CBSE Class 8 Mathematics Fractions in Disguise Short Answer Questions with Answers

Short Answer Questions

Question 1. Express each of the following fractions as a percentage. Show the step you use in each case.
(i) \( \frac{7}{20} \)
(ii) \( \frac{3}{8} \)
Answer:
To change a fraction into a percentage, multiply it by 100.
(i) \( \frac{7}{20} \) × 100 = 35, so \( \frac{7}{20} \) = 35%. (Equivalent fraction method: \( \frac{7}{20} \) = \( \frac{7 × 5}{20 × 5} \) = \( \frac{35}{100} \).)
(ii) \( \frac{3}{8} \) × 100 = \( \frac{300}{8} \) = 37.5, so \( \frac{3}{8} \) = 37.5%.
In simple words: Multiply each fraction by 100: 7/20 is 35% and 3/8 is 37.5%.

Teacher's Note:
1. When the denominator does not divide 100, just multiply by 100.
2. A percentage can be a decimal like 37.5%.

Question 2. A sports channel reports that 65% of the seats in a stadium were filled.
(i) Write 65% as a fraction in its simplest form.
(ii) Write 65% as a decimal.
Answer:
A percentage is a fraction with denominator 100, so 65% = \( \frac{65}{100} \).
(i) \( \frac{65}{100} \) = \( \frac{13}{20} \) (dividing the numerator and the denominator by 5).
(ii) \( \frac{65}{100} \) = 0.65.
So the same proportion is written as 65%, \( \frac{13}{20} \) and 0.65.
In simple words: 65% is 65/100, which simplifies to 13/20 and equals 0.65.

Teacher's Note:
1. Fraction, decimal and percentage are three forms of one value.
2. Divide by 100 to get the decimal form.

Question 3. There are 40 students in Class 8-A and 24 of them are girls. Find the percentage of girls and the percentage of boys in the class. Check that the two percentages add up to 100%.
Answer:
Girls = \( \frac{24}{40} \) × 100 = 60%.
Boys = 40 − 24 = 16, so boys = \( \frac{16}{40} \) × 100 = 40%.
Check: 60% + 40% = 100%, because the two groups together make the whole class.
In simple words: Girls are 60% and boys are 40%, which together make the whole class, 100%.

Teacher's Note:
1. Find the number of boys first.
2. Parts of a whole always add to 100%.

Question 4. Find the following values and write one sentence explaining the method you used.
(i) 15% of 240
(ii) An average human body is about 60% water by weight. Find the weight of water in a person of weight 55 kg.
Answer:
(i) 15% of 240 = \( \frac{15}{100} \) × 240 = 36. I multiplied the quantity by the fraction \( \frac{15}{100} \).
(ii) 60% of 55 = \( \frac{60}{100} \) × 55 = 0.6 × 55 = 33. So the body contains about 33 kg of water. Here I used the decimal form 0.6 of 60%.
In simple words: To find a percentage of a quantity, multiply the quantity by the percentage written as a fraction or decimal.

Teacher's Note:
1. y% of z = (y/100) × z.
2. 15% can also be done as 10% + 5% = 24 + 12.

Question 5. Answer briefly.
(i) What does the word ‘per cent’ mean, and from which Latin phrase does it come?
(ii) Name the ancient Indian text in which the idea of ‘per hundred’ appears in connection with rates of interest.
Answer:
(i) ‘Per cent’ means “out of a hundred” or “by the hundred”. It comes from the Latin phrase per centum.
(ii) The idea of ‘per hundred’ appears as early as the 4th century BCE in Kautilya’s Arthashastra, which describes fair, commercial and other rates of interest in panas per month per cent.
In simple words: Per cent means out of a hundred (from per centum), and Kautilya’s Arthashastra used this idea for interest rates.

Teacher's Note:
1. The idea of percentages is more than 2000 years old.
2. The Arthashastra fixed interest ‘per month per cent’.

Question 6. A shop sells a shirt with a marked price of ₹700 at a discount of 20%. Find the discount amount and the selling price of the shirt.
Answer:
A discount of 20% is a reduction of 20% of the marked price.
Discount = \( \frac{20}{100} \) × 700 = ₹140.
Selling price = 700 − 140 = ₹560.
Check by another route: the customer pays 80% of ₹700 = 0.8 × 700 = ₹560.
In simple words: 20% of ₹700 is ₹140 off, so the shirt sells for ₹560.

Teacher's Note:
1. Discount is always taken on the marked price.
2. SP = 80% of MP gives the answer in one step.

Question 7. The price of a notebook rose from ₹40 to ₹46 in one year. Find the percentage increase in the price, and state clearly which amount you used as the base.
Answer:
Increase in price = 46 − 40 = ₹6.
Percentage increase = \( \frac{\text{increase}}{\text{original price}} \) × 100 = \( \frac{6}{40} \) × 100 = 15%.
The base used is the original price ₹40, not the new price ₹46, because a change is always compared with the amount we started from.
In simple words: The ₹6 rise on the original ₹40 is a 15% increase.

Teacher's Note:
1. The base is the starting value.
2. Using ₹46 as base would give a wrong answer of about 13%.

Question 8. Find 15% of 320 mentally, without multiplying by \( \frac{15}{100} \). Write down the two percentages you used and how you combined them.
Answer:
I used 10% and 5%.
10% of 320 = 32 (one tenth).
5% is half of 10%, so 5% of 320 = 16.
15% = 10% + 5%, so 15% of 320 = 32 + 16 = 48.
This works because \( \frac{10}{100} \)y + \( \frac{5}{100} \)y = \( \frac{15}{100} \)y.
In simple words: 10% of 320 is 32 and 5% is 16, so 15% is 32 + 16 = 48.

Teacher's Note:
1. 10% is the easiest starting point for mental work.
2. Percentages of the same quantity can be added.

Question 9. Rehana scored 38 marks out of 50 in a Hindi test and 52 marks out of 70 in a Maths test. In which test did she perform better? Justify your answer by converting both scores to percentages (round to two decimal places where needed).
Answer:
The two tests have different maximum marks, so the raw scores cannot be compared directly. Convert both to percentages.
Hindi: \( \frac{38}{50} \) × 100 = 76%.
Maths: \( \frac{52}{70} \) × 100 ≈ 74.29% (to two decimal places).
Since 76% > 74.29%, Rehana performed better in the Hindi test.
In simple words: Hindi is 76% and Maths about 74.29%, so she did better in Hindi.

Teacher's Note:
1. Bigger raw marks do not mean better performance.
2. Percentages compare on the same base of 100.

Question 10. 35% of a number is 91.
(i) Find the number.
(ii) Using your answer, find 60% of the same number.
Answer:
Let the number be n.
(i) \( \frac{35}{100} \) × n = 91, so n = 91 × \( \frac{100}{35} \) = 260.
(ii) 60% of 260 = \( \frac{60}{100} \) × 260 = 156.
Check: 35% of 260 = 0.35 × 260 = 91. Correct.
In simple words: The number is 260, and 60% of it is 156.

Teacher's Note:
1. Find the whole first, then any other percentage of it.
2. Shortcut: 1% is 91 ÷ 35 = 2.6, so 60% is 156.

Question 11. In a mortar mixture, sand and cement are mixed in the ratio 3 : 2.
(i) What percentage of the mixture is sand and what percentage is cement?
(ii) How much cement is needed to prepare 40 kg of this mixture?
Answer:
Sand : cement = 3 : 2, so the mixture has 3 + 2 = 5 equal parts.
(i) Sand = \( \frac{3}{5} \) × 100 = 60%; cement = \( \frac{2}{5} \) × 100 = 40%.
(ii) Cement in 40 kg = \( \frac{2}{5} \) × 40 = 16 kg (and sand = 24 kg).
In simple words: Sand is 60% and cement 40% of the mix, so 40 kg needs 16 kg of cement.

Teacher's Note:
1. Total the ratio parts to get the whole.
2. Check: 16 + 24 = 40 kg.

Question 12. Ravi deposits ₹8000 in a fixed deposit at 9% per annum for 3 years. The interest is paid out every year and is not added back to the deposit. Find the total interest he earns and the total amount he receives at the end of 3 years.
Answer:
The interest is paid out each year, so the principal stays ₹8000 throughout.
Interest for one year = 0.09 × 8000 = ₹720.
Interest for 3 years = 3 × 720 = ₹2160.
Amount received = p(1 + rt) = 8000 + 2160 = ₹10,160.
In simple words: He earns ₹720 each year, ₹2160 in all, and gets ₹10,160 back.

Teacher's Note:
1. Without compounding, the yearly interest stays the same.
2. Check: 8000 × (1 + 0.27) = 10,160.

Question 13. A motorbike bought for ₹60,000 loses 10% of its value every year. Find its value after 2 years. Explain why the loss in the second year is not the same amount as in the first year.
Answer:
Each year the value becomes 90% of the value at the start of that year.
After 1 year: 0.9 × 60000 = ₹54,000 (fall of ₹6000).
After 2 years: 0.9 × 54000 = ₹48,600 (fall of ₹5400).
The fall is smaller in the second year because 10% is taken on the reduced value ₹54,000, not on ₹60,000.
In simple words: The bike is worth ₹48,600 after 2 years; the second loss is smaller because it is 10% of a smaller value.

Teacher's Note:
1. Depreciation works like compounding in reverse.
2. 60000 × 0.9 × 0.9 = 48,600.

Question 14. A trader buys a table fan for ₹250.
(i) If he sells it for ₹310, find his profit percentage.
(ii) If instead he sells it for ₹225, find his loss percentage.
Answer:
Profit and loss percentages are always taken on the cost price ₹250.
(i) Profit = 310 − 250 = ₹60, so profit % = \( \frac{60}{250} \) × 100 = 24%.
(ii) Loss = 250 − 225 = ₹25, so loss % = \( \frac{25}{250} \) × 100 = 10%.
In simple words: Selling at ₹310 gives a 24% profit and selling at ₹225 gives a 10% loss.

Teacher's Note:
1. Use the cost price as the base in both parts.
2. First decide whether it is a profit or a loss.

Question 15. A bakery has a daily sales target of ₹4000.
(i) On Monday the sales were ₹5600. What percentage of the target was achieved?
(ii) On Tuesday the bakery achieved only 85% of the target. Find Tuesday’s sales.
Answer:
(i) Percentage of target = \( \frac{5600}{4000} \) × 100 = 140%. A value above 100% means the target was crossed, here by 40%.
(ii) Tuesday’s sales = \( \frac{85}{100} \) × 4000 = ₹3400, which is ₹600 short of the target.
In simple words: Monday reached 140% of the target, and Tuesday’s sales were ₹3400.

Teacher's Note:
1. A percentage can be greater than 100.
2. The target is the base in both parts.

Question 16. Find 18% of 50 and 50% of 18. Are the two answers equal? Using algebra, show that x% of y is always equal to y% of x.
Answer:
18% of 50 = \( \frac{18}{100} \) × 50 = 9 and 50% of 18 = \( \frac{50}{100} \) × 18 = 9. The answers are equal.
In general, x% of y = \( \frac{x}{100} \) × y = \( \frac{xy}{100} \) and y% of x = \( \frac{y}{100} \) × x = \( \frac{xy}{100} \).
Both simplify to the same expression, so x% of y always equals y% of x.
In simple words: Both are xy/100, so swapping the numbers gives the same answer, like 18% of 50 = 50% of 18 = 9.

Teacher's Note:
1. Use this swap to make mental calculations easier.
2. Multiplication order does not change a product.

Question 17. The price of a bag was ₹500. The shopkeeper first increased it by 20% and later reduced the increased price by 20%.
(i) Find the final price of the bag.
(ii) Is the final price equal to the original price? Find the overall percentage change and name the base you used.
Answer:
(i) After the rise: 500 × 1.2 = ₹600. After the fall: 600 × 0.8 = ₹480.
(ii) No. The final price ₹480 is not equal to ₹500. Overall change = 480 − 500 = −20, so the percentage change = \( \frac{20}{500} \) × 100 = 4% decrease, measured on the original price ₹500.
The rise of 20% was on ₹500, but the fall of 20% was on the bigger amount ₹600, so the fall is larger.
In simple words: The bag ends at ₹480, a 4% drop from ₹500, because the second 20% is taken on ₹600.

Teacher's Note:
1. Equal rise and fall percentages do not cancel.
2. 1.2 × 0.8 = 0.96, so 4% is lost.

Question 18. A showroom advertises “25% + 10% off” on a jacket priced at ₹1200, while a rival shop offers a flat 35% off on the same jacket.
(i) Find the price a customer pays in each shop.
(ii) Find the single discount percentage that is equivalent to “25% + 10% off”.
Answer:
(i) Showroom: 25% off gives 1200 × 0.75 = ₹900; then 10% off gives 900 × 0.9 = ₹810.
Rival shop: 1200 × 0.65 = ₹780. The rival shop is cheaper by ₹30.
(ii) Total reduction at the showroom = 1200 − 810 = ₹390, so the equivalent single discount = \( \frac{390}{1200} \) × 100 = 32.5%, not 35%, because the 10% is taken on the already reduced price.
In simple words: The showroom charges ₹810 and the rival ₹780; “25% + 10%” is really a 32.5% discount.

Teacher's Note:
1. Successive discounts multiply: 0.75 × 0.9 = 0.675.
2. They are always less than the sum of the two.

Question 19. A shopkeeper marks every item at 40% above its cost price and then allows a discount of 25% on the marked price. Taking the cost price as x, show by algebra whether he makes a profit or a loss, and find the percentage.
Answer:
Let the cost price be x.
Marked price = x + 40% of x = 1.4x.
Selling price = 1.4x × (1 − 0.25) = 1.4x × 0.75 = 1.05x.
Since 1.05x > x, he makes a profit. Profit = 1.05x − x = 0.05x, so profit % = \( \frac{0.05x}{x} \) × 100 = 5% of the cost price.
In simple words: He sells at 1.05 times his cost, which is a 5% profit.

Teacher's Note:
1. The markup and discount have different bases.
2. Compare the final multiplier with 1.

Question 20. A sum of ₹25,000 is to be invested for 2 years. Bank A offers 10% per annum without compounding, while Bank B offers 8% per annum compounded annually. Find the amount received from each bank and state which bank is better and by how much.
Answer:
Bank A (no compounding): amount = p(1 + rt) = 25,000 × (1 + 0.10 × 2) = 25,000 × 1.2 = ₹30,000.
Bank B (compounded annually): amount = p(1 + r)^t = 25,000 × 1.08 × 1.08 = 25,000 × 1.1664 = ₹29,160.
Bank A gives 30,000 − 29,160 = ₹840 more, so Bank A is better by ₹840.
Compounding helps, but 8% compounded for only 2 years cannot catch up with 10% simple interest.
In simple words: Bank A pays ₹30,000 and Bank B ₹29,160, so Bank A is better by ₹840.

Teacher's Note:
1. Compounding helps more over long periods.
2. A higher rate can beat compounding over short periods.