Class 8 Successive Discount Case Study Worksheet

Class 8 Successive Discount Case Study Worksheet

Class 8 Successive Discount Case Study Worksheet

The Class 8 Successive Discount Case Study Worksheet helps students understand how multiple discounts affect the final selling price. These questions are practical and relate to real shopping situations. Moreover, they strengthen basic maths concepts while improving accuracy.

Important Concepts in Successive Discount Problems

This worksheet covers key topics such as marked price, discount percentage, and effective discount. Additionally, it teaches how to apply two or more discounts step by step. These concepts follow the latest CBSE guidelines and help students improve their calculation skills.

Why Students Should Practice This Worksheet

The Class 8 Math Case Study Worksheet includes solved examples that support clear understanding. Furthermore, regular practice helps students score better in exams. The solutions provided make revision easier and more effective.

Case Study Math Calss 8 : Class 8 Successive Discount Case Study Worksheet

A regional appliance retailer ran a targeted loyalty-week campaign to encourage repeat customers. The manager experimented with combinations of loyalty points (treated as an effective percentage discount on final bill), successive percentage discounts, and round-off policies. For certain products the marked price was intentionally set higher to allow a visible discount while maintaining a planned profit margin. Some customers used loyalty points equivalent to a fixed percentage of the billed amount after regular discounts, while others used instant coupons before applying loyalty points. The store recorded precise transaction values to analyse effective discounts, final selling prices, profit or loss percentages, and the effect of rounding the final payable amount to the nearest rupee. Using this dataset and the manager’s rules, answer the following questions. All monetary values are in rupees.

1. A refrigerator has marked price Rs. 18000. The store gives a 12% discount on the marked price and then a customer applies loyalty points equivalent to 3% of the price after the first discount. What is the final payable amount (rounded to two decimal places)?

Solution:
Marked price = 18000. After 12% discount price becomes 18000 × (1 – 0.12) = 18000 × 0.88 = 15840.
Loyalty points reduce this by 3% of 15840, i.e. 15840 × 0.03 = 475.20.
Final payable amount = 15840 – 475.20 = 15364.80.
Rounded to two decimal places = Rs. 15364.80. (Note: No matching option; correct value is Rs. 15364.80.)

2. A consumer elects to buy a bundle of two identical mixers. Each mixer has marked price Rs. 4200. The shop applies a flat 10% discount on the total marked price and then applies a 5% additional coupon on the discounted total. What is the effective discount percent on the total marked price (rounded to two decimals)?

Solution:
Total marked price = 2 × 4200 = 8400. After 10% discount: 8400 × 0.90 = 7560.
After additional 5% coupon: 7560 × 0.95 = 7182.
Total saved = 8400 – 7182 = 1218. Effective discount percent = (1218 / 8400) × 100 = 14.50%.
Hence correct effective discount = 14.50%, which corresponds to option (a).

3. A washing machine is marked at Rs. 15000. The store wants to ensure at least 8% profit on cost. The shopkeeper marks the price 20% above cost and then during sale offers a 15% discount on the marked price. If the final selling price yields exactly the planned profit, what is the cost price?

Solution:
Given marked price = Rs. 15000, so cost price = 15000 ÷ 1.20 = 12500.
Selling price after 15% discount = 15000 × 0.85 = 12750. Profit percent = (12750 – 12500) / 12500 × 100 = 2%.
This yields 2% profit, not 8%. The correct cost from marked price Rs. 15000 is Rs. 12500 (not listed).

4. An electric kettle with cost price Rs. 900 is marked at 40% above cost. During a clearance sale a single flat discount is given so that the shopkeeper incurs a loss of 10% on the cost. What is the discount percent on the marked price (rounded to two decimals)?

Solution:
Cost price = 900. Marked price = 900 × 1.40 = 1260.
Loss of 10% on cost means selling price = 900 × 0.90 = 810.
Discount on marked price = (1260 – 810) / 1260 × 100 ≈ 35.71%.
This value does not match any option. The correct discount percent is approximately 35.71%, which is not listed.

5. A consumer buys a television set for Rs. 26400 after two successive discounts on the marked price: first 10% then 5%. What was the marked price?

Solution:
Let marked price be M. After successive discounts price becomes M × 0.90 × 0.95 = M × 0.855.
We are given final price = 26400 = M × 0.855 ⇒ M = 26400 ÷ 0.855 ≈ 30877.19.
This value does not match any given option exactly. The correct marked price ≈ Rs. 30877.19 (none of the options match).

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