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: \(\dfrac{5}{25}\times 100\% = 20\%\).
3. Discount and Sale Price
Discount is the reduction on Marked Price (MP). Sale Price (SP) = MP − Discount.
Discount \(= \) MP − SP.
Discount percent \(= \dfrac{\text{Discount}}{\text{MP}} \times 100\%.\)
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: \(CI = A - P.\)
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)
\(\text{Percent} = \dfrac{\text{part}}{\text{whole}}\times100\% \) |
\(SP = MP - \text{Discount}\) |
\(MP = \dfrac{SP}{1 - \text{Discount\%}/100}\)
\(A = P\left(1+\frac{R}{100}\right)^n\) |
\(V = P\left(1-\frac{R}{100}\right)^n\)
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.)
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.)
