Here is a tiny formula that, once you remember it, gives you the answer to an entire family of JEE arithmetic questions in your head. Whenever a quantity rises by X\% and then falls by X\% — or falls by X\% and then rises by X\% — the result is always below the starting value, by an amount that is exactly:

\text{loss} \;=\; \left(\dfrac{X}{100}\right)^{2} \times \text{original}

Not X\%. Not zero. A quadratic-sized gap.

For X = 10\%, you lose 1\% of the original. For X = 20\%, you lose 4\%. For X = 50\%, you lose a full 25\%. The loss grows as the square of the percentage, not linearly with it — and it always points downward, regardless of whether the rise or the fall comes first.

The one-line derivation

A rise of X\% is a multiplication by 1 + \tfrac{X}{100}. A fall of X\% is a multiplication by 1 - \tfrac{X}{100}. Whichever order you do them in, the net multiplier is the product:

\left(1 + \tfrac{X}{100}\right)\left(1 - \tfrac{X}{100}\right) \;=\; 1 - \left(\dfrac{X}{100}\right)^{2}

This is just the difference of squares identity (1 + a)(1 - a) = 1 - a^{2}, with a = X/100.

Why the cross-terms cancel: when you multiply (1 + a)(1 - a) you get 1 - a + a - a^2, and the two middle terms (-a and +a) exactly kill each other. This cancellation is the algebraic source of the "almost zero" net change — but "-a^2" survives, and because a^2 > 0 for any nonzero a, the survivor is always negative. That is why you always lose.

The net multiplier is 1 - (X/100)^{2}, which is always less than 1 (for X \ne 0). So the price is always below its starting value after the two changes, by exactly the square of X/100 as a fraction of the original.

Numbers make it concrete

Start with ₹1000 and walk through three cases.

The loss grows fast. Double X from 10\% to 20\% and the loss quadruples from 1\% to 4\%. Triple X from 10\% to 30\% and the loss climbs by a factor of nine, from 1\% to 9\%. That is the quadratic shape.

Loss as a quadratic function of the percentage changeA graph whose horizontal axis is the percentage value X from zero up to sixty percent, and whose vertical axis shows the fractional loss from zero to forty percent. A smooth upward-opening parabola rises slowly near zero, picks up speed, and by X equals sixty percent the curve has reached thirty-six percent loss. A draggable red point on the curve shows both the current X value and the loss for that X in a readout at the top. At X equals twenty percent the loss is four percent, and at X equals fifty percent the loss is twenty-five percent. 0% 20% 40% 60% (X) 0 20% 40% loss ↔ drag to change X
The loss after a "$+X\%$ then $-X\%$" pair grows as $(X/100)^{2}$ — a parabola. Small $X$ costs almost nothing (at $X = 10\%$ you lose only $1\%$); large $X$ costs a lot (at $X = 50\%$ you lose $25\%$). The quadratic curve explains why volatility is so expensive for big swings and nearly free for small ones.

Why the order doesn't matter

Whether the rise or the fall goes first, the two multipliers are the same two numbers — (1 + X/100) and (1 - X/100) — and multiplication is commutative. So:

(1 + \tfrac{X}{100})(1 - \tfrac{X}{100}) \;=\; (1 - \tfrac{X}{100})(1 + \tfrac{X}{100})

Both orderings give 1 - (X/100)^2, exactly the same loss. A stock that drops 20\% and then recovers 20\% ends at 96\% of its starting value — the same place it would end at if it had first gone up 20\% and then down 20\%.

Why the asymmetry exists at all

If you ever wonder "shouldn't +X\% and -X\% cancel?", the answer is: they would cancel, if both acted on the same base. But they don't. The +X\% acts on the original price; the -X\% acts on the inflated price (which is larger). The -X\% therefore subtracts more rupees than the +X\% added. The asymmetry of bases is the geometric source of the -(X/100)^2 residue.

You can see this viscerally: +20\% on ₹1000 adds ₹200; but -20\% on the resulting ₹1200 removes ₹240. Same percentage, different base, net loss ₹40.

The three cases you should memorise

Three values of X come up often enough in JEE and real life that knowing the loss by heart is a time-saver:

For other values, just square X/100 on the spot. X = 15\%? (0.15)^2 = 0.0225 = 2.25\%. X = 30\%? (0.30)^2 = 0.09 = 9\%.

The generalisation — rise by X and fall by Y

If the two percentage changes are different, the formula stretches. A rise of X\% followed by a fall of Y\% gives a net multiplier

\left(1 + \tfrac{X}{100}\right)\left(1 - \tfrac{Y}{100}\right) \;=\; 1 + \tfrac{X - Y}{100} - \tfrac{XY}{10{,}000}

The linear term X - Y is the naive "add percentages" guess; the quadratic term -XY/10{,}000 is the correction that multiplicative composition forces you to subtract. When X = Y, the linear term is zero and only the quadratic correction survives — which is how you get the pure -(X/100)^2 result above.

So the quadratic loss of the symmetric case is not a strange accident. It is the universal "correction term" that shows up whenever you stack two percentage changes, stripped of its linear companion by the symmetry.

Reflex checklist

When a problem chains a +X\% with a -X\%:

  1. Do not write zero. The answer is never zero.
  2. Compute (X/100)^2 in your head. That is the fractional loss.
  3. Subtract from the original to get the final value.
  4. If the percentages are not equal — say +X\% then -Y\% — use the full product (1 + X/100)(1 - Y/100) instead.

One formula — \text{loss} = (X/100)^{2} \times \text{original} — closes off an entire class of trap questions. Remember the parabola, and remember it always points downward.

Related: Percentages and Ratios · Price Up 20% Then Down 20% — Why You End Up Below Where You Started · See Successive Percentage Changes → Multiply Factors, Never Add Them · Plus 20% Then Minus 20%: Why You End Up Below Where You Started