Up Twenty Per Cent Then Down Twenty Leaves You Worse Off

Lower than it started — by exactly $4$ per cent.

Take $\text{₹}250$. A rise of $20\%$ takes it to $\text{₹}300$. A fall of $20\%$ then applies to **$300$**, not to the original $250$, and $20\%$ of $300$ is $\text{₹}60$ — so the price becomes $\text{₹}240$.

The rise added $\text{₹}50$ and the fall removed $\text{₹}60$, because the two percentages were taken of different amounts. Net change:

$$\frac{250-240}{250}\times 100=4\%\text{ decrease}$$

Percentages cannot simply be added and subtracted, and this page covers the second part of the ICSE Class 8 Mathematics chapter on percentage: change, successive change, and the problems both are used for.

Multiply the two multipliers together. Each change applies to whatever the value was at that moment, not to the original.

Worked example 1 — the opening case in full. A price of $\text{₹}250$ rises by $20\%$ and then falls by $20\%$.

$$250\times 1.20\times 0.80=250\times 0.96=\text{₹}240$$

The combined multiplier is $1.20\times 0.80=0.96$, which is a **$4\%$ decrease** overall — and not the zero change that adding $+20$ and $-20$ would suggest.

The single equivalent change. For two successive changes of $a\%$ and $b\%$, taking a decrease as negative:

$$\text{net percentage change}=a+b+\frac{ab}{100}$$

Checking it on the case above, with $a=20$ and $b=-20$:

$$20+(-20)+\frac{20\times (-20)}{100}=0-4=-4\%$$

The same $4\%$ decrease. The extra term $\frac{ab}{100}$ is exactly the part that plain addition misses, and it is why successive percentages never simply add.

Worked example 2 — two increases. A salary of $\text{₹}20000$ rises by $10\%$ and then by $10\%$ again.

$$20000\times 1.10\times 1.10=20000\times 1.21=\text{₹}24200$$

The net rise is $21\%$, not $20\%$. By the formula: $10+10+\frac{100}{100}=21\%$.

Worked example 3 — two decreases. A value of $\text{₹}5000$ falls by $10\%$ and then by $20\%$.

$$5000\times 0.90\times 0.80=5000\times 0.72=\text{₹}3600$$

The net fall is $28\%$, not $30\%$. By the formula: $-10-20+\frac{200}{100}=-28\%$.

The order does not matter. $250\times 1.2\times 0.8$ and $250\times 0.8\times 1.2$ both give $\text{₹}240$, because multiplication is commutative. So increased then decreased and decreased then increased end at the same place — though neither returns to the start.

Why two opposite changes always lose. For a rise of $r\%$ followed by a fall of $r\%$, the combined multiplier is

$$\left(1+\frac{r}{100}\right)\left(1-\frac{r}{100}\right)=1-\frac{r^2}{10000}$$

which is **always less than $1$** whatever $r$ is. So equal opposite percentage changes always leave you worse off, and the loss is ${\left(\frac{r}{10}\right)}^2$ per cent — giving $4\%$ for $r=20$, as found.

All three are repeated percentage changes, so each is a multiplier raised to a power.

Population growth. A population of $25000$ increases at $4\%$ per year. Find it after two years.

Each year multiplies by $1.04$, so two years multiply by $1.04^2$:

$$25000\times 1.04^2=25000\times 1.0816=27040$$

Notice the second year's increase is larger than the first, because it is $4\%$ of a bigger population. The first year adds $1000$; the second adds $1040$.

Depreciation — the fall in value of a machine or vehicle with use and age. A machine costing $\text{₹}80000$ depreciates at $10\%$ per year. Find its value after two years.

Each year multiplies by $0.90$:

$$80000\times 0.90^2=80000\times 0.81=\text{₹}64800$$

Again the reductions are unequal — $\text{₹}8000$ in the first year and $\text{₹}7200$ in the second, since the second $10\%$ is taken of a smaller value.

**The general form for $n$ periods:**

$$\text{final}=\text{initial}\times {\left(1\pm \frac{r}{100}\right)}^n$$

with plus for growth and minus for depreciation.

Marks in an examination. A student scores $486$ out of $600$. Find the percentage.

$$\frac{486}{600}\times 100=81\%$$

And the reverse. A student needs $40\%$ to pass a paper of $600$ marks and scores $210$. By how many marks did they fail?

Pass mark $=\frac{40}{100}\times 600=240$, so

$$240-210=30\text{ marks short}$$

Worked example — a common examination question type. A student scores $30\%$ and fails by $40$ marks, while the pass mark is $35\%$. Find the total marks.

The $40$ marks correspond to the difference $35\%-30\%=5\%$, so

$$\text{total}=\frac{40}{5}\times 100=800\text{ marks}$$

Checking: $30\%$ of $800=240$ scored, $35\%$ of $800=280$ needed, and $280-240=40$. Correct.

Why growth is not simple multiplication by the number of years. Two years at $4\%$ is not $8\%$ — it is $1.04^2=1.0816$, or $8.16\%$. The extra $0.16\%$ is the growth on the first year's growth, which is the same $\frac{ab}{100}$ term from the previous section appearing again.

Identify which quantity stays fixed while the total changes, and percentages become straightforward.

Mixtures — the key is that adding water does not change the amount of acid.

Worked example 1. $40\text{L}$ of a solution contains $15\%$ acid. If $10\text{L}$ of water is added, what is the new percentage of acid?

$$\text{acid}=\frac{15}{100}\times 40=6\text{L}$$

Adding water leaves the acid unchanged at $6\text{L}$ while the total becomes $40+10=50\text{L}$:

$$\text{new percentage}=\frac{6}{50}\times 100=12\%$$

So the concentration falls from $15\%$ to $12\%$ with no acid removed at all. Tracking the absolute amount of acid rather than its percentage is what makes this work.

Income, expenditure and savings — the three are linked by one equation:

$$\text{income}=\text{expenditure}+\text{savings}$$

So if expenditure is $75\%$ of income, savings must be the remaining $25\%$.

Worked example 2. A man's income is $\text{₹}32000$ per month. He spends $75\%$ and saves the rest. Find his savings.

$$\text{expenditure}=\frac{75}{100}\times 32000=\text{₹}24000$$

$$\text{savings}=32000-24000=\text{₹}8000$$

Or directly, savings are $25\%$ of $32000=\text{₹}8000$.

Worked example 3 — the harder version. His income then rises by $10\%$ and his expenditure by $8\%$. Find the percentage increase in his savings.

$$\text{new income}=32000\times 1.10=\text{₹}35200$$

$$\text{new expenditure}=24000\times 1.08=\text{₹}25920$$

$$\text{new savings}=35200-25920=\text{₹}9280$$

$$\text{increase in savings}=9280-8000=\text{₹}1280$$

$$\text{percentage increase}=\frac{1280}{8000}\times 100=16\%$$

**Why savings rose by $16\%$ when income rose by only $10\%$. Savings are what is left over, so they absorb the whole of the gap between the two increases. A small difference between the income rise and the expenditure rise produces a large percentage change in the much smaller savings figure.

This is the real lesson of the problem. A percentage change in a small quantity means much less in absolute terms than the same percentage of a large one — and three percentages taken of three different bases can never be added or compared directly. Working in rupees** first and converting to a percentage only at the end is what keeps these problems straight.