Download CBSE Class 8 Mathematics Assignments
Explore structured practice materials through the CBSE Class 8 Mathematics Comparing Quantities Assignment Set 08. Tailored for Class 8 learners, utilizing these Mathematics assignments ensures thorough preparation and strengthens foundational knowledge before final CBSE evaluations.
Access CBSE School Assignments and Answers
View or download the dedicated CBSE Class 8 Mathematics Comparing Quantities Assignment Set 08 resource below. Engaging with these assignments under focused study conditions ensures continuous academic progress and mastery of the 2026-27 curriculum.
Profit and Loss
Question 1. A retailer buys a watch for Rs 335. By paying Rs 15 more, he replaces its strap by a costlier strap. If he sells the watch for Rs 400. Find his gain or loss percent.
Answer: Cost price of the watch = Rs 335
Additional overhead charges (extra strap) = Rs 15
Total Cost Price (C.P.) = \(335 + 15 =\) Rs 350
Selling Price (S.P.) = Rs 400
Since \(\text{S.P.} > \text{C.P.}\), there is a profit.
\(\text{Profit} = \text{S.P.} - \text{C.P.} = 400 - 350 =\) Rs 50
\(\text{Profit percent} = \left( \frac{\text{Profit}}{\text{C.P.}} \right) \times 100 = \frac{50}{350} \times 100 = \frac{1}{7} \times 100 = \frac{100}{7}\% = 14\frac{2}{7}\%\)
In simple words: Add extra expenses to the original cost price to find the total cost, then calculate the profit percentage.
Exam Tip: Always include overhead charges (repairs, transport, or extra parts) in the total cost price before calculating profit or loss.
Question 2. A shopkeeper looses 20% by selling a bicycle for Rs 1536. Find the cost price of the bicycle.
Answer: Selling Price (S.P.) = Rs 1536
Loss percent = \(20\%\)
Let the Cost Price (C.P.) be \(x\).
\(\text{S.P.} = \text{C.P.} \times \left( \frac{100 - \text{Loss}\%}{100} \right)\)
\(1536 = x \times \left( \frac{100 - 20}{100} \right)\)
\(1536 = x \times \frac{80}{100}\)
\(x = \frac{1536 \times 100}{80}\)
\(x = 192 \times 10 =\) Rs 1920
In simple words: When loss is given, subtract it from 100% to find the selling price percentage relative to the cost price.
Exam Tip: Formula to find C.P. when S.P. and loss percent are given: \(\text{C.P.} = \frac{\text{S.P.} \times 100}{100 - \text{Loss}\%}\).
Question 3. Anil purchased an old type-writer for Rs 1200 and spent Rs 200 on repairs. He sold it for Rs 1680. Find his profit and loss, and express it as a %
Answer: Purchase price = Rs 1200
Repair cost = Rs 200
Total Cost Price (C.P.) = \(1200 + 200 =\) Rs 1400
Selling Price (S.P.) = Rs 1680
Since \(\text{S.P.} > \text{C.P.}\), there is a profit.
\(\text{Profit} = \text{S.P.} - \text{C.P.} = 1680 - 1400 =\) Rs 280
\(\text{Profit percent} = \left( \frac{280}{1400} \right) \times 100 = \frac{280}{14} = 20\%\)
In simple words: Add repair costs to the purchase price to get the total cost, then find the profit and convert it into a percentage.
Exam Tip: Repairs and overheads are always added to the initial buying price to establish the true cost price.
Question 4. By selling an article for Rs 630, a person gains 5%. Find the cost price of the article.
Answer: Selling Price (S.P.) = Rs 630
Gain percent = \(5\%\)
Let the Cost Price (C.P.) be \(x\).
\(\text{S.P.} = \text{C.P.} \times \left( \frac{100 + \text{Gain}\%}{100} \right)\)
\(630 = x \times \left( \frac{100 + 5}{100} \right)\)
\(630 = x \times \frac{105}{100}\)
\(x = \frac{630 \times 100}{105}\)
\(x = 6 \times 100 =\) Rs 600
In simple words: Use the selling price and gain percentage formula in reverse to isolate and find the original cost price.
Exam Tip: Formula for C.P. when S.P. and gain percent are given: \(\text{C.P.} = \frac{\text{S.P.} \times 100}{100 + \text{Gain}\%}\).
Question 5. If the C.P. of 15 chairs is equal to the S.P. of 10 chairs, find the gain percent
Answer: Let the Cost Price of 1 chair be Rs 1.
Cost Price of 15 chairs = Rs 15
Given that C.P. of 15 chairs = S.P. of 10 chairs.
Therefore, Selling Price of 10 chairs = Rs 15
Cost Price of 10 chairs = \(10 \times 1 =\) Rs 10
Selling Price of 10 chairs = Rs 15
Since \(\text{S.P.} > \text{C.P.}\), there is a profit.
\(\text{Profit} = \text{S.P.} - \text{C.P.} = 15 - 10 =\) Rs 5
\(\text{Gain percent} = \left( \frac{\text{Profit}}{\text{C.P. of 10 chairs}} \right) \times 100 = \frac{5}{10} \times 100 = 50\%\)
In simple words: Assume a unit price for the items, equate the total amounts for a common quantity, and find the profit percentage.
Exam Tip: For quantity-based problems, always calculate the C.P. and S.P. for the exact same number of items.
Question 6. Toffees are bought 8 for a rupee and sold 6 for a rupee. Find the profit percent.
Answer: Let the number of toffees bought and sold be the LCM of 8 and 6, which is 24.
Cost Price of 8 toffees = Re 1
Cost Price of 24 toffees = \(\frac{1}{8} \times 24 =\) Rs 3
Selling Price of 6 toffees = Re 1
Selling Price of 24 toffees = \(\frac{1}{6} \times 24 =\) Rs 4
Total Cost Price for 24 toffees = Rs 3
Total Selling Price for 24 toffees = Rs 4
\(\text{Profit} = 4 - 3 =\) Rs 1
\(\text{Profit percent} = \left( \frac{1}{3} \right) \times 100 = \frac{100}{3}\% = 33\frac{1}{3}\%\)
In simple words: Take the LCM of the quantities to find the common number of items, calculate total cost and selling price, then find the profit.
Exam Tip: Finding the LCM of quantities makes calculation fraction-free and much easier.
Question 7. Anil purchased an article for Rs 4000 and sold it to Arun for a profit of 25%. Arun sold it to Ajay at a loss of 5%. For how much did Ajay buy it?
Answer: Cost price for Anil = Rs 4000
Anil sells to Arun at a profit of \(25\%\):
Selling Price for Anil (Cost Price for Arun) = \(4000 \times \left( \frac{100 + 25}{100} \right)\)
\(= 4000 \times \frac{125}{100} = 40 \times 125 =\) Rs 5000
Arun sells to Ajay at a loss of \(5\%\):
Selling Price for Arun (Cost Price for Ajay) = \(5000 \times \left( \frac{100 - 5}{100} \right)\)
\(= 5000 \times \frac{95}{100} = 50 \times 95 =\) Rs 4750
In simple words: Calculate successive selling prices step-by-step from one person to the next.
Exam Tip: In chain transactions, the selling price of the previous person always becomes the cost price for the next person.
Question 8. A man sold two watches for Rs 480 each. He gains 20% on one and loses 20% on the other. By how much he gains or loses in the whole transaction.
Answer: Selling Price of each watch = Rs 480
Total Selling Price of two watches = \(480 + 480 =\) Rs 960
For the first watch (Gain of \(20\%\)):
\(\text{C.P.}_1 = \frac{\text{S.P.} \times 100}{100 + \text{Gain}\%} = \frac{480 \times 100}{120} = 4 \times 100 =\) Rs 400
For the second watch (Loss of \(20\%\)):
\(\text{C.P.}_2 = \frac{\text{S.P.} \times 100}{100 - \text{Loss}\%} = \frac{480 \times 100}{80} = 6 \times 100 =\) Rs 600
Total Cost Price of both watches = \(400 + 600 =\) Rs 1000
Total Selling Price = Rs 960
Since \(\text{C.P.} > \text{S.P.}\), there is a total loss.
\(\text{Loss} = \text{Total C.P.} - \text{Total S.P.} = 1000 - 960 =\) Rs 40
\(\text{Loss percent} = \left( \frac{40}{1000} \right) \times 100 = 4\%\)
In simple words: Find the individual cost price for each watch separately, add them up, and compare with the total selling price.
Exam Tip: When the selling price and equal percentage of profit and loss are given, there is always a net loss calculated by \(\text{Loss}\% = \left(\frac{\text{Common}\%}{10}\right)^2\).
Question 9. A fruit seller buys oranges for Rs 12 a dozen and sells them for Rs 18 a dozen, Find his profit per cent
Answer: Cost Price of 1 dozen oranges = Rs 12
Selling Price of 1 dozen oranges = Rs 18
\(\text{Profit} = \text{S.P.} - \text{C.P.} = 18 - 12 =\) Rs 6
\(\text{Profit percent} = \left( \frac{\text{Profit}}{\text{C.P.}} \right) \times 100 = \frac{6}{12} \times 100 = \frac{1}{2} \times 100 = 50\%\)
In simple words: Since both buying and selling prices are given for the same quantity (one dozen), calculate the profit directly.
Exam Tip: Always verify that the quantities are identical before calculating profit or loss.
Question 10. If by selling a saree for Rs 200, a shopkeeper suffers a loss of 20%. for what amount should he sell it as to gain a profit of 10%.
Answer: Selling Price (S.P.) = Rs 200
Loss percent = \(20\%\)
First, find the Cost Price (C.P.):
\(\text{C.P.} = \frac{\text{S.P.} \times 100}{100 - \text{Loss}\%} = \frac{200 \times 100}{100 - 20} = \frac{20000}{80} =\) Rs 250
Now, find the new Selling Price to gain a profit of \(10\%\):
Desired Profit percent = \(10\%\)
\(\text{New S.P.} = \text{C.P.} \times \left( \frac{100 + \text{Profit}\%}{100} \right)\)
\(\text{New S.P.} = 250 \times \left( \frac{100 + 10}{100} \right)\)
\(\text{New S.P.} = 250 \times \frac{110}{100} = 25 \times 11 =\) Rs 275
In simple words: First find the cost price using the initial loss condition, then calculate the new selling price using the desired profit percentage.
Exam Tip: You can combine this into a single step: \(\text{New S.P.} = \text{Old S.P.} \times \left( \frac{100 + \text{New Profit}\%}{100 - \text{Old Loss}\%} \right)\).
Free study material for Mathematics
Chapter 08 Comparing Quantities Printable Assignments & Solutions for Class 8 Mathematics
Chapter Practice Questions for Class 8 Mathematics
Explore reliable practice questions for Chapter 08 Comparing Quantities tailored for Class 8 learners. Use these structured worksheets to evaluate preparedness and strengthen core problem-solving skills.
Maximize Exam Scores with Chapter Practice Sets
- Syllabus Compliance: Sets reflect current CBSE evaluation criteria and official marking frameworks.
- Multi-Format Practice: Includes varied problem types designed to deepen comprehension across all sub-topics.
- Time Management: Routine practice optimizes pacing to finish school examinations comfortably within schedule.
Effective Strategy for Class 8 Mathematics Assignments
- Textbook Review: Always study the core NCERT book for Class 8 Mathematics prior to beginning the assignment.
- Independent Attempt: Solve Chapter 08 Comparing Quantities questions on your own initially before cross-checking with expert solutions.
- Error Tracking: Record challenging concepts in a dedicated notebook and practice online MCQ tests for revision.
FAQs
You can download free PDF assignments for Class 8 Mathematics Chapter 08 Comparing Quantities from StudiesToday.com. These practice sheets have been updated for the 2026-27 session covering all concepts from latest NCERT textbook.
Yes, our teachers have given solutions for all questions in the Class 8 Mathematics Chapter 08 Comparing Quantities assignments. This will help you to understand step-by-step methodology to get full marks in school tests and exams.
Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 08 Comparing Quantities.
Practicing topicw wise assignments will help Class 8 students understand every sub-topic of Chapter 08 Comparing Quantities. Daily practice will improve speed, accuracy and answering competency-based questions.
Yes, all printable assignments for Class 8 Mathematics Chapter 08 Comparing Quantities are available for free download in mobile-friendly PDF format.