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

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

The picture above is the whole D4 map: each rung uses only the tools below it. Refer back to it whenever you feel lost — you are always somewhere on this ladder.

Level 1 — Recognition

Here you only need to spot which formula the words are pointing at and plug in. No traps of base yet.

Problem 1.1

Convert $\frac{7}{20}$ to a percentage.

Recall Solution 1.1

WHAT we want: a "parts per 100" number equal to $\frac{7}{20}$ . WHY multiply by 100: a percent is the fraction scaled so its denominator becomes 100. $\frac{7}{20} \times 100 = \frac{700}{20} = 35%.$ Check: $35% = 0.35$ and $7 \div 20 = 0.35$ ✓.

Problem 1.2

Write $62%$ as a decimal and as a fraction in lowest terms.

Recall Solution 1.2

Decimal: percent → decimal means divide by 100: $62% = \frac{62}{100} = 0.62$ . Fraction: $\frac{62}{100}$ ; divide top and bottom by their common factor $2$ : $\frac{62}{100} = \frac{31}{50} .$ Check: $31 \div 50 = 0.62$ ✓.

Problem 1.3

Find $30%$ of $150$ .

Recall Solution 1.3

"OF means multiply": $30% of 150 = \frac{30}{100} \times 150 = 0.30 \times 150 = 45.$ Feynman check: $10%$ of $150 = 15$ , so $30% = 3 \times 15 = 45$ ✓. The answer is a quantity ( $45$ ), not a percent.

Level 2 — Application

Now you must choose the base (the "of" number) correctly and run the formula.

Problem 2.1

$27$ is what percent of $75$ ?

Recall Solution 2.1

Identify: the "is"-number is $A = 27$ , the "of"-number (base) is $B = 75$ . WHY $\frac{A}{B}$ : we solve $\frac{p}{100} = \frac{27}{75}$ for $p$ . $p = \frac{27}{75} \times 100 = 0.36 \times 100 = 36%.$ Reverse-check: $36%$ of $75 = 0.36 \times 75 = 27$ ✓.

Problem 2.2

A jacket costs $₹1200$ . A discount of $18%$ is applied. What is the sale price?

Recall Solution 2.2

Discount amount: $18%$ of $1200 = \frac{18}{100} \times 1200 = 216$ . Sale price: original minus discount: $1200 - 216 = ₹984$ . Multiplier shortcut: a $18%$ cut multiplies by $(1 - 0.18) = 0.82$ , so $1200 \times 0.82 = 984$ ✓. Same answer, fewer steps.

Problem 2.3

A number increases from $80$ to $92$ . Find the percentage increase.

Recall Solution 2.3

"New minus Old, over Old": $% change = \frac{92 - 80}{80} \times 100 = \frac{12}{80} \times 100 = 15%.$ The result is positive, so it is a $15%$ increase. Check: $80 \times 1.15 = 92$ ✓.

Level 3 — Analysis

Here the base shifts between steps, and you must reason about why the answers aren't symmetric.

Problem 3.1

A price rises from $₹400$ to $₹500$ , then later falls from $₹500$ back to $₹400$ . Find the percentage rise and the percentage fall, and explain why they differ.

Recall Solution 3.1

Rise: base is the starting value $400$ . $\frac{500 - 400}{400} \times 100 = \frac{100}{400} \times 100 = 25%.$ Fall: now the journey starts at $500$ , so the base is $500$ . $\frac{400 - 500}{500} \times 100 = \frac{- 100}{500} \times 100 = - 20%,$ a $20%$ decrease. WHY different: the absolute jump ( $100$ ) is the same, but $%$ change divides by the starting value. Going up, we divide $100$ by the smaller $400$ ; coming down, we divide $100$ by the larger $500$ . Bigger base ⇒ smaller percent. See the figure below.

Problem 3.2

A salary is increased by $20%$ , then the new salary is decreased by $20%$ . What is the net percentage change?

Recall Solution 3.2

Multiplier method (the safe way across different bases):

$+ 20%$ multiplies by $1.20$ .
$- 20%$ multiplies by $0.80$ . $V_{n} = V_{o} \times 1.20 \times 0.80 = V_{o} \times 0.96.$ A multiplier of $0.96$ means the final value is $96%$ of the original, i.e. a net $4%$ decrease. WHY not $0%$ : the $- 20%$ acts on the raised salary $1.2 V_{o}$ , which is bigger, so it removes more than the $+ 20%$ added.

Problem 3.3

After a $25%$ increase a quantity becomes $150$ . What was the original value?

Recall Solution 3.3

Set up the multiplier equation: $V_{o} \times 1.25 = 150$ . WHY $1.25$ : a $25%$ rise means final $= (1 + 0.25) \times$ original. $V_{o} = \frac{150}{1.25} = 120.$ Check: $120 \times 1.25 = 150$ ✓. Trap avoided: the original is not $150 - 25% of 150 = 112.5$ — that would take $25%$ off the wrong base (the new value, not the old).

Level 4 — Synthesis

Now several ideas combine in one problem — reverse percentages, successive changes, and profit/loss thinking.

Problem 4.1

A shopkeeper marks a book up by $40%$ over the cost price, then gives a $25%$ discount on the marked price. If the cost price is $₹200$ , find the final selling price and the shopkeeper's overall profit percentage.

Recall Solution 4.1

Step 1 — markup: marked price $= 200 \times 1.40 = 280$ . Step 2 — discount: selling price $= 280 \times 0.75 = 210$ . Step 3 — overall change vs cost price (base = cost price $200$ ): $\frac{210 - 200}{200} \times 100 = \frac{10}{200} \times 100 = 5%.$ So the final selling price is $₹210$ and the overall profit is $5%$ . This links to Profit and Loss — profit $%$ is just $%$ change of the cost price. Multiplier check: $1.40 \times 0.75 = 1.05$ , a $5%$ gain ✓.

Problem 4.2

The population of a town grows by $10%$ in year 1 and by $20%$ in year 2. If it starts at $50, 000$ , what is the population after two years, and what single percentage increase would produce the same result?

Recall Solution 4.2

Two-year multiplier: $1.10 \times 1.20 = 1.32$ . Final population: $50, 000 \times 1.32 = 66, 000$ . Equivalent single increase: the combined multiplier is $1.32$ , which is $(1 + 0.32)$ , so a single $32%$ increase. WHY not $10 + 20 = 30%$ : the year-2 $20%$ acts on the already-grown population, adding extra on top — the "interest on interest" effect. This is exactly the seed of Simple and Compound Interest and Exponential Growth and Decay.

Problem 4.3

A student scores $18$ out of $25$ in Test A and $27$ out of $40$ in Test B. Which performance is better, and by how many percentage points?

Recall Solution 4.3

Convert both to a common measuring stick (percent): $Test A = \frac{18}{25} \times 100 = 72%, Test B = \frac{27}{40} \times 100 = 67.5%.$ Compare: Test A is higher. The gap is $72 - 67.5 = 4.5$ percentage points. WHY percent here: the two tests have different totals ( $25$ vs $40$ ); percentages give both the same $100$ -scale so they are directly comparable — the core reason percentages exist. (Relates to Fractions and Decimals and Ratio and Proportion.)

Level 5 — Mastery

These require full modelling — set up unknowns, chain multipliers, and interpret.

Problem 5.1

After two successive equal percentage discounts, a $₹1000$ item sells for $₹810$ . Find the single discount rate applied each time.

Recall Solution 5.1

Set up: let each discount be $r %$ , so each step multiplies by $(1 - \frac{r}{100})$ . Call this multiplier $m$ . Equation: $1000 \times m^{2} = 810$ , so $m^{2} = 0.81$ . WHY square root: two identical multipliers give $m \times m = m^{2}$ ; to undo squaring we take the square root. $m = 0.81 = 0.9.$ Interpret: $m = 1 - \frac{r}{100} = 0.9 \Rightarrow \frac{r}{100} = 0.1 \Rightarrow r = 10%.$ Each discount is $10%$ . Check: $1000 \times 0.9 \times 0.9 = 810$ ✓.

Problem 5.2

A quantity is increased by $50%$ . By what percentage must it now be decreased to return exactly to its original value?

Recall Solution 5.2

After the rise: value is $1.5 V_{o}$ . We need a fall multiplier $f$ so that $1.5 V_{o} \times f = V_{o}$ , giving $f = \frac{1}{1.5} = \frac{2}{3} \approx 0.6 \overline{6} .$ Convert to a percentage decrease: $f = 1 - \frac{r}{100}$ , so $\frac{r}{100} = 1 - \frac{2}{3} = \frac{1}{3} \Rightarrow r = 33. \overline{3} %.$ WHY not $50%$ : the decrease is measured against the raised value $1.5 V_{o}$ , a larger base, so a smaller percent is enough. Undo of $+ 50%$ is a $33.3%$ cut, not $50%$ . Check: $1.5 \times \frac{2}{3} = 1$ ✓ (exactly back to start).

Problem 5.3

Sugar price rises by $25%$ . A family wants to keep its total sugar spending unchanged. By what percentage must it cut its sugar consumption?

Recall Solution 5.3

Model spending as $price \times quantity$ . Call the reduction multiplier on quantity $q$ . Keep spending fixed: $(1.25 P) \times (q Q) = P \times Q$ , so $1.25 q = 1$ . $q = \frac{1}{1.25} = 0.8.$ Interpret: $q = 1 - \frac{r}{100} = 0.8 \Rightarrow r = 20%.$ The family must cut consumption by $20%$ . WHY the asymmetry ( $+ 25%$ price ↔ $- 20%$ quantity): the cut acts on the quantity side while the rise acts on the price side; balancing the product requires reciprocal multipliers, and $\frac{1}{1.25} = 0.8$ is a $20%$ drop, not a $25%$ one.

Recall One-line self-test: state the master rule for every chained-change problem

Turn each percentage change into a multiplier, multiply them, then read the final multiplier as $(1 + \frac{net}{100})$ . ::: A net multiplier $m$ means net change $= (m - 1) \times 100%$ ; e.g. $m = 0.96 \Rightarrow - 4%$ , $m = 1.32 \Rightarrow + 32%$ .

Connections

Fractions and Decimals — every percent is a fraction/decimal in disguise (Problems 1.1–1.2, 4.3).
Ratio and Proportion — comparing scores puts different totals on one scale (Problem 4.3).
Simple and Compound Interest — successive multipliers are compounding (Problem 4.2).
Profit and Loss — profit $%$ is $%$ change of cost price (Problem 4.1).
Exponential Growth and Decay — repeated equal multipliers $m^{n}$ (Problems 4.2, 5.1).