Comparing Quantities

Every day you compare quantities without calling it mathematics. A shirt is "20% off," a bank pays "8% interest," a shopkeeper adds "5% tax" to a bill, a city's population "grows by 5% a year." Behind each of these ordinary sentences sits one idea: a ratio turned into a percentage, and a percentage turned into a rule for finding a new number from an old one. This chapter, Comparing Quantities, builds that toolkit, so that "roughly cheaper" becomes an exact rupee amount and "growing fast" becomes a number you can predict years ahead. We begin with the oldest tool, the ratio, and its cousin, the percentage, moved between using the unitary method. From there the chapter builds outward to money problems that use exactly this idea: discount on a marked price, sales tax and GST added to a bill, and the reverse problem of stripping a tax back out of a price already paid. The second half turns to interest. Simple interest treats the principal as fixed; compound interest lets last year's interest join the principal and earn interest of its own — which is why a bank deposit, a growing population and a depreciating car all obey the same formula, $A = P(1+R/100)^n$ , with only a sign choice between growth and decay. By the end you should convert confidently between ratios, fractions and percentages, calculate discount, sales tax, VAT and GST in both directions, distinguish simple from compound interest and explain why compound interest is always at least as large, derive and apply the compound interest formula to money, population and depreciation, and spot the classic trap of adding percentages that should instead be multiplied. Comparing quantities is not a side skill for word problems; it is how numbers describe change in the real world.

Key takeaways

A ratio compares two quantities of the same kind by division; a percentage is a ratio expressed with 100 as the denominator, $\frac{\text{Part}}{\text{Whole}} \times 100$ .
The unitary method — find the value for one unit, then scale — solves almost every ratio and percentage problem.
Discount is a reduction on the Marked Price (MP): Discount = MP minus Sale Price, and Sale Price = MP times $\left(1 - \frac{\text{Discount \%}}{100}\right)$ .
Sales tax, VAT and GST are charged on the selling price and added to the bill: Bill Amount = Cost times $\left(1 + \frac{\text{Tax \%}}{100}\right)$ , and the relation runs backwards to strip a tax out of a tax-included price.
Simple interest keeps the principal fixed every year; compound interest adds each year's interest to the principal first, so it grows faster.
The compound interest formula, built by compounding year after year, is $A = P\left(1+\frac{R}{100}\right)^n$ , with Compound Interest CI = A minus P.
The same formula with a minus sign, $A = P\left(1-\frac{R}{100}\right)^n$ , models depreciation — the steady percentage fall in the value of machinery or vehicles.
Population and bacterial growth are compounding processes too: each period's growth acts on the previous period's total, not on the original amount.
Two successive percentage changes never simply add; a 10% rise followed by a 10% fall leaves you below the start, since the fall acts on a larger base.
Key idea: ratios become percentages by the unitary method, and every quantity changing by a fixed percentage repeatedly — money, population or value — follows the single formula A = P(1 ± R/100)^n.

The core insight

Ratios become percentages by the unitary method, and every quantity that changes by a fixed percentage repeatedly — interest, population, or depreciating value — grows or shrinks according to the single formula $A = P(1 \pm R/100)^n$ .

What this chapter covers

This chapter builds one comparison tool after another, each resting on the one before it. It opens by recalling ratios and percentages and the unitary method that converts between them — the engine used throughout the chapter. It applies that engine to shopping: marked price and discount, then a mental-math shortcut for estimating a percentage of a bill. Next comes sales tax, VAT and GST, including the reverse problem of finding the original price from a tax-included price. The chapter then turns to money over time: it recalls simple interest, contrasts it with compound interest, and derives the general compound interest formula year by year. Finally it shows the same formula describes population growth, bacterial growth, and, with a sign flip, depreciation — the steady percentage loss of value in machinery or vehicles.

The core ideas

Ratios and percentages: the language of comparison

A ratio compares two quantities of the same kind by division. A basket with 20 apples and 5 oranges has oranges to apples in the ratio 5 : 20, which simplifies to 1 : 4. A percentage is a ratio written with 100 as its denominator, answering "how many out of every hundred?"

\text{Percentage} = \frac{\text{Part}}{\text{Whole}} \times 100. \qquad (1)

where: Part and Whole must share a unit, and the ratio is simplified before scaling to 100. For the basket, oranges are $\frac{5}{25}\times 100 = 20\%$ of all the fruit, so apples are the remaining $80\%$ , since the two shares of a whole always add to $100\%$ . The unitary method — find what one unit is worth, then multiply — is the workhorse behind every conversion here. In the chapter's class-picnic problem, girls are $60\%$ of a class and number 18; treating $60\%$ as 18 students gives one percent as $18/60$ students, so the whole class is $\frac{100}{60}\times 18 = 30$ students, leaving $30-18=12$ boys, and a girls-to-boys ratio of $18:12 = 3:2$ . The same problem finds a cost per head: a refreshment bill of ₹4280 plus a return transport charge of ₹1320 gives total expenses of ₹5600, shared among 32 people (18 girls, 12 boys, 2 teachers) for ₹175 per person.

Marked price, discount and sale price

Shops advertise a Marked Price (MP) and often sell at a lower Sale Price (SP) to attract buyers. The difference is the discount, always calculated as a percentage of the marked price, never of the sale price.

\text{Discount} = \text{MP} - \text{SP}, \qquad \text{Sale Price} = \text{MP}\left(1 - \frac{\text{Discount \%}}{100}\right). \qquad (2)

where: MP is the listed price, SP is what the customer pays, and Discount % is found by the unitary method. For an item marked ₹840 and sold for ₹714, the discount is ₹126, so on ₹100 the discount is $\frac{126}{840}\times 100 = 15\%$ . Running it the other way: a frock listed at ₹220 with a $20\%$ discount has discount $\frac{20}{100}\times 220 = ₹44$ , so the sale price is ₹176 — or, in one step, $80\%$ of ₹220. Buyers also incur overhead expenses, added to the buying price to give the true Cost Price: CP = Buying Price + Overhead Expenses.

Estimating percentages quickly

For a bill of ₹577.80 with a $15\%$ discount, round the bill to ₹580; $10\%$ of ₹580 is ₹58; half of that, $5\%$ , is ₹29; adding gives about ₹87 off, so the amount to pay is roughly ₹493 — close enough for a quick decision at the counter.

Sales tax, VAT and GST

A tax on the sale of goods or services is calculated on the selling price and added to the bill — the opposite direction to a discount.

\text{Bill Amount} = \text{Cost of item}\left(1 + \frac{\text{Tax \%}}{100}\right). \qquad (3)

where: Tax % is Sales Tax, VAT, or, since 1 July 2017 in India, GST (Goods and Services Tax), which replaced the older sales tax and VAT system and is levied on the supply of goods or services or both. Roller skates costing ₹450 with $5\%$ sales tax cost $₹450 + ₹22.50 = ₹472.50$ billed. The reverse problem strips tax from a tax-included price by the unitary method on a base of ₹100: an air cooler bought for ₹3300 including $10\%$ VAT had an original price of $\frac{100}{110}\times 3300 = ₹3000$ , and an article bought for ₹784 including $12\%$ GST had an original price of $\frac{100}{112}\times 784 = ₹700$ .

Simple interest, revisited

Simple Interest (SI) is calculated on a principal that never changes: $\text{SI} = \frac{P \times R \times T}{100}$ , and the amount to repay is $A = P + \text{SI}$ . For ₹10,000 at $15\%$ per annum for 2 years, one year's interest is ₹1500, so two years give SI = ₹3000 and amount ₹13,000.

Compound interest: interest on interest

In real banking, interest is calculated on last year's amount, not the original sum — this is Compound Interest (C.I.). Follow ₹20,000 borrowed by Heena for 2 years at $8\%$ per annum. Year one's interest is $\frac{20000\times8}{100}=₹1600$ , giving amount ₹21,600, which becomes the principal for year two. Year two's interest is $\frac{21600\times8}{100}=₹1728$ , giving amount ₹23,328. Total CI is $₹1600+₹1728=₹3328$ — ₹128 more than two years of simple interest (₹3200), because the second year's interest was partly earned on the first year's interest. The chapter's own ₹100-at- $10\%$ table shows the gap widening: simple interest gives ₹110, ₹120, ₹130 at years one to three, while compound interest gives ₹110, ₹121, ₹133.10.

Figure

A grouped bar chart comparing the value of 100 rupees under simple interest and under compound interest at 10 percent per annum at the end of year 0 1 2 and 3 with the compound interest bars pulling higher each year to reach 133 point 10 against 130 for simple interest

The compound interest formula

Working out compound interest year by year is reliable but slow. Compounding a principal $P$ annually at rate $R\%$ for $n$ years multiplies it by the same factor each year, giving

A = P\left(1+\frac{R}{100}\right)^{n}, \qquad \text{CI} = A - P. \qquad (4)

where: $P$ is the principal, $R$ the annual rate, $n$ the number of years, $A$ the final amount. For ₹12,600 at $10\%$ for 2 years, $A = 12600\times\frac{11}{10}\times\frac{11}{10} = ₹15{,}246$ , so $\text{CI} = ₹2646$ .

Growth and decay: population and depreciation

Anything changing by a fixed percentage repeatedly — a growing population, a rising price, a depreciating machine — follows the same rule, since each period's change acts on the previous total, exactly like compounding.

A = P\left(1+\frac{R}{100}\right)^{n} \ \text{(growth)}, \qquad A = P\left(1-\frac{R}{100}\right)^{n} \ \text{(decay).} \qquad (5)

where: the sign inside the bracket is the only difference; $P$ is now a population, count or value, and $R\%$ the steady rate per period. A city of 20,000 growing at $5\%$ per annum reaches $20000\times\left(\frac{21}{20}\right)^3 = 23{,}152.5$ , about $23{,}153$ , after 3 years. A TV bought for ₹21,000, depreciating $5\%$ in one year, is worth $21000\times\frac{19}{20} = ₹19{,}950$ .

Figure

A flow diagram showing a marked price of 220 rupees passing through a 20 percent discount arrow to 176 rupees and then through a 5 percent tax arrow to a final 184 point 80 rupees, illustrating that successive percentage changes multiply along the chain rather than simply adding

Derivation ADeriving the compound interest formula from year-by-year compounding

The goal is to show, without assuming the formula in advance, that compounding a principal $P$ at a fixed rate $R\%$ per annum for $n$ years must give $A = P\left(1+\frac{R}{100}\right)^n$ . The one fact used repeatedly is that each year's simple interest is calculated only on that year's principal, and that year's amount then becomes next year's principal.

Step 1. Find the amount after year one. Let $P_1 = P$ . Interest earned is $\text{SI}_1 = \frac{P_1 R}{100}$ , so the amount is $A_1 = P_1 + \text{SI}_1 = P\left(1+\frac{R}{100}\right)$ . This becomes the principal for year two, $P_2 = P\left(1+\frac{R}{100}\right)$ .

Step 2. Find the amount after year two using $P_2$ . Interest is $\text{SI}_2 = P_2\times\frac{R}{100}$ , so $A_2 = P_2\left(1+\frac{R}{100}\right) = P\left(1+\frac{R}{100}\right)^2$ , factoring the common bracket out of both terms.

Step 3. Spot the pattern. Each year multiplies the current amount by the same factor $\left(1+\frac{R}{100}\right)$ , since every year repeats the identical calculation on a larger base. After $n$ years the factor has been applied $n$ times: $A = P\left(1+\frac{R}{100}\right)^n$ .

Step 4. Recover CI and check. Since $\text{CI} = A - P$ , and with Heena's numbers $P=20000$ , $R=8$ , $n=2$ : $A = 20000(1.08)^2 = 23{,}328$ and $\text{CI} = 3328$ , matching the year-by-year calculation exactly.

Go deeper · Derivation B · 1 minThe combined effect of two successive percentage changes

Derivation BThe combined effect of two successive percentage changes

The goal is to find a single rule for what happens when a quantity changes by one percentage and then a second — a discount followed by a tax, or a rise followed by a fall. The common mistake is to add the two percentages; this derivation shows why that fails and what the correct combined change actually is.

Step 1. Write each change as a multiplier. A change of $x\%$ turns $Q$ into $Q\left(1+\frac{x}{100}\right)$ ; a discount uses a negative $x$ , a tax a positive $x$ .

Step 2. Apply the second change of $y\%$ to the already-changed quantity, not the original $Q$ : the result is $Q\left(1+\frac{x}{100}\right)\left(1+\frac{y}{100}\right)$ . The two factors multiply because the second change acts on the first change's result.

Step 3. Expand to find the net percentage. $\left(1+\frac{x}{100}\right)\left(1+\frac{y}{100}\right) = 1 + \frac{x+y+\frac{xy}{100}}{100}$ , so $\text{net}\% = x + y + \frac{xy}{100}$ — the sum of the two percentages plus a correction $\frac{xy}{100}$ that naive addition loses.

Step 4. Check with numbers. A $15\%$ discount then $5\%$ tax: $x=-15$ , $y=5$ , giving $\text{net}\% = -15+5+\frac{(-15)(5)}{100} = -10.75\%$ , not $-10\%$ . A $10\%$ rise then $10\%$ fall: $x=10$ , $y=-10$ , giving $\text{net}\% = -1\%$ , not zero, because the fall acts on the larger, already-risen base — the same compounding idea behind the CI formula, applied to two unequal opposite steps.

Worked examples

Board level - Discount followed by sales tax on a marked price (extends NCERT Example 3 and Example 4)

A frock is marked at ₹220. The shop gives a discount of 20% and then charges 5% sales tax on the discounted price. Find the final amount paid.

Step 1. Sale price after discount: $80\%$ of ₹220, since a $20\%$ discount means paying $80\%$ of MP. Sale price $= \frac{80}{100}\times 220 = ₹176$ , matching the chapter's Example 3.

Step 2. Apply $5\%$ tax to this sale price, not the marked price: tax $= \frac{5}{100}\times 176 = ₹8.80$ .

Step 3. Add tax to the discounted price: $₹176 + ₹8.80 = ₹184.80$ . Answer: the customer pays ₹184.80. This is not $220\times0.85=₹187$ ; using $\text{net}\% = -20+5+\frac{(-20)(5)}{100}=-16\%$ , $220\times(1-0.16)=₹184.80$ , confirming it.

Board level - Compound interest and amount over three years (extends NCERT Example 8)

Find the compound interest on ₹12,600 for 3 years at 10% per annum, compounded annually, and the amount at the end of 3 years.

Step 1. Identify $P=₹12{,}600$ , $R=10$ , $n=3$ .

Step 2. Apply the formula: $A = 12600\times\left(\frac{11}{10}\right)^3 = 12600\times\frac{1331}{1000}$ .

Step 3. Simplify: $A = ₹16{,}770.60$ . Answer: amount ₹16,770.60, and CI = A minus P = ₹4,170.60. This extends the chapter's 2-year answer of ₹15,246 by one more year.

Higher-order thinking (HOTS / olympiad-style) - A population that grows then declines (extends NCERT Example 9 and Example 10)

A town of 20,000 grows at 5% per annum for 3 years, as in the chapter's Example 9, then falls by 5% in a fourth year. Find the population after the fourth year and explain why it is not simply back to 20,000.

Step 1. After 3 years of $5\%$ growth: $A = 20000\times\left(\frac{21}{20}\right)^3 = 23{,}152.5$ , about $23{,}153$ .

Step 2. Apply the $5\%$ decline to this larger population, not to 20,000: $23{,}153\times\frac{19}{20} \approx 21{,}995$ .

Step 3. Compare with 20,000. Answer: the population is about 21,995, not 20,000. Three unequal-sized growth steps compound on ever-larger bases, while the single decline removes only one-twentieth of the already-inflated total — the same asymmetry as in the successive-percentage derivation. Timing note: allow about four minutes. Reuse the chapter's own Example 9 result for the 3-year growth rather than re-deriving it, then apply one decay step to that result — never to the original 20,000. Examiners design multi-step compounding questions precisely to catch students who quietly reset to the starting value partway through.

Common traps

Mistakes here are rarely about the formulas; they are about which number becomes the base for the next percentage. Discount is a percentage of the marked price, never the sale price; tax is a percentage of the price actually charged, never the marked price. In compound interest, the commonest slip is using the original principal every year instead of the previous year's amount, quietly turning compound interest into simple interest. Two successive percentage changes must never be added directly; the correct combined change is $x+y+\frac{xy}{100}$ . Depreciation is sometimes solved with the growth formula by mistake — check whether the quantity is rising or falling before choosing the sign. Finally, ratio quantities must share a unit first; comparing 5 m to 10 km without converting to metres gives a meaningless ratio.

You should be able to…

Convert confidently between a ratio, a fraction and a percentage.
Use the unitary method to solve any ratio or percentage word problem.
Calculate discount, discount percentage, marked price or sale price when any two are given.
Estimate a percentage of a bill quickly using the 10%-then-half trick.
Calculate sales tax, VAT or GST in both directions, including stripping tax from a tax-included price.
Distinguish simple from compound interest, and explain why compound interest grows faster.
State and apply $A = P\left(1+\frac{R}{100}\right)^n$ , and recover CI = A minus P.
Apply the same formula, with a sign flip, to depreciation and to population or bacterial growth.
Combine two successive percentage changes using $x+y+\frac{xy}{100}$ , never by simple addition.
Identify the correct base for a given percentage before doing any arithmetic.

Ask the tutor: If you would like, I can work through a few more successive-discount-and-tax or multi-year compound interest problems with you step by step.

Why is compound interest always more than simple interest for the same principal, rate and time beyond one year?

For the first year the two are identical, since there is nothing yet to compound. From the second year onward, compound interest is calculated on the previous year's amount, which already includes earlier interest, while simple interest keeps calculating on the same original principal. So compound interest earns interest on interest every year after the first, and this extra amount grows larger each year — visible in the chapter's own ₹100 example, where three years give ₹130 under simple interest but ₹133.10 under compound interest.

When I calculate a discount, why do I always use the marked price as the base, and not the sale price?

Discount is defined as a reduction on the marked price, so the shopkeeper's discount percentage is a fraction of the price originally listed, not of whatever price you eventually pay. Basing the discount percentage on the sale price instead would give the same rupee discount a different, larger percentage, which would not match what the shop actually advertised. Always convert word by word: "20% discount" means 20% of the marked price is subtracted, giving 80% of the marked price as the sale price.

If GST replaced sales tax and VAT, do I need to learn all three separately for this chapter?

The calculation is identical for all three — a percentage of the price is added to the bill — so no separate formulas are needed. Sales tax and VAT were the older Indian tax systems; GST, introduced from 1 July 2017, is the current system levied on the supply of goods or services or both, replacing the earlier taxes. You may still see "sales tax" or "VAT" as a general name for any such added percentage, solved exactly like a GST problem: Bill Amount equals Cost times $\left(1+\frac{\text{Tax}\%}{100}\right)$ .

How do I know whether to use the plus sign or the minus sign in the compound interest style formula?

Ask whether the quantity is getting bigger or smaller over time. If it is growing — interest, a rising population, multiplying bacteria — use $A = P\left(1+\frac{R}{100}\right)^n$ . If it is shrinking — a depreciating machine, a declining population — use $A = P\left(1-\frac{R}{100}\right)^n$ . Both are repeated percentage changes on the previous total, not the original amount; only the direction decides the sign.