Overview
Introduction: "Comparing Quantities" is a practical chapter that teaches students how to measure and compare amounts using ratios, fractions, decimals and especially percentages. It develops tools to express how one quantity relates to another and how quantities change — increases, decreases or undergo repeated (successive) changes. Importance: Percentages and related ideas are used daily (discounts, taxes, interest, profit/loss, population growth, data interpretation). Mastery of this chapter builds numerical sense, improves problem solving and prepares students for algebraic manipulation and finance-related topics. Key themes: converting between fractions, decimals and percentages; understanding percent as ‘per hundred’; calculating percentage increase and decrease; successive percentage changes; using percentage to find original or final quantities; and practical applications such as discounts, profit & loss, tax/GST and interest (simple and compound). What the student will learn: Students will learn to convert among different representations, compute and interpret percentages, set up and solve word problems using percentage and ratios, handle successive percentage changes…
Learning Objectives
- Define percentage and express a given fraction or decimal as a percentage and vice versa
- Convert between fractions, decimals and percentages fluently for application in problems
- Explain percentage increase and percentage decrease and state their underlying formulas
- Calculate percentage increase or decrease for single-step and successive change situations
- Apply the concept of percentage to find the original (base) value when final value and percent change are given
- Solve problems on discounts: compute discount amount, discount percentage, marked price and selling price
- Compute profit, loss, profit percentage and loss percentage given cost price and selling price
- Determine final payable amount by applying taxes or additional charges (e.g., GST) expressed as percentages
Topics in this chapter
8 topics · tap a topic title to jump straight to it.
Ratio and Proportion (Revision)
Ratio compares two quantities of the same kind. If two quantities are a and b (b ≠ 0), their ratio is written as a:b or a/b. A ratio can be simplified to its lowest terms by dividing both terms by their greatest common divisor (GCD).
Equivalent ratios are ratios that have equal fractional value: a:b = c:d if a/b = c/d. To get equivalent ratios multiply or divide both terms by the same nonzero number.
Proportion is an equality of two ratios. a:b = c:d is a proportion. In any proportion the product of extremes equals the product of means (cross-multiplication): a × d = b × c.
Types and ideas to remember:
- Simplest form: divide both terms by GCD.
- Unitary method: find the value of one unit then scale up or down.
- Direct proportion: y ∝ x (y = kx). Graph is a straight line through origin; ratio y/x is constant.
- Inverse proportion: y ∝ 1/x (xy = k). As x increases y decreases; graph is a rectangular hyperbola.
- Continued proportion: a:b = b:c means b^2 = ac.
How to solve typical problems:
- Use cross-multiplication for a proportion: a:b = c:d ⇒ ad = bc, solve the unknown.
- Use unitary method for sharing, price, speed: find per-unit quantity then multiply by needed units.
- For inverse problems (more workers → less time), use xy = constant (workers × days = constant).
Common classroom applications: sharing amounts in a given ratio, map scale conversions, mixing solutions, recipe adjustments, speed/worker problems, percentage conversions (percent = ratio to 100).
Tips: always check units, reduce ratios to simplest form to compare, draw double-number-line or bar models for clarity, and verify by plugging back into the original ratio or proportion.
- 1) Simplify 42:56. GCD(42,56)=14 → 42:56 = (42÷14):(56÷14) = 3:4.
- 2) Solve 5:8 = x:32. Cross-multiply: 5×32 = 8×x ⇒ 160 = 8x ⇒ x = 20.
- 3) Share Rs 600 in ratio 3:2:5. Sum parts = 3+2+5 = 10. One part = 600÷10 = 60. Shares: 3×60=180, 2×60=120, 5×60=300.
- 4) Direct proportion (price): 4 kg apples cost Rs 120. Price per kg = 120÷4 = 30. Cost of 7 kg = 7×30 = Rs 210.
- 5) Inverse proportion (work): 6 men finish work in 15 days. How many days for 10 men? Assume work constant: men×days = constant ⇒ 6×15 = 10×d ⇒ d = (6×15)/10 = 9 days.
- 6) Continued proportion: 2 : x = x : 8 ⇒ x^2 = 2×8 = 16 ⇒ x = 4 (positive value for ratio).
- Ratio: a:b = a/b
- Simplest form: divide both terms by gcd(a,b)
- Equivalent ratios: a:b = (ka):(kb) for k ≠ 0
- Proportion: a:b = c:d ⇒ a×d = b×c (cross-multiplication)
- Continued proportion: a:b = b:c ⇒ b^2 = a×c
- Direct proportion: y ∝ x ⇒ y = kx (k = y/x constant)
Percent (Per Cent)
Meaning: "Percent" (written %) means "per hundred". 1% = 1/100. Percent tells how many parts out of 100.
Converting between fraction, decimal and percent:
- Fraction to percent: multiply by 100. Example: 3/5 = (3/5)×100% = 60%.
- Decimal to percent: multiply by 100. Example: 0.25 = 25%.
- Percent to decimal: divide by 100. Example: 15% = 0.15.
Finding percent of a quantity: To find P% of a number N, compute (P/100) × N.
Finding the whole when a percent is given: If P% of x = y, then x = y × (100/P).
Percent change (increase / decrease): Percent change = (Change / Original) × 100. Use positive sign for increase, negative for decrease.
Successive percentages: Apply each percent as a multiplication factor. An increase of a% → multiply by (1 + a/100). A decrease of a% → multiply by (1 − a/100). For successive changes multiply the factors.
Important distinction: A change of 2 percentage points (e.g., from 5% to 7%) is not the same as a 40% increase: percentage points = difference in % values (2 points); percent change = (2/5)×100 = 40%.
Short methods and tips:
- Use the 100-square (grid of 100 small squares) to visualise percentages quickly.
- Convert to fractions with denominator 100 for quick comparison.
- For repeated percent increase/decrease use multiplication factors rather than adding/subtracting percentages directly.
- Example 1: What is 30% of 200? Solution: (30/100)×200 = 0.30×200 = 60.
- Example 2: What percent is 45 of 150? Solution: (45/150)×100 = 0.3×100 = 30%.
- Example 3: 20% of a number is 50. Find the number. Solution: number = 50×(100/20) = 50×5 = 250.
- Example 4: A price rises from ₹200 to ₹250. Find percent increase. Change = 50. Percent increase = (50/200)×100 = 25%.
- Example 5: Successive change — price increases by 10% then by 20%. Net factor = 1.10×1.20 = 1.32 → net increase 32%.
- Percent as part of whole: Percent = (Part / Whole) × 100
- Part from percent: Part = (Percent / 100) × Whole
- Whole from part: Whole = Part × (100 / Percent)
- Percent change: Percent change = (Change / Original) × 100
- Successive change factor: Final = Initial × (1 ± p1/100) × (1 ± p2/100) ...
- Convert: Percent → Decimal: divide by 100. Decimal → Percent: multiply by 100.
Increase and Decrease of Quantities (Percentage Change)
Percentage change measures how much a quantity grows (increase) or shrinks (decrease) relative to its original value, expressed as a percentage. It is useful for comparing changes across different contexts like prices, population, marks or salaries.
Key ideas
- Increase: When a new value is greater than the original, the percentage increase = (increase ÷ original) × 100.
- Decrease: When a new value is less than the original, the percentage decrease = (decrease ÷ original) × 100.
- Direct computation of new value: For an increase of r%, new = original × (1 + r/100). For a decrease of r%, new = original × (1 − r/100).
- Reverse problem: If you know the new value after a change, the original = new ÷ (1 ± r/100) where + is for increase, − for decrease.
- Successive changes (compound effect): If a quantity changes by r1% then r2%, the overall multiplier = (1 ± r1/100) × (1 ± r2/100). The net percentage change = (overall multiplier − 1) × 100.
Important note: Equal percentage increase and decrease do not cancel out. For example, a 20% increase followed by a 20% decrease does not return to the original value because the base values differ.
Worked mini-example inside explanation: An article priced at ₹250 is increased to ₹300. Increase = 300 − 250 = ₹50. Percentage increase = (50 ÷ 250) × 100 = 20%. Alternatively, new = 250 × (1 + 20/100) = 250 × 1.2 = 300.
- 1) Price increase: A bag costs ₹250 and its price rises to ₹300. Increase = 300 − 250 = ₹50. Percentage increase = (50/250)×100 = 20%.
- 2) Discount (decrease): A shirt marked ₹1200 is sold at 25% discount. Discount amount = 25% of 1200 = 0.25×1200 = ₹300. Selling price = 1200 − 300 = ₹900.
- 3) Reverse problem: After a 20% decrease the price becomes ₹800. Original price = 800 ÷ (1 − 20/100) = 800 ÷ 0.8 = ₹1000.
- 4) Successive changes: An item first increases by 10% then decreases by 20%. Overall multiplier = 1.10 × 0.80 = 0.88. Net change = (0.88 − 1)×100 = −12%. So overall a 12% decrease.
- 5) Compound growth example: A town's population grows 5% each year for 2 years. New population factor = 1.05 × 1.05 = 1.1025. Net increase = 10.25% (not exactly 10%).
- Percentage increase = (Increase ÷ Original) × 100
- Percentage decrease = (Decrease ÷ Original) × 100
- New value after r% increase = Original × (1 + r/100)
- New value after r% decrease = Original × (1 − r/100)
- Original = New ÷ (1 + r/100) (if r% was an increase)
- Original = New ÷ (1 − r/100) (if r% was a decrease)
Successive Percentage Change
Successive percentage change means applying two or more percentage increases or decreases one after the other on an original quantity. Instead of adding or subtracting the percentages directly, each change is applied to the new value obtained after the previous change.
Key idea: convert each percentage change to a multiplication factor and multiply the factors. For a change of +p%, use the factor (1 + p/100). For a change of −p% (a decrease), use the factor (1 − p/100). If changes p, q, r ... are applied successively, the final value = original value × (1 + p/100) × (1 + q/100) × (1 + r/100) ...
Net percentage change = (product of factors − 1) × 100%. If the product > 1, there is a net increase; if < 1, a net decrease.
Important notes:
- Order does not affect the final result because multiplication is commutative: applying +10% then −20% gives the same final value as −20% then +10%.
- Two equal opposite percent changes do not cancel: +10% then −10% does not return to the original value. Example: 1.1 × 0.9 = 0.99 → a 1% net decrease.
- Percentage points are different from percent change. Do not confuse them.
- Example 1 — Increase 10% then 20%: Start with 100. Factor for +10% = 1.10, for +20% = 1.20. Final = 100 × 1.10 × 1.20 = 132. Net change = (132/100 − 1) × 100% = 32% increase.
- Example 2 — Increase 10% then decrease 10%: Start with 100. Factors = 1.10 and 0.90. Final = 100 × 1.10 × 0.90 = 99. Net change = (99/100 − 1) × 100% = −1% (a 1% decrease).
- Example 3 — Decrease 25% then increase 12%: Start with 100. Factors = 0.75 and 1.12. Final = 100 × 0.75 × 1.12 = 84. Net change = −16% (price falls from 100 to 84).
- Example 4 — Same percent applied n times: A value increases by 10% twice. Factor = (1 + 10/100)^2 = 1.1^2 = 1.21. Net increase = 21%. If repeated n times, final factor = (1 + p/100)^n.
- Factor for +p% = 1 + p/100. For −p% = 1 − p/100.
- Final value after successive changes p, q, r... = Original × (1 + p/100) × (1 + q/100) × (1 + r/100) ...
- Net percentage change (%) = [ (product of factors) − 1 ] × 100
- If same percentage p is applied n times: Net change (%) = [ (1 + p/100)^n − 1 ] × 100
Profit, Loss and Discount (Applications of Percent)
Basic terms
Cost Price (CP): the price at which an article is bought.
Selling Price (SP): the price at which it is sold.
Marked Price (MP): the printed price on an article (also called list price).
Profit and Loss
If SP > CP, there is a Profit = SP - CP. If SP < CP, there is a Loss = CP - SP.
Profit and loss are usually expressed as a percentage of CP:
- Profit% = (Profit / CP) × 100
- Loss% = (Loss / CP) × 100
Useful rearrangements:
- SP = CP + Profit = CP × (1 + profit%/100)
- SP = CP - Loss = CP × (1 - loss%/100)
- CP = SP / (1 + profit%/100) (when profit% is given)
- CP = SP / (1 - loss%/100) (when loss% is given)
Discount
Discount = MP - SP. Discount% is usually calculated on MP:
- Discount% = (Discount / MP) × 100
- SP after a discount d% on MP: SP = MP × (1 - d/100)
Combining markup (on CP) and discount (on MP)
If an article is marked up by m% on CP (so MP = CP × (1 + m/100)) and then sold at a discount d% on MP, the final SP is
SP = CP × (1 + m/100) × (1 - d/100). Therefore net profit% = [ (1 + m/100)(1 - d/100) - 1 ] × 100.
Key points for students
- Always note which base percentage is taken on: profit/loss are on CP; discount is on MP.
- When percentages are small, you can use multipliers: multiply CP by (1 ± percent/100) to get SP, MP, or CP.
- For successive discounts d1% and d2% on the same MP, effective multiplier is (1 - d1/100)(1 - d2/100).
- 1) Profit example: A shopkeeper buys a toy for Rs. 450 and sells it for Rs. 540. Find profit and profit%. Solution: Profit = SP - CP = 540 - 450 = Rs. 90. Profit% = (90/450)×100 = 20%.
- 2) Loss example: A book is bought for Rs. 150 and sold for Rs. 120. Find loss and loss%. Solution: Loss = 150 - 120 = Rs. 30. Loss% = (30/150)×100 = 20%.
- 3) Discount example: A jacket is marked Rs. 800 and sold at 25% discount. Find selling price and discount amount. Solution: Discount = 25% of 800 = 0.25×800 = Rs. 200. SP = 800 - 200 = Rs. 600.
- 4) Finding CP from SP and profit%: An article is sold for Rs. 264 after a profit of 10%. Find its cost price. Solution: SP = CP×(1+10/100)=1.1×CP. So CP = SP/1.1 = 264/1.1 = Rs. 240.
- 5) Markup and discount combined: An article costs Rs. 100. It is marked 20% above cost and later sold at 10% discount on the marked price. Find the final profit%. Solution: MP = 100×1.20 = Rs. 120. SP = 120×0.90 = Rs. 108. Profit = 108 - 100 = Rs. 8. Profit% = (8/100)×100 = 8%.
- Profit = SP - CP
- Loss = CP - SP
- Profit% = (Profit / CP) × 100
- Loss% = (Loss / CP) × 100
- SP = CP + Profit = CP × (1 + profit%/100)
- SP = CP - Loss = CP × (1 - loss%/100)
Simple Interest
What is Simple Interest?
Simple Interest (SI) is the interest calculated only on the original principal amount (the initial sum of money) for the entire period of the loan or investment. It does not consider interest on interest.
Formula and meaning of symbols
SI = (P × R × T) / 100
where P = principal (initial amount), R = rate of interest per annum (in %), T = time (in years). The total amount A to be paid or received at the end of the period is A = P + SI.
Important points
- Rate R is usually given per annum. If time is given in months, convert to years: T (in years) = months / 12.
- SI is directly proportional to P, R and T individually. Doubling any one (keeping others fixed) doubles the SI.
- Simple interest gives a linear growth of interest with time (no compounding).
How to solve problems (step-by-step)
- Identify P, R and T (convert units if needed).
- Substitute into SI = (P × R × T)/100 and compute SI.
- If asked for the total amount, compute A = P + SI.
- Example 1 — Loan interest: Ria borrows ₹6,000 at 5% per annum for 3 years. SI = (6000 × 5 × 3)/100 = (6000 × 15)/100 = 900. Total amount = 6000 + 900 = ₹6,900.
- Example 2 — Short deposit: Amit deposits ₹10,000 for 6 months at 8% per annum. Convert time: T = 6/12 = 0.5 years. SI = (10000 × 8 × 0.5)/100 = (10000 × 4)/100 = 400. Amount = ₹10,400.
- Example 3 — Find missing value: A sum lent at 6% per annum for 4 years yields ₹720 as simple interest. Find the principal. Use P = (SI × 100) / (R × T) = (720 × 100) / (6 × 4) = 72000 / 24 = ₹3,000.
- SI = (P × R × T) / 100
- Amount A = P + SI
- P = (SI × 100) / (R × T)
- R = (SI × 100) / (P × T)
- T = (SI × 100) / (P × R)
- When time in months: T(years) = months / 12 (use in SI formula)
Compound Interest (Introductory)
What is Compound Interest?
Compound interest is the interest calculated on the original principal and also on the interest that accumulates each period. In other words, each time interest is added to the principal, the new total becomes the base for calculating interest in the next period. This causes the amount to grow faster than with simple interest.
Key idea: Interest earns interest.
Basic terms:
- Principal (P) — the initial amount of money invested or borrowed.
- Rate (r) — annual interest rate in percent.
- Time (n) — number of years (periods) the money is invested or borrowed.
- Amount (A) — total money after n years (principal + interest).
- Compound Interest (CI) — the difference between the amount and the principal: CI = A − P.
How it differs from Simple Interest
With simple interest (SI), interest is computed only on the original principal every year. The amount with SI grows linearly. With compound interest, the growth is exponential because interest each year is added to the principal for the next year's calculation.
When is compounding done? Usually compounding is done yearly (annual compounding) at the end of each year. For introductory Class 8 level we normally consider annual compounding. (Later you learn about semi-annual, quarterly or monthly compounding.)
- Example 1 — Annual compounding (simple numbers): If Rs. 1000 is invested at 10% per annum compounded annually for 2 years, amount A = 1000 × (1 + 10/100)^2 = 1000 × (1.1)^2 = 1000 × 1.21 = Rs. 1210. Compound interest CI = 1210 − 1000 = Rs. 210.
- Example 2 — Three years: If Rs. 5000 is invested at 8% per annum compounded annually for 3 years, A = 5000 × (1 + 8/100)^3 = 5000 × (1.08)^3 = 5000 × 1.259712 = Rs. 6298.56 (approx). CI = 6298.56 − 5000 = Rs. 1298.56 (approx).
- Example 3 — Compare SI and CI for same P, r, n: P = 1000, r = 10%, n = 2 years. SI amount = 1000 × (1 + 2×10/100) = 1000 × 1.20 = Rs. 1200 (SI). CI amount = 1000 × (1.1)^2 = Rs. 1210 (CI). Difference = 10. This shows CI gives a slightly higher amount because interest in year 2 is calculated on interest earned in year 1 as well.
- Amount after n years (annual compounding): A = P × (1 + r/100)^n
- Compound Interest: CI = A − P = P × [ (1 + r/100)^n − 1 ]
- If interest is compounded m times a year (general formula): A = P × (1 + r/(100m))^(m n). (For Class 8, m = 1 is usually used.)
Applications and Word Problems
What this topic covers
Applications and Word Problems in the chapter Comparing Quantities teach how to use percentages, percentage increase/decrease, successive changes, and related ideas (discount, profit & loss, mark-up) to solve real-life problems. The main skill is to identify the base (reference) quantity, convert percent to a multiplier, and apply algebra or arithmetic carefully.
Key ideas & methods
- Always decide what is the original/base quantity (the denominator for percentage).
- Convert percent to decimal/multiplier: p% = p/100 and the multiplier for an increase of p% is (1 + p/100); for a decrease it is (1 − p/100).
- For successive changes, multiply the respective multipliers (changes are not simply added).
- To find the original when you know the final after a percent change, divide by the multiplier: original = final / (1 ± p/100).
- For profit and loss: profit% = (Profit / Cost Price) × 100, loss% = (Loss / Cost Price) × 100. For discounts: Selling Price = Marked Price × (1 − discount%).
Strategy for word problems
- Read carefully and identify what is asked (new value, original value, percentage, or comparison).
- Identify the base (original) quantity and whether change is increase or decrease.
- Write the multiplier or formula and compute step by step (for successive steps multiply multipliers).
- Check units and reasonableness (e.g., after a 10% increase value should be larger than original).
Common pitfalls
- Mixing up base: percent change is always relative to the original (or specified) base.
- Adding successive percentages directly — instead multiply factors.
- Confusing percentage points (difference of percentages) with percent change (relative change).
- 1) Single percentage increase: A bike priced at Rs. 12,000 is increased by 15%. Find new price. (Multiplier = 1 + 15/100 = 1.15; New price = 12000 × 1.15 = Rs. 13,800.)
- 2) Successive percentage changes: A watch increases 10% in year 1 and then 20% in year 2. Net increase? (Net multiplier = 1.10 × 1.20 = 1.32 → 32% net increase.)
- 3) Finding original from final: After a 25% discount, a shirt sells for Rs. 1,125. What was the marked price? (Selling price = MP × (1 − 0.25) = 0.75 MP. So MP = 1125 ÷ 0.75 = Rs. 1,500.)
- 4) Profit percent: A shopkeeper buys an item for Rs. 480 and sells at Rs. 600. Find profit% (Profit = 120; Profit% = (120/480) × 100 = 25%.)
- 5) Successive discount: Two successive discounts of 10% and 20% on an article marked Rs. 2,000. Final price? (Multiplier = 0.90 × 0.80 = 0.72; Final price = 2000 × 0.72 = Rs. 1,440.)
- 6) Comparing quantities: A is 25% more than B. If B = 400, what percent of A is B? (A = 1.25 × B = 500. B as percent of A = (400/500) × 100 = 80%.)
- Percent change = ((new − original) / original) × 100
- New value after p% increase = original × (1 + p/100)
- New value after p% decrease = original × (1 − p/100)
- Successive changes: final = original × Π(1 ± p_i/100) for each change p_i
- Original from final after p% change: original = final / (1 ± p/100)
- Profit% = (Selling Price − Cost Price) / Cost Price × 100
Key Concepts
- Ratio
- A comparison of two quantities by division, written a:b or a/b, showing how many times one value contains the other.
- Simplest Form of a Ratio
- A ratio expressed so that the two terms have no common factor other than 1.
- Equivalent Ratios
- Two or more ratios that express the same relationship; obtained by multiplying or dividing both terms of a ratio by the same nonzero number.
- Proportion
- An equality of two ratios; written as a:b = c:d meaning a/b = c/d.
- Percentage
- A way of expressing a number as a fraction of 100; marked by the symbol % (percent means 'per hundred').
- Percentage Increase
- The percent by which a quantity increases compared to its original (base) value: (increase/original) × 100%.
- Percentage Decrease
- The percent by which a quantity decreases compared to its original value: (decrease/original) × 100%.
- Successive Percentage
- Applying percentages one after another on a quantity; percentages compound multiplicatively, not additively.
- Conversion between Fraction, Decimal and Percentage
- Methods to change representations: percent = fraction×100, decimal = percent/100, fraction = percent/100.
- Cost Price (CP)
- The price at which an item is purchased (the seller's cost).
- Selling Price (SP)
- The price at which an item is sold to a buyer.
- Marked Price (MP)
- The price printed or labeled on an item (list price) before any discounts.
- Discount
- A reduction given on the marked price; discount = MP − SP (when SP is after discount).
- Discount Rate
- The percentage of the marked price reduced as discount: (discount/MP) × 100%.
- Profit
- When SP > CP, the seller gains: Profit = SP − CP.
- Loss
- When SP < CP, the seller incurs a loss: Loss = CP − SP.
- Profit Percent
- Profit expressed as a percentage of cost price: (Profit/CP) × 100%.
- Loss Percent
- Loss expressed as a percentage of cost price: (Loss/CP) × 100%.
- Tax (e.g., GST) and Final Price
- A government-imposed charge on goods (tax rate%). Final price = price + tax on that price. GST example: tax applied to selling price or marked price as specified.
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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Which of the following is the correct formula for percentage increase? / निम्नलिखित में से प्रतिशत वृद्धि का सही सूत्र कौन-सा है? (a) (Decrease / Original) × 100 / (कमी / मूल मान) × 100 (b) (Increase / Original) × 100 / (वृद्धि / मूल मान) × 100 (c) (Increase / New Value) × 100 / (वृद्धि / नया मान) × 100 (d) (Original / Increase) × 100 / (मूल मान / वृद्धि) × 100
Show answer
(b) — Percentage increase is always calculated relative to the original (base) value: (Increase / Original) × 100. / प्रतिशत वृद्धि हमेशा मूल (आधार) मान के सापेक्ष निकाली जाती है: (वृद्धि / मूल मान) × 100।
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A shopkeeper buys a pen for ₹40 and sells it for ₹50. What is the profit percentage? / एक दुकानदार ₹40 में कलम खरीदकर ₹50 में बेचता है। लाभ प्रतिशत क्या है? (a) 20% / 20% (b) 25% / 25% (c) 10% / 10% (d) 15% / 15%
Show answer
(b) — Profit = ₹50 − ₹40 = ₹10. Profit% = (10/40) × 100 = 25%. The base for profit% is always CP. / लाभ = ₹50 − ₹40 = ₹10। लाभ% = (10/40) × 100 = 25%। लाभ% का आधार हमेशा क्रय मूल्य होता है।
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A jacket is marked at ₹1200. After a 20% discount, what is the selling price? / एक जैकेट का अंकित मूल्य ₹1200 है। 20% की छूट के बाद विक्रय मूल्य क्या होगा? (a) ₹980 / ₹980 (b) ₹960 / ₹960 (c) ₹1000 / ₹1000 (d) ₹1440 / ₹1440
Show answer
(b) — SP = MP × (1 − 20/100) = 1200 × 0.80 = ₹960. Discount = 20% of MP reduces price by ₹240. / SP = MP × (1 − 20/100) = 1200 × 0.80 = ₹960। छूट = अंकित मूल्य का 20% = ₹240 घटाने पर।
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Simple Interest is calculated using the formula SI = _____ / साधारण ब्याज का सूत्र SI = _____ है।
Show answer
SI = (P × R × T) / 100 / SI = (मूलधन × दर × समय) / 100 — P is principal, R is annual rate (%), T is time in years. The total amount paid is A = P + SI. / P = मूलधन, R = वार्षिक दर (%), T = समय वर्षों में। कुल धनराशि A = P + SI।
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When successive percentage changes of +10% and −10% are applied to a quantity, the net result is _____. / जब किसी राशि पर क्रमशः +10% और −10% का परिवर्तन लागू होता है, तो कुल परिणाम _____ होता है।
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A net decrease of 1% / 1% की कुल कमी — The combined factor is 1.10 × 0.90 = 0.99, so the result is 1% less than the original. Equal opposite percentages do NOT cancel out. / संयुक्त गुणांक 1.10 × 0.90 = 0.99 है, अतः मूल से 1% कम। समान विपरीत प्रतिशत एक-दूसरे को रद्द नहीं करते।
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True or False: Profit percentage is always calculated on the Selling Price. / सत्य या असत्य: लाभ प्रतिशत हमेशा विक्रय मूल्य पर निकाला जाता है।
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False / असत्य — Profit (and loss) percentage is always calculated on the Cost Price (CP), not on the Selling Price. / लाभ (और हानि) प्रतिशत हमेशा क्रय मूल्य (CP) पर निकाला जाता है, विक्रय मूल्य पर नहीं।
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A sum of ₹8000 is invested at a compound interest rate of 10% per annum for 2 years. What is the amount at the end of 2 years? / ₹8000 की राशि 10% वार्षिक दर पर 2 वर्ष के लिए चक्रवृद्धि ब्याज पर लगाई जाती है। 2 वर्ष बाद राशि क्या होगी?
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A = 8000 × (1 + 10/100)² = 8000 × 1.1 × 1.1 = 8000 × 1.21 = ₹9680. / A = 8000 × (1 + 10/100)² = 8000 × 1.21 = ₹9680। — The formula A = P(1 + r/100)^n gives the compound amount. / सूत्र A = P(1 + r/100)^n से चक्रवृद्धि राशि निकाली जाती है।
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After a 25% increase, a price becomes ₹500. What was the original price? / 25% वृद्धि के बाद कोई मूल्य ₹500 हो जाता है। मूल मूल्य क्या था?
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Original price = 500 ÷ (1 + 25/100) = 500 ÷ 1.25 = ₹400. / मूल मूल्य = 500 ÷ (1 + 25/100) = 500 ÷ 1.25 = ₹400। — To find the original, divide the new value by the multiplier (1 + r/100). / मूल मान जानने के लिए नए मान को गुणांक (1 + r/100) से भाग दें।
Related Laws & Principles
Explore allFoundational laws & principles behind this chapter. Each one opens a full page — what it says, why it matters, five practice questions and the mistakes to avoid.