CBSE Class 8 Mathematics Comparing Quantities Worksheet Set 04

8. Comparing Quantities

Question 1. When 5% sale tax is added on the purchase of a bedsheet of Rs 300, find the buying price or the cost price of the bedsheet.
Answer: Marked price of bedsheet = Rs 300
Sales tax rate = 5%
Sales tax amount = \(5\% \text{ of Rs } 300 = \frac{5}{100} \times 300 = \text{Rs } 15\)
Buying price (Cost price) = Marked price + Sales tax = \(300 + 15 = \text{Rs } 315\)
In simple words: Add the 5% tax amount calculated on the original price to find the total buying price.

Exam Tip: Always add sales tax to the marked price to find the total amount paid by the customer.

Question 2. Find the ratio of the speed of a car 60 km/hr to the speed of a scooter 30 km/hr.
Answer: Speed of car = 60 km/hr
Speed of scooter = 30 km/hr
Ratio = \(\frac{60}{30} = \frac{2}{1} = 2:1\)
In simple words: Divide both quantities by their highest common factor to simplify the ratio.

Exam Tip: Ensure both quantities are in the same units before finding the ratio.

Question 3. Find the ratio of 5 m to 10 km.
Answer: First quantity = 5 m
Second quantity = 10 km = \(10 \times 1000\text{ m} = 10000\text{ m}\)
Ratio = \(\frac{5}{10000} = \frac{1}{2000} = 1:2000\)
In simple words: Convert kilometres to metres first since ratios require identical units, then simplify.

Exam Tip: Never calculate ratios between different units directly; always convert them to the same unit.

Question 4. Convert the ratio 4 : 5 to percentage.
Answer: Given ratio = \(4:5 = \frac{4}{5}\)
Percentage = \(\frac{4}{5} \times 100\% = 4 \times 20\% = 80\%\)
In simple words: Multiply the fractional form of the ratio by 100 to convert it into a percentage.

Exam Tip: To convert any fraction or ratio to a percentage, simply multiply by 100%.

Question 5. 76% of 50 students are good in English. How many students are not good in English?
Answer: Total number of students = 50
Percentage of students not good in English = \(100\% - 76\% = 24\%\)
Number of students not good in English = \(24\% \text{ of } 50 = \frac{24}{100} \times 50 = \frac{24}{2} = 12\)
In simple words: Find the percentage of students who are weak in English and calculate that percentage out of 50.

Question 6. A table fan is bought for Rs 560 and expenses of Rs 40 are made on its repairs. It is sold at a profit of 5%. Find the selling price.
Answer: Cost price (C.P.) = Rs 560
Overhead expenses = Rs 40
Total Cost Price = \(560 + 40 = \text{Rs } 600\)
Profit percentage = 5%
Profit = \(5\% \text{ of Rs } 600 = \frac{5}{100} \times 600 = \text{Rs } 30\)
Selling Price (S.P.) = Total C.P. + Profit = \(600 + 30 = \text{Rs } 630\)
In simple words: Add repair costs to the purchase price to get the total cost, then add the 5% profit.

Exam Tip: Always include overhead charges (repairs, transport) in the Cost Price before calculating profit or loss.

Question 7. A man got 10% increase in his salary. If his original salary was Rs 30,000, find his new salary.
Answer: Original salary = Rs 30,000
Increase percentage = 10%
Increase amount = \(10\% \text{ of Rs } 30,000 = \frac{10}{100} \times 30,000 = \text{Rs } 3,000\)
New salary = Original salary + Increase amount = \(30,000 + 3,000 = \text{Rs } 33,000\)
In simple words: Calculate 10% of the original salary and add it to the original amount.

Exam Tip: Alternatively, calculate 110% of the original salary directly to get the new salary in one step.

Question 8. Find the amount on a principal of Rs 2000 for 2 years at 10% per annum compounded annually. Also find the compound interest.
Answer: Principal (\(P\)) = Rs 2000
Rate (\(R\)) = 10% per annum
Time (\(n\)) = 2 years
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n\)
\(A = 2000 \left(1 + \frac{10}{100}\right)^2 = 2000 \left(\frac{11}{10}\right)^2\)
\(A = 2000 \times \frac{11}{10} \times \frac{11}{10} = 20 \times 121 = \text{Rs } 2420\)
Compound Interest (C.I.) = \(A - P = 2420 - 2000 = \text{Rs } 420\)
In simple words: Use the compound amount formula to find the total amount, then subtract the principal to get the interest.

Exam Tip: Memorize the compound amount formula \(A = P(1 + R/100)^n\) for quick calculations.

Question 9. An item marked at Rs 640 is sold for 560. What is the discount and discount%?
Answer: Marked Price (M.P.) = Rs 640
Selling Price (S.P.) = Rs 560
Discount = Marked Price - Selling Price = \(640 - 560 = \text{Rs } 80\)
Discount% = \(\frac{\text{Discount}}{\text{M.P.}} \times 100\% = \frac{80}{640} \times 100\%\)
\(\text{Discount}\% = \frac{1}{8} \times 100\% = 12.5\%\)
In simple words: Subtract selling price from marked price to find discount money, then find what percentage it is of the marked price.

Exam Tip: Discount percentage is always calculated on the Marked Price, never on the Cost Price.

Question 10. The price of a car was Rs 2,60,000 last year. It has increased 15% this year. What is the new price?
Answer: Original price = Rs 2,60,000
Increase percentage = 15%
Increase in price = \(15\% \text{ of Rs } 2,60,000 = \frac{15}{100} \times 2,60,000 = \text{Rs } 39,000\)
New price = Original price + Increase = \(2,60,000 + 39,000 = \text{Rs } 2,99,000\)
In simple words: Find 15% of last year's price and add it to get this year's price.

Question 11. A table is sold at Rs 5000 after allowing a discount of 12.5%. Find its marked price.
Answer: Selling Price (S.P.) = Rs 5000
Discount% = 12.5%
Let the Marked Price be \(x\).
\(\text{S.P.} = \text{M.P.} - \text{Discount}\)
\(5000 = x - 12.5\% \text{ of } x = x \left(1 - \frac{12.5}{100}\right) = x \left(\frac{87.5}{100}\right)\)
\(x = \frac{5000 \times 100}{87.5} = \frac{500000}{87.5} = \text{Rs } 5714.29\)
In simple words: Set up an equation where selling price is 87.5% of the marked price to solve for marked price.

Exam Tip: When given selling price and discount percentage, never calculate discount on S.P.; always work backward from M.P.

Question 12. Ram had Rs 400 left after spending 60% of his money. How much did he have in the beginning?
Answer: Let Ram's total initial money be \(x\).
Percentage of money spent = 60%
Percentage of money left = \(100\% - 60\% = 40\%\)
According to the question, \(40\% \text{ of } x = 400\)
\(\frac{40}{100} \times x = 400\)
\(x = \frac{400 \times 100}{40} = \text{Rs } 1000\)
In simple words: Since 40% of his money equals Rs 400, divide 400 by 40% to find the total initial amount.

Exam Tip: Total percentage is always 100%. Subtract the spent percentage to find the remaining percentage.

Question 13. Shalu bought a water cooler for Rs 1100 including a tax of 10%. Find the price of the water cooler before the tax was added.
Answer: Price including tax = Rs 1100
Tax rate = 10%
Let the original price before tax be \(x\).
\(x + 10\% \text{ of } x = 1100\)
\(1.10x = 1100\)
\(x = \frac{1100}{1.10} = \text{Rs } 1000\)
In simple words: The given price is 110% of the original price. Divide by 1.10 to remove the tax.

Question 14. On selling a fan for Rs 810, Sunil gains 8%. For how much did he purchase it?
Answer: Selling Price (S.P.) = Rs 810
Gain% = 8%
Let the Cost Price (C.P.) be \(x\).
\(\text{S.P.} = \text{C.P.} + \text{Gain}\)
\(810 = x + 8\% \text{ of } x = x \left(1 + \frac{8}{100}\right) = 1.08x\)
\(x = \frac{810}{1.08} = \text{Rs } 750\)
In simple words: Since there is a gain, selling price is 108% of cost price. Divide selling price by 1.08 to find cost price.

Exam Tip: Formula shortcut: \(\text{C.P.} = \frac{100}{100 + \text{Gain}\%} \times \text{S.P.}\)

Question 15. A scooter was bought at Rs 32,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Answer: Initial value = Rs 32,000
Depreciation rate = 8% per annum
Depreciation amount = \(8\% \text{ of Rs } 32,000 = \frac{8}{100} \times 32000 = \text{Rs } 2560\)
Value after one year = Initial value - Depreciation amount
Value after one year = \(32,000 - 2560 = \text{Rs } 29,440\)
In simple words: Calculate 8% depreciation and subtract it from the initial cost of the scooter.

Exam Tip: Depreciation means reduction in value, so always subtract the depreciation amount from the principal.

Question 16. Find the compound interest on Rs 1,60,000 for 2 years at 10% per annum when compounded semi-annually.
Answer: Principal (\(P\)) = Rs 1,60,000
Annual Rate = 10% per annum
Semi-annual Rate (\(R\)) = \(\frac{10\%}{2} = 5\%\) per half-year
Time (\(t\)) = 2 years
Number of semi-annual periods (\(n\)) = \(2 \times 2 = 4\) half-years
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n\)
\(A = 160000 \left(1 + \frac{5}{100}\right)^4 = 160000 \left(\frac{21}{20}\right)^4\)
\(A = 160000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} = 160000 \times \frac{194481}{160000} = \text{Rs } 194,481\)
Compound Interest = \(A - P = 194,481 - 160,000 = \text{Rs } 34,481\)
In simple words: Halve the annual interest rate and double the number of years when compounding semi-annually.

Exam Tip: For half-yearly compounding, always use \(R/2\) for rate and \(2t\) for time periods.

Question 17. If the cost price of 10 greeting cards is equal to the selling price of 8 greeting cards, find the gain or loss%.
Answer: Let the cost price of 1 greeting card be Re 1.
Cost Price of 10 greeting cards = Rs 10
Given that Selling Price of 8 greeting cards = Cost Price of 10 greeting cards = Rs 10
Selling Price of 10 greeting cards = \(\frac{10}{8} \times 10 = \text{Rs } 12.50\)
Profit = S.P. - C.P. = \(12.50 - 10 = \text{Rs } 2.50\)
Gain% = \(\frac{\text{Profit}}{\text{C.P.}} \times 100\% = \frac{2.50}{10} \times 100\% = 25\%\)
In simple words: Assume a cost price for one item, find total cost and selling price for a common quantity, then calculate profit percentage.

Exam Tip: When selling fewer items than bought for the same total money, there is always a net profit.

Question 18. Rishi bought a wrist watch for Rs 2200 and sold it for Rs 1980. Find his gain or loss%.
Answer: Cost Price (C.P.) = Rs 2200
Selling Price (S.P.) = Rs 1980
Since C.P. > S.P., there is a loss.
Loss = C.P. - S.P. = \(2200 - 1980 = \text{Rs } 220\)
Loss% = \(\frac{\text{Loss}}{\text{C.P.}} \times 100\% = \frac{220}{2200} \times 100\% = 10\%\)
In simple words: Subtract selling price from cost price to find loss, then express it as a percentage of the cost price.

Exam Tip: Loss percentage is always evaluated with respect to the Cost Price.

Question 19. A cricket team won 10 matches out of the total matches they played. If their win percentage was 40, then how many matches did they play in all?
Answer: Let the total number of matches played be \(x\).
Number of matches won = 10
Win percentage = 40%
\(40\% \text{ of } x = 10\)
\(\frac{40}{100} \times x = 10\)
\(x = \frac{10 \times 100}{40} = \frac{1000}{40} = 25\)
In simple words: Set 40% of total matches equal to 10 and solve for the total number of matches.

Exam Tip: Use algebraic variables for unknown totals in percentage word problems.

Question 20. A shopkeeper offers his customers 10% discount and still makes a profit of 26%. What is the actual cost of an article marked Rs 280?
Answer: Marked Price (M.P.) = Rs 280
Discount = 10%
Selling Price (S.P.) = \(\text{M.P.} - 10\% \text{ of M.P.} = 280 - \frac{10}{100} \times 280 = 280 - 28 = \text{Rs } 252\)
Profit = 26%
Let Cost Price (C.P.) be \(x\).
\(\text{S.P.} = \text{C.P.} + 26\% \text{ of C.P.} = 1.26x\)
\(252 = 1.26x\)
\(x = \frac{252}{1.26} = \text{Rs } 200\)
In simple words: First find the selling price using the discount and marked price, then use the profit percentage to find the cost price.

Exam Tip: Selling Price is the bridge connecting Marked Price and Cost Price.

Question 21. A person sells an article for Rs 550, gaining 1/10 of its C.P. Find gain%.
Answer: Selling Price (S.P.) = Rs 550
Gain = \(\frac{1}{10} \text{ of C.P.}\)
We know that \(\text{S.P.} = \text{C.P.} + \text{Gain}\)
\(550 = \text{C.P.} + \frac{1}{10}\text{C.P.} = \frac{11}{10}\text{C.P.}\)
\(\text{C.P.} = \frac{550 \times 10}{11} = \text{Rs } 500\)
Gain = \(\frac{1}{10} \times 500 = \text{Rs } 50\)
Gain% = \(\frac{\text{Gain}}{\text{C.P.}} \times 100\% = \frac{50}{500} \times 100\% = 10\%\)
In simple words: Express gain as a fraction of cost price, find the cost price, and then calculate the gain percentage.

Exam Tip: When gain is given as a fraction of C.P., combine terms algebraically to find C.P. easily.

Question 22. Rahul bought an air conditioner for Rs 22,000 including a tax of 10%. Find the price of the air conditioner before VAT was added.
Answer: Price including VAT = Rs 22,000
VAT rate = 10%
Let price before VAT be \(x\).
\(x + 10\% \text{ of } x = 22,000\)
\(1.10x = 22,000\)
\(x = \frac{22,000}{1.10} = \text{Rs } 20,000\)
In simple words: Divide the total tax-inclusive price by 1.10 to find the original list price.

Exam Tip: Tax-inclusive prices always require dividing by \((1 + \text{tax rate})\) to revert to the base price.

Question 23. Harish sold a bicycle at 8% gain. Had it been sold for Rs 75 more, the gain would have been 14%. Find the cost price of the bicycle.
Answer: Let the cost price of the bicycle be \(x\).
Initial selling price at 8% gain = \(x + 8\% \text{ of } x = 1.08x\)
New selling price at 14% gain = \(x + 14\% \text{ of } x = 1.14x\)
Difference between two selling prices = Rs 75
\(1.14x - 1.08x = 75\)
\(0.06x = 75\)
\(x = \frac{75}{0.06} = \frac{7500}{6} = \text{Rs } 1250\)
In simple words: The difference in profit percentages (14% - 8% = 6%) corresponds directly to the extra money of Rs 75.

Question 24. Shruti borrows Rs 12,000 at 10% per annum for 3 years at simple interest and Shalini borrows the same amount for the same period at 8% per annum compounded annually. Who pays more interest and by how much?
Answer: For Shruti (Simple Interest):
\(P = 12000, R = 10\%, T = 3\)
\(\text{S.I.} = \frac{P \times R \times T}{100} = \frac{12000 \times 10 \times 3}{100} = \text{Rs } 3600\)
For Shalini (Compound Interest):
\(P = 12000, R = 8\%, n = 3\)
\(A = P \left(1 + \frac{R}{100}\right)^n = 12000 \left(1 + \frac{8}{100}\right)^3 = 12000 \left(\frac{27}{25}\right)^3\)
\(A = 12000 \times \frac{19683}{15625} = \text{Rs } 15,116.16\)
\(\text{C.I.} = A - P = 15,116.16 - 12,000 = \text{Rs } 3,116.16\)
Comparing interests: Shruti pays Rs 3600 and Shalini pays Rs 3116.16.
Shruti pays more interest by \(3600 - 3116.16 = \text{Rs } 483.84\).
In simple words: Calculate simple interest for Shruti and compound interest for Shalini separately, then compare their values.

Exam Tip: Always compute simple and compound interest using their respective formulas step-by-step.

Question 25. Mr. Kashiv sold his old chair at a loss of 15%. If he had sold it for Rs. 800 more, he would have received a profit of 5%. Find the cost price of the table.
Answer: Let the cost price of the chair (table) be \(x\).
Initial selling price at 15% loss = \(x - 15\% \text{ of } x = 0.85x\)
New selling price at 5% profit = \(x + 5\% \text{ of } x = 1.05x\)
Difference in selling prices = Rs 800
\(1.05x - 0.85x = 800\)
\(0.20x = 800\)
\(x = \frac{800}{0.20} = \text{Rs } 4000\)
In simple words: The total percentage gap between 15% loss and 5% profit is 20%, which equals Rs 800.

Exam Tip: When moving from a loss to a profit, add the loss percentage and profit percentage to find the total percentage difference.

Question 26. Rakesh bought a watch for Rs. 800 and sold it for Rs. 1000. Mukesh bought a car for Rs. 4,00,000 and sold it for Rs. 4,20,000. Who made a better sale, Rakesh or Mukesh?
Answer: For Rakesh:
\(\text{C.P.} = \text{Rs } 800, \text{S.P.} = \text{Rs } 1000\)
\(\text{Profit} = 1000 - 800 = \text{Rs } 200\)
\(\text{Profit}\% = \frac{200}{800} \times 100\% = 25\%\)
For Mukesh:
\(\text{C.P.} = \text{Rs } 4,00,000, \text{S.P.} = \text{Rs } 4,20,000\)
\(\text{Profit} = 4,20,000 - 4,00,000 = \text{Rs } 20,000\)
\(\text{Profit}\% = \frac{20,000}{4,00,000} \times 100\% = 5\%\)
Conclusion: Rakesh made a better sale because his profit percentage (25%) is much higher than Mukesh's profit percentage (5%).
In simple words: Compare profit percentages rather than absolute profit amounts to determine who made a better sale.

Exam Tip: Better business deals are always judged by comparing profit percentages, not absolute money values.

Question 27. A washing machine was sold for Rs. 5760 after giving successive discounts of 15% and 10% respectively. What was the marked price?
Answer: Let the Marked Price be \(x\).
First discount = 15%
Price after first discount = \(x \left(1 - \frac{15}{100}\right) = 0.85x\)
Second discount = 10% on discounted price
Final Selling Price = \(0.85x \left(1 - \frac{10}{100}\right) = 0.85x \times 0.90 = 0.765x\)
Given Final S.P. = Rs 5760
\(0.765x = 5760\)
\(x = \frac{5760}{0.765} = \text{Rs } 7529.41\)
In simple words: Apply successive discounts one after the other to the marked price to find the final selling price equation.

Exam Tip: Successive discounts are applied multiplicatively, not by adding percentages together.

Question 28. The marked price of a television is 25% more than the cost price. It is sold at a discount of 10%. If the marked price is Rs. 12,000, then find the cost price, selling price and the profit. What will be the percentage of profit?
Answer: Marked Price (M.P.) = Rs 12,000
Given M.P. is 25% more than C.P.:
\(\text{M.P.} = \text{C.P.} + 25\% \text{ of C.P.} = 1.25 \times \text{C.P.}\)
\(12,000 = 1.25 \times \text{C.P.}\)
\(\text{C.P.} = \frac{12,000}{1.25} = \text{Rs } 9600\)
Discount = 10% on M.P.
\(\text{Selling Price (S.P.)} = \text{M.P.} - 10\% \text{ of M.P.} = 12,000 - 1200 = \text{Rs } 10,800\)
Profit = S.P. - C.P. = \(10,800 - 9600 = \text{Rs } 1200\)
Profit% = \(\frac{\text{Profit}}{\text{C.P.}} \times 100\% = \frac{1200}{9600} \times 100\% = 12.5\%\)
In simple words: Find cost price from marked price, calculate selling price using discount, then find profit and profit percentage.

Exam Tip: Break multi-part questions into sequential steps: C.P. first, then S.P., followed by profit and percentage.

Question 29. In how much time will the simple interest on Rs. 3,500 at the rate of 6% per annum be the same as simple interest on Rs. 4,000 at 10.5% per annum for 4 years?
Answer: For the second case:
\(P_2 = 4000, R_2 = 10.5\%, T_2 = 4 \text{ years}\)
\(\text{Simple Interest} = \frac{P_2 \times R_2 \times T_2}{100} = \frac{4000 \times 10.5 \times 4}{100} = 40 \times 42 = \text{Rs } 1680\)
For the first case:
\(P_1 = 3500, R_1 = 6\%, T_1 = ?\)
Given that S.I. is the same (Rs 1680):
\(1680 = \frac{3500 \times 6 \times T_1}{100}\)
\(1680 = 210 \times T_1\)
\(T_1 = \frac{1680}{210} = 8 \text{ years}\)
In simple words: Calculate the interest in the second scenario and use it to find the unknown time period in the first scenario.

Exam Tip: Equate the simple interest formulas when problems state that two interests are equal.

Question 30. A man sold two of his buffaloes for Rs 20,000 each. On one, he made a profit of 5% and on the other, he lost 10%. Find his overall profit% or loss%.
Answer: For Buffalo 1 (Profit):
\(\text{S.P.} = \text{Rs } 20,000, \text{Profit} = 5\%\)
\(\text{C.P.}_1 = \frac{100}{100 + 5} \times 20,000 = \frac{100}{105} \times 20,000 = \text{Rs } 19,047.62\)
For Buffalo 2 (Loss):
\(\text{S.P.} = \text{Rs } 20,000, \text{Loss} = 10\%\)
\(\text{C.P.}_2 = \frac{100}{100 - 10} \times 20,000 = \frac{100}{90} \times 20,000 = \text{Rs } 22,222.22\)
Total S.P. = \(20,000 + 20,000 = \text{Rs } 40,000\)
Total C.P. = \(19,047.62 + 22,222.22 = \text{Rs } 41,269.84\)
Since Total C.P. > Total S.P., there is an overall loss.
Loss = \(41,269.84 - 40,000 = \text{Rs } 1,269.84\)
Overall Loss% = \(\frac{1269.84}{41269.84} \times 100\% \approx 3.08\%\)
In simple words: Calculate the individual cost price for each buffalo, sum them up, and compare with total selling price to find overall loss percentage.

Exam Tip: Never average profit and loss percentages; always find individual cost prices and total amounts.

Question 31. Find the amount and compound interest on Rs 20,000 for 1.5 years at 10% per annum compounded half yearly. Would this interest be more than the interest he gets if it was compounded annually?
Answer: Principal (\(P\)) = Rs 20,000
Annual Rate = 10%
Case 1: Compounded half-yearly:
Half-yearly Rate = \(10\% / 2 = 5\%\)
Time = \(1.5 \text{ years} = 3 \text{ half-years}\)
\(A = 20000 \left(1 + \frac{5}{100}\right)^3 = 20000 \left(\frac{21}{20}\right)^3 = 20000 \times \frac{9261}{8000} = \text{Rs } 23,152.50\)
\(\text{C.I. (Half-yearly)} = 23,152.50 - 20,000 = \text{Rs } 3,152.50\)
Case 2: Compounded annually for 1.5 years:
Amount for 1 year = \(20000 \left(1 + \frac{10}{100}\right)^1 = 20000 \times 1.1 = \text{Rs } 22,000\)
Interest for the remaining half year (0.5 year) at 10% p.a. = \(\frac{22000 \times 10 \times 0.5}{100} = \text{Rs } 1100\)
Total Amount (Annually) = \(22000 + 1100 = \text{Rs } 23,100\)
\(\text{C.I. (Annually)} = 23,100 - 20,000 = \text{Rs } 3,100\)
Comparison: Interest compounded half-yearly (Rs 3,152.50) is more than compounded annually (Rs 3,100) by Rs 52.50.
In simple words: Compounding half-yearly yields more interest because interest is added more frequently.

Exam Tip: More frequent compounding periods always result in higher compound interest returns.

Question 32. If 60% people speak Spanish, 30% speak English and the remaining speak other language, then what percent of the people speak other languages? If the total population is 50 lakhs, find the exact number of people who speak each type of language?
Answer: Percentage of people speaking Spanish = 60%
Percentage of people speaking English = 30%
Percentage of people speaking other languages = \(100\% - (60\% + 30\%) = 100\% - 90\% = 10\%\)
Total population = 50 lakhs = 5,000,000
Number of people speaking Spanish = \(60\% \text{ of } 50 \text{ lakhs} = \frac{60}{100} \times 50 = 30 \text{ lakhs}\) (or 3,000,000)
Number of people speaking English = \(30\% \text{ of } 50 \text{ lakhs} = \frac{30}{100} \times 50 = 15 \text{ lakhs}\) (or 1,500,000)
Number of people speaking other languages = \(10\% \text{ of } 50 \text{ lakhs} = \frac{10}{100} \times 50 = 5 \text{ lakhs}\) (or 500,000)
In simple words: Subtract the known percentages from 100% to find the percentage for other languages, then calculate counts for each category.

Exam Tip: Always verify that the sum of all percentage shares equals 100% before calculating actual numbers.

Most Important Questions

Question 1. Ratio is a means of comparing quantities of similar kinds. (T/F)
Answer: True
In simple words: Ratios compare two quantities of the same unit and kind.

Exam Tip: Ratios have no units because the units cancel out during division.

Question 2. Is 2:4 = 1:2?
Answer: Yes, 2:4 is equal to 1:2.
In simple words: Simplifying 2:4 by dividing both numbers by 2 gives 1:2.

Exam Tip: Equivalent ratios can be found by multiplying or dividing both terms by the same non-zero number.

Question 3. Percentages are a means of comparison. (T/F)
Answer: True
In simple words: Percentages express numbers as fractions out of 100 to make comparison easy.

Exam Tip: 'Percent' literally means 'per hundred'.

Question 4. Increase in price = old price + increase (True/False)
Answer: False
In simple words: New price equals old price plus increase, whereas increase alone is just the extra amount.

Question 5. Discount = _______ - _______
Answer: Discount = Marked Price - Selling Price
In simple words: Discount is the price reduction given on the marked price.

Exam Tip: Marked Price is always greater than or equal to Selling Price during discounts.

Question 6. Find the discount price of a skirt with a price of Rs 200 and a discount of 25%. What will be the sale price.
Answer: Marked Price = Rs 200
Discount percentage = 25%
Discount amount = \(25\% \text{ of Rs } 200 = \frac{25}{100} \times 200 = \text{Rs } 50\)
Sale Price = Marked Price - Discount amount = \(200 - 50 = \text{Rs } 150\)
In simple words: Calculate 25% of Rs 200 and subtract it from the marked price to get the sale price.

Exam Tip: Sale price is the final price paid after subtracting the discount from the marked price.

Question 7. What will be the price of petrol if its price now is Rs 50 and the hike is expected to be 12%?
Answer: Original price = Rs 50
Hike percentage = 12%
Increase amount = \(12\% \text{ of Rs } 50 = \frac{12}{100} \times 50 = \text{Rs } 6\)
New price of petrol = Original price + Increase = \(50 + 6 = \text{Rs } 56\)
In simple words: Find 12% of Rs 50 and add it to the original price.

Exam Tip: Price hikes represent percentage increases added to the base price.

Question 8. What will be the price of rice if its price now is Rs 25/kg and the reduction is expected to be 8 %?
Answer: Original price = Rs 25/kg
Reduction percentage = 8%
Reduction amount = \(8\% \text{ of Rs } 25 = \frac{8}{100} \times 25 = \text{Rs } 2\)
New price of rice = Original price - Reduction = \(25 - 2 = \text{Rs } 23/\text{kg}\)
In simple words: Calculate 8% of Rs 25 and subtract it to find the reduced price.

Exam Tip: Reductions and discounts are percentage decreases subtracted from the original value.

Question 9. Convert into percentage: - a) 3 : 4
Answer: Given ratio = \(3:4 = \frac{3}{4}\)
Percentage = \(\frac{3}{4} \times 100\% = 3 \times 25\% = 75\%\)
In simple words: Multiply the fractional form by 100 to convert a ratio into percentage.

Exam Tip: Always multiply by 100% when converting fractions or ratios to percentages.

Question 10. What percent of the total distance of 100km is 22km?
Answer: Total distance = 100 km
Part distance = 22 km
Percentage = \(\frac{22}{100} \times 100\% = 22\%\)
In simple words: Put the part over the whole and multiply by 100 to find the percentage.

Question 11. If M.P = Rs 650 and discount = 10% Find Sale Price
Answer: Marked Price (M.P.) = Rs 650
Discount percentage = 10%
Discount amount = \(10\% \text{ of Rs } 650 = \frac{10}{100} \times 650 = \text{Rs } 65\)
Sale Price = Marked Price - Discount = \(650 - 65 = \text{Rs } 585\)
In simple words: Calculate 10% of the marked price and subtract it to get the sale price.

Exam Tip: Sale Price = Marked Price \(\times (1 - \text{Discount}\%/100)\)

Question 12. A fan of M.P of Rs 1500 is sold for Rs 1080 after allowing a certain discount. Find the rate of discount
Answer: Marked Price (M.P.) = Rs 1500
Selling Price (S.P.) = Rs 1080
Discount = M.P. - S.P. = \(1500 - 1080 = \text{Rs } 420\)
Discount% = \(\frac{\text{Discount}}{\text{M.P.}} \times 100\% = \frac{420}{1500} \times 100\% = \frac{42}{15}\% = 28\%\)
In simple words: Find the monetary discount by subtracting selling price from marked price, then compute its percentage on the marked price.

Exam Tip: Discount percentage is always calculated with respect to the Marked Price.

Question 13. A shirt with marked price Rs 1500 was sold to a customer for Rs 1080. Find the rate of discount allowed on the shirt.
Answer: Marked Price (M.P.) = Rs 1500
Selling Price (S.P.) = Rs 1080
Discount = \(1500 - 1080 = \text{Rs } 420\)
Discount Rate% = \(\frac{420}{1500} \times 100\% = 28\%\)
In simple words: This is identical to the previous question; find the discount amount and express it as a percentage of the marked price.

Exam Tip: Always double check calculations by applying the discount percentage back to the marked price.

Question 14. A team won 6 matches out of total that they played. If their winning percentage is 60 percent, what is the total number of matches they played? How many did they loose?
Answer: Let total matches played be \(x\).
Matches won = 6
Winning percentage = 60%
\(60\% \text{ of } x = 6\)
\(\frac{60}{100} \times x = 6\)
\(x = \frac{6 \times 100}{60} = 10 \text{ matches in total}\)
Number of matches lost = Total matches - Matches won = \(10 - 6 = 4 \text{ matches}\)
In simple words: Calculate total matches using the winning percentage equation, then subtract wins from total to find losses.

Exam Tip: Total matches equal the sum of matches won, lost, and drawn.

Question 15. What price should Kavita mark on an item, which costs her Rs 2200 so as to gain 26% after allowing a discount of 12%
Answer: Cost Price (C.P.) = Rs 2200
Gain = 26%
Selling Price (S.P.) = \(\text{C.P.} + 26\% \text{ of C.P.} = 2200 + \frac{26}{100} \times 2200 = 2200 + 572 = \text{Rs } 2772\)
Let Marked Price be \(x\).
Discount = 12% on Marked Price
\(\text{S.P.} = \text{M.P.} - 12\% \text{ of M.P.} = 0.88x\)
\(2772 = 0.88x\)
\(x = \frac{2772}{0.88} = \text{Rs } 3150\)
In simple words: Calculate selling price from cost price and profit first, then work backward using the discount to find the marked price.

Exam Tip: Selling Price acts as the key link between Cost Price/Profit and Marked Price/Discount.

Question 16. A cycle merchant allows 25% discount on the marked price of the cycle and still makes a profit of 20%. If he gains Rs 360 over the sale of one cycle find M.P of the cycle.
Answer: Profit amount = Rs 360
Profit percentage = 20%
Let Cost Price (C.P.) be \(x\).
\(20\% \text{ of } x = 360\)
\(\frac{20}{100} \times x = 360 \Rightarrow x = \frac{360 \times 100}{20} = \text{Rs } 1800 \text{ (C.P.)}\)
Selling Price (S.P.) = \(\text{C.P.} + \text{Profit} = 1800 + 360 = \text{Rs } 2160\)
Let Marked Price be \(y\).
Discount = 25% on Marked Price
\(\text{S.P.} = y - 25\% \text{ of } y = 0.75y\)
\(2160 = 0.75y\)
\(y = \frac{2160}{0.75} = \text{Rs } 2880\)
In simple words: Find cost price from profit amount, calculate selling price, and then use the 25% discount to find the marked price.

Exam Tip: Proceed step-by-step: Profit -> Cost Price -> Selling Price -> Marked Price.

Question 17. A shopkeeper offers his customers 10% discount and still makes a profit of 26%. What is the actual cost to his of an article marked Rs 280.
Answer: Marked Price (M.P.) = Rs 280
Discount = 10%
Selling Price (S.P.) = \(280 - 10\% \text{ of } 280 = 280 - 28 = \text{Rs } 252\)
Profit = 26%
Let Cost Price (C.P.) be \(x\).
\(\text{S.P.} = \text{C.P.} + 26\% \text{ of C.P.} = 1.26x\)
\(252 = 1.26x \Rightarrow x = \frac{252}{1.26} = \text{Rs } 200\)
In simple words: Compute selling price from marked price and discount, then determine cost price using the profit percentage.

Question 18. 0.6 % can be expressed as
Answer: \(0.6\% = \frac{0.6}{100} = \frac{6}{1000} = 0.006\) (or as a fraction \(\frac{3}{500}\))
In simple words: Remove the percentage sign by dividing the number by 100.

Exam Tip: To convert any percentage to a decimal, divide by 100 or shift the decimal point two places to the left.

Question 19. Gain% = \(\frac{\text{Gain}}{\text{S.P.}} \times 100\%\) [Note: Standard formula is \(\frac{\text{Gain}}{\text{C.P.}} \times 100\%\)]
Answer: Formula: Gain% = \(\frac{\text{Gain}}{\text{C.P.}} \times 100\%\)
In simple words: Profit percentage is always calculated on the basis of Cost Price.

Exam Tip: Never divide profit by selling price unless specifically instructed by a unique problem statement.

Question 20. Discount = \(\frac{\text{M.P.} \times \text{Discount}\%}{100}\)
Answer: Formula: Discount = \(\text{M.P.} \times \frac{\text{Discount}\%}{100}\)
In simple words: Calculate discount amount by taking the discount percentage of the marked price.

Exam Tip: This formula gives the direct money value of the discount.

Question 21. Loss% = \(\frac{(\text{C.P.} - \text{S.P.})}{\text{C.P.}} \times 100\%\)
Answer: Formula: Loss% = \(\frac{\text{Loss}}{\text{C.P.}} \times 100\% = \frac{\text{C.P.} - \text{S.P.}}{\text{C.P.}} \times 100\%\)
In simple words: Loss percentage is the ratio of loss amount to cost price expressed as a percentage.

Exam Tip: Both profit and loss percentages are always calculated relative to the Cost Price.

Question 22. If M.P = Rs x and discount = y%. Find S.P
Answer: Selling Price (S.P.) = \(x - \left(\frac{y}{100} \times x\right) = x \left(1 - \frac{y}{100}\right) = \frac{x(100 - y)}{100}\)
In simple words: Subtract the discount amount from the marked price to get the selling price formula.

Exam Tip: Remember this algebraic expression for quick calculations involving unknown marked prices.

Question 23. If C.P = Rs X and loss = A% Find S.P.
Answer: Selling Price (S.P.) = \(X - \left(\frac{A}{100} \times X\right) = X \left(1 - \frac{A}{100}\right) = \frac{X(100 - A)}{100}\)
In simple words: Subtract the loss amount from the cost price to obtain the selling price.

Exam Tip: For loss, multiply C.P. by \((100 - \text{Loss}\%)/100\).

Question 24. If S.P = Rs. Y and Gain = B% find C.P
Answer: Cost Price (C.P.) = \(\frac{100}{100 + B} \times Y\)
In simple words: Divide the selling price by \((1 + \text{Gain}\%/100)\) to find the cost price.

Exam Tip: Use this formula when working backward from selling price to cost price during a profit.

Question 25. Total Bill amount = Bill amount + Sales Tax% of (bill amount) (T/F)
Answer: True
In simple words: Total bill includes the base cost plus the sales tax calculated on that cost.

Exam Tip: Sales tax is always charged on the taxable bill amount.

Question 26. Price including VAT = _________.
Answer: Original Price + VAT amount (or Original Price \(\times \left(1 + \frac{\text{VAT}\%}{100}\right)\))
In simple words: Price including tax is the base price plus the added value-added tax.

Question 27. To calculate the compound interest with the help of simple interest the Principal Amount of second year = _____________
Answer: Principal + Simple Interest of the first year (Amount at the end of the first year)
In simple words: In compound interest, the principal for the next year becomes the total amount of the previous year.

Exam Tip: This principle is why compound interest grows faster than simple interest.

Question 28. If Principal Amount = P, Rate of Interest = r%, Time period = n years. Than compound Interest will be = ___________
Answer: \(P\left(1 + \frac{r}{100}\right)^n - P\)
In simple words: Compound interest is total amount minus the initial principal.

Exam Tip: Always subtract \(P\) from the compound amount formula to get the interest only.

Question 29. The Difference between the original principal and last amount is said to be ___________
Answer: Compound Interest (or Total Interest)
In simple words: The extra money earned over the principal is the total interest.

Question 30. The time period after which interest is added each time to form a new principal is said to be __________
Answer: Conversion period
In simple words: Conversion period is the interval after which interest is compounded.

Question 31. What is the formula to calculate Simple interest?
Answer: Simple Interest (S.I.) = \(\frac{P \times R \times T}{100}\)
In simple words: Multiply principal, rate, and time, then divide by 100.

Exam Tip: Here \(P\) is principal, \(R\) is rate per cent per annum, and \(T\) is time in years.

Question 32. Write the formula for computing price including Sales Tax.
Answer: Price including tax = \(\text{Original Price} + (\text{Sales Tax}\% \text{ of Original Price})\)
In simple words: Add the calculated sales tax to the original item price.

Exam Tip: Useful for finding final consumer bills.

Question 33. A man purchased an article for Rs 1000. He sold it by gaining two-fifth of its cost price. Find his S.P. of article.
Answer: Cost Price (C.P.) = Rs 1000
Gain = \(\frac{2}{5} \text{ of C.P.} = \frac{2}{5} \times 1000 = \text{Rs } 400\)
Selling Price (S.P.) = C.P. + Gain = \(1000 + 400 = \text{Rs } 1400\)
In simple words: Calculate two-fifths of Rs 1000 and add it to the cost price to get the selling price.

Exam Tip: Fractional gains are calculated directly on the cost price.

Question 34. C.P = Rs. 225, overhead charges = 15, S.P = Rs 300. Find loss or profit percent
Answer: Initial C.P. = Rs 225
Overhead charges = Rs 15
Total Cost Price = \(225 + 15 = \text{Rs } 240\)
Selling Price (S.P.) = Rs 300
Since S.P. > Total C.P., there is a profit.
Profit = S.P. - Total C.P. = \(300 - 240 = \text{Rs } 60\)
Profit% = \(\frac{\text{Profit}}{\text{Total C.P.}} \times 100\% = \frac{60}{240} \times 100\% = \frac{1}{4} \times 100\% = 25\%\)
In simple words: Add overhead expenses to get total cost price, find the profit, and calculate profit percentage.

Exam Tip: Always include overhead charges in the cost price before computing profit or loss.

Question 35. A fruit seller purchased 6 oranges for a rupee and sold them at the rate of Rs 2.50 per dozen, find his gain or loss%.
Answer: Cost of 6 oranges = Re 1 (Rs 1)
Cost of 12 oranges (1 dozen) = \(1 \times 2 = \text{Rs } 2\)
Selling Price of 1 dozen oranges = Rs 2.50
Since S.P. > C.P., there is a profit.
Profit = \(2.50 - 2.00 = \text{Rs } 0.50\)
Profit% = \(\frac{\text{Profit}}{\text{C.P.}} \times 100\% = \frac{0.50}{2.00} \times 100\% = \frac{1}{4} \times 100\% = 25\%\)
In simple words: Make the quantity of oranges purchased and sold equal (1 dozen) to compare cost and selling prices.

Exam Tip: Always equalize the number of articles when solving purchase-sale ratio word problems.

Question 36. Find the Simple interest of Rs 8500 for 2 years at 8% p.a.
Answer: Principal (\(P\)) = Rs 8500
Time (\(T\)) = 2 years
Rate (\(R\)) = 8% p.a.
Simple Interest (S.I.) = \(\frac{P \times R \times T}{100} = \frac{8500 \times 8 \times 2}{100} = 85 \times 16 = \text{Rs } 1360\)
In simple words: Substitute principal, rate, and time into the simple interest formula.

Exam Tip: Double check multiplication steps to avoid arithmetic errors.

Question 37. Find the Compound interest on Rs 5000 for 1 yr at 8% p.a. compounded annually.
Answer: Principal (\(P\)) = Rs 5000
Rate (\(R\)) = 8% p.a.
Time (\(n\)) = 1 year
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n = 5000 \left(1 + \frac{8}{100}\right)^1 = 5000 \times \frac{108}{100} = 50 \times 108 = \text{Rs } 5400\)
Compound Interest (C.I.) = \(A - P = 5400 - 5000 = \text{Rs } 400\)
In simple words: For 1 year, compound interest equals simple interest because compounding occurs only once.

Exam Tip: For \(n = 1\), compound interest and simple interest are always identical.

Question 38. Find the amount on Rs 16000 for 3 yrs at 5% p.a. compounded annually.
Answer: Principal (\(P\)) = Rs 16000
Time (\(n\)) = 3 years
Rate (\(R\)) = 5% p.a.
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n = 16000 \left(1 + \frac{5}{100}\right)^3 = 16000 \left(\frac{21}{20}\right)^3\)
\(A = 16000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} = 16000 \times \frac{9261}{8000} = 2 \times 9261 = \text{Rs } 18,522\)
In simple words: Apply the compound amount formula for 3 years with 5% annual interest.

Exam Tip: Cancel out common factors in fractions carefully before multiplying large numbers.

Question 39. The sale price of an article including the sales tax is Rs 616. The rate of sales tax is 10%. If the shopkeeper has made a profit of 12% then find the cost price.
Answer: Price including sales tax = Rs 616
Sales tax rate = 10%
Let marked/list price before tax be \(x\).
\(x + 10\% \text{ of } x = 616 \Rightarrow 1.10x = 616 \Rightarrow x = \frac{616}{1.10} = \text{Rs } 560\) (This is the selling price before tax / list price)
Since the shopkeeper makes a profit of 12% on C.P.:
Let Cost Price be \(C\).
\(C + 12\% \text{ of } C = 560 \Rightarrow 1.12C = 560 \Rightarrow C = \frac{560}{1.12} = \text{Rs } 500\)
In simple words: First remove the sales tax to find the selling price, then use the 12% profit to find the cost price.

Exam Tip: Sales tax is added to selling price, while profit is calculated on cost price.

Question 40. Find the Compound Interest on Rs 6400 for 2yrs compounded annually at 7\(\frac{1}{2}\)% p.a.
Answer: Principal (\(P\)) = Rs 6400
Time (\(n\)) = 2 years
Rate (\(R\)) = \(7\frac{1}{2}\% = \frac{15}{2}\%\) p.a.
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n = 6400 \left(1 + \frac{15}{200}\right)^2 = 6400 \left(\frac{215}{200}\right)^2 = 6400 \left(\frac{43}{40}\right)^2\)
\(A = 6400 \times \frac{43}{40} \times \frac{43}{40} = 6400 \times \frac{1849}{1600} = 4 \times 1849 = \text{Rs } 7396\)
Compound Interest = \(A - P = 7396 - 6400 = \text{Rs } 996\)
In simple words: Convert the mixed fraction rate into an improper fraction and apply the compound amount formula.

Exam Tip: Simplify fractions like \(215/200\) to \(43/40\) before squaring to make calculations easier.

Question 41. Find the amount of Rs 12000 after 2yrs, compounded annually, the rate of interest being 5% p.a. during the 1st yr and 6% p.a. during the 2nd yr.
Answer: Principal = Rs 12000
First year rate (\(R_1\)) = 5%
Second year rate (\(R_2\)) = 6%
Amount after 2 years (\(A\)) = \(P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right)\)
\(A = 12000 \left(1 + \frac{5}{100}\right) \left(1 + \frac{6}{100}\right) = 12000 \times \frac{105}{100} \times \frac{106}{100}\)
\(A = 12000 \times \frac{21}{20} \times \frac{53}{50} = \frac{12000 \times 1113}{1000} = 12 \times 1113 = \text{Rs } 13,356\)
In simple words: When annual rates differ each year, multiply the principal by each year's growth factor successively.

Exam Tip: This method works whenever annual interest rates change year by year.

Question 42. If the C.P of 10 chairs be equal to S.P of 16 chairs. Find gain or loss percent
Answer: Let C.P. of 1 chair be Re 1.
C.P. of 10 chairs = Rs 10
Given S.P. of 16 chairs = C.P. of 10 chairs = Rs 10
S.P. of 10 chairs = \(\frac{10}{16} \times 10 = \text{Rs } 6.25\)
Since C.P. > S.P. (for 10 chairs, C.P. is 10 and S.P. is 6.25), there is a loss.
Alternatively, considering 16 chairs: C.P. of 16 chairs = Rs 16, S.P. of 16 chairs = Rs 10.
Loss = \(16 - 10 = \text{Rs } 6\)
Loss% = \(\frac{\text{Loss}}{\text{C.P.}} \times 100\% = \frac{6}{16} \times 100\% = \frac{3}{8} \times 100\% = 37.5\%\)
In simple words: Assume cost price for 16 items, find total cost and selling price, then calculate loss percentage.

Exam Tip: Equate quantities to a common number (like 16 chairs) to easily determine profit or loss.

Question 43. Sabena bought 16 dozen-ball pens and sold them at a loss equal to S.P of 8 ball pens. Find her loss%
Answer: Total pens = 16 dozen = \(16 \times 12 = 192\) pens.
Let S.P. of 1 pen be Re 1.
S.P. of 192 pens = Rs 192.
Loss = S.P. of 8 pens = \(8 \times 1 = \text{Rs } 8\).
C.P. = S.P. + Loss = \(192 + 8 = \text{Rs } 200\).
Loss% = \(\frac{\text{Loss}}{\text{C.P.}} \times 100\% = \frac{8}{200} \times 100\% = 4\%\)
In simple words: Express selling price and loss in terms of a single pen's price, find total cost price, and compute loss percentage.

Exam Tip: When loss is given in terms of S.P., use \(\text{C.P.} = \text{S.P.} + \text{Loss}\).

Question 44. Find the difference between the compound interest and the simple interest on Rs 32000 at 12% p.a. for 3 yrs.
Answer: Principal (\(P\)) = Rs 32000, Rate (\(R\)) = 12% p.a., Time (\(T\)) = 3 years.
Simple Interest (S.I.) = \(\frac{P \times R \times T}{100} = \frac{32000 \times 12 \times 3}{100} = 320 \times 36 = \text{Rs } 11,520\)
Compound Interest (C.I.):
\(A = P \left(1 + \frac{R}{100}\right)^3 = 32000 \left(1 + \frac{12}{100}\right)^3 = 32000 \left(\frac{112}{100}\right)^3 = 32000 \left(\frac{28}{25}\right)^3\)
\(A = 32000 \times \frac{21952}{15625} = \frac{70246400}{15625} = \text{Rs } 44,957.696\)
\(\text{C.I.} = A - P = 44,957.696 - 32,000 = \text{Rs } 12,957.696\)
Difference = \(\text{C.I.} - \text{S.I.} = 12,957.696 - 11,520 = \text{Rs } 1,437.70\)
In simple words: Calculate simple interest and compound interest separately for 3 years and subtract them.

Exam Tip: For 3 years, the direct formula for C.I. - S.I. difference is \(P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right)\).

Question 45. A charity purchased 100 blankets at Rs 2000 each. They found that 10 blankets were defective and they sold these at Rs 1200 each. At what rate should they sell the remaining blankets so as to gain 14% on the whole?
Answer: Total blankets purchased = 100
Cost price per blanket = Rs 2000
Total Cost Price = \(100 \times 2000 = \text{Rs } 200,000\)
Desired total profit = 14% on the whole
Total Selling Price required = \(\text{Total C.P.} + 14\% \text{ of Total C.P.} = 200,000 + \frac{14}{100} \times 200,000 = 200,000 + 28,000 = \text{Rs } 228,000\)
Defective blankets = 10, sold at Rs 1200 each.
Money realized from defective blankets = \(10 \times 1200 = \text{Rs } 12,000\)
Remaining good blankets = \(100 - 10 = 90\)
Money needed from remaining blankets = \(228,000 - 12,000 = \text{Rs } 216,000\)
Selling price per remaining blanket = \(\frac{216,000}{90} = \text{Rs } 2400\)
In simple words: Calculate total target selling price for a 14% overall gain, subtract money earned from defective blankets, and divide the remainder by the number of good blankets.

Question 46. Rajan bought two buffaloes for Rs 30000. By selling one at a loss of 15% and other at a gain of 19%, he found that selling price of both buffaloes is the same. Fine the C.P of each.
Answer: Let C.P. of the first buffalo be \(x\).
C.P. of the second buffalo = \(30,000 - x\)
First buffalo sold at 15% loss, so S.P. = \(x - 15\% \text{ of } x = 0.85x\)
Second buffalo sold at 19% gain, so S.P. = \((30,000 - x) + 19\% \text{ of } (30,000 - x) = 1.19(30,000 - x)\)
Since selling prices are equal:
\(0.85x = 1.19(30,000 - x)\)
\(0.85x = 35,700 - 1.19x\)
\(0.85x + 1.19x = 35,700\)
\(2.04x = 35,700\)
\(x = \frac{35,700}{2.04} = \text{Rs } 17,500\)
C.P. of first buffalo = Rs 17,500
C.P. of second buffalo = \(30,000 - 17,500 = \text{Rs } 12,500\)
In simple words: Set up an equation equating the selling price expressions of both buffaloes and solve for the cost price of the first.

Exam Tip: When total C.P. and equal S.P. are given, use a single variable \(x\) for one cost price and \(\text{Total} - x\) for the other.

Question 47. A man sold two articles at Rs 25920 each. These were sold at 8% gain and 4% loss respectively. Find the gain or loss present in the whole transaction.
Answer: Selling price of each article = Rs 25920.
Total Selling Price = \(25920 + 25920 = \text{Rs } 51,840\)
For Article 1 (8% gain):
\(\text{C.P.}_1 = \frac{100}{100 + 8} \times 25,920 = \frac{100}{108} \times 25,920 = \text{Rs } 24,000\)
For Article 2 (4% loss):
\(\text{C.P.}_2 = \frac{100}{100 - 4} \times 25,920 = \frac{100}{96} \times 25,920 = \text{Rs } 27,000\)
Total Cost Price = \(24,000 + 27,000 = \text{Rs } 51,000\)
Since Total S.P. (Rs 51,840) > Total C.P. (Rs 51,000), there is an overall profit.
Profit = \(51,840 - 51,000 = \text{Rs } 840\)
Overall Profit% = \(\frac{840}{51,000} \times 100\% \approx 1.65\%\)
In simple words: Find individual cost prices for both articles, add them to get total cost, and compare with total selling price.

Exam Tip: Never mix up overall profit/loss with individual percentages; always compute absolute total C.P. and S.P.

Question 48. In how much time will a sum of Rs 1000 amount to Rs 1331.20 at 10% p.a. Compound interest?
Answer: Principal (\(P\)) = Rs 1000, Amount (\(A\)) = Rs 1331.20, Rate (\(R\)) = 10% p.a.
\(A = P \left(1 + \frac{R}{100}\right)^n\)
\(1331.20 = 1000 \left(1 + \frac{10}{100}\right)^n\)
\(\frac{1331.20}{1000} = \left(\frac{11}{10}\right)^n\)
\(\frac{13312}{10000} = \left(\frac{11}{10}\right)^n \Rightarrow \frac{1664}{1250} = \dots\) wait, let us simplify properly:
\(\frac{1331.2}{1000} = \frac{13312}{10000} = \frac{1331}{1000} = \left(\frac{11}{10}\right)^3\)
Therefore, \(\left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^n\)
Time (\(n\)) = 3 years.
In simple words: Express the amount-to-principal ratio as a power of \((1 + R/100)\) to find the time exponent \(n\).

Question 49. At what rate % p.a. will a sum of Rs 7500 amount to Rs 8427 in 2 yrs, compounded annually?
Answer: Principal (\(P\)) = Rs 7500, Amount (\(A\)) = Rs 8427, Time (\(n\)) = 2 years.
\(A = P \left(1 + \frac{R}{100}\right)^n\)
\(8427 = 7500 \left(1 + \frac{R}{100}\right)^2\)
\(\frac{8427}{7500} = \left(1 + \frac{R}{100}\right)^2\)
Divide numerator and denominator by 3:
\(\frac{2809}{2500} = \left(1 + \frac{R}{100}\right)^2\)
Notice that \(2809 = 53^2\) and \(2500 = 50^2\):
\(\left(\frac{53}{50}\right)^2 = \left(1 + \frac{R}{100}\right)^2\)
Taking square root on both sides:
\(\frac{53}{50} = 1 + \frac{R}{100}\)
\(\frac{53}{50} - 1 = \frac{R}{100} \Rightarrow \frac{3}{50} = \frac{R}{100}\)
\(R = \frac{3 \times 100}{50} = 6\%\) p.a.
In simple words: Take the ratio of amount to principal, equate it to the squared binomial term, and solve for rate \(R\).

Exam Tip: Recognizing square numbers (like 2500, 2809) makes solving rate problems much faster.

Question 50. Semi annual means ________.
Answer: Twice a year (every 6 months)
In simple words: Semi-annual events occur every six months.

Exam Tip: For semi-annual compounding, interest is calculated twice a year.

Question 51. Quarterly means every ______ months in a year.
Answer: 3 months
In simple words: A quarter of a year equals 3 months, occurring 4 times a year.

Exam Tip: Quarterly compounding means 4 conversion periods per year.

Question 52. There are ______ conversion periods when the interest is computed half-yearly.
Answer: 2 conversion periods in a year
In simple words: Half-yearly computation divides the year into 2 periods.

Exam Tip: Number of conversion periods = \(2 \times \text{Number of years}\).

Question 53. There are ______ conversion periods when the interest is computed quarterly.
Answer: 4 conversion periods in a year
In simple words: Quarterly computation divides the year into 4 periods.

Exam Tip: Number of quarterly conversion periods = \(4 \times \text{Number of years}\).

Question 54. What is the formula for finding increased population during n years?
Answer: Population after \(n\) years = \(P \left(1 + \frac{R}{100}\right)^n\) (where \(P\) is initial population, \(R\) is growth rate %)
In simple words: Population growth follows the exact same formula as compound interest amount.

Exam Tip: Use positive sign for population growth and negative sign for population decline.

Question 55. If the growth rate of a company’s employees is r% every year and there head count in 2004 was x million, find their head count in 2009?
Answer: Initial head count (\(P\)) = \(x\) million
Time period (\(n\)) = \(2009 - 2004 = 5\) years
Growth rate = \(r\%\) per year
Head count in 2009 = \(x \left(1 + \frac{r}{100}\right)^5\) million
In simple words: Apply the compound growth formula for 5 years with rate \(r\%\).

Exam Tip: Count the number of years carefully from 2004 to 2009 to get \(n = 5\).

Question 56. If the value of a car depreciates by 6 % each year and its current value is 3 lakhs, find its value the next yr.
Answer: Current value = 3 lakhs = Rs 300,000
Depreciation rate = 6% per year
Depreciation amount = \(6\% \text{ of } 300,000 = \frac{6}{100} \times 300,000 = \text{Rs } 18,000\)
Value next year = Current value - Depreciation = \(300,000 - 18,000 = \text{Rs } 282,000\) (or 2.82 lakhs)
In simple words: Calculate 6% depreciation on 3 lakhs and subtract it from the current value.

Exam Tip: Depreciation always decreases the principal value year after year.

Question 57. When the interest is calculated semi-annually, the rate of interest of each conversion period is _______ & the time period gets ______.
Answer: halved (\(R/2\)) & doubled (\(2n\))
In simple words: Semi-annual calculation cuts the annual rate in half and doubles the number of years into half-year periods.

Exam Tip: Always make these two adjustments whenever semi-annual compounding is mentioned.

Question 58. When the interest is calculated quarterly, the rate of interest of each conversion period becomes _______ & the time period becomes ______.
Answer: divided by 4 (\(R/4\)) & multiplied by 4 (\(4n\))
In simple words: Quarterly compounding divides the annual rate by 4 and multiplies the number of years by 4.

Exam Tip: Remember: Quarterly = 4 quarters per year.

Question 59. Compound Interest formula is applicable in following cases (T/F).
a) To calculate Increase/Decrease in population.
b) Calculating value of an item increasing/decreasing over a number of years.
Answer: True (for both a and b)
In simple words: The compound growth/decay formula applies to any exponential increase or decrease scenario.

Exam Tip: Population growth and depreciation are practical applications of compound interest formulas.

Question 60. What will be the compound interest on Rs 1000 at the rate of 8% per annum for 1 and \(\frac{1}{2}\) years when interest is compounded quarterly.
Answer: Principal (\(P\)) = Rs 1000
Annual Rate = 8%
Quarterly Rate (\(R\)) = \(\frac{8\%}{4} = 2\%\) per quarter
Time = \(1\frac{1}{2} \text{ years} = \frac{3}{2} \text{ years}\)
Number of quarterly periods (\(n\)) = \(4 \times \frac{3}{2} = 6\) quarters
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n = 1000 \left(1 + \frac{2}{100}\right)^6 = 1000 \left(\frac{51}{50}\right)^6\)
\(A = 1000 \times (1.02)^6 = 1000 \times 1.1261624 = \text{Rs } 1,126.16\)
Compound Interest = \(A - P = 1126.16 - 1000 = \text{Rs } 126.16\)
In simple words: Divide annual rate by 4 and multiply years by 4 to get quarterly rate and conversion periods, then apply the formula.

Exam Tip: For quarterly compounding, divide rate by 4 and multiply time in years by 4.

Question 61. Find the compound interest on Rs 14,000 at the rate of 12% per annum for 1 and \(\frac{1}{2}\) years when interest is compounded semi-annually.
Answer: Principal (\(P\)) = Rs 14,000
Annual Rate = 12%
Semi-annual Rate (\(R\)) = \(\frac{12\%}{2} = 6\%\) per half-year
Time = \(1\frac{1}{2} \text{ years} = 3\) half-years
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n = 14000 \left(1 + \frac{6}{100}\right)^3 = 14000 \left(\frac{53}{50}\right)^3\)
\(A = 14000 \times \frac{148877}{125000} = \text{Rs } 16,674.22\)
Compound Interest = \(A - P = 16,674.22 - 14,000 = \text{Rs } 2,674.22\)
In simple words: Halve the 12% rate to 6% and double the 1.5 years to 3 half-yearly periods.

Exam Tip: Semi-annual compounding means 2 compounding cycles per year.

Question 62. Rakesh bought a scooter for Rs 42,000. If its value decreases by 8% every year, find its value at the end of 2 years.
Answer: Initial value (\(P\)) = Rs 42,000
Depreciation rate (\(R\)) = 8% per year
Time (\(n\)) = 2 years
Value after 2 years (\(A\)) = \(P \left(1 - \frac{R}{100}\right)^n = 42000 \left(1 - \frac{8}{100}\right)^2 = 42000 \left(\frac{92}{100}\right)^2\)
\(A = 42000 \times 0.92 \times 0.92 = 42000 \times 0.8464 = \text{Rs } 35,548.80\)
In simple words: Use the compound depreciation formula with a minus sign for the 2-year decrease in value.

Exam Tip: For decreasing values (depreciation), use \((1 - R/100)\) instead of \((1 + R/100)\).

Question 63. Find the amount to be paid on a loan of Rs 50,000 for 3 years at 12% per annum compounded semi-annually.
Answer: Principal (\(P\)) = Rs 50,000
Annual Rate = 12%
Semi-annual Rate (\(R\)) = \(\frac{12\%}{2} = 6\%\) per half-year
Time = 3 years
Number of semi-annual periods (\(n\)) = \(3 \times 2 = 6\) half-years
Amount (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n = 50000 \left(1 + \frac{6}{100}\right)^6 = 50000 \left(\frac{53}{50}\right)^6\)
\(A = 50000 \times (1.06)^6 = 50000 \times 1.418519 = \text{Rs } 70,925.95\)
In simple words: Compute semi-annual rate as 6% and conversion periods as 6, then apply the compound amount formula.

Question 64. The bacteria in our hands grow at the rate of 20% every hour. What will be their count after 3 hrs if their initial count is 5 lakhs.
Answer: Initial count (\(P\)) = 5 lakhs = 500,000
Growth rate (\(R\)) = 20% per hour
Time (\(n\)) = 3 hours
Bacteria count after 3 hours (\(A\)) = \(P \left(1 + \frac{R}{100}\right)^n = 500,000 \left(1 + \frac{20}{100}\right)^3 = 500,000 \left(\frac{6}{5}\right)^3\)
\(A = 500,000 \times \frac{216}{125} = 4000 \times 216 = 864,000\) (or 8.64 lakhs)
In simple words: Apply compound growth formula with 20% growth rate over 3 hourly periods.

Exam Tip: Growth problems in biology follow the exact compound interest exponential model.

Question 65. The population of the world in 2008 is 6 billion. It increases at a rate of 1.5% every year. What will be the population 2 years from now? Also estimate what would have been the population in 2006?
Answer: Part 1: Population 2 years from 2008 (i.e., in 2010):
Initial population in 2008 (\(P\)) = 6 billion
Rate (\(R\)) = 1.5% per year
Time (\(n\)) = 2 years
Population in 2010 = \(P \left(1 + \frac{R}{100}\right)^n = 6 \left(1 + \frac{1.5}{100}\right)^2 = 6 \left(\frac{101.5}{100}\right)^2\)
\(\text{Population in 2010} = 6 \times (1.015)^2 = 6 \times 1.030225 = 6.18135 \text{ billion}\)
Part 2: Estimated population 2 years before 2008 (i.e., in 2006):
Let population in 2006 be \(P_{2006}\).
\(6 = P_{2006} \left(1 + \frac{1.5}{100}\right)^2\)
\(P_{2006} = \frac{6}{(1.015)^2} = \frac{6}{1.030225} \approx 5.824 \text{ billion}\)
In simple words: Multiply current population by growth factor squared for future population, and divide by growth factor squared to find past population.

Exam Tip: To calculate past population, treat the past value as principal and present value as amount, working backward through division.