Ratio and Proportion | ICSE Class 8 Maths Notes

This note covers ratios and percentages, the unitary method, profit and loss, overhead expenses, discounts, tax calculations, simple interest, annual and half-yearly compound interest, direct and inverse variation, and time and work.

How do ratios, proportions and percentages compare quantities?

A ratio compares quantities by division. The order of comparison matters: oranges to apples compares the number of oranges with the number of apples. A proportion states that two ratios are equal. Equivalent ratios express the same comparison using different numbers.

The colon in a ratio means “is to”; a slash indicates division. The signs =, +, − and × mean equals, addition, subtraction and multiplication respectively. For example, 5 : 20 = 5/20 = 1/4 = 1 : 4. Dividing both numbers by the same non-zero number preserves the ratio.

Percentage means a comparison per hundred; the symbol % means “per cent”.

Which quantity is the comparison based on?

Worked example 1. A basket contains 20 apples and 5 oranges. Find the ratio of oranges to apples and the percentage of all the fruits that are oranges.

Answer: Oranges to apples = 5 : 20 = 1 : 4. The total is 20 + 5 = 25 fruits. Oranges form 5/25 of the total, so their percentage is (5/25) × 100% = 20%.

In the example, oranges are one-fourth as numerous as apples, but one-fifth of all the fruits. The second quantity, or base of the comparison, has changed.

Apples to oranges is 20 : 5 = 4 : 1. Apples make up 80% of all the fruits. Reversing a ratio reverses the comparison, while changing a part-to-part comparison into a part-to-total comparison changes the denominator, the number below the fraction bar.

How are fractions converted into percentages?

Express the required part as a fraction of the chosen whole, then multiply by 100%. Conversely, a percentage is written as a fraction by placing its numerical value over 100. Simplify that fraction if possible. Keep the meaning of the whole unchanged throughout.

Note: When comparing measurements of the same kind, express them in the same unit before simplifying their ratio. Write the requested quantities in their stated order.

How does the unitary method solve percentage problems?

The unitary method first finds the value corresponding to one unit, then uses it to find the required value. A unit can mean one person, one metre, or one percentage point, depending on the question. Name the quantity before deciding whether to multiply or divide.

How can a whole be recovered from a percentage?

Worked example 2. In a class going on a picnic, 18 girls form 60% of all the students. Find the total number of students and the ratio of girls to boys.

Answer: If 60% represents 18 students, 1% represents 18/60 students. Thus 100% represents (18/60) × 100 = 30 students. There are 30 − 18 = 12 boys. Girls to boys = 18 : 12 = 3 : 2.

An intermediate unitary value need not be a whole number, even when the final answer counts people. It is a calculation step, not a claim that a fraction of a person attends.

Check the result against the original information. Taking 60% of 30 gives 18. The ratio 3 : 2 compares girls with boys; it does not say that girls form three-halves of the whole class. The whole class includes both groups.

How is a percentage of a journey found?

For a picnic site 55 kilometres away, a first stop 22 kilometres from the school represents (22/55) × 100% = 40% of the outward journey. A kilometre, abbreviated km, is a unit of distance. The remaining percentage is 100% − 40% = 60%.

These calculations use the outward distance as the whole. If a question asks for transport costs for a return journey, both directions must instead be counted. Identify the whole from the wording; do not transfer a percentage calculation to a different whole without checking.

How are profit, loss and overhead expenses calculated?

Cost price, abbreviated CP, is the total cost of an article to the seller. Overhead expenses are additional expenses incurred after buying the article and included in its cost price. Therefore, cost price = buying price + overhead expenses.

Selling price, abbreviated SP, is the price at which the seller sells the article. A profit, also called a gain, is the excess of selling price over cost price. A loss is the excess of cost price over selling price.

Result: Profit and loss percentages use cost price

Profit = selling price − cost price; loss = cost price − selling price. Profit percentage = (profit / cost price) × 100%, and loss percentage = (loss / cost price) × 100%. Use the formula matching the transaction.

If selling price equals cost price, there is neither profit nor loss. Include any stated overhead expenses before calculating the difference or its percentage. A comparison based only on buying price would omit part of the seller's total cost.

Why must quantities match before prices are compared?

Worked example 3. Lemons are bought at ₹60 a dozen and sold at ₹40 for 10 lemons. Find the loss percentage. The symbol ₹ means rupees, and a dozen means 12 items.

Answer: Cost of one lemon = ₹60/12 = ₹5. Cost of 10 lemons = ₹50. Selling price of those 10 lemons = ₹40. Loss = ₹50 − ₹40 = ₹10. Loss percentage = (10/50) × 100% = 20%.

The two quoted prices initially refer to different numbers of lemons. Comparing ₹60 directly with ₹40 would give the wrong loss. The unitary method makes both prices refer to 10 lemons before the percentage formula is used.

As a check, a 20% loss on a dozen costing ₹60 is ₹12, leaving a selling price of ₹48 per dozen. That is ₹4 per lemon, which agrees with ₹40 for 10 lemons. This check returns to the original selling rate.

How do discounts differ from profit and loss?

A discount is a reduction in an article's marked price. The marked price, abbreviated MP, is its listed price before discount. Discount is generally offered to attract customers or promote sales. The sale price is the price after the discount has been deducted.

Result: Discount percentage uses marked price

Discount = marked price − sale price. Discount percentage = (discount / marked price) × 100%. If the percentage is supplied, discount = (discount percentage / 100) × marked price. Subtract this amount from the marked price to obtain the sale price.

Worked example 4. An item marked at ₹840 sells for ₹714. Find the discount and its percentage.

Answer: Discount = ₹840 − ₹714 = ₹126. The base is the marked price, so discount percentage = (126/840) × 100% = 15%. Dividing by ₹714 would use the wrong base.

A discount does not by itself establish a seller's loss. Discount compares sale price with marked price; profit or loss compares selling price with cost price. These are different comparisons, even when they concern the same article.

How can a discount and a profit occur together?

Worked example 5. An article marked at ₹280 is sold at a discount of 20%, yet gives a profit of 12%. Find its cost price.

Answer: Discount = (20/100) × ₹280 = ₹56. Selling price = ₹280 − ₹56 = ₹224. A 12% profit makes selling price 112% of cost price. Therefore cost price = ₹224 × 100/112 = ₹200.

Here the sale price is below the marked price but above the cost price. The first percentage uses ₹280; the second uses ₹200. When several percentages occur in one problem, attach the correct base to each calculation instead of treating them as interchangeable.

How do tax calculations change the amount paid?

In a sales tax calculation, tax is an amount charged on a sale and added to the selling price. For these problems, use the tax percentage stated in the question. The bill amount is the price together with the tax added to it.

How is tax added to a stated price?

Worked example 6. A pair of roller skates costs ₹450 before sales tax. The stated tax rate is 5%. Find the bill amount.

Answer: Tax = (5/100) × ₹450 = ₹22.50. Bill amount = ₹450 + ₹22.50 = ₹472.50. The price before tax is the base of the percentage calculation.

The bill in this calculation represents 105% of the price before tax: the original 100% plus the stated 5%. This relationship helps distinguish adding a tax from finding a price when the tax has already been included.

How is the price before tax recovered?

A tax-inclusive price already contains the tax. For a stated tax rate of 10%, every ₹100 of price before tax corresponds to ₹110 including tax. To recover the original price, work backwards through that relationship.

Worked example 7. An air cooler costs ₹3,300 including value added tax of 10%. Find its price before tax. Value added tax, abbreviated VAT, is the tax named in this calculation.

Answer: The inclusive price represents 110% of the original price. Price before tax = ₹3,300 × 100/110 = ₹3,000. The tax included is ₹3,300 − ₹3,000 = ₹300.

Taking 10% off the inclusive price would use a different base. The percentage applies to the price before tax, not to the final amount. Read “including tax” carefully, and identify whether the question asks for the base price, the tax, or the total bill.

How does simple interest depend on principal, rate and time?

Interest is money paid for the use of money, such as money borrowed or deposited. The principal is the original sum borrowed or deposited. Simple interest is calculated on that original principal for the entire stated time.

Let P denote principal, R the numerical annual percentage rate, and T time in years. Let SI denote simple interest and A the amount, meaning principal plus interest. “Per annum” means per year.

How is the formula obtained by the unitary method?

At R% per annum, the interest on ₹100 for one year is ₹R. The interest on principal P for one year is PR/100, where adjacent letters indicate multiplication. Multiplying this yearly interest by T gives the interest for T years.

SI = PRT/100

A = P + SI

Worked example 8. A sum of ₹10,000 is borrowed at 15% per annum simple interest for two years. Find the interest and the amount to be repaid.

Answer: Yearly interest = ₹10,000 × 15/100 = ₹1,500. Interest for two years = ₹1,500 × 2 = ₹3,000. Amount = ₹10,000 + ₹3,000 = ₹13,000.

What stays unchanged from year to year?

The calculation continues to use ₹10,000 as principal in the second year. The interest earned or owed in the first year is not added to the base for calculating simple interest. At an unchanged rate, equal periods therefore produce equal simple interest.

Distinguish the interest from the amount in the final sentence. The interest is the extra money; the amount includes the original principal as well. Writing only the interest when the amount is requested leaves out the original sum.

How is compound interest built up year by year?

With compound interest, interest is added to the principal at the end of each conversion period. A conversion period is the interval after which interest is added to form a new principal. With annual compounding, this interval is one year.

Let CI denote compound interest. Each year's closing amount becomes the next year's principal. The next interest calculation therefore includes interest already added. For the annual calculations here, work through up to three years.

How does the principal change?

Worked example 9. ₹20,000 is borrowed for two years at 8% per annum compounded annually. Find the amount and compound interest, and compare with simple interest at the same rate.

Answer: First-year interest = ₹20,000 × 8/100 = ₹1,600; amount = ₹21,600. Second-year interest = ₹21,600 × 8/100 = ₹1,728; final amount = ₹23,328. Compound interest = ₹3,328. Simple interest = ₹20,000 × 8 × 2/100 = ₹3,200, so compound interest is ₹128 more.

For ₹100 at 10% per annum, the following values show the different patterns. Each principal entry is the sum used to calculate that year's interest; each amount entry is the total accumulated by that year-end.

What explains the difference?

The first year's interest is the same under both methods. Later, compound interest uses the increased principal, while simple interest continues to use ₹100. After three years, the interest totals are ₹30 and ₹33.10 respectively. Compare like principals, rates and times.

How do patterns give the compound interest formula?

Adding one year's interest at R% multiplies the opening principal by the growth factor 1 + R/100. A growth factor is the number by which a quantity is multiplied to obtain its increased value. Repeating annual compounding repeats this multiplication.

Result: Annual compounding repeats the same factor

The superscripts ² and ³ indicate two and three repeated factors respectively.

1. After one year, the amount is P + PR/100 = P(1 + R/100).
2. After two years, multiply the first amount by the same factor, giving P(1 + R/100)².
3. After three years, multiply again, giving P(1 + R/100)³.

More generally, let n be the number of annual conversion periods. The superscript ⁿ means that the factor is used n times. Then the amount is A = P(1 + R/100)ⁿ, and CI = A − P. Use a constant annual rate throughout this pattern.

Worked example 10. Find compound interest on ₹12,600 for two years at 10% per annum compounded annually.

Answer: A = ₹12,600 × (1 + 10/100)² = ₹12,600 × 11/10 × 11/10 = ₹15,246. Therefore CI = ₹15,246 − ₹12,600 = ₹2,646.

What changes with half-yearly compounding?

Half-yearly compounding adds interest every six months. There are two conversion periods in one year. Halve the annual percentage rate for each period and count the half-years. The superscript ²ⁿ means 2 × n repeated factors. The amount for n years is A = P(1 + R/200)²ⁿ; use no more than three half-yearly steps here.

Worked example 11. ₹1,600 is lent at 5% per annum compounded half-yearly for one year. Find the amount.

Answer: Each half-year uses 2.5%. After the first half-year, amount = ₹1,600 × 1.025 = ₹1,640. After the second, amount = ₹1,640 × 1.025 = ₹1,681.

The annual rate remains 5%; 2.5% is the rate used for each six-month calculation. Both the rate per period and the number of periods must match the chosen compounding interval. Altering just one gives a different calculation.

How can direct variation be recognised and used?

Direct variation, or direct proportion, occurs when two quantities change together while the ratio of their corresponding values stays constant. A constant is a value that remains unchanged in the relationship. An increase in both quantities alone does not establish direct proportion.

Let x and y represent the two positive quantities being compared, and let k be their constant ratio. Direct proportion can be written x/y = k, or x = ky. “Corresponding” means belonging to the same case or observation.

Property: Corresponding ratios remain equal

Let x₁ and y₁ represent a first pair of corresponding values, and x₂ and y₂ a second pair. The small numerals distinguish the pairs. In direct proportion, x₁/y₁ = x₂/y₂. Keep the same kind of quantity above each fraction bar.

Worked example 12. Five metres of cloth cost ₹210. Find the costs of 2, 4, 10 and 13 metres of the same cloth at the same price per metre. A metre, abbreviated m, is a unit of length.

Answer: One metre costs ₹210/5 = ₹42. The required costs are 2 × ₹42 = ₹84; 4 × ₹42 = ₹168; 10 × ₹42 = ₹420; and 13 × ₹42 = ₹546.

Here doubling the cloth length doubles its cost because the price per metre is unchanged. Dividing cost by length gives the same result for every pair. The unitary method and the equal-ratio method express the same relationship.

Which conditions must stay the same?

For travel at a uniform speed, meaning an unchanged distance covered per unit time, distance is directly proportional to time. Changing the speed would change the relationship. Likewise, a fixed price per metre is essential to the cloth calculation.

Before using direct proportion, identify the quantities and explain what remains fixed. Then calculate the missing value and check that the direction of change makes sense. A longer length at the same price per metre should cost more, not less.

How does inverse variation differ from direct variation?

Inverse variation, or inverse proportion, occurs when one positive quantity increases and the other decreases proportionately, keeping their product constant. A product is the result of multiplication. Opposite directions of change alone are insufficient; the constant-product condition must also hold.

Property: Corresponding products remain equal

Using x and y for the two quantities and k for the constant product, inverse proportion is written xy = k. For corresponding pairs, x₁y₁ = x₂y₂. Thus multiplying one quantity by a factor divides the other by that same factor.

Consider a fixed budget of ₹6,000 for books. The following table gives the price of each book and the number that can be bought. The fixed budget is the condition that makes the relationship inverse.

For example, 40 × 150 = 50 × 120 = 6,000. More expensive books mean fewer books can be bought with the same money. Direct proportion would instead require a constant ratio, which is a different test.

How are fixed food provisions shared over time?

Worked example 13. A hostel has food for 100 students for 20 days. If 25 more students join, how long will it last, assuming unchanged food consumption per student per day?

Answer: The new total is 100 + 25 = 125 students. Students × days remains constant. Required days = (100 × 20)/125 = 16 days.

The increase in students is 25, but the new number receiving food is 125. Use the total in the equation. The answer is fewer than 20 days, as expected when the same provisions serve more students at the same daily rate.

State the condition about consumption: without it, the number of students would not determine how long the food lasts. The fixed product describes a specific relationship, not every situation in which one quantity rises while another falls.

How are simple time and work problems solved?

For a fixed task, workers of equal and unchanged efficiency take less time when more workers work together. Efficiency here means the amount of work completed by a worker in a given time. Under these conditions, the number of workers and completion time vary inversely.

How is the constant amount of work represented?

The product of the number of workers and their working time remains constant for the same task. Keep time units consistent, and assume each worker contributes at the same rate throughout. The fixed task must be identical in both cases.

Worked example 14. Fifteen workers build a wall in 48 hours. How many workers of the same efficiency are needed to build the same wall in 30 hours, working at unchanged rates?

Answer: The constant product is 15 × 48 = 720 worker-hours, where a worker-hour is one worker's work for one hour. Required workers = 720/30 = 24 workers. The shorter time requires more workers.

How does the same method apply to pipes?

Worked example 15. Six identical pipes fill a tank in 1 hour 20 minutes. How long will five such pipes take, each delivering water at the same constant rate, with no leakage?

Answer: One hour equals 60 minutes, so the original time is 80 minutes. Required time = (6 × 80)/5 = 96 minutes, or 1 hour 36 minutes.

Here the tank capacity is unchanged, and the pipe rates are unchanged. Fewer pipes give a smaller combined filling rate, so a longer time is needed. Converting the original time into minutes makes the multiplication consistent.

1. Identify the fixed task or quantity and the two quantities that vary.
2. Check the stated rates or efficiencies and convert times to a common unit.
3. Use equal products for inverse proportion to calculate the unknown quantity.
4. State the answer with its unit and check whether the increase or decrease is sensible.

The same reasoning links the wall, tank and food problems. In each case, the product stays fixed because the total requirement is fixed. Make that connection before calculating, rather than deciding from the numbers alone.

Glossary

• Ratio — A comparison made by dividing one quantity by another, keeping their order clear.
• Proportion — A statement that two ratios express the same comparison and are equal.
• Percentage — A comparison per hundred, written using the symbol % after its numerical value.
• Unitary method — Finding a value for one unit before finding the value for the required quantity.
• Cost price — The seller's total cost, including the buying price and any stated overhead expenses.
• Overhead expenses — Additional expenses incurred after buying an article and included in its cost price.
• Marked price — The listed price of an article before a discount is deducted from it.
• Discount — A reduction from marked price, calculated using marked price as the percentage base.
• Principal — The original sum borrowed or deposited on which interest calculations begin.
• Simple interest — Interest calculated on the original principal throughout the stated period of time.
• Compound interest — Interest accumulated when each conversion period's interest is added to form the next principal.
• Conversion period — The interval after which interest is added to form a new principal.
• Direct variation — A relationship in which corresponding values of two quantities have a constant ratio.
• Inverse variation — A relationship in which corresponding values of two quantities have a constant product.

Common errors and misconceptions

• Misconception: The ratio of oranges to apples is also the fraction of all fruits that are oranges. Correct: The first comparison uses apples as its base; the second uses all the fruits.
• Misconception: Profit percentage and discount percentage use the same base. Correct: Profit percentage uses cost price, including overhead expenses; discount percentage uses marked price.
• Misconception: A discount means the seller makes a loss. Correct: Discount reduces marked price, but the resulting selling price can still exceed cost price.
• Misconception: Subtract the tax percentage from a tax-inclusive price to recover the original price. Correct: Relate the inclusive price to 100% plus the stated tax percentage, then work backwards.
• Misconception: Compound interest uses the original principal every year. Correct: Each closing amount becomes the principal for the following conversion period.
• Misconception: Half-yearly compounding uses the full annual rate every six months. Correct: Use half the annual rate per half-year and count the half-yearly periods.
• Misconception: Quantities increasing together must be directly proportional. Correct: Their corresponding ratios must remain constant; for inverse proportion, test corresponding products instead.
• Misconception: More workers require more time for the same task. Correct: With equal, unchanged efficiency, more workers require less time, and workers multiplied by time stays constant.

Exam-style questions with model answers

Q1. A basket contains 20 apples and 5 oranges. Find the ratio of oranges to apples and the percentage of all fruits that are oranges. [2 marks]

1. Oranges to apples = 5 : 20 = 1 : 4, keeping the requested order of comparison.
2. Total fruits = 25, so the percentage of oranges is (5/25) × 100% = 20%.

Q2. Eighteen girls form 60% of the students in a class. Find the total number of students, the number of boys and the ratio of girls to boys. [3 marks]

1. The whole class represents 100%. Therefore the total number of students is 18 × 100/60 = 30 students.
2. Subtract the 18 girls from the total of 30 students. The number of boys is 30 − 18 = 12.
3. Girls to boys is 18 : 12. Dividing both numbers by 6 gives the required ratio, 3 : 2.

Q3. Lemons are bought at ₹60 a dozen, where a dozen means 12, and sold at ₹40 for 10 lemons. Find the loss percentage. [3 marks]

1. Compare equal quantities. One lemon costs ₹60/12 = ₹5, so the cost price of 10 lemons is ₹50.
2. Those 10 lemons sell for ₹40, which is less than their cost price. The loss is ₹50 − ₹40 = ₹10.
3. Loss percentage uses cost price as its base. Therefore loss percentage = (10/50) × 100% = 20%, calculated for the same 10 lemons.

Q4. An article marked at ₹280 is sold at a discount of 20%. The seller makes a profit of 12% on cost price. Find the discount, selling price and cost price. [4 marks]

1. Use marked price for the discount calculation: discount = (20/100) × ₹280 = ₹56.
2. Subtract this reduction from the marked price. Selling price = ₹280 − ₹56 = ₹224.
3. A profit of 12% means that the selling price represents 112% of cost price, since the cost itself represents 100%.
4. Hence cost price = ₹224 × 100/112 = ₹200. This is below the selling price, consistent with a profit.

Q5. An air cooler costs ₹3,300 including value added tax of 10% on its price before tax. Find the price before tax and the tax included. [3 marks]

1. The price before tax represents 100%, while the tax-inclusive price represents 110%. The given ₹3,300 therefore corresponds to 110%.
2. Recover the base price by the unitary method: price before tax = ₹3,300 × 100/110 = ₹3,000.
3. The tax included is ₹3,300 − ₹3,000 = ₹300. This is 10% of ₹3,000, confirming that the percentage uses the price before tax.

Q6. A loan of ₹20,000 runs for two years at 8% per annum compounded annually. Find each year's interest, the final amount and total compound interest. Compare the compound interest with simple interest on the same principal at the same rate for the same time. [5 marks]

1. First-year interest is 8% of the original ₹20,000: ₹20,000 × 8/100 = ₹1,600. This interest is added at the end of the year.
2. The first closing amount is ₹20,000 + ₹1,600 = ₹21,600. Annual compounding makes this the principal for the second year.
3. Second-year interest is ₹21,600 × 8/100 = ₹1,728. The larger interest follows from using the increased principal, while the rate stays unchanged.
4. Final amount = ₹21,600 + ₹1,728 = ₹23,328. Total compound interest = ₹23,328 − ₹20,000 = ₹3,328.
5. Simple interest = ₹20,000 × 8 × 2/100 = ₹3,200. Thus compound interest exceeds simple interest by ₹3,328 − ₹3,200 = ₹128.

Q7. ₹1,600 is lent at 5% per annum compounded half-yearly for one year. Calculate the final amount, showing the rate per period and both compounding steps. [4 marks]

1. Half-yearly compounding gives two six-month periods in one year. The rate per period is half of 5%, which is 2.5%.
2. First-period interest = ₹1,600 × 2.5/100 = ₹40, so the amount after six months is ₹1,640.
3. Use ₹1,640 as the new principal. Second-period interest = ₹1,640 × 2.5/100 = ₹41.
4. The final amount is ₹1,640 + ₹41 = ₹1,681, including the original principal and interest from both periods.

Q8. Five metres of cloth cost ₹210. At the same price per metre, find the cost of 13 metres. Explain the type of proportion used. [3 marks]

1. Cloth length and cost vary directly because the price per metre is unchanged. Their corresponding ratio therefore remains constant.
2. Use the unitary method to find the cost of one metre: ₹210/5 = ₹42 per metre.
3. Thirteen metres cost 13 × ₹42 = ₹546. This is greater than ₹210, as expected for a longer length bought at the same price per metre.

Q9. Fifteen workers build a wall in 48 hours. How many workers are needed to build the same wall in 30 hours? Assume all workers have equal efficiency and work at unchanged rates. [3 marks]

1. For the same wall and equal, unchanged efficiency, the number of workers and the required time are inversely proportional.
2. The constant product is workers × hours = 15 × 48 = 720 worker-hours. Use this same product for the new arrangement.
3. Required workers = 720/30 = 24. More workers than the original 15 are needed because the permitted completion time is shorter.

Key takeaways

• A ratio compares quantities in a stated order; changing the order or the base changes the meaning of the comparison.
• The unitary method finds the value for one unit first, then uses that value to obtain the required quantity.
• Include stated overhead expenses in cost price, and compare equal quantities before calculating profit or loss percentages.
• Discount percentage uses marked price; profit and loss percentages use cost price. Identify the base for each percentage separately.
• A tax-inclusive price represents the original 100% plus tax. Work backwards through that relationship to recover the original price.
• Simple interest uses the original principal throughout; compound interest uses the previous conversion period's closing amount as the new principal.
• Half-yearly compounding requires half the annual percentage rate per period and two conversion periods for each year.
• Direct proportion preserves corresponding ratios, whereas inverse proportion preserves corresponding products. State the conditions that keep the relationship valid.

Test yourself

What does the colon mean in the ratio 1 : 4?

It means “is to”, so the ratio is read as “one is to four”.

Why is the oranges-to-apples ratio different from the fraction of all fruits that are oranges?

The ratio uses the number of apples as its base; the fraction uses the total number of fruits.

What must be added to buying price before profit is calculated?

Add any stated overhead expenses to obtain the full cost price used in the profit calculation.

Can a seller offer a discount and still make a profit?

Yes. The discounted selling price can remain above cost price even though it is below marked price.

Why does a price including 10% tax represent 110%?

The original price represents 100%, and adding tax of 10% of that price gives 110%.

What becomes the principal for the second year of annual compound interest?

The amount at the end of the first year becomes the principal for the second year.

What rate applies each half-year when the annual rate is 5%?

The rate is 2.5% per half-year, and one year contains two such conversion periods.

What numerical test distinguishes direct from inverse proportion?

Direct proportion has equal corresponding ratios; inverse proportion has equal corresponding products under the stated conditions.