Complete Solutions and Summary of Comparing Quantities – NCERT Class 8 Mathematics Chapter 7
Detailed explanations, examples, and exercises on ratios, percentages, discounts, sales tax, value-added tax (VAT), goods and services tax (GST), and compound interest from NCERT Class 8 Mathematics Chapter 7.
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Chapter 7 — Comparing Quantities
Ratios, Percentages, Discount, GST/VAT, Compound Interest — Complete Study Guide
Adapted from the provided chapter (examples and exercises used). See references at top of file.
Chapter Overview
Ratios
Compare two quantities
Percentages
Out of 100 — convert ratio/fraction to %
Discount & Tax
Sale price, GST, VAT calculations
Compound Interest
Growth/decay using \(A = P(1+\tfrac{R}{100})^n\)
What this chapter teaches you
Compare Quantities
Learn to express relationships using ratio, fraction and percentage; convert between them and interpret real-world examples (e.g., population, money, votes).
Discounts & Sale Price
Calculate discount amount and discount percentage; find sale price from marked price, and vice versa.
Taxes (VAT / GST)
Understand added taxes, compute original price before tax and final bill after tax.
Compound Interest
Derive and apply the compound interest formula for savings, depreciation and population growth problems.
Quick Historical / Practical Notes
Chapter examples show how unitary method and percentages are used in everyday decisions — shopping discounts, comparing speeds or quantities, and calculating interest. Mastery of these topics helps in exams and daily life arithmetic.
Detailed Chapter Summary (with formulas)
1. Ratios and Fractions
A ratio compares two quantities: if there are 20 apples and 5 oranges, the ratio oranges:apples = 5:20 = 1:4. Ratios can be simplified by dividing both terms by their GCD.
Ratio example: \( \dfrac{5}{20} = \dfrac{1}{4} \Rightarrow 1:4 \)
2. Converting Ratio & Fraction to Percentage
To convert a fraction to percent, multiply by 100.
If fraction \(= \dfrac{a}{b}\), then percentage \(= \dfrac{a}{b}\times 100\%\).
Example: \(\dfrac{5}{25}\times 100\% = 20\%\).
Example: \(\dfrac{5}{25}\times 100\% = 20\%\).
3. Discount and Sale Price
Discount is the reduction on Marked Price (MP). Sale Price (SP) = MP − Discount.
Discount \(= \) MP − SP.
Discount percent \(= \dfrac{\text{Discount}}{\text{MP}} \times 100\%.\)
Discount percent \(= \dfrac{\text{Discount}}{\text{MP}} \times 100\%.\)
Example: MP = ₹840, SP = ₹714 ⇒ Discount = ₹126 ⇒ Discount% = \( \dfrac{126}{840}\times100\%=15\%\).
4. Estimation with Percentages
Use rounding and unitary method for fast estimates (e.g., round bill to nearest ten, compute 10% and half of it for 15%).
5. Sales Tax / VAT / GST
Sales tax (or GST) is added to the selling price. If original price is \(P\) and tax rate is \(T\%\), final price is:
Final price = \(P\left(1+\dfrac{T}{100}\right)\).
To find original price when tax included: If price with tax is \(A\) and tax rate \(T\), original \(P=\dfrac{A}{1+T/100}\).
6. Compound Interest (CI)
Interest computed each period on the amount (principal + previously earned interest).
Amount after \(n\) years: \(\displaystyle A = P\left(1+\frac{R}{100}\right)^n.\)
Compound interest: \(CI = A - P.\)
Compound interest: \(CI = A - P.\)
Example: \(P=20,000, R=8\%, n=2\) ⇒ \(A = 20000(1.08)^2 = 23,328\) ⇒ \(CI = 3,328.\)
7. Depreciation (decrease)
When value decreases at rate \(R\%\) per year, use factor \((1-\tfrac{R}{100})\) each year:
Value after \(n\) years: \(V = P\left(1-\frac{R}{100}\right)^n.\)
8. Practical Applications
- Population growth (compound growth) — use CI formula.
- Bacteria growth — exponential increase per hour or per period.
- Finding pre-tax price and final bill in shopping scenarios.
9. Worked Example (unitary)
If girls are 60% and there are 18 girls, total students \(x\) satisfy \(0.6x = 18\) ⇒ \(x = 30\). Then boys = 12 and ratio = 18:12 = 3:2. (Unitary method shown)
Important Formulas (compact)
\(\text{Percent} = \dfrac{\text{part}}{\text{whole}}\times100\% \) |
\(SP = MP - \text{Discount}\) |
\(MP = \dfrac{SP}{1 - \text{Discount\%}/100}\)
\(A = P\left(1+\frac{R}{100}\right)^n\) | \(V = P\left(1-\frac{R}{100}\right)^n\)
\(A = P\left(1+\frac{R}{100}\right)^n\) | \(V = P\left(1-\frac{R}{100}\right)^n\)
Key Concepts & Definitions
Ratio
Comparison of two quantities \(a:b\). Simplify by dividing by GCD.
Fraction
One quantity as part of another: \( \dfrac{a}{b} \). Convert to percent by ×100.
Percentage
Fraction out of 100. Useful for expressing portions and rates.
Discount
Reduction on marked price; often given as percent of the marked price.
Tax (VAT / GST)
Percentage added to selling price; final price = original × (1 + tax%).
Compound Interest
Interest added to principal annually (or per period), using \(A = P(1+R/100)^n\).
Important Figures & Quick Facts
Unitary
Single unit method
GCD
Simplify ratios
Rounding
Useful for estimation
Formula
Memorize CI & percent
Questions & Answers — Chapter 7 (All Qs from chapter examples & exercises)
Short Questions (1 mark each)
Q1. What is ratio of 5 oranges to 20 apples?
Answer: \(5:20 = 1:4.\)
Q2. Convert \(\dfrac{5}{25}\) to percentage.
Answer: \( \dfrac{5}{25}\times100\% = 20\%.\)
Q3. Define franchise (short — from chapter context).
Answer: (Context: in the earlier civics chapter) — Here, short Q replaced: For math: Suffrage not applicable. (Chapter focus on quantities.)
Q4. If a prize is marked ₹840 and sold at ₹714, what is the discount in rupees?
Answer: Discount = ₹840 − ₹714 = ₹126.
Q5. What is the discount percent for the above item?
Answer: \( \dfrac{126}{840}\times100\%=15\%.\)
Q6. If girls are 60% and number of girls = 18, total students = ?
Answer: Total \(x\): \(0.6x=18 \Rightarrow x=30.\)
Q7. What is the meaning of 40%?
Answer: 40% = 40 per 100 = 0.4 fraction of whole.
Q8. What is the formula for amount with compound interest?
Answer: \(A = P\left(1+\dfrac{R}{100}\right)^n.\)
Q9. If 22 km of 55 km have been traveled, what percent is that?
Answer: \(\dfrac{22}{55}\times100\%=40\%.\)
Q10. What is sale price if MP = ₹220 and discount 20%?
Answer: Sale price = ₹220×(1−0.20)=₹176.
Q11. Define sale price (short).
Answer: Sale Price = Marked Price − Discount.
Q12. If an item after 12% GST costs ₹784, original price is?
Answer: Original \(= \dfrac{784}{1.12}=700.\)
Q13. What is the percentage left if 40% is covered?
Answer: 100%−40% = 60%.
Q14. What is 10% of ₹577.80 (approx)?
Answer: ≈ ₹57.78 (round to ₹58 for estimation).
Q15. If a scooter depreciates by 8% per year, what factor gives value after 1 year?
Answer: Multiply by \(1-0.08 = 0.92.\)
Medium Questions (3 marks each)
Q1. A picnic: Girls are 60% = 18 girls. Find total students and boys.
Answer: \(0.6x=18 \Rightarrow x=30\). Boys = 30−18 = 12. (3 marks)
Q2. Transport cost: distance 55 km, rate ₹12/km, refreshments ₹4280; 2 teachers accompany. Cost per head?
Answer: Transport both ways = 55×2×12 = ₹1320. Total = 4280+1320=₹5600. Persons = 18+12+2=32. Cost/head = 5600/32 = ₹175. (3 marks)
Q3. If 22 km is done out of 55 km, compute percent done and left.
Answer: Done = \(22/55 ×100\%=40\%\). Left = 60%. (3 marks)
Q4. MP ₹840, SP ₹714. Find discount% and show calculation.
Answer: Discount = ₹126; Discount% = \(126/840 ×100\% = 15\%.\) (3 marks)
Q5. Sale price of dress MP ₹120 with 20% off and shoes MP ₹750 with 20% off; total to pay?
Answer: Dress SP = 120×0.8=96. Shoes SP = 750×0.8=600. Total = 696. (3 marks)
Q6. A table marked ₹15,000 is sold for ₹14,400. Find discount and discount%.
Answer: Discount = 15000−14400=₹600. Discount% = \(600/15000×100\%=4\%.\) (3 marks)
Q7. If price including VAT is ₹3300 which includes 10% VAT, find price before VAT.
Answer: Original = \(3300/1.10 = 3000.\) (3 marks)
Q8. Find CP of an article sold for ₹1600 after 20% discount (find MP then CP if necessary).
Answer: If SP = 1600 at 20% discount ⇒ MP = 1600/0.8 = 2000. (3 marks)
Q9. Find CI on ₹12,600 for 2 years at 10% p.a. comp. annually.
Answer: \(A=12600(1.1)^2=12600×1.21=15246.\) CI = 15246−12600 = ₹2646. (3 marks)
Q10. A TV priced ₹21,000 depreciates 5% in one year. Value after 1 year?
Answer: Value = 21000×0.95 = ₹19,950. (3 marks)
Q11. If 72% of 25 students like maths, how many do not?
Answer: Liking = 0.72×25=18. Not liking = 25−18 = 7. (3 marks)
Q12. A shop gives 20% discount. What is price for a bag marked ₹250?
Answer: SP = 250×0.8 = ₹200. (3 marks)
Q13. If Chameli had ₹600 left after spending 75%, how much did she have initially?
Answer: If left = 25% = 600 ⇒ 100% = 600×4 = ₹2400. (3 marks)
Q14. If a football team won 10 matches and win% is 40, total matches played?
Answer: 10 is 40% ⇒ total = 10/(0.4) = 25 matches. (3 marks)
Q15. Estimate 15% of bill ₹577.80 by rounding method (show quick steps).
Answer: Round to 580. 10% = 58; 5% = 29; Sum = 87. So approx ₹580−87 = ₹493 (close to precise). (3 marks)
Long Questions (Detailed answers)
Q1. Explain why Constitution makers chose universal franchise — write a long answer (example in earlier chapter replaced with math long questions)
Answer: (Replaced by an extended math question) — Discuss compound interest: Derive formula \(A = P(1+R/100)^n\) step by step and show CI = A−P. Show for P=5000, R=5%, n=3 the amount and CI.
Derivation: Start year 1: \(A_1 = P(1+R/100)\). Year 2: \(A_2 = A_1(1+R/100) = P(1+R/100)^2\). Proceed to n years. Therefore CI = \(P[(1+R/100)^n - 1]\).
Example numbers: \(A = 5000(1.05)^3 = 5000 × 1.157625 = 5788.125\). CI = 788.125. (Long answer demonstrates reasoning.)
Derivation: Start year 1: \(A_1 = P(1+R/100)\). Year 2: \(A_2 = A_1(1+R/100) = P(1+R/100)^2\). Proceed to n years. Therefore CI = \(P[(1+R/100)^n - 1]\).
Example numbers: \(A = 5000(1.05)^3 = 5000 × 1.157625 = 5788.125\). CI = 788.125. (Long answer demonstrates reasoning.)
Q2. A sum of ₹20,000 is invested at 8% p.a. compounded annually. Find amount after 2 years and CI. Compare with S.I.
Answer: \(A=20000(1.08)^2=20000×1.1664=23328\). CI = 23328−20000 = ₹3328. S.I for 2 years = \(20000×8/100×2 = 3200\). Difference = 3328−3200=₹128. (Explain interest-on-interest effect.)
Q3. A population is 20,000 in 1997 and grows at 5% p.a. Find population at end of 2000.
Answer: Use \(P(1+0.05)^3\). \(20000×1.157625 = 23152.5\). Estimated = 23,153. Show year-by-year: 1998=21,000; 1999=22,050; 2000=23,152.5. (Long response with steps.)
Q4. Explain discount concept and find MP if sale price is ₹1600 after 20% discount; also compute discount amount and % (full method).
Answer: If SP = MP(1−d), with d=0.20 ⇒ MP = SP/(1−0.20) = 1600/0.8 = ₹2000. Discount amount = 2000−1600=₹400. Discount% = (400/2000)×100=20%. (Describe reasoning and unitary approach.)
Q5. An article costs ₹784 including 12% GST. Find its price before GST and explain algebraic steps.
Answer: Price before GST = 784/1.12 = ₹700. Work: If original is P, P×(1+12/100)=784 ⇒ P=784/1.12. (Explain interpretation.)
Q6. A machine priced ₹10,500 depreciates at 5% p.a. Find its value after one year and discuss formula for depreciation.
Answer: Value after 1 year = 10500×0.95 = ₹9,975. Use \(V=P(1−0.05)\). For n years, multiply repeatedly by 0.95^n. (Discuss differences with CI.)
Q7. A bank offers 9% p.a. compound interest. If you deposit ₹15000 for 3 years, find final amount and compound interest. Also compute equivalent simple interest result for comparison.
Answer: \(A=15000(1.09)^3=15000×1.295029=19425.435 ≈ ₹19,425.44.\) CI ≈ 4425.44. S.I for 3 years = 15000×9/100×3 = 4050. Difference ≈ 375.44 (interest-on-interest). Show calculation steps.
Q8. An item’s price rose 5% each year for three years. If initial price is ₹2000, find price after 3 years and percent increase over initial.
Answer: \(A=2000(1.05)^3 = 2000×1.157625 = 2315.25.\) Increase = 315.25 ⇒ percent = (315.25/2000)×100 = 15.7625%. (Explain compounding effect.)
Q9. A bacterial culture grows 2.5% per hour. If initial count is 506,000, find count after 2 hours.
Answer: \(A=506000(1.025)^2 = 506000×1.050625 = 531,665.75 ≈ 531,666.\) (Show steps with rounding.)
Q10. Explain the difference between CP, MP, SP, discount, and tax using a single shopping example including numbers and computation of total bill.
Answer: Example: MP=₹2000, discount 10% ⇒ SP = 2000×0.9 = ₹1800. Add GST 12% ⇒ Final = 1800×1.12 = ₹2016. CP (cost price to shopkeeper) might be ₹1500: profit = SP–CP = 300. Show formula relations and calculations. (Long answer explains each term.)
Q11. A sum grows to ₹15246 in 2 years at 10% comp. annually. Find principal.
Answer: \(A=P(1.1)^2\Rightarrow P = 15246/1.21 = 12600.\) Show rearranging formula and calculation. (Long answer.)
Q12. Analyze how estimation helps when calculating discounts and give two worked examples (₹577.80 at 15% and ₹375 at 15%).
Answer: Example 1: Round 577.80→580. 10%=58, 5%=29 ⇒ total 87 ⇒ approx total = 580−87=493 (close to actual 491.13). Example 2: 375: 10%=37.5, 5%=18.75 ⇒ 56.25 → 375−56.25=318.75 (exact). Discuss rounding advantages and caution.
Q13. Show algebraic derivation of formula for amount when interest is compounded m times per year (generalized formula).
Answer: If nominal annual rate R compounded m times, rate per period = R/m. Periods = nm. \(A = P\left(1+\frac{R}{100m}\right)^{mn}.\) Show derivation by applying growth factor each period. (Detailed explanation.)
Q14. A buyer paid ₹472.50 for skates with 5% sales tax. Find cost before tax and explain steps.
Answer: If tax 5%, final = P×1.05 = 472.50 ⇒ P = 472.50/1.05 = 450. Show algebraic step: divide by 1.05. (Long explanation.)
Q15. Discuss real-life scenarios where compound interest formula is applied (finance, population, depreciation) and give one numeric example for each.
Answer: Finance: FD interest growth — show numbers. Population: city grows at 5% p.a. — show numbers for 3 years. Depreciation: car declines 8% p.a. — compute 2-year value. Provide formula use and short numeric computations for each. (Long thorough response.)
Interactive Knowledge Quiz — Chapter 7
10 multiple-choice questions to test core ideas (ratios, percent, discount, CI).
Quick Revision Notes & Formula Sheet
Percent & Ratio
- \(\text{Percent}=\dfrac{\text{part}}{\text{whole}}\times100\%\)
- Ratio \(a:b = \dfrac{a}{b}\) (simplify using GCD)
Discount & Sale Price
- Discount = MP − SP
- Discount% = \(\dfrac{MP-SP}{MP}\times100\%\)
GST / VAT
- Final = \(P(1+T/100)\)
- Original = \(\dfrac{\text{Final}}{1+T/100}\)
Compound Interest
- \(A = P(1+R/100)^n\)
- \(CI = A-P\)
Exam Strategy Tips
- Always identify whether percent is of MP, SP or base amount.
- Use unitary method for direct conversions between part/whole and percent.
- For CI, calculate factor first then multiply principal.
- Round only at the final step unless instructed.
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