This MCQ module is based on: Finding Percentage of a Quantity
TOPIC 2 OF 23
Finding Percentage of a Quantity
🎓 Class 8
Mathematics
CBSE
Theory
Ch 1 — Percentages
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Finding Percentage of a Quantity
Targeting Class 8 level in General Mathematics, with Basic difficulty.
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1.2 Percentage of Some Quantity
Example 1: Madhu and Madhav's Biscuits
Madhu and Madhav each ate biscuits of a different variety. Madhu's biscuits had 25% sugar, while Madhav's had 35% sugar. Can you tell who ate more sugar?
Percentages represent fractional quantities or proportions. It would be inappropriate to compare just the percentages when they are referring to different quantities or wholes. If Madhu ate 120 g of biscuits and Madhav ate 95 g:
Madhu's sugar: \(25\% \times 120 = \frac{25}{100} \times 120 = 30\) g
Madhav's sugar: \(35\% \times 95 = \frac{35}{100} \times 95 = 33.25\) g
Madhav ate more sugar (33.25 g vs 30 g) even though his percentage was higher — the base quantities matter!
Key Insight
A higher percentage does not always mean a larger amount. It depends on what the percentage is of — the base quantity matters. Always check: "Percentage of WHAT?"
Example 5: Delhi to Agra Cycling
A cyclist cycles from Delhi to Agra and completes 40% of the journey. If he has covered 92 km, how many more kilometres does he have to travel to reach Agra?
Solution (4 Methods):
Method 1: 40% is 92 km, so 20% = 46 km. 60% = 3 × 46 = 138 km.
Method 2: If 40% = 92, total \(d = 92 \times \frac{100}{40} = 230\) km. Remaining = 230 − 92 = 138 km.
Method 3: \(\frac{40}{100} = \frac{92}{d}\). Cross-multiply: \(40d = 9200\), so \(d = 230\). Remaining = 138 km.
Method 4: If \(x\) is remaining, then total = \(92 + x\). \(\frac{40}{100} \times (92 + x) = 92\). Solving: \(x = 138\) km.
Method 2: If 40% = 92, total \(d = 92 \times \frac{100}{40} = 230\) km. Remaining = 230 − 92 = 138 km.
Method 3: \(\frac{40}{100} = \frac{92}{d}\). Cross-multiply: \(40d = 9200\), so \(d = 230\). Remaining = 138 km.
Method 4: If \(x\) is remaining, then total = \(92 + x\). \(\frac{40}{100} \times (92 + x) = 92\). Solving: \(x = 138\) km.
Percentages Greater than 100
Can there be percentages with a value more than 100? What could it mean when a percentage is greater than 100%? Let us explore.
Example 6: Kishanlal's Garment Shop
Kishanlal recently opened a garment shop. He aims to achieve a daily sales of at least ₹5000. The sales on the first 2 days were ₹2000 and ₹3500. What percentage of his target did he achieve?
Solution:
Day 1: \(\frac{2000}{5000} \times 100 = 40\%\) — below target.
Day 2: \(\frac{3500}{5000} \times 100 = 70\%\) — still below target.
On Day 7, he achieved 150% of his target → Sales = \(\frac{150}{100} \times 5000 = ₹7500\).
On Day 8, he achieved 210% → Sales = \(\frac{210}{100} \times 5000 = ₹10,500\).
150% means 1.5 times the target. 210% means 2.1 times. A percentage > 100% means the actual value exceeds the base/reference quantity.
Day 2: \(\frac{3500}{5000} \times 100 = 70\%\) — still below target.
On Day 7, he achieved 150% of his target → Sales = \(\frac{150}{100} \times 5000 = ₹7500\).
On Day 8, he achieved 210% → Sales = \(\frac{210}{100} \times 5000 = ₹10,500\).
150% means 1.5 times the target. 210% means 2.1 times. A percentage > 100% means the actual value exceeds the base/reference quantity.
Percentages > 100%
When a percentage is greater than 100%, it means the quantity is more than the whole/reference. For example:• 150% of target = 1.5 × target (50% more than the target)
• 200% = double the original
• 250% = 2.5 times the original
Example 7: Farmer's Harvest
A farmer harvested 260 kg of wheat last year. This year, they harvested 650 kg. What percentage of last year's harvest is this year's?
\(\frac{650}{260} \times 100 = 250\%\). This means this year's harvest is 2.5 times last year's harvest.
Figure it Out (Section 1.2)
Q1. Find the missing numbers using the bar model. The first has been worked out.
Using proportional reasoning with the bar models, find each missing value by setting up: \(\frac{\text{percentage}}{100} = \frac{\text{value}}{\text{total}}\) and solving for the unknown.
Q2-Q8. Solve these percentage problems (from the NCERT Figure it Out on pages 12-13):
Q2. Nandini has 25 marbles, 15 are white. What % are white?
Q3. 15 of 80 students walk to school. What %?
Q4. Match runners to approximate % of race completed.
Q5. Compare: 50% vs 5%, \(\frac{3}{10}\) vs 50%, \(\frac{3}{4}\) vs 61%, 30% vs \(\frac{3}{5}\).
Q6-Q8. Various proportion questions.
These were solved in Part 1. Key results: Q2 = 60%, Q3 = 18.75%, Q5: >, <, >, <.
Q9. Workers in a coffee plantation take 18 days to pick coffee berries in 20% of the plantation. How many days will they take to complete the entire plantation?
20% takes 18 days. So 100% takes \(\frac{18}{20} \times 100 = 90\) days (assuming the rate of work stays the same).
Q10. A badminton coach plans training sessions with warm up : play : cool down = 10% : 80% : 10%. If the total training is 90 minutes, how long should each activity be?
Warm up: \(10\% \times 90 = 9\) min. Play: \(80\% \times 90 = 72\) min. Cool down: \(10\% \times 90 = 9\) min. Check: 9 + 72 + 9 = 90 ✓
Q11. An estimated 90% of the world's population lives in the Northern Hemisphere. Find the approximate number of people living in the Northern Hemisphere based on this year's worldwide population.
World population (2025) ≈ 8.2 billion. Northern Hemisphere: \(90\% \times 8.2 = 7.38\) billion ≈ 7.4 billion people.
Q12. A recipe for halwa for 4 people has: Rava 40%, Sugar 40%, Ghee 20%.
(i) To make halwa for 8 people, what is the proportion of each ingredient?
(ii) If the total weight of ingredients is 2 kg, how much rava, sugar and ghee are present?
(i) To make halwa for 8 people, what is the proportion of each ingredient?
(ii) If the total weight of ingredients is 2 kg, how much rava, sugar and ghee are present?
(i) The proportions stay the same: Rava 40%, Sugar 40%, Ghee 20%. But the total quantity doubles.
(ii) Rava: \(40\% \times 2 = 0.8\) kg = 800 g. Sugar: \(40\% \times 2 = 0.8\) kg. Ghee: \(20\% \times 2 = 0.4\) kg = 400 g.
(ii) Rava: \(40\% \times 2 = 0.8\) kg = 800 g. Sugar: \(40\% \times 2 = 0.8\) kg. Ghee: \(20\% \times 2 = 0.4\) kg = 400 g.
1.3 Using Percentages
To Compare Proportions
Example 1: Eesha's Test Scores
Eesha scored 42 marks out of 50 on an English test and 70 marks out of 80 on a Science test. She lost only 8 marks in English but 10 marks in Science. She thinks she has done better at English. Reema does not agree. Who do you think is correct?
Solution:
We cannot compare raw marks because the maximum marks are different. Convert to percentages:
English: \(\frac{42}{50} \times 100 = 84\%\)
Science: \(\frac{70}{80} \times 100 = 87.5\%\)
The Science score (87.5%) is higher than the English score (84%). So Eesha actually scored better on the Science test. Reema was right!
English: \(\frac{42}{50} \times 100 = 84\%\)
Science: \(\frac{70}{80} \times 100 = 87.5\%\)
The Science score (87.5%) is higher than the English score (84%). So Eesha actually scored better on the Science test. Reema was right!
Percentage Increase and Decrease
Formulas
Percentage Increase: \(\text{Percentage increase} = \frac{\text{amount of increase}}{\text{original amount or base}} \times 100\)Percentage Decrease: \(\text{Percentage decrease} = \frac{\text{amount of decrease}}{\text{original amount or base}} \times 100\)
Example: Theatre Footfall
The average footfall in a theatre before COVID was 160. Now it is just 100. The decrease in the footfall is 60.
\(\text{Percentage decrease} = \frac{60}{160} \times 100 = 37.5\%\)
Profit and Loss
You may have the experience of buying something — snacks, groceries, clothes, toys. Very often, the shopkeeper quotes a price and after some bargaining, the customer pays the negotiated amount.
Key Terms
Cost Price (CP): The price the shopkeeper paid to purchase the item.Marked Price (MP): The price quoted by the shopkeeper (sometimes the MRP).
Selling Price (SP): The price the customer pays after any discount.
Profit = SP − CP (when SP > CP)
Loss = CP − SP (when CP > SP)
Discount = MP − SP
Manufacturing Unit
CP: ₹200
MP: ₹255
SP: ₹203
MP: ₹255
SP: ₹203
→
Wholesale Store
CP: ₹203
MP: ₹310
SP: ₹300
MP: ₹310
SP: ₹300
→
Retail Store
CP: ₹300
MP: ₹480
SP: ₹430
MP: ₹480
SP: ₹430
→
Customer
Pays: ₹430
The journey of a sweater: CP, MP, and SP at each step in the supply chain
🔵 At each stage, the selling price becomes the cost price for the next buyer. Each seller makes a profit (SP > CP). The customer pays the final SP.
Activity: Percentage Detective in a Shop
L4 Analyse
Challenge: Next time you go to a shop, look at the MRP on product labels and the final selling price. Can you calculate the discount percentage?
- Pick any 3 items with visible MRP and discounted price.
- For each item, calculate: Discount = MRP − SP, and Discount % = \(\frac{\text{Discount}}{\text{MRP}} \times 100\).
- Which item has the highest discount percentage?
- Is a 50% discount on a ₹100 item better or worse than a 30% discount on a ₹200 item? (Hint: compare the actual savings!)
Item: T-shirt, MRP = ₹800, SP = ₹600. Discount = ₹200. Discount % = \(\frac{200}{800} \times 100 = 25\%\).
50% of ₹100 = ₹50 saved. 30% of ₹200 = ₹60 saved. The 30% discount saves more in absolute terms!
Competency-Based Questions
Scenario: Sneha buys strawberries from her farm at ₹80/kg. She sells some at ₹90/kg (fresh) and some at ₹60/kg (slightly damaged due to rain). She sells 40 kg fresh and 10 kg damaged.
Q1. What is Sneha's overall profit or loss percentage?
L3 ApplyAnswer: (b) 8% profit.
Total CP = \(50 \times 80 = ₹4000\). Revenue = \(40 \times 90 + 10 \times 60 = 3600 + 600 = ₹4200\). Profit = ₹200. Profit % = \(\frac{200}{4000} \times 100 = 5\%\). Wait — let me recalculate. Actually: CP = 50 × 80 = 4000. SP = 3600 + 600 = 4200. Profit = 200. Profit % = 200/4000 × 100 = 5%. So the correct answer is actually not listed exactly — closest is between (b) and (c). Let's recheck: the answer is 5% profit.
Total CP = \(50 \times 80 = ₹4000\). Revenue = \(40 \times 90 + 10 \times 60 = 3600 + 600 = ₹4200\). Profit = ₹200. Profit % = \(\frac{200}{4000} \times 100 = 5\%\). Wait — let me recalculate. Actually: CP = 50 × 80 = 4000. SP = 3600 + 600 = 4200. Profit = 200. Profit % = 200/4000 × 100 = 5%. So the correct answer is actually not listed exactly — closest is between (b) and (c). Let's recheck: the answer is 5% profit.
Q2. If Sneha could sell all 50 kg at the fresh price (₹90/kg), what would be her profit percentage? Analyse the impact of the damaged stock.
L4 AnalyseAnswer: All fresh: SP = \(50 \times 90 = ₹4500\). Profit = 4500 − 4000 = ₹500. Profit % = \(\frac{500}{4000} \times 100 = 12.5\%\). With damaged stock: profit was 5%. The 10 kg of damaged stock reduced her profit by \(12.5 - 5 = 7.5\) percentage points. The rain damage cost her ₹300 in profit (₹500 − ₹200).
Q3. Sneha's neighbour sells at MRP ₹100/kg with a "10% discount." Is this the same as selling at ₹90/kg? Evaluate whether the neighbour makes more or less profit than Sneha (fresh stock).
L5 EvaluateAnswer: 10% discount on MRP ₹100 = ₹10 off. SP = ₹90/kg — exactly the same as Sneha's fresh price. If the neighbour also buys at ₹80/kg, both make the same profit per kg (₹10, i.e., 12.5%). The discount is just a different way of presenting the same final price.
Q4. Create a pricing strategy for Sneha where she can make at least 15% overall profit even if 20% of her stock gets damaged (sold at ₹60/kg). What should the fresh selling price be?
L6 CreateAnswer: CP = 50 × 80 = ₹4000. Target: 15% profit → SP total ≥ ₹4600.
Damaged (20% = 10 kg at ₹60): ₹600. Fresh (80% = 40 kg at price \(p\)): \(40p\).
\(40p + 600 \ge 4600\) → \(40p \ge 4000\) → \(p \ge 100\).
Sneha needs to sell fresh strawberries at ₹100/kg or more to guarantee 15% overall profit.
Damaged (20% = 10 kg at ₹60): ₹600. Fresh (80% = 40 kg at price \(p\)): \(40p\).
\(40p + 600 \ge 4600\) → \(40p \ge 4000\) → \(p \ge 100\).
Sneha needs to sell fresh strawberries at ₹100/kg or more to guarantee 15% overall profit.
Assertion–Reason Questions
Assertion (A): If a shopkeeper buys at ₹500 and sells at ₹600, the profit percentage is 20%.
Reason (R): Profit % = \(\frac{\text{Profit}}{\text{CP}} \times 100\).
Reason (R): Profit % = \(\frac{\text{Profit}}{\text{CP}} \times 100\).
Answer: (a) — Profit = 600 − 500 = ₹100. Profit % = \(\frac{100}{500} \times 100 = 20\%\). Both true, R explains A.
Assertion (A): A 250% increase means the new value is 2.5 times the original.
Reason (R): Percentage increase is calculated on the base/original value.
Reason (R): Percentage increase is calculated on the base/original value.
Answer: (d) — A is false: a 250% increase means the new value = original + 250% of original = 3.5 times the original (not 2.5). If the value is 250% of the original, then it's 2.5 times. But a 250% increase is different from being 250%. R is true.
Assertion (A): Discount is always calculated on the marked price (MRP), not the cost price.
Reason (R): The customer sees only the marked price and the selling price, not the cost price.
Reason (R): The customer sees only the marked price and the selling price, not the cost price.
Answer: (a) — Both true. Discount = MP − SP, and discount % is based on MP. R explains why: customers know MP (it's on the price tag) but not CP.
Frequently Asked Questions — Chapter 1
What is Finding Percentage of a Quantity in NCERT Class 8 Mathematics?
Finding Percentage of a Quantity is a key concept covered in NCERT Class 8 Mathematics, Chapter 1: Chapter 1. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Finding Percentage of a Quantity step by step?
To solve problems on Finding Percentage of a Quantity, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Chapter 1?
The essential formulas of Chapter 1 (Chapter 1) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Finding Percentage of a Quantity important for the Class 8 board exam?
Finding Percentage of a Quantity is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Finding Percentage of a Quantity?
Common mistakes in Finding Percentage of a Quantity include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Finding Percentage of a Quantity?
End-of-chapter NCERT exercises for Finding Percentage of a Quantity cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
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