Percentages — finding %, % of a quantity, % increase - decrease: Visual walkthrough

Step 1 — Draw the starting amount as a bar

WHAT. We pick a starting value and draw it as a horizontal bar. Call the starting value $V_{o}$ where the little " $o$ " just means "old / original". To make the pictures concrete we take $V_{o} = 100$ (rupees, grams, anything — the units don't matter for percentages).

WHY a bar? Because a percentage is a share of a length. If the whole bar is "100%", then any slice of it is easy to see as a fraction of the whole. Length is the friendliest picture of "part of a whole".

PICTURE. The full lavender bar below is $V_{o}$ . Its whole length is the "100%" that every later percentage will be measured against.

Figure — Percentages — finding %, % of a quantity, % increase - decrease

Step 2 — What does "increase by 10%" do to the bar?

WHAT. We add a piece equal to $10%$ of the bar onto its right end.

WHY multiply by $1.10$ ? " $10%$ of $V_{o}$ " means $\frac{10}{100} \times V_{o} = 0.10 V_{o}$ . The new length is the old bar plus this piece:

$V_{1} = whole starting bar V_{o} + the added 10% slice 0.10 V_{o} = V_{o} (keep all of V_{o} 1 + add a tenth 0.10) = 1.10 V_{o}$

Each symbol: $V_{1}$ is the value after step one (the raised value); the $1$ inside the bracket says "keep the whole original", the $0.10$ says "attach one extra tenth". Pulling out $V_{o}$ turns adding into a single multiply by $1.10$ — the multiplier form from the parent note.

PICTURE. The mint slice is the new $10%$ . Crucially, its length is a tenth of the original bar.

With $V_{o} = 100$ : $V_{1} = 1.10 \times 100 = 110$ .

Step 3 — Now cut 10% off — but off of WHAT?

WHAT. We remove $10%$ from the raised bar $V_{1}$ , not the original.

WHY this is the whole trick. A percentage always eats its base (Step 1's definition). The second $10%$ arrives after the increase, so its base is the new, longer bar $V_{1} = 1.10 V_{o}$ — a bigger stick than $V_{o}$ . A tenth of a bigger stick is a bigger chunk than a tenth of the original.

$V_{2} = V_{1} - a tenth of the raised bar 0.10 V_{1} = V_{1} (1 - 0.10) = 0.90 V_{1}$

Here $V_{2}$ is the value after both changes; the $- 0.10$ says "shave off a tenth", and the base being shaved is $V_{1}$ , not $V_{o}$ .

PICTURE. Compare the coral chunk (the $10%$ we now remove) with the mint chunk we added in Step 2. The coral one is visibly longer — that gap is the entire mystery.

Step 4 — Multiply the multipliers

WHAT. Chain the two operations into one number.

WHY multiply, not add? Each step scaled the bar (Step 2 by $1.10$ , Step 3 by $0.90$ ). Doing one scaling then another means the second scaling acts on the already-scaled length — that is exactly what multiplication of factors means. You never add multipliers, because $1.10$ and $0.90$ live on different bases.

$V_{2} = Step 3 shrink 0.90 \times Step 2 grow 1.10 \times V_{o} = 0.99 V_{o}$

The order doesn't matter ( $0.90 \times 1.10 = 1.10 \times 0.90$ ), which is why "+10% then −10%" and "−10% then +10%" give the same answer.

PICTURE. A number line of multipliers: start at $1$ , jump right by the factor $1.10$ , then jump left by the factor $0.90$ , landing at $0.99$ — just short of home.

With $V_{o} = 100$ : $V_{2} = 0.99 \times 100 = 99$ .

Step 5 — Point at the missing 1%

WHAT. We line up the original bar and the final bar and find the shortfall.

WHY. The final bar $V_{2} = 0.99 V_{o}$ is $0.01 V_{o}$ shorter than the original. That is a $1%$ loss measured against $V_{o}$ — exactly the parent note's claim.

$V_{o} - V_{2} = V_{o} - 0.99 V_{o} = 0.01 V_{o} = 1% of V_{o}$

PICTURE. The thin butter-yellow sliver is the whole $1%$ loss. It is precisely the overlap gap: the coral chunk removed in Step 3 stuck out past the mint chunk added in Step 2 — and that overhang is $0.10 \times 0.10 V_{o} = 0.01 V_{o}$ .

Step 6 — All cases: does this always lose? What about other $r$ ?

WHAT. We test the formula $V_{2} = (1 - \frac{r ^{2}}{10 0 ^{2}}) V_{o}$ across signs and edges so no scenario surprises you.

WHY. A good derivation must cover every input, not just $r = 10$ .

Define the loss fraction $L (r) = \frac{r ^{2}}{10 0 ^{2}}$ — the tidy name for "what fraction of the original you end up short after an up-then-down of rate $r$ ". It is a pure number: multiply it by $V_{o}$ to get the actual money lost, or by $100$ to read it as a percent. Reading it aloud: $L (r)$ answers "for a swing of $r$ percent each way, what share of my starting bar vanishes?"

$r = 0$ (do nothing): $L (0) = 0$ , so $V_{2} = V_{o}$ . No change, no loss. ✓ (degenerate case.)
Any nonzero $r$ , up-then-down: $L (r) = \frac{r ^{2}}{10 0 ^{2}} > 0$ always, so $V_{2} < V_{o}$ — always a loss. Bigger $r$ , bigger loss (it grows like $r^{2}$ ): $r = 50$ gives $1.50 \times 0.50 = 0.75$ , a $25%$ loss, not $0$ .
Down-then-up (−r% then +r%): same factors, same product $(1 - \frac{r ^{2}}{10 0 ^{2}})$ — still a loss, same size. Cutting first then raising doesn't save you.
$r = 100$ (up $100%$ then down $100%$ ): $2.00 \times 0.00 = 0$ . The bar vanishes — a total wipe-out. That's the largest $r$ that makes sense for a "down" step (you can't remove more than $100%$ ).

PICTURE. The loss curve $L (r) = \frac{r ^{2}}{10 0 ^{2}}$ : flat and tiny near $r = 0$ , curving up as a parabola, reaching $100%$ loss at $r = 100$ .

The one-picture summary

The whole story in one frame: start ( $V_{o}$ ), grow by a tenth of $V_{o}$ (mint), then shrink by a tenth of the taller bar (coral) — the coral overhang leaves you the butter-yellow $1%$ short.

Recall Feynman retelling — say it like a story

Picture a rubber band $100$ cm long. You stretch it $10%$ — now it's $110$ cm, you added $10$ cm. Then you shrink it by $10%$ . But $10%$ of $110$ is $11$ cm, not $10$ ! So you pull off $11$ cm and land at $99$ cm. You added ten, took away eleven, so you're one short — that's the $1%$ loss. The catch is always the same: the second percent is a slice of a bigger thing, so it's a bigger slice. Multiply the two multipliers, $1.1 \times 0.9 = 0.99$ , and the $0.01$ gap is really $10%$ of the $10%$ you added — a percent of a percent. Do it with any number $r$ and the shortfall is $\frac{r ^{2}}{10 0 ^{2}}$ : tiny for small wiggles, brutal for big ones. Up-then-down or down-then-up, it never comes back to zero except when you did nothing at all.

Recall Quick self-check

Why is up-10%-then-down-10% a loss and not zero? ::: The down-10% acts on the raised bar (110), removing 11, while you only added 10 — net $- 1$ . What single number captures both changes? ::: The product of multipliers $1.10 \times 0.90 = 0.99$ . General shortfall for $\pm r %$ ? ::: $L (r) = \frac{r ^{2}}{10 0 ^{2}}$ of the original, i.e. loss $= (\frac{r}{100})^{2}$ . Does order (up-first vs down-first) matter? ::: No — multiplication commutes, same $0.99$ . When is there truly no loss? ::: Only when $r = 0$ (you did nothing).

Connections

Fractions and Decimals — every multiplier $1.10, 0.90$ is a fraction-of-the-whole made decimal.
Ratio and Proportion — each step scales the bar by a fixed ratio.
Simple and Compound Interest — chaining multipliers is exactly compounding; here two factors, there $n$ .
Profit and Loss — the $1%$ gap is a loss on the original cost.
Exponential Growth and Decay — repeated multipliers are the seed of exponential change.
↑ Parent: Percentages