Quick Revision

Simple Interest: P, R and T

$\text{SI} = \frac{P \times R \times T}{100}$ and amount = P + SI. Any one of P, R and T comes from the rest: $R = \frac{100 \times \text{SI}}{P \times T}$ , and P and T the same way.
₹4,800 at 6.25% for 3 years: ₹300 a year, so SI = ₹900.
Months: divide by 12. Days: divide by 365, counting the last day but not the first.
Rate per month × 12 = rate per annum.
Interest = $\frac{1}{n}$ of the sum means $R \times T = \frac{100}{n}$ .

Amounts after two times: one year's interest = difference ÷ gap in years, and the sum = earlier amount − its interest.
₹6,840 after 3 years and ₹7,560 after 5: ₹360 a year, sum = ₹5,760, rate = 6.25%.
n times in T years: the interest is (n − 1) times the sum, so $R = \frac{100(n - 1)}{T}$ .
Time for m times = $T \times \frac{m - 1}{n - 1}$ . Doubles in 8 years: 4 times in 24.
Trap: scale the interest, not the amount.

Changing rates: add rate × years for each stretch. 5% for 2 years, then 8% for 3: 10 + 24 = 34%.
Two parts at two rates with a known total interest: put all at the lower rate; extra interest ÷ rate gap = the part at the higher rate.
Equal interest: parts in the ratio $\frac{1}{R_1 T_1} : \frac{1}{R_2 T_2}$ .
Equal amounts: ratio $\frac{1}{100 + R_1 T_1} : \frac{1}{100 + R_2 T_2}$ .
Borrow low, lend high: yearly gain = sum × rate gap.

Part repaid midway: amount so far − payment = the new principal, and interest runs on that.
Equal annual instalments x that clear a debt A due in n years: each instalment earns SI until the due date, so $x = \frac{100A}{100n + \frac{R n (n - 1)}{2}}$ .
A debt of ₹6,615 due in 3 years at 5%: $x = \frac{661500}{315}$ = ₹2,100.
For a loan P taken now, first find $A = P\left(1 + \frac{nR}{100}\right)$ .
Trap: dividing the debt by n forgets that early payments earn interest.

$A = P\left(1 + \frac{R}{100}\right)^n$ and CI = A − P. Each year's interest is on the amount at the start of that year.
Turn the rate into a fraction: 5% is $\frac{21}{20}$ , 6.25% is $\frac{17}{16}$ , 8% is $\frac{27}{25}$ , 10% is $\frac{11}{10}$ , 16% is $\frac{29}{25}$ and 20% is $\frac{6}{5}$ .
₹5,120 at 6.25% for 2 years: $5120 \times \frac{289}{256}$ = ₹5,780, so CI = ₹660.
Choose numbers the denominator divides and the working stays in whole rupees.
For one year, CI = SI.

2 years at R%: $2R + \frac{R^2}{100}$ percent. 10% gives 21%, 6% gives 12.36% and 30% gives 69%.
3 years at R%: $3R + \frac{3R^2}{100} + \frac{R^3}{10000}$ percent. 10% gives 33.1% and 30% gives 90 + 27 + 2.7 = 119.7%.
₹9,000 at 6% for 2 years: 12.36% of 9,000 = ₹1,112.40.
Two different rates a% and b%: $a + b + \frac{ab}{100}$ percent.
Going back: the sum = CI ÷ the combined percent.

Half-yearly: half the rate, twice the periods. Quarterly: a quarter of the rate, four times the periods.
₹10,000 at 12% compounded half-yearly for 1 year: $10000 \times 1.06^2$ = ₹11,236, so CI = ₹1,236 (yearly gives ₹1,200).
Effective yearly rate: 12% half-yearly is 12.36%, and 8% quarterly is about 8.24%.
9 months quarterly is 3 periods; $1\frac{1}{2}$ years half-yearly is also 3 periods.
Trap: using the full yearly rate for each half-year.

A different rate each year: multiply the multipliers. 8% and then 12.5%: 1.08 × 1.125 = 1.215, so 21.5% in 2 years.
A fraction of a year: compound the whole years, then add SI for the fraction on the last amount.
₹12,500 at 8% for 2 years 3 months: 12,500 → 13,500 → 14,580, then 2% of 14,580 = ₹291.60, so the amount is ₹14,871.60.
Trap: working out the fraction on the original sum.
Growth and depreciation work the same way, with a minus sign for depreciation.

Amounts after consecutive years: rate = $\frac{\text{later} - \text{earlier}}{\text{earlier}} \times 100$ . ₹9,000 and ₹9,720 give 8%.
Amounts after n and m years: $\left(1 + \frac{R}{100}\right)^{m - n} = \frac{A_m}{A_n}$ .
Then the sum = the earlier amount ÷ the multiplier, once for each year back.
Rate from CI: (sum + CI) ÷ sum is a perfect square for 2 years or a cube for 3. 1.1236 = $1.06^2$ gives 6%.
Trap: dividing by the later amount.

At CI money multiplies. If it becomes k times in t years, it becomes $k^2$ times in 2t years and $k^3$ times in 3t years.
Triples in 9 years: 9 times in 18 and 27 times in 27.
16 times in 40 years means 2 times in 10, so 8 times takes 30 years.
At SI the time grows in a straight line: doubling in 9 years means 4 times in 27 years.
Trap: using the SI rule in a CI question.

2 years: CI − SI = $\frac{P R^2}{100^2}$ , the interest on the first year's interest.
3 years: CI − SI = $\frac{P R^2 (300 + R)}{100^3}$ .
₹4,000 at 15%: ₹90 for 2 years, and $\frac{4000 \times 225 \times 315}{1000000}$ = ₹283.50 for 3 years.
Sum from the 2-year difference: $P = \text{difference} \times \left(\frac{100}{R}\right)^2$ .
Ratio of the 3-year difference to the 2-year difference = $\frac{300 + R}{100}$ .

2-year CI and SI given: $R = \frac{2(\text{CI} - \text{SI})}{\text{SI}} \times 100$ , and then the sum = one year's SI ÷ R%.
CI ₹2,163.20 and SI ₹2,080: one year's SI is ₹1,040, the extra ₹83.20 is 8% of it, so R = 8% and the sum = ₹13,000.
SI given, CI wanted: CI = SI + R% of one year's SI. SI of ₹960 for 2 years at 8% gives CI = 960 + 38.40 = ₹998.40.
Trap: taking R% of the full 2-year SI.

Two parts lent for $t_1$ and $t_2$ years at R% with equal amounts: part for fewer years : part for more years = $\left(1 + \frac{R}{100}\right)^{t_2 - t_1}$ : 1.
₹3,300 lent for 2 and 3 years at 20%: ratio 6 : 5, so ₹1,800 and ₹1,500.
Ages: the younger child's money grows for longer, so the younger child gets the smaller share.
Three parts: each part is the one before it divided by the yearly multiplier.

Bring each instalment back to today: $P = \frac{x}{1 + r} + \frac{x}{(1 + r)^2} + \cdots$ , with r = R ÷ 100.
2 instalments at 20%: $P = x\left(\frac{5}{6} + \frac{25}{36}\right) = \frac{55x}{36}$ . Instalments of ₹3,600 clear a loan of ₹5,500.
To check, run the loan forward: add a year's interest, take off the payment and repeat. It must end at zero.
Trap: dividing the final amount by the number of instalments.