Extra Questions for Class 8 Maths Chapter 8: Fractions in Disguise (Percentages) – HOTS

Original HOTS-level (Higher Order Thinking Skills) practice questions for Class 8 Maths Chapter 8 (Fractions in Disguise — Percentages), genuinely harder than the textbook exercise, using fresh numbers not found in the NCERT book. Full worked solutions included. These Class 8 Mathematics Chapter 8 important questions are handy for last-minute exam practice.

Last Updated: September 23, 2026

Q1 (Reverse percentage). A jacket’s price was reduced twice: first by 20%, then the reduced price by a further 10%, to reach ₹1,440. Find the original price.
Solution: Original × 0.8 × 0.9 = 1440 ⇒ Original × 0.72 = 1440 ⇒ Original = ₹2,000.

Q2 (General proof). Prove that increasing a quantity by a% and then decreasing the result by a% always gives a net decrease, and find that decrease in terms of a.
Solution: Overall factor = (1+a/100)(1−a/100) = 1 − a²/100², which is always less than 1 for a ≠ 0. So there’s always a net decrease of a²/100 percent. Example: a = 20 ⇒ decrease = 400/100 = 4%.

Q3 (Compound growth, spot-the-error). A town’s population is 20,00,000 and grows at 8% per year. A student claims that after 2 years, the population will be exactly 2 × (first year’s growth) more than the starting figure of 21,60,000. Check this claim with real numbers.
Solution: After 1 year: 20,00,000 × 1.08 = 21,60,000 (growth = 1,60,000). After 2 years: 21,60,000 × 1.08 = 23,32,800 (total growth over 2 years = 3,32,800). Twice the first year’s growth = 3,20,000. Since 3,32,800 ≠ 3,20,000, the student’s claim is false — compounding always adds a bit extra (here, 12,800 extra) beyond simply doubling the first year’s growth.

Q4 (Assertion–Reason). Assertion (A): If the cost price of 8 articles equals the selling price of 5 articles, the profit percentage is 60%. Reason (R): Profit % is always calculated as (Profit ÷ CP) × 100.
Solution: Let CP of 1 article = ₹1, so CP of 8 = ₹8 = SP of 5 ⇒ SP of 1 = 8/5 = 1.6. Profit per article = 0.6, so Profit% = 60%. A is true. R is also a true, correct general statement that explains A. Both A and R are true, and R is the correct explanation of A.

Q5 (Multi-step word problem). A trader marks goods 40% above cost price and allows two successive discounts of 10% and 5% during a sale. Find his actual profit percentage.
Solution: MP = CP × 1.40. SP = MP × 0.90 × 0.95 = CP × 1.40 × 0.855 = CP × 1.197. Profit = 19.7%.

Q6 (Reverse-engineering GST). The price printed on a bill after adding 18% GST is ₹11,800. Find the pre-GST price and the GST amount.
Solution: Pre-GST price = 11800 ÷ 1.18 = ₹10,000. GST amount = 11800 − 10000 = ₹1,800.

Q7 (General proof – order independence). Show that increasing a quantity first by p% then by q% gives the same final result as increasing it first by q% then by p%, and find the single equivalent overall percentage increase.
Solution: Final value = P(1+p/100)(1+q/100); since multiplication is commutative, the order doesn’t matter. Expanding: overall increase = p + q + pq/100 percent.

Q8 (Compound vs simple interest crossover). At what whole number of years will the compound interest (annually compounded) on ₹15,000 at 10% p.a. first exceed the simple interest on the same sum and rate by more than ₹500?
Solution: SI at year t = 1500t. CI at year t = 15000[(1.1)^t − 1]. t=3: CI=4965, SI=4500, difference=465 (not yet over 500). t=4: CI≈6966.15, SI=6000, difference≈966. The difference first exceeds ₹500 in year 4.

Q9 (Combined percentage – area). The length and breadth of a rectangular field are both increased by 25%. By what percentage does the area increase? If the original area was 4,800 m², find the new area.
Solution: New area factor = 1.25 × 1.25 = 1.5625, an increase of 56.25%. New area = 4800 × 1.5625 = 7,500 m².

Q10 (Chained discount word problem). A shopkeeper wants an effective discount of 30% on a ₹2,000 item using two successive discounts, the first being 20%. Find the required second discount, and the final selling price.
Solution: Overall factor needed = 0.70. First discount factor = 0.80. Second factor = 0.70 ÷ 0.80 = 0.875, so the second discount = 12.5%. Final SP = 2000 × 0.70 = ₹1,400.

More on This Chapter

See also: NCERT Solutions for Class 8 Maths Chapter 8, Class 8 Maths NCERT Book and the Class 8 Maths Formulas Handbook.

Frequently Asked Questions

If a shopkeeper offers a discount of a certain percentage on the marked price, how is the actual selling price calculated?
The discount amount is calculated by finding the given percentage of the marked price, and this discount amount is then subtracted from the marked price to arrive at the final selling price.

How is percentage increase or decrease calculated when comparing an old value to a new value?
Percentage change is found by dividing the difference between the new and old value by the old value, then multiplying the result by 100 to express the change as a percentage.

Written by Satish

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