Percentages, Profit and Loss, Discount and Interest

Percentages sit under half the quantitative questions in any placement test: profit and loss, discount, simple and compound interest, population change, successive changes. If you master a handful of identities and the habit of choosing a convenient base (usually 100), most of these take under 40 seconds. Every worked answer below is checked by computation.

1. Percentage basics

$x\%$ of $N$ is $\frac{x}{100}N$. Percentage change $=\frac{\text{new}-\text{old}}{\text{old}}\times 100$. The base is always the original quantity, which is where most mistakes happen.

Useful equivalents: $\frac{1}{2}=50\%$, $\frac{1}{3}=33\frac{1}{3}\%$, $\frac{1}{4}=25\%$, $\frac{1}{5}=20\%$, $\frac{1}{6}=16\frac{2}{3}\%$, $\frac{1}{7}=14\frac{2}{7}\%$, $\frac{1}{8}=12\frac{1}{2}\%$, $\frac{1}{9}=11\frac{1}{9}\%$, $\frac{1}{10}=10\%$, $\frac{1}{11}=9\frac{1}{11}\%$, $\frac{1}{12}=8\frac{1}{3}\%$.

The asymmetry of increase and decrease. If a quantity rises by $x\%$, to return to the original it must fall by $\frac{100x}{100+x}\%$. If it falls by $x\%$, it must rise by $\frac{100x}{100-x}\%$.

def fall_needed(rise):
    return 100 * rise / (100 + rise)

def rise_needed(fall):
    return 100 * fall / (100 - fall)

assert abs(fall_needed(25) - 20) < 1e-12              # +25 % then -20 % returns to the start
assert abs(rise_needed(20) - 25) < 1e-12
assert abs(rise_needed(50) - 100) < 1e-12             # halving needs a 100 % rise to restore
assert round(100 * 1.25 * 0.8, 6) == 100

Successive percentage changes

Two successive changes of $a\%$ and $b\%$ equal one net change of $a+b+\frac{ab}{100}\%$ (use signs: a decrease is negative).

Example 1. A price rises 20% and then falls 20%. Net change? $20-20+\frac{(20)(-20)}{100}=-4\%$. A 4% loss.

Example 2. Price up 10%, then up 20%, then down 25%. Multiply factors: $1.10\times 1.20\times 0.75=0.99$, a 1% decrease.

def net_change(*changes):
    factor = 1.0
    for c in changes:
        factor *= 1 + c / 100
    return (factor - 1) * 100

assert abs(net_change(20, -20) + 4) < 1e-9
assert abs(net_change(10, 20, -25) + 1) < 1e-9
assert abs(net_change(20, 20) - 44) < 1e-9             # two 20 % rises are a 44 % rise, not 40 %

Percentage with population, depreciation and "A is x% more than B"

• If A is $x\%$ more than B, then B is $\frac{100x}{100+x}\%$ less than A.
• Population growing at $r\%$ for $n$ years: $P(1+r/100{)}^n$. Depreciation: $P(1-r/100{)}^n$.

Example 3. A town's population is 50,000 and grows 10% per year. Population after 2 years? $50000\times 1.1^2=60500$.

Example 4. A machine bought for Rs 80,000 depreciates 15% a year. Value after 2 years? $80000\times 0.85^2=57800$.

assert round(50000 * 1.1 ** 2) == 60500
assert round(80000 * 0.85 ** 2) == 57800
assert abs(fall_needed(40) - 28.571428) < 1e-5          # if A is 40 % more than B, B is 28.57 % less than A

Election and exam style percentages

Example 5. In an election between two candidates, the winner gets 55% of the votes and wins by 2,400 votes. Total votes? Margin = $55\%-45\%=10\%$ of the total, so total $=24000$.

Example 6. In an exam, 70% pass in maths and 80% in English; 10% fail both. Percentage passing both? Pass at least one $=90\%$. Both $=70+80-90=60\%$.

assert 2400 / ((55 - 45) / 100) == 24000
assert 70 + 80 - (100 - 10) == 60

2. Profit, loss and discount

Definitions. Cost price (CP), selling price (SP), marked price (MP).

• Profit $=\text{SP}-\text{CP}$; loss $=\text{CP}-\text{SP}$.
• Profit % $=\frac{\text{profit}}{\text{CP}}\times 100$; loss % likewise. Always on cost price unless stated.
• Discount $=\text{MP}-\text{SP}$; discount % $=\frac{\text{discount}}{\text{MP}}\times 100$. Always on marked price.
• $\text{SP}=\text{CP}\left(1+\frac{p}{100}\right)$ for profit $p\%$, and $\text{SP}=\text{MP}\left(1-\frac{d}{100}\right)$ for discount $d\%$.

Example 7. An article bought for Rs 800 is sold at a 25% profit. SP? $800\times 1.25=1000$.

Example 8. An article is marked at Rs 1,500 and sold for Rs 1,275. Discount %? Discount $=225$; $225/1500=15\%$.

Example 9. A trader marks goods 40% above cost and allows a 25% discount. Profit or loss %? Take CP = 100: MP = 140; SP = $140\times 0.75=105$. 5% profit.

Example 10. Marked price Rs 2,000, two successive discounts of 10% and 20%. Selling price? $2000\times 0.9\times 0.8=1440$. Equivalent single discount $=10+20-\frac{10\cdot 20}{100}=28\%$.

assert 800 * 1.25 == 1000
assert (1500 - 1275) / 1500 * 100 == 15
assert round(100 * 1.40 * 0.75, 6) == 105
assert round(2000 * 0.9 * 0.8) == 1440 and abs(net_change(-10, -20) + 28) < 1e-9

Standard patterns

Same SP, one gain and one loss of $x\%$. Always a loss of $\frac{x^2}{100}\%$. Two articles sold at Rs 1,200 each, one at 20% gain and one at 20% loss. Overall? Loss $=20^2/100=4\%$.

cp1 = 1200 / 1.2
cp2 = 1200 / 0.8
total_cp, total_sp = cp1 + cp2, 2400
assert round((total_cp - total_sp) / total_cp * 100, 6) == 4.0       # an overall loss of 4 %
assert abs(((1000 + 1500) - 2400) / 2500 * 100 - 4.0) < 1e-9

Profit when SP = x times CP. If SP is 1.2 times CP the profit is 20%.

False weights. A shopkeeper sells at cost price but uses a 900 g weight for 1 kg. Gain $=\frac{1000-900}{900}\times 100=11.11\%$.

assert round((1000 - 900) / 900 * 100, 2) == 11.11

Marking up to allow a discount and still profit $p\%$. If CP = 100, desired SP $=100+p$; MP $=\frac{(100+p)\times 100}{100-d}$. How much above cost should goods be marked so that after a 20% discount the profit is 20%? $\text{MP}=\frac{120\cdot 100}{80}=150$, so 50% above cost.

assert 120 * 100 / 80 == 150

Cost price from loss or gain. A man sells an article at a 15% loss. Had he sold it for Rs 240 more he would have gained 5%. CP? The difference between the two selling prices is $20\%$ of CP, so CP $=240/0.20=1200$.

cp = 240 / 0.20
assert cp == 1200 and round(cp * 1.05 - cp * 0.85) == 240

Dishonest gain on mixed items and mark-up on SP

If profit is $p\%$ of the selling price, the cost is $(100-p)\%$ of the selling price. A 20% profit on SP means a markup on cost of $\frac{20}{80}=25\%$.

assert 20 / (100 - 20) * 100 == 25

3. Simple interest

$SI=\frac{P\cdot R\cdot T}{100},A=P+SI$

with $T$ in years and $R$ the rate per annum.

Example 11. SI on Rs 5,000 at 8% per annum for 3 years: $\frac{5000\cdot 8\cdot 3}{100}=1200$.

Example 12. At what rate does a sum triple in 10 years at simple interest? Interest $=2P$ in 10 years: $\frac{PR\cdot 10}{100}=2P$, so $R=20\%$.

Example 13. A sum amounts to Rs 6,200 in 2 years and Rs 7,100 in 4 years at simple interest. Find the principal and rate. Interest for 2 years $=900$, so for 1 year $=450$. Principal $=6200-900=5300$. Rate $=\frac{450}{5300}\times 100\approx 8.49\%$.

assert 5000 * 8 * 3 / 100 == 1200
assert 100 * 2 / 10 == 20.0
principal = 6200 - 2 * (7100 - 6200) / 2
assert principal == 5300 and round(450 / 5300 * 100, 2) == 8.49

4. Compound interest

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

For half-yearly compounding use rate $R/2$ and $2n$ periods; for quarterly, $R/4$ and $4n$.

Example 14. CI on Rs 10,000 at 10% per annum for 2 years: $10000\times 1.21-10000=2100$. (SI would be 2,000: the extra 100 is interest on the first year's interest.)

Example 15. Rs 8,000 at 10% per annum compounded half-yearly for 1 year. $8000\times 1.05^2=8820$. CI $=820$.

def compound(p, rate, years, per_year=1):
    return p * (1 + rate / 100 / per_year) ** (years * per_year)

assert round(compound(10000, 10, 2) - 10000) == 2100
assert round(compound(8000, 10, 1, per_year=2)) == 8820
assert round(compound(8000, 10, 1, per_year=2) - 8000) == 820

Difference between CI and SI

• For 2 years: $CI-SI=P{\left(\frac{R}{100}\right)}^2$.
• For 3 years: $CI-SI=P{\left(\frac{R}{100}\right)}^2\left(3+\frac{R}{100}\right)$.

Example 16. The difference between CI and SI on a sum for 2 years at 5% is Rs 25. Find the sum. $P\times 0.0025=25$, so $P=10000$.

assert round(25 / (5 / 100) ** 2) == 10000
p, r = 10000, 5
ci2 = compound(p, r, 2) - p
si2 = p * r * 2 / 100
assert round(ci2 - si2, 6) == 25
ci3 = compound(p, r, 3) - p
assert round(ci3 - p * r * 3 / 100, 6) == round(p * (r / 100) ** 2 * (3 + r / 100), 6)

Doubling time (rule of 72, approximation). At $r\%$ compound interest, money doubles in about $72/r$ years.

import math
assert abs(math.log(2) / math.log(1.08) - 72 / 8) < 0.3     # 9.01 years exactly versus 9 by the rule of 72

Equal instalments. A loan repaid in equal annual instalments: each instalment $x$ satisfies $\frac{x}{1+r}+\frac{x}{(1+r{)}^2}=P$ for 2 years, where $r=R/100$.

Example 17. Rs 5,500 is borrowed at 10% compounded annually and repaid in two equal annual instalments. Each instalment? $x\left(\frac{1}{1.1}+\frac{1}{1.21}\right)=5500$, so $x\cdot \frac{2.1}{1.21}=5500$ and $x=3169.05$ (to the paisa, 3,169.05).

x = 5500 / (1 / 1.1 + 1 / 1.1 ** 2)
assert round(x, 2) == 3169.05
assert round(x / 1.1 + x / 1.21, 6) == 5500

5. Quick percentage patterns

• Price and consumption: if a price rises $x\%$, to keep expenditure the same consumption must fall $\frac{100x}{100+x}\%$.
• Salary: expenses as a fraction of income; savings $=$ income $-$ expenses.
• Area of a rectangle when length rises $a\%$ and breadth $b\%$: net change $a+b+\frac{ab}{100}\%$. If the sides both increase 10%, the area rises 21%.
• A number reduced by $x\%$ gives $N(1-x/100)$; a number increased then decreased by the same percent always ends below the start.

Example 18. The price of sugar rises 25%. By what percent should a family reduce consumption to keep spending unchanged? $\frac{100\cdot 25}{125}=20\%$.

Example 19. A rectangle's length rises 20% and its breadth falls 10%. Change in area? $20-10-\frac{20\cdot 10}{100}=8\%$ increase.

assert fall_needed(25) == 20
assert abs(net_change(20, -10) - 8) < 1e-9
assert abs(net_change(10, 10) - 21) < 1e-9

6. Common traps

• Choosing the wrong base: profit on CP, discount on MP, percentage change on the original.
• Adding successive percentages ($20\%+20\%\ne 40\%$).
• Using simple interest in a compound question, or forgetting to halve the rate and double the periods for half-yearly compounding.
• "A is 25% more than B" versus "B is 25% less than A": not equivalent.
• Rounding mid-way: keep fractions until the end.
• Mixing time units (months versus years) in interest questions.

7. Practice set with answers

1. A number is first increased by 30% and then decreased by 30%. Net change?
2. A shopkeeper marks goods 25% above cost and gives a 12% discount. Profit %?
3. An article is sold at a 10% loss. Had it been sold for Rs 280 more, the gain would have been 8%. Find the cost price.
4. Find CI on Rs 12,500 for 2 years at 8% per annum, compounded annually.
5. A sum becomes Rs 7,260 in 2 years at 10% per annum compound interest. Find the sum.
6. The price of rice rises by 20%. A family wants to keep the same expenditure. By what percent must it cut consumption?
7. Rs 6,000 lent at simple interest for 3 years yields Rs 1,260. Find the rate.
8. A sells to B at a 20% profit; B sells to C at a 25% profit. C pays Rs 1,500. What did A pay?

assert abs(net_change(30, -30) + 9) < 1e-9
assert abs(1.25 * 0.88 * 100 - 100 - 10) < 1e-9
assert round(280 / (0.08 + 0.10)) == 1556                                  # a difference of 18 % of CP is Rs 280 (nearest rupee)
assert round(compound(12500, 8, 2) - 12500) == 2080
assert round(7260 / 1.1 ** 2) == 6000
assert round(fall_needed(20), 2) == 16.67
assert 1260 * 100 / (6000 * 3) == 7.0
assert round(1500 / (1.2 * 1.25)) == 1000

Answers: 1) 9% decrease; 2) 10% profit; 3) about Rs 1,556 (the exact value is $280/0.18=1555.56$); 4) Rs 2,080; 5) Rs 6,000; 6) $16\frac{2}{3}\%$; 7) 7% per annum; 8) Rs 1,000.