%. If an option for the difference looks like a full year's interest, it is the trap.

SI and CI on the Same Sum

When to use it: the SI and the CI on one sum for 2 years at one rate are both given, or one is given and the other is asked.

Fast way: one year's SI is half the 2-year SI. The gap CI − SI is R% of that one year's interest. So

$$R = \frac{2(\text{CI} - \text{SI})}{\text{SI}} \times 100 \qquad P = \frac{\text{SI}^2}{4(\text{CI} - \text{SI})}$$

For example: for 2 years the SI is ₹6,600 and the CI is ₹6,996. One year's SI = ₹3,300, and the gap of ₹396 is 12% of 3,300, so R = 12%. P = 3,300 ÷ 0.12 = ₹27,500. The formula agrees: $\frac{6600^2}{4 \times 396}$ = 27,500.

From SI to CI

SI for 2 years at 8% is ₹1,440. One year's SI is ₹720, so CI = 1,440 + 8% of 720 = 1,440 + 57.60 = ₹1,497.60.

From CI to SI

CI for 2 years at 10% is ₹1,680. That is 21% of P, and the SI is 20% of P, so SI = 1,680 × $\frac{20}{21}$ = ₹1,600.

Different times

SI for 3 years is ₹2,520 and CI for 2 years is ₹1,722 at the same rate. One year's SI is ₹840, so 2 years' SI is ₹1,680. The extra ₹42 is R% of 840, so R = 5% and P = 840 ÷ 0.05 = ₹16,800.

Where it fails:

• Dividing the gap by the whole SI. $\frac{396}{6600}$ = 6% is exactly half the true rate, and it is nearly always among the options.
• Comparing CI for 2 years with SI for some other time before finding one year's SI.
• For 3 years the gap is not R% of one year's SI. Use the 3-year difference formula instead.

[SSC Tip] For 2 years, CI ÷ SI = $1 + \frac{R}{200}$. A ratio of 1.06 means R = 12%, with no other working.

Equal Amounts and Instalments

Equal amounts under CI

When to use it: a sum is split into parts lent for different times at the same CI rate, and the parts grow to equal amounts.

Fast way: the part lent for fewer years must be bigger by the growth factor for the extra years. If the times differ by k years, shorter part : longer part = $\left(1 + \frac{R}{100}\right)^k$ : 1.

For example: ₹41,000 is split between two sisters aged 17 and 15 so that each gets the same amount at 18, at 25% CI. The times are 1 and 3 years, 2 years apart. Elder : younger = $\left(\frac{5}{4}\right)^2$ : 1 = 25 : 16, which is 41 parts of ₹1,000. So ₹25,000 and ₹16,000. Both grow to ₹31,250.

Equal instalments under CI: the ratio method

An instalment paid later clears less of today's loan. At 10%, call the instalment 121 units. Paid after 1 year it clears 110 units of loan, and paid after 2 years it clears 100. So the loan is 210 units and instalment : loan = 121 : 210. For three instalments, call the instalment 1,331 units: it clears 1,210 + 1,100 + 1,000 = 3,310 units.

For example: ₹8,400 at 10% CI is repaid in two equal yearly instalments. 210 units = 8,400, so 1 unit = 40 and the instalment = 121 × 40 = ₹4,840. Check forward: 8,400 grows to 9,240, less 4,840 leaves 4,400, which grows to exactly 4,840.

At 20% the units are 36 for the instalment against 30 + 25 = 55 for the loan. For three instalments it is 216 against 180 + 150 + 125 = 455.

Equal instalments under SI

For a debt of ₹A due after n years at R% SI, paid in n equal yearly instalments of x:

$$x = \frac{100A}{100n + \frac{R n(n - 1)}{2}}$$

For example: ₹9,460 due in 4 years at 5%. The bottom is 400 + 5 × 6 = 430, so x = $\frac{946000}{430}$ = ₹2,200. If the question gives the sum borrowed today instead, add its SI for n years first to get A.

Where it fails:

• Splitting the final amount evenly. 8,400 × 1.21 ÷ 2 = ₹5,082 is the classic wrong option.
• Under SI each instalment earns interest only for the years left after it is paid, not for the full term.
• In an equal-amount split, giving the bigger share to the part that runs longer.

Run any CI instalment answer forward once as a check. It takes about 10 seconds.

Option Elimination

When to use it: before the full working, or when you are short of time.

Fast way: run a few quick checks on the four options.

• CI beats SI after year 1. For 1 year with yearly compounding they are equal, and for 2 years CI is above SI by only $\frac{R^2}{100}$% of P.
• The amount is more than the principal, and the interest is less than the amount. An option that mixes the two up is out.
• Rates make neat factors. For 2 years, amount ÷ P should be the square of a simple fraction. ₹3,600 growing to ₹4,356 gives $\frac{4356}{3600} = \frac{121}{100} = \left(\frac{11}{10}\right)^2$, so R = 10%.
• SI grows in a straight line. Twice the time at SI means twice the interest. At CI it is more than twice.

For example: the CI on ₹19,000 at 10% for 2 years is (a) ₹3,800 (b) ₹3,990 (c) ₹22,990 (d) ₹3,819.

(c) is the amount, so it is out. (a) is the SI, and CI must be more. The extra is 10% of one year's ₹1,900, which is ₹190, so the answer is (b) ₹3,990. (d) adds 1% of ₹1,900 instead of 1% of ₹19,000.

Another one: the CI on a sum for 2 years at 5% is ₹1,435. The SI for the same time and rate is (a) ₹1,470 (b) ₹1,435 (c) ₹1,400 (d) ₹1,330.

The SI must be less than the CI, so (a) and (b) are out. The gap is 5% of one year's SI, which is about 5% of 700 = ₹35, so the answer is (c) ₹1,400. Check: CI is 10.25% of P and SI is 10%, so SI = 1,435 × $\frac{10}{10.25}$ = 1,400.

Where it fails:

• Half-yearly compounding makes CI bigger than SI even in the first year.
• When two options both pass the checks, do the working.

[SSC Tip] Trap options come from a handful of slips: the SI given as the CI, the amount given as the interest, 2R without the $\frac{R^2}{100}$, the rate halved without doubling the periods, and (CI − SI) divided by the whole SI. When your answer equals one of these figures, redo that one step.

Solved Examples

Every answer below uses the quickest route and gives a rough time.

Question 1

Find the simple interest on ₹8,400 at 7.5% a year for 3 years 4 months.

Answer. One year's interest is 7.5% of 8,400 = ₹630. 3 years 4 months is $3\frac{1}{3}$ years, so the SI is 630 × $\frac{10}{3}$ = ₹2,100. Don't read 3 years 4 months as 3.4 years; that gives ₹2,142.

Time check: about 12 seconds.

Question 2

A sum at simple interest amounts to ₹10,384 in 2 years and to ₹12,760 in 5 years. Find the rate.

Answer. The gap of ₹2,376 is 3 years' interest, so one year's SI is ₹792. The sum is 10,384 − 2 × 792 = ₹8,800, and the rate is $\frac{792}{8800} \times 100$ = 9%.

Time check: about 20 seconds.

Question 3

At simple interest a sum becomes 2.5 times itself in 12 years. In how many years will it become 4 times itself?

Answer. Becoming 2.5 times means 1.5 times the sum as interest in 12 years. Becoming 4 times needs 3 times the sum as interest, which is double that, so it takes 24 years. Scaling by 4 ÷ 2.5 gives 19.2 years, which is wrong. (The rate is 12.5%, but you don't need it.)

Time check: about 8 seconds.

Question 4

₹36,000 is lent in two parts, one at 8% and the other at 12.5% simple interest. The total interest after 2 years is ₹6,840. How much was lent at 8%?

Answer. The yearly interest is ₹3,420, so the average rate is $\frac{3420}{36000} \times 100$ = 9.5%. Alligation: 8% part : 12.5% part = (12.5 − 9.5) : (9.5 − 8) = 3 : 1.5 = 2 : 1. So ₹24,000 was lent at 8%. Check: 1,920 + 1,500 = 3,420.

Time check: about 25 seconds.

Question 5

₹19,500 is divided between two sisters aged 14 and 16 so that, at 6% simple interest, each gets the same amount when she turns 20. Find the younger sister's share.

Answer. The younger sister's money runs 6 years (100 + 36 = 136) and the elder's runs 4 years (100 + 24 = 124). Shares go as $\frac{1}{136} : \frac{1}{124}$ = 124 : 136 = 31 : 34. That is 65 parts of ₹300, so the younger sister gets ₹9,300. Check: 9,300 × 1.36 = 12,648 = 10,200 × 1.24.

Time check: about 25 seconds.

Question 6

Find the compound interest on ₹6,250 at 16% a year for 2 years, compounded yearly.

Answer. Two years at 16% is 32 + 2.56 = 34.56%, and 34.56% of 6,250 = ₹2,160.

Time check: about 12 seconds.

Question 7

Find the compound interest on ₹20,480 for 2 years at 6.25% a year.

Answer. 6.25% is $\frac{1}{16}$, so the multiplier is $\frac{17}{16}$. The amount is 20,480 × $\frac{289}{256}$ = 80 × 289 = ₹23,120, so the CI is ₹2,640.

Time check: about 12 seconds.

Question 8

Find the compound interest on ₹31,250 for 9 months at 16% a year, compounded quarterly.

Answer. Quarterly means 4% a quarter, and 9 months is 3 quarters. 4% is $\frac{26}{25}$, so the amount is 31,250 × $\frac{17576}{15625}$ = 2 × 17,576 = ₹35,152. The CI is ₹3,902. Make sure you use 4%, not 16%, for each quarter.

Time check: about 20 seconds.

Question 9

A sum at compound interest amounts to ₹10,580 in 2 years and ₹12,167 in 3 years. Find the rate and the sum.

Answer. The gap is ₹1,587, and $\frac{1587}{10580}$ = 0.15, so R = 15%. 15% is $\frac{23}{20}$, so P = 10,580 × $\frac{400}{529}$ = ₹8,000.

Time check: about 20 seconds.

Question 10

A sum at compound interest becomes 3 times itself in 4 years. In how many years will it become 81 times itself?

Answer. 81 = $3^4$, so it takes 4 × 4 = 16 years.

Time check: about 6 seconds.

Question 11

The difference between the compound interest and the simple interest on a sum for 2 years at 6% is ₹72. Find the sum.

Answer. P = 72 × $\left(\frac{100}{6}\right)^2$ = 72 × $\frac{10000}{36}$ = ₹20,000. Check: 0.36% of 20,000 is 72.

Time check: about 10 seconds.

Question 12

The difference between CI and SI on a sum for 3 years at 10% is ₹186. Find the sum.

Answer. At 10% the 3-year difference is 3.1% of P, so P = 186 ÷ 0.031 = ₹6,000. Using the 2-year form (1% of P) by mistake gives ₹18,600.

Time check: about 10 seconds.

Question 13

On a certain sum, the SI for 2 years is ₹1,960 and the CI for 2 years at the same rate is ₹2,038.40. Find the rate and the sum.

Answer. One year's SI is ₹980. The gap of ₹78.40 is R% of 980, so R = $\frac{78.40}{980} \times 100$ = 8%, and P = 980 ÷ 0.08 = ₹12,250. Dividing 78.40 by the full 1,960 gives 4%, half the right rate.

Time check: about 20 seconds.

Question 14

A loan is repaid in two equal yearly instalments of ₹7,056 each at 5% compound interest. Find the loan.

Answer. 5% is $\frac{21}{20}$. Call the instalment 441 units. It clears 420 units of loan after 1 year and 400 units after 2 years, so the loan is 820 units. 441 units = ₹7,056, so 1 unit = ₹16 and the loan is ₹13,120. Check: 13,120 grows to 13,776, less 7,056 leaves 6,720, which grows to exactly 7,056.

Time check: about 20 seconds.

Question 15

What equal yearly instalment will clear a debt of ₹13,800 due in 4 years at 10% simple interest?

Answer. x = $\frac{100 \times 13800}{400 + \frac{10 \times 4 \times 3}{2}}$ = $\frac{1380000}{460}$ = ₹3,000. Check: the four instalments earn 30%, 20%, 10% and 0% till the end, so 3,000 × 4.6 = 13,800.

Time check: about 20 seconds.