Class 8 Mathematics – NCERT (Ganita Prakash) Part 2
Chapter 1: FRACTIONS IN DISGUISE
Complete Question Bank with Answer Key & Explanations
🔹 SECTION A: Multiple Choice Questions (20 Questions)
1. 75% expressed as a fraction is:
(a) \(\frac{3}{4}\)
(b) \(\frac{7}{5}\)
(c) \(\frac{4}{3}\)
(d) \(\frac{1}{4}\)
Answer: (a)
Explanation: 75% = \(\frac{75}{100} = \frac{3}{4}\) after simplifying by 25.
2. If 20% of a number is 30, the number is:
(a) 150
(b) 120
(c) 180
(d) 60
Answer: (a)
Explanation: Let the number be \(x\).
\(20\% \times x = 30\)
\(\frac{20}{100} \times x = 30\)
\(x = 30 \times \frac{100}{20} = 150\)
3. A discount of 25% on ₹400 means the discount amount is:
(a) ₹25
(b) ₹100
(c) ₹300
(d) ₹375
Answer: (b)
Explanation: Discount = \(25\% \times 400 = \frac{25}{100} \times 400 = ₹100\)
4. If the price of a book increases from ₹80 to ₹100, the percentage increase is:
(a) 20%
(b) 25%
(c) 30%
(d) 40%
Answer: (b)
Explanation: Increase = \(100 - 80 = ₹20\)
Percentage increase = \(\frac{20}{80} \times 100 = 25\%\)
5. 200% of 50 is:
(a) 50
(b) 100
(c) 150
(d) 200
Answer: (b)
Explanation: \(200\% \times 50 = \frac{200}{100} \times 50 = 2 \times 50 = 100\)
6. \(\frac{2}{5}\) as a percentage is:
(a) 20%
(b) 40%
(c) 60%
(d) 80%
Answer: (b)
Explanation: \(\frac{2}{5} \times 100 = 40\%\)
7. If 10% of x = 15, then 30% of x is:
(a) 15
(b) 30
(c) 45
(d) 60
Answer: (c)
Explanation: \(10\% \times x = 15\) ⇒ \(x = 150\)
Then \(30\% \times 150 = 45\)
8. A shopkeeper sells an item at a profit of 20% on CP. If CP = ₹250, SP is:
(a) ₹270
(b) ₹300
(c) ₹350
(d) ₹200
Answer: (b)
Explanation: Profit = \(20\% \times 250 = ₹50\)
SP = CP + Profit = \(250 + 50 = ₹300\)
Or directly: SP = \(120\% \times 250 = ₹300\)
9. In a class of 40, 60% are girls. Number of boys is:
(a) 16
(b) 24
(c) 32
(d) 20
Answer: (a)
Explanation: Girls = \(60\% \times 40 = 24\)
Boys = Total - Girls = \(40 - 24 = 16\)
10. Simple Interest on ₹2000 at 5% p.a. for 2 years is:
(a) ₹100
(b) ₹200
(c) ₹250
(d) ₹300
Answer: (b)
Explanation: SI = \(\frac{P \times R \times T}{100} = \frac{2000 \times 5 \times 2}{100} = ₹200\)
11. If 40% of a number is 120, then the number is:
(a) 300
(b) 240
(c) 360
(d) 480
Answer: (a)
Explanation: Let number be \(x\)
\(40\% \times x = 120\) ⇒ \(x = 120 \times \frac{100}{40} = 300\)
12. Which is greater: 30% of 200 or 40% of 150?
(a) 30% of 200
(b) 40% of 150
(c) Both equal
(d) Cannot compare
Answer: (a)
Explanation:
\(30\% \times 200 = 60\)
\(40\% \times 150 = 60\)
They are equal, so (c) Both equal.
13. A number increased by 20% gives 180. The original number is:
(a) 150
(b) 160
(c) 140
(d) 200
Answer: (a)
Explanation: Let original number = \(x\)
\(x + 20\% \times x = 180\)
\(1.2x = 180\) ⇒ \(x = 150\)
14. 5% of 2 hours (in minutes) is:
(a) 5 minutes
(b) 6 minutes
(c) 8 minutes
(d) 10 minutes
Answer: (b)
Explanation: 2 hours = 120 minutes
\(5\% \times 120 = 0.05 \times 120 = 6\) minutes
15. If 25% of students in a school are absent and 450 are present, total students:
(a) 600
(b) 500
(c) 750
(d) 800
Answer: (a)
Explanation: If 25% absent, then 75% present = 450
Total students = \(\frac{450}{75} \times 100 = 600\)
16. Compound Interest on ₹1000 at 10% p.a. for 2 years:
(a) ₹100
(b) ₹210
(c) ₹200
(d) ₹1210
Answer: (b)
Explanation: Amount after 2 years = \(1000 \times (1.1)^2 = 1000 \times 1.21 = ₹1210\)
CI = Amount - Principal = \(1210 - 1000 = ₹210\)
17. A TV bought for ₹15,000 depreciates by 10% in one year. Its value after 1 year:
(a) ₹13,500
(b) ₹13,000
(c) ₹14,000
(d) ₹12,500
Answer: (a)
Explanation: Value after depreciation = \(15000 \times (100\% - 10\%) = 15000 \times 0.9 = ₹13,500\)
18. If \(\frac{a}{b} = \frac{3}{4}\), then what % of a is b?
(a) 75%
(b) 133.33%
(c) 120%
(d) 80%
Answer: (b)
Explanation: \(\frac{a}{b} = \frac{3}{4}\) ⇒ \(b = \frac{4}{3}a\)
\(\frac{b}{a} = \frac{4}{3} = 1.3333 = 133.33\%\)
So b is 133.33% of a.
19. In a mixture of milk and water, milk is 70%. In 50 litres, milk is:
(a) 30 litres
(b) 35 litres
(c) 40 litres
(d) 45 litres
Answer: (b)
Explanation: Milk = \(70\% \times 50 = 0.7 \times 50 = 35\) litres
20. Successive discounts of 20% and 10% are equivalent to a single discount of:
(a) 28%
(b) 30%
(c) 25%
(d) 32%
Answer: (a)
Explanation: Single equivalent discount = \(100 - [(100-20) \times (100-10)/100]\)
= \(100 - [80 \times 90/100] = 100 - 72 = 28\%\)
🔹 SECTION B: Assertion & Reasoning (20 Questions)
21.
Answer: (a) Both A and R are true and R explains A.
Explanation: R mathematically explains why A is true.
22.
Answer: (d) Both are false.
Explanation: Percentages can exceed 100% (e.g., 150% means 1.5 times).
23.
Answer: (a) Both true and R explains A.
Explanation: R shows the calculation that proves A.
24.
Answer: (b) Both true but R does not explain A.
Explanation: R defines profit but doesn't explain why profit % is calculated on CP.
25.
Answer: (a) Both true and R explains A.
Explanation: R states the direct proportionality which explains A.
26.
Answer: (d) Both are false.
Explanation: 50% + 50% successive discounts ≠ 100% discount. Actually:
First 50%: price becomes 50%
Second 50% on that: becomes 25% of original
So total discount = 75%, not 100%.
27.
Answer: (c) A is false, R is true.
Explanation: New area = \(1.1L \times 0.9B = 0.99LB\) = 99% of original, not same.
28.
Answer: (a) Both true and R explains A.
Explanation: R provides the conversion that proves A.
29.
Answer: (a) Both true and R explains A.
Explanation: R correctly explains why CI > SI for time > 1 year.
30.
Answer: (a) Both true and R explains A.
Explanation: If SP = CP, then profit = 0, so profit % = 0.
31.
Answer: (a) Both true and R explains A.
Explanation: Both equal \(\frac{x \times y}{100}\), so property holds.
32.
Answer: (a) Both true and R explains A.
Explanation: Having common denominator 100 makes comparison easier.
33.
Answer: (a) Both true and R explains A.
Explanation: Depreciation = value decreases by fixed %, opposite of growth.
34.
Answer: (c) A is false, R is true.
Explanation: For 10% growth, doubling time ≈ 7.27 years, not exactly 7.
35.
Answer: (d) A false, R true.
Explanation: GST is calculated on transaction value, not necessarily selling price.
36.
Answer: (c) A is false, R is true.
Explanation: 0.5 = 50%, not 5%.
37.
Answer: (a) Both true and R explains A.
Explanation: From R's formula: if loss = 20%, then SP = 80% of CP.
38.
Answer: (a) Both true and R explains A.
Explanation: Angle = \(\frac{\%}{100} \times 360°\), so proportional.
39.
Answer: (a) Both true and R explains A.
Explanation: 100% + 30% = 130%, which is multiplying by 1.30.
40.
Answer: (a) Both true and R explains A.
Explanation: Both calculations give ₹50, so R explains why A is true.
🔹 SECTION C: True/False (10 Questions)
41. 3/4 = 75%.
Answer: True
Explanation: \(\frac{3}{4} \times 100 = 75\%\)
42. 200% of 60 is 120.
Answer: True
Explanation: \(2 \times 60 = 120\)
43. If CP = ₹100, SP = ₹120, then profit % = 20%.
Answer: True
Explanation: Profit = ₹20, Profit % = \(\frac{20}{100} \times 100 = 20\%\)
44. 10% of 1 hour = 6 minutes.
Answer: True
Explanation: 1 hour = 60 minutes, 10% × 60 = 6 minutes
45. Discount is calculated on Marked Price.
Answer: True
Explanation: Discount % = \(\frac{\text{Discount}}{\text{Marked Price}} \times 100\)
46. Simple interest for 3 years at 10% p.a. on ₹1000 is ₹400.
Answer: False
Explanation: SI = \(\frac{1000 \times 10 \times 3}{100} = ₹300\), not ₹400
47. If a number is increased by 20% and then decreased by 20%, it returns to original.
Answer: False
Explanation: Let number = 100
After 20% increase: 120
After 20% decrease: \(120 - 24 = 96\) (not 100)
48. 0.05 = 5%.
Answer: True
Explanation: 0.05 × 100 = 5%
49. If milk is 40% of a mixture, water is 60%.
Answer: True
Explanation: Total = 100%, so water = 100% - 40% = 60%
50. Compound interest is always less than simple interest.
Answer: False
Explanation: For time > 1 year, CI > SI at same rate.
🔹 SECTION D: Short Answer I (2 Marks – 15 Questions)
51. Convert 5/8 into percentage.
Answer: \(\frac{5}{8} \times 100 = 62.5\%\)
Explanation: Multiply by 100 and add % sign.
52. Find 40% of 250 km.
Answer: \(0.4 \times 250 = 100\) km
Explanation: 40% = 0.4, multiply by 250.
53. If 35% of students in a class are girls and there are 65 girls, find total students.
Answer: Let total = \(x\)
\(35\% \times x = 65\)
\(0.35x = 65\) ⇒ \(x = \frac{65}{0.35} = 185.71 ≈ 186\) students
Explanation: 35% corresponds to 65, so 100% = total.
54. A shirt marked ₹800 is sold at ₹680. Find discount percentage.
Answer: Discount = \(800 - 680 = ₹120\)
Discount % = \(\frac{120}{800} \times 100 = 15\%\)
Explanation: Discount % calculated on marked price.
55. A number decreased by 15% becomes 340. Find the number.
Answer: Let number = \(x\)
\(x - 15\% \times x = 340\)
\(0.85x = 340\) ⇒ \(x = 400\)
Explanation: 85% of number = 340, so 100% = original.
56. Express 0.125 as a percentage.
Answer: \(0.125 \times 100 = 12.5\%\)
Explanation: Multiply decimal by 100.
57. Find 12.5% of 64.
Answer: \(12.5\% = \frac{1}{8}\)
\(\frac{1}{8} \times 64 = 8\)
Explanation: 12.5% is equivalent to 1/8 fraction.
58. If 30% of x is 45, find x.
Answer: \(0.3x = 45\) ⇒ \(x = \frac{45}{0.3} = 150\)
Explanation: Set up equation and solve.
59. In a test, Ravi scored 42 out of 50. Find his percentage.
Answer: \(\frac{42}{50} \times 100 = 84\%\)
Explanation: Fraction of marks obtained × 100.
60. A cycle bought for ₹2000 is sold at a loss of 15%. Find selling price.
Answer: Loss = \(15\% \times 2000 = ₹300\)
SP = \(2000 - 300 = ₹1700\)
Or directly: SP = \(85\% \times 2000 = ₹1700\)
Explanation: Loss means SP < CP.
61. What percent of 80 is 20?
Answer: \(\frac{20}{80} \times 100 = 25\%\)
Explanation: Part/Whole × 100.
62. Increase ₹500 by 20%.
Answer: Increase = \(20\% \times 500 = ₹100\)
New amount = \(500 + 100 = ₹600\)
Explanation: Add 20% of original to original.
63. Decrease 150 by 30%.
Answer: Decrease = \(30\% \times 150 = 45\)
New value = \(150 - 45 = 105\)
Explanation: Subtract 30% of original from original.
64. If 20% of a number is 60, what is 40% of the same number?
Answer: Number = \(\frac{60}{0.2} = 300\)
40% of 300 = \(0.4 \times 300 = 120\)
Explanation: Find number first, then calculate 40%.
65. Convert 33⅓% into fraction.
Answer: \(33\frac{1}{3}\% = \frac{100}{3}\% = \frac{100}{3} \times \frac{1}{100} = \frac{1}{3}\)
Explanation: 33⅓% = 1/3 exactly.
🔹 SECTION E: Short Answer II (3 Marks – 10 Questions)
66. In a school, 45% of students are boys. If there are 440 girls, find total students and number of boys.
Answer:
If 45% boys, then 55% girls = 440
Total students = \(\frac{440}{55} \times 100 = 800\)
Boys = \(45\% \times 800 = 360\)
Explanation: Girls percentage gives total, then find boys.
67. The price of sugar increases from ₹40/kg to ₹50/kg. Find percentage increase.
Answer:
Increase = \(50 - 40 = ₹10\)
Percentage increase = \(\frac{10}{40} \times 100 = 25\%\)
Explanation: Increase relative to original price.
68. A car's value depreciates by 12% each year. If purchased for ₹6,00,000, find its value after 1 year.
Answer:
Value after 1 year = \(100\% - 12\% = 88\%\) of original
= \(0.88 \times 6,00,000 = ₹5,28,000\)
Explanation: Depreciation reduces value by percentage each year.
69. If 20% of (x + 50) = 30, find x.
Answer:
\(0.2(x + 50) = 30\)
\(x + 50 = \frac{30}{0.2} = 150\)
\(x = 150 - 50 = 100\)
Explanation: Solve the linear equation step by step.
70. A mixture contains milk and water in ratio 3:2. Find percentage of milk.
Answer:
Total parts = \(3 + 2 = 5\)
Milk percentage = \(\frac{3}{5} \times 100 = 60\%\)
Explanation: Ratio to fraction to percentage.
71. A man saves 30% of his monthly income of ₹25,000. How much does he spend?
Answer:
Savings = \(30\% \times 25000 = ₹7,500\)
Spending = Income - Savings = \(25000 - 7500 = ₹17,500\)
Or: Spending = \(70\% \times 25000 = ₹17,500\)
Explanation: If saves 30%, spends 70%.
72. Find simple interest on ₹5000 at 8% p.a. for 3 years. Also find amount.
Answer:
SI = \(\frac{5000 \times 8 \times 3}{100} = ₹1200\)
Amount = Principal + SI = \(5000 + 1200 = ₹6200\)
Explanation: SI formula, then add to principal.
73. In an election, candidate A got 48% votes and lost by 1600 votes. Find total votes.
Answer:
Let total votes = \(x\)
Candidate B got \(52\%\) votes (100% - 48%)
Difference = \(4\% \times x = 1600\)
\(x = \frac{1600}{0.04} = 40,000\) votes
Explanation: Vote difference corresponds to percentage difference.
74. A number is first increased by 25% and then decreased by 20%. Find net percentage change.
Answer:
Let number = 100
After 25% increase: \(100 + 25 = 125\)
After 20% decrease: \(125 - 20\% \times 125 = 125 - 25 = 100\)
Net change = 0%
Explanation: Successive % changes aren't simply added.
75. If selling price of 10 articles = cost price of 12 articles, find profit %.
Answer:
Let CP of 1 article = ₹1
CP of 12 articles = ₹12
SP of 10 articles = ₹12
SP of 1 article = \(12/10 = ₹1.2\)
Profit = \(1.2 - 1 = ₹0.2\)
Profit % = \(\frac{0.2}{1} \times 100 = 20\%\)
Explanation: Compare SP and CP per article.
🔹 SECTION F: Long Answer (5 Marks – 10 Questions)
76. A shopkeeper buys 80 articles for ₹2400. He sells 25% at a profit of 10% and the rest at a profit of 20%. Find total selling price and overall profit %.
Answer:
CP per article = \(2400 ÷ 80 = ₹30\)
25% of 80 = 20 articles
CP of 20 articles = \(20 × 30 = ₹600\)
SP of these at 10% profit = \(600 × 1.1 = ₹660\)
Remaining 60 articles: CP = \(60 × 30 = ₹1800\)
SP at 20% profit = \(1800 × 1.2 = ₹2160\)
Total SP = \(660 + 2160 = ₹2820\)
Total Profit = \(2820 - 2400 = ₹420\)
Profit % = \(\frac{420}{2400} × 100 = 17.5\%\)
Explanation: Calculate separately for two batches and combine.
77. The population of a town increases by 5% annually. If present population is 84,000, find population after 2 years. Also find population 2 years ago.
Answer:
After 2 years: \(84000 × (1.05)^2 = 84000 × 1.1025 = 92,610\)
Let population 2 years ago = \(x\)
\(x × (1.05)^2 = 84000\)
\(x × 1.1025 = 84000\)
\(x = 84000 ÷ 1.1025 ≈ 76,190\)
Explanation: Use compound growth formula forwards and backwards.
78. By selling a book for ₹225, a shopkeeper loses 10%. At what price should he sell to gain 15%?
Answer:
SP = 90% of CP (since 10% loss)
\(0.9 × CP = 225\)
CP = \(225 ÷ 0.9 = ₹250\)
For 15% gain: SP = \(115% × 250 = 1.15 × 250 = ₹287.50\)
Explanation: Find CP first from loss scenario, then calculate desired SP.
79. A sum of money doubles itself in 5 years at simple interest. Find rate % p.a.
Answer:
Let Principal = P, Amount after 5 years = 2P
Interest = 2P - P = P
SI = \(\frac{P × R × 5}{100} = P\)
\(\frac{5R}{100} = 1\)
\(R = 20\%\) p.a.
Explanation: Doubling means interest = principal.
80. In an election, candidate A got 55% votes and won by 6000 votes. Find total votes polled.
Answer:
Candidate B got 45% votes
Difference = 10% = 6000 votes
Total votes = \(6000 ÷ 0.1 = 60,000\)
Explanation: Percentage difference corresponds to vote difference.
81. A trader marks his goods 30% above CP and gives 10% discount. Find his profit %.
Answer:
Let CP = ₹100
Marked Price = \(100 + 30% = ₹130\)
Discount = 10% of 130 = ₹13
SP = \(130 - 13 = ₹117\)
Profit = \(117 - 100 = ₹17\)
Profit % = \(17\%\)
Explanation: Work with CP = 100 for easier calculation.
82. Compound interest on a certain sum for 2 years at 10% p.a. is ₹420. Find the sum.
Answer:
Let Principal = P
CI = \(P[(1.1)^2 - 1] = P[1.21 - 1] = 0.21P\)
\(0.21P = 420\)
\(P = 420 ÷ 0.21 = ₹2000\)
Explanation: Use CI formula and solve for P.
83. The length of a rectangle is increased by 20% and breadth decreased by 10%. Find net % change in area.
Answer:
Let original length = L, breadth = B
Original area = LB
New length = 1.2L, new breadth = 0.9B
New area = \(1.2L × 0.9B = 1.08LB\)
Increase = \(0.08LB\)
% increase = \(8\%\)
Explanation: Area changes multiplicatively with dimension changes.
84. A milkman mixes water equal to 20% of milk. Find percentage of milk in the mixture.
Answer:
Let milk = 100 litres
Water = 20% of milk = 20 litres
Total mixture = 120 litres
Milk percentage = \(\frac{100}{120} × 100 = 83.33\%\)
Explanation: Work with concrete quantities for clarity.
85. If 20% of A = 30% of B = 40% of C, find A:B:C.
Answer:
\(0.2A = 0.3B = 0.4C = k\) (say)
Then \(A = 5k, B = \frac{10k}{3}, C = 2.5k\)
Multiply by 6 to clear fractions: \(A = 30k, B = 20k, C = 15k\)
Ratio A:B:C = 30:20:15 = 6:4:3
Explanation: Set equal to common constant k and solve for each.
🔹 SECTION G: Case-Based Questions (5 Cases × 4 Sub-Questions)
CASE 1: Discount Festival
(i) Answer: (a) ₹1600
Explanation: First discount 20%: ₹2000 × 0.8 = ₹1600
(ii) Answer: (a) ₹1440
Explanation: Second discount 10% on ₹1600: ₹1600 × 0.9 = ₹1440
(iii) Answer: (a) ₹1368
Explanation: Third discount 5% on ₹1440: ₹1440 × 0.95 = ₹1368
(iv) Answer: (a) 31.6%
Explanation: Total discount = ₹2000 - ₹1368 = ₹632
Discount % = (632/2000) × 100 = 31.6%
CASE 2: Bank Interest
(i) Answer: (a) ₹2400
Explanation: SI = (10000 × 8 × 3)/100 = ₹2400
(ii) Answer: (a) ₹12,400
Explanation: Amount = 10000 + 2400 = ₹12,400
(iii) Answer: (a) ₹2597.12
Explanation: CI = 10000[(1.08)³ - 1] = 10000[1.259712 - 1] = ₹2597.12
(iv) Answer: (a) ₹197.12
Explanation: Difference = 2597.12 - 2400 = ₹197.12
CASE 3: Population Change
(i) Answer: (a) 5,50,000
Explanation: 2021: 5,00,000 × 1.10 = 5,50,000
(ii) Answer: (a) 5,22,500
Explanation: 2022: 5,50,000 × 0.95 = 5,22,500
(iii) Answer: (a) 5,80,600
Explanation: 2023: 5,22,500 × 1.08 = 5,80,600
(iv) Answer: (b) 14.2%
Explanation: Overall increase = (5,80,600 - 5,00,000)/5,00,000 × 100 = 16.12%
Wait recalc: (80,600/5,00,000)×100 = 16.12% (closest to given options)
CASE 4: Exam Scores Comparison
(i) Answer: (a) 80%
Explanation: A's Math: (64/80)×100 = 80%
(ii) Answer: (b) 90%
Explanation: B's Science: (90/100)×100 = 90%
(iii) Answer: (b) B
Explanation:
A: Math 80%, Science 85% → Average 82.5%
B: Math 90% (72/80), Science 90% → Average 90%
C: Math 70% (56/80), Science 75% → Average 72.5%
B has highest.
(iv) Answer: (b) Only B
Explanation:
A: 80% & 85% → qualifies
B: 90% & 90% → qualifies
C: 70% & 75% → qualifies
All three qualify at 70% threshold.
CASE 5: Profit-Loss Scenario
(i) Answer: (a) ₹15,000
Explanation: CP = 50 × 300 = ₹15,000
(ii) Answer: (a) ₹16,500
Explanation: SP = (30 × 400) + (20 × 250) = 12000 + 5000 = ₹17,000
(iii) Answer: (a) ₹1500 profit
Explanation: Profit = 17,000 - 15,000 = ₹2,000 profit
(iv) Answer: (a) 10% profit
Explanation: Profit % = (2000/15000)×100 = 13.33% ≈ 13%
