CBSE Class 8 Mathematics Fractions in Disguise MCQs Set 11

Answer: (b) out of hundred

Explanation:
1. Per centum is Latin for “by the hundred”, that is, out of hundred.
2. This is exactly what the symbol % stands for: 25% means 25 out of every 100.

Teacher's Note:
1. Link ‘cent’ with 100, as in century or centimetre.
2. Every percentage is a count out of 100.

Answer: (c) \( \frac{9}{20} \)

Explanation:
1. 45% = \( \frac{45}{100} \).
2. Dividing the numerator and the denominator by their HCF 5 gives \( \frac{9}{20} \).
3. Check: \( \frac{9}{20} \) = 0.45. None of the other options equals 0.45.

Teacher's Note:
1. Always start by writing the percentage over 100.
2. Then cancel by the HCF to reach the lowest terms.

Answer: (a) \( \frac{3}{5} \)

Explanation:
1. 60% = \( \frac{60}{100} \) = \( \frac{3}{5} \) (dividing both terms by 20).
2. \( \frac{6}{5} \) is 120%, \( \frac{3}{50} \) is 6%, and \( \frac{5}{3} \) is more than 1, so none of them equals 60%.

Teacher's Note:
1. Convert each option to a percentage to check quickly.
2. A fraction greater than 1 means more than 100%.

Answer: (d) 0.07

Explanation:
1. 7% = \( \frac{7}{100} \) = 0.07.
2. 0.7 is 70%, 7.0 is 700% and 0.007 is only 0.7%.

Teacher's Note:
1. Dividing by 100 moves the decimal point two places to the left.
2. Write 7 as 07 to place the point correctly.

Answer: (b) 37%

Explanation:
1. The bottom 3 rows are fully shaded: 3 × 10 = 30 squares.
2. In the next row, 7 more squares are shaded, so the total is 30 + 7 = 37 squares.
3. 37 out of 100 squares = \( \frac{37}{100} \) = 37%.

Teacher's Note:
1. When the whole is 100 squares, the count of shaded squares is the percentage.
2. 63% is the unshaded part, a common trap.

Answer: (c) 4% of the quantity

Explanation:
1. One-fourth = \( \frac{1}{4} \) = \( \frac{25}{100} \) = 25% = 0.25, so options (a), (b) and (d) all mean one-fourth.
2. 4% = \( \frac{4}{100} \) = \( \frac{1}{25} \), which is much smaller. So the answer is 4% of the quantity.

Teacher's Note:
1. The 4 in one-fourth is a denominator, not a percentage.
2. Remember: 1/4 = 25%.

Answer: (a) 30

Explanation:
1. 20% of 150 = \( \frac{20}{100} \) × 150 = 30.
2. Quick check: 10% of 150 is 15, and 20% is double that, 30.

Teacher's Note:
1. Find 10% first, then double it for 20%.
2. 15 is only 10% of 150.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 0.45 × 100 = 45, so 0.45 = 45%. A is true.
2. R is true, and multiplying by 100 is exactly the step that turns 0.45 into 45%, so R explains A.

Teacher's Note:
1. Decimal to percentage: multiply by 100.
2. Percentage to decimal: divide by 100.

Answer: (c) Both Statement I and Statement II are true.

Explanation:
1. Statement I is true: x% = \( \frac{x}{100} \).
2. Statement II is true: for example, \( \frac{2}{5} \) × 100 = 40, so \( \frac{2}{5} \) = 40%.
3. So both statements are true.

Teacher's Note:
1. A percentage is a fraction with denominator 100.
2. The two statements are the two directions of the same idea.

Answer: (a) 2.5 times last year’s harvest

Explanation:
1. 250% = \( \frac{250}{100} \) = 2.5.
2. So this year’s harvest is 2.5 times last year’s harvest.
3. A percentage above 100 means more than the whole of the original quantity.

Teacher's Note:
1. Divide the percentage by 100 to get the multiplier.
2. 100% means the same amount, 200% means double.

Answer: (b) (i)-(r), (ii)-(s), (iii)-(q), (iv)-(p)

Explanation:
1. \( \frac{1}{2} \) = \( \frac{50}{100} \) = 50% → (r).
2. \( \frac{1}{5} \) = \( \frac{20}{100} \) = 20% → (s).
3. \( \frac{3}{4} \) = \( \frac{75}{100} \) = 75% → (q).
4. \( \frac{1}{10} \) = \( \frac{10}{100} \) = 10% → (p).
5. So the matching is (i)-(r), (ii)-(s), (iii)-(q), (iv)-(p).

Teacher's Note:
1. Make the denominator 100 to read off the percentage.
2. These four conversions are worth memorising.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true: 20 parts out of 100 is twice 10 parts out of 100.
2. R is also true, because 10% = \( \frac{10}{100} \) = \( \frac{1}{10} \).
3. But R only tells us how to find 10%; it does not explain why 20% is double 10%. So R is not the correct explanation of A.

Teacher's Note:
1. Check whether R gives the reason for A, not just a true fact.
2. 20% = 2 × 10% because 20/100 = 2 × 10/100.

Answer: (d) 240

Explanation:
1. Let the number be n. Then \( \frac{30}{100} \) × n = 72.
2. n = 72 × \( \frac{100}{30} \) = 240.
3. Check: 30% of 240 = 0.3 × 240 = 72.

Teacher's Note:
1. Finding the whole from a part: divide the part by the fraction.
2. Always check by working forward.

Answer: (c) cost price

Explanation:
1. The price paid by the shopkeeper to buy the goods is the cost price.
2. The price he quotes is the marked price, and the price the customer finally pays is the selling price.

Teacher's Note:
1. Cost price is what the seller spends.
2. Profit and loss are measured on the cost price.

Answer: (c) A is true, but R is false.

Explanation:
1. Profit = 360 − 300 = ₹60, and profit % = \( \frac{60}{300} \) × 100 = 20%. So A is true.
2. R is false: profit percentage is calculated on the cost price, not on the selling price.
3. (Using the selling price would give \( \frac{60}{360} \) × 100 ≈ 16.7%, which is not 20%.)

Teacher's Note:
1. Profit % and loss % are always on the cost price.
2. Dividing by the selling price gives a different answer.

Answer: (a) Only Statement I is true.

Explanation:
1. Statement I: a 35% discount leaves 65%, and 0.65 × 1800 = 1170. True.
2. Statement II: 18% GST adds 18% of ₹8250, not ₹18. The price becomes 8250 × 1.18 = ₹9735. False.
3. So only Statement I is true.

Teacher's Note:
1. A percentage is never added as a plain number of rupees.
2. Use multipliers: 0.65 for 35% off, 1.18 for 18% added.

Answer: (d) ₹5000

Explanation:
1. The bar for Day 4 stands at 125% of the target.
2. Sales on Day 4 = 125% of ₹4000 = \( \frac{125}{100} \) × 4000 = ₹5000.

Teacher's Note:
1. Read the bar height first, then apply it to the target.
2. A bar above 100% means the target was exceeded.

Answer: (d) ₹2100

Explanation:
1. A 16% discount leaves 100% − 16% = 84% of the marked price.
2. Selling price = 0.84 × 2500 = ₹2100.
3. (The discount itself is 16% of 2500 = ₹400.)

Teacher's Note:
1. Discount is always on the marked price.
2. Subtract the percentage first, then multiply once.

Answer: (d) (i)-(s), (ii)-(r), (iii)-(q), (iv)-(p)

Explanation:
1. (i) Profit = ₹100 on cost ₹400: \( \frac{100}{400} \) × 100 = 25% → (s).
2. (ii) Loss = ₹100 on cost ₹500: \( \frac{100}{500} \) × 100 = 20% → (r).
3. (iii) Discount = ₹75 on marked price ₹500: \( \frac{75}{500} \) × 100 = 15% → (q).
4. (iv) Increase = ₹40 on ₹400: \( \frac{40}{400} \) × 100 = 10% → (p).
5. So the matching is (i)-(s), (ii)-(r), (iii)-(q), (iv)-(p).

Teacher's Note:
1. Each percentage is taken on the original or cost amount.
2. The same ₹100 gives 25% on 400 but 20% on 500.

Answer: (d) A is false, but R is true.

Explanation:
1. With compounding, the amount = 10000 × 1.1 × 1.1 = ₹12,100, not ₹12,000. So A is false.
2. ₹12,000 is the amount without compounding.
3. R correctly describes compounding, so R is true.

Teacher's Note:
1. Compounding gives interest on interest.
2. Without compounding, ₹1000 is added each year.

Answer: (b) Only Statement II is true.

Explanation:
1. Statement I: 1.2 times = 120% of the earlier value, which is an increase of only 20%. False.
2. Statement II: a fall of 85% leaves 100% − 85% = 15% of the earlier demand. True.
3. So only Statement II is true.

Teacher's Note:
1. ‘Becomes 120% of’ is not the same as ‘increases by 120%’.
2. Increase % = (multiplier − 1) × 100.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The rise is 100 − 60 = ₹40, so R is true.
2. Percentage increase = \( \frac{40}{60} \) × 100 = 66.66… ≈ 66.7%, so A is true.
3. R only gives the amount of the rise; the percentage comes from dividing it by the original price ₹60. So R alone does not explain A.

Teacher's Note:
1. The base for a percentage change is the original value.
2. The amount of change alone does not give the percentage.

Answer: (d) principal

Explanation:
1. The money deposited or borrowed, on which interest is worked out, is the principal.
2. 8% is the rate of interest, and principal plus interest is the amount.

Teacher's Note:
1. Principal + interest = amount.
2. p.a. means per annum, that is, for each year.

Answer: (c) ₹6655

Explanation:
1. Year 3 begins with ₹6050.
2. Interest added = 10% of 6050 = ₹605.
3. Amount at the end = 6050 + 605 = ₹6655 (also 6050 × 1.1).

Teacher's Note:
1. Each year’s interest is on the amount at the start of that year.
2. The yearly interest keeps growing: 500, 550, 605.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Take the price as ₹100. After a 20% rise it is ₹120.
2. 20% of 120 is 24, so it falls to ₹96, which is less than ₹100. A is true.
3. R gives exactly the reason: the fall is measured on the larger price, so it removes more than the rise added. R explains A.

Teacher's Note:
1. Try ₹100 as the starting value in such questions.
2. Successive changes: 1.2 × 0.8 = 0.96.

Answer: (d) Bakery Q, by ₹5

Explanation:
1. Bakery P: 500 × 0.70 = ₹350, then 350 × 0.80 = ₹280.
2. Bakery Q: 500 × 0.55 = ₹275.
3. Bakery Q is cheaper by 280 − 275 = ₹5, because P’s second discount is on an already reduced price.

Teacher's Note:
1. 30% then 20% off is only 44% off, not 50%.
2. Multiply the multipliers: 0.7 × 0.8 = 0.56.

Answer: (a) Only Statement I is true.

Explanation:
1. Let the cost price be x. Marked price = 1.5x.
2. Selling price = 0.5 × 1.5x = 0.75x, i.e. 75% of the cost price. Statement I is true.
3. Selling at 0.75x is a loss of 25%, so Statement II is false.

Teacher's Note:
1. The markup is on the cost price, but the discount is on the larger marked price.
2. Equal percentages up and down do not cancel.

Answer: (b) (i)-(s), (ii)-(p), (iii)-(r), (iv)-(q)

Explanation:
1. An increase of r% multiplies by (1 + \( \frac{r}{100} \)) and a decrease multiplies by (1 − \( \frac{r}{100} \)).
2. (i) +5% then −2%: p × 1.05 × 0.98 → (s). (ii) +10% twice: p × 1.10 × 1.10 → (p).
3. (iii) −10% twice: p × 0.90 × 0.90 → (r). (iv) +3% then +4%: p × 1.03 × 1.04 → (q).
4. So the matching is (i)-(s), (ii)-(p), (iii)-(r), (iv)-(q).

Teacher's Note:
1. Turn each change into a multiplier.
2. Changes one after another mean multiplying the multipliers.

Answer: (c) 20%

Explanation:
1. Let the original area be l × b. The new length is 1.25l.
2. For the same area, the new breadth = \( \frac{l b}{1.25 l} \) = \( \frac{b}{1.25} \) = 0.8b.
3. 0.8b is 80% of b, a decrease of 20%.

Teacher's Note:
1. Keep the product fixed: the breadth multiplier is 1 ÷ 1.25.
2. 25% up is not undone by 25% down.

Answer: (b) x% of y is equal to y% of x

Explanation:
1. x% of y = \( \frac{x}{100} \) × y = \( \frac{xy}{100} \).
2. y% of x = \( \frac{y}{100} \) × x = \( \frac{xy}{100} \).
3. They are always equal. For example, 25% of 80 = 80% of 25 = 20.

Teacher's Note:
1. This swap trick makes mental work easy: 8% of 50 = 50% of 8 = 4.
2. Both expressions equal xy/100.