Percentages — finding %, % of a quantity, % increase - decrease: Worked examples

The scenario matrix

Every cell below is one class of problem. The right column names the example that covers it.

#	Case class	What makes it tricky	Covered by
C1	A of B, part smaller than whole	plain, warm-up	Ex 1
C2	A of B, part larger than whole → answer $> 100%$	you can have "150% of"	Ex 2
C3	x% of Q with a real-world word wrapper (tax/discount)	keep the units, don't return a %	Ex 3
C4	% increase (new $>$ old, base = old)	sign positive	Ex 4
C5	% decrease & the asymmetry (up ≠ down)	base changes	Ex 4
C6	Find the base / original from a % change	reverse of C4, divide not multiply	Ex 5
C7	Successive changes (+% then −%)	multiply multipliers, never add	Ex 6
C8	Zero / degenerate inputs ( $A = 0$ , $V_{o} = 0$ , $B = 0$ )	$0%$ , $- 100%$ , and undefined	Ex 7
C9	Limiting behaviour (as the fall approaches the whole)	max possible decrease is $- 100%$	Ex 8
C10	Exam twist: percent-of-a-percent / point vs percent	"20% off then 10% off ≠ 30% off"; percentage points	Ex 9

Ten cells, nine examples (C4 and C5 share one). Let's fill them all.

Worked example Ex 2 — Cell

C2: answer bigger than 100% This year a bakery sold $360$ cakes; last year it sold $240$ . This year's sales are what percent of last year's?

Forecast: more cakes than last year → the answer must exceed $100%$ . (A percent can go past 100 whenever the part beats the whole.)

Base $B = 240$ (last year, the "of" number); part $A = 360$ (this year). Why this step? The phrase is "this year is what % of last year", so last year is the base.
$p = \frac{360}{240} \times 100$ . Why this step? Same formula as C1 — nothing forbids $A > B$ .
$\frac{360}{240} = 1.5$ , so $p = 1.5 \times 100 = 150%$ .

Verify: $150%$ of $240 = 1.5 \times 240 = 360$ ✓. Reading check: " $150%$ of" simply means "one-and-a-half times", which lines up with 360 being $1.5 \times 240$ .

Worked example Ex 3 — Cell

C3: % of a quantity, keep the units A jacket is listed at ₹ $2500$ . There is a $30%$ discount, then $18%$ GST is added on the discounted price. What do you pay?

Forecast: guess whether the final price is above or below ₹ $2500$ . (30% off is a big cut, 18% tax is smaller — expect below.)

Discount amount $= 30%$ of $2500 = \frac{30}{100} \times 2500 = ₹750$ . Why this step? "of" = multiply; we want an amount in rupees, not a percent.
Discounted price $= 2500 - 750 = ₹1750$ . Why this step? Discount is money removed from the base.
GST $= 18%$ of the new base $1750 = 0.18 \times 1750 = ₹315$ . Why this step? Tax is charged on what you actually owe (the discounted price), not the original list.
Total paid $= 1750 + 315 = ₹2065$ .

Verify: units are rupees throughout (never a bare percent) ✓. Sanity: $2065 < 2500$ , matching the forecast. Multiplier form as a cross-check: $2500 \times 0.70 \times 1.18 = 2065$ ✓. This "chain of multipliers" idea powers Profit and Loss markups.

Worked example Ex 4 — Cells

C4 & C5: increase, decrease, and why they differ A stock moves $250 \to 300$ , then later $300 \to 250$ . Find the % increase, then the % decrease, and explain the mismatch.

Forecast: same $50$ -rupee jump both ways — will the two percents be equal? (Trap: they won't.)

Up: change $= 300 - 250 = + 50$ , base = old $= 250$ . $% = \frac{50}{250} \times 100 = + 20%$ . Why this step? $%$ change measures the jump against where you started — the old value.
Down: change $= 250 - 300 = - 50$ , base = old $= 300$ . $% = \frac{- 50}{300} \times 100 = - 16. \overline{6} % \approx - 16.7%$ . Why this step? Now the starting value is 300, so the same 50 is a smaller slice.
Why the asymmetry — look at the figure: same arrow length, different-sized base bars.

Verify: up-multiplier $1.20$ , down-multiplier $1 - 0.1667 = 0.8333$ ; $1.20 \times 0.8333 = 1.0000$ ✓ (they take you exactly back). So the trip is a round trip in value, but $+ 20%$ vs $- 16.7%$ in percent — because the bases differ.

Worked example Ex 5 — Cell

C6: find the ORIGINAL from a % change After a $25%$ pay rise, Meera earns ₹ $60000$ . What was her salary before the rise?

Forecast: the "before" number is smaller than 60000 — but is it $60000 - 25%$ ? (No! That's the classic trap.)

Model with the multiplier: new $=$ old $\times (1 + \frac{r}{100})$ , so $60000 = V_{o} \times 1.25$ . Why this step? A $25%$ rise multiplies the old salary by $1.25$ ; the old value is the unknown base.
Solve for the old value by dividing (undo the multiply): $V_{o} = \frac{60000}{1.25} = ₹48000$ . Why this step? Division reverses the $\times 1.25$ . We must divide, not subtract 25%.
Contrast the trap: $60000 - 25%$ of $60000 = 60000 - 15000 = 45000$ — wrong, because 25% here would be of the new salary, not the old.

Verify: grow it back — $48000 \times 1.25 = 60000$ ✓. And $48000 < 60000$ ✓. This "divide by the multiplier to reverse growth" is exactly how you strip interest in Simple and Compound Interest.

Worked example Ex 6 — Cell

C7: successive changes (multiply, never add) A ₹ $1000$ item is marked up $20%$ for the festival, then put on a $20%$ off sale. Net change?

Forecast: up 20% then down 20% — back to ₹1000? (Feels like it. It isn't.)

Turn each change into a multiplier: up $20% \Rightarrow \times 1.20$ ; down $20% \Rightarrow \times 0.80$ . Why this step? Multipliers compose cleanly; raw percents don't, because each acts on a different base.
Chain them: final $= 1000 \times 1.20 \times 0.80 = 1000 \times 0.96 = ₹960$ . Why this step? The $0.80$ acts on the already-raised ₹1200, a bigger base than the original.
Net multiplier $0.96 \Rightarrow$ net change $= (0.96 - 1) \times 100 = - 4%$ .

Verify: $1200 \times 0.80 = 960$ and $\frac{960 - 1000}{1000} \times 100 = - 4%$ ✓. General rule: $(1 + a) (1 - a) = 1 - a^{2}$ , always a loss when the raw percents are equal. This compounding of multipliers is the seed of Exponential Growth and Decay.

Worked example Ex 7 — Cell

C8: zero and degenerate inputs Handle three edge cases carefully. (a) In a jar of $80$ marbles, $0$ are red. What % are red? (b) A shop had $40$ mangoes; today it has $0$ . What is the % change? (c) A startup had $0$ users last month and $500$ this month. What is the % growth?

Forecast: which one has no answer at all? (Hold that thought for (c).)

(a) $A = 0, B = 80$ : $p = \frac{0}{80} \times 100 = 0%$ . Why this step? Zero part of any non-zero whole is genuinely $0%$ — perfectly defined.
(b) old $= 40$ , new $= 0$ : $% = \frac{0 - 40}{40} \times 100 = - 100%$ . Why this step? Losing everything is exactly a $100%$ decrease — the multiplier is $\times 0$ .
(c) old $= 0$ , new $= 500$ : $% = \frac{500 - 0}{0} \times 100$ . The base is $0$ → division by zero → undefined. Why this step? Percent change compares to the starting amount; if you started at nothing, "how many times bigger" has no numeric answer. We report "growth from zero — percentage undefined", or state the raw increase ( $+ 500$ users).

Verify: (a) $0%$ of $80 = 0$ marbles ✓. (b) multiplier $1 + \frac{- 100}{100} = 0$ , and $40 \times 0 = 0$ ✓. (c) any base of $0$ makes $\frac{change}{0}$ undefined — no plug-back exists, which is the correct "answer" ✓.

Worked example Ex 8 — Cell

C9: the limiting value of a decrease A tank drains. Its water level falls by $x %$ . As losses grow, what is the most it can fall by, and what happens near that limit?

Forecast: guess the biggest possible percentage decrease. (Can a level drop by $150%$ ?)

Multiplier after a fall of $x %$ is $(1 - \frac{x}{100})$ ; the amount left is old $\times$ this. Why this step? We track the surviving fraction, which must stay $\geq 0$ (you can't have negative water).
Empty tank ⇒ multiplier $= 0$ ⇒ $1 - \frac{x}{100} = 0 \Rightarrow x = 100$ . So the maximum decrease is $100%$ . Why this step? At $100%$ nothing remains; you cannot lose more than everything.
Limiting behaviour: as $x \to 10 0^{-}$ , the remaining amount $\to 0^{+}$ — you approach empty but each extra percent removes less absolute water (a shrinking base). A " $110%$ decrease" is meaningless for a physical quantity.

Verify: for a $100$ -litre tank, a $99%$ fall leaves $1$ L; a $99.9%$ fall leaves $0.1$ L; the leftover $= 100 \times (1 - x /100) \to 0$ as $x \to 100$ ✓. Contrast increases, which have no upper cap (a price can rise $500%$ ), matching Ex 2.

Worked example Ex 9 — Cell

C10: the exam twist (percent of a percent, and "points") A vaccine's effectiveness rose from $60%$ to $75%$ . A newspaper writes two headlines: (i) "effectiveness up 15 percentage points", (ii) "effectiveness up 25 percent". Which is right?

Forecast: can both be correct at once? (Yes — they measure different things.)

Percentage points = the plain subtraction of the two percentages: $75% - 60% = 15$ points. Why this step? "Points" compares two percent numbers directly, with no re-basing.
Percent change = the rise measured against the old percent as base: $\frac{75 - 60}{60} \times 100 = \frac{15}{60} \times 100 = 25%$ . Why this step? Here the base is the old value $60$ , so we're asking "how much bigger, relative to before".
So both headlines are true and not contradictory: $+ 15$ points and $+ 25%$ .

Verify: grow the old value by 25%: $60 \times 1.25 = 75$ ✓, and $75 - 60 = 15$ points ✓.

Recall Rapid scenario check (say the cell out loud)

Match each phrasing to its cell and formula. "39 out of 50 is what %?" ::: C1 · $\frac{39}{50} \times 100 = 78%$ "This is 130% of last year" ::: C2 · answer exceeds 100%, base = last year "18% GST on ₹500" ::: C3 · $\frac{18}{100} \times 500 = ₹90$ , a rupee amount "After −25%, find the original" ::: C6 · divide by $0.75$ , don't subtract "+20% then −20%" ::: C7 · $\times 1.2 \times 0.8 = 0.96 \Rightarrow - 4%$ "Fell from 40 to 0" ::: C8 · exactly $- 100%$ "Grew from 0 to 500" ::: C8 · base zero ⇒ undefined "60% to 75%: points vs percent?" ::: C10 · +15 points and +25%

Connections

Fractions and Decimals — every ratio $\frac{A}{B}$ here is first a fraction/decimal.
Ratio and Proportion — "A of B" and "find the base" are proportion solves.
Simple and Compound Interest — reversing growth (Ex 5) = stripping interest; chaining multipliers (Ex 6) = compounding.
Profit and Loss — the discount/tax chain of Ex 3 is a cost→selling price markup.
Exponential Growth and Decay — successive multipliers (Ex 6) and the $\to 0$ limit (Ex 8).

Worked examples — Percentages — finding %, % of a quantity, % increase - decrease

The scenario matrix

Connections

Concept Map