Maths · Suggested Grade 8
Successive Discounts: Class 8 Worked Shopping Problems
20% off, then another 10% off, does not mean 30% off. The second discount is calculated on a smaller price.
Original explanations and practice by BananaLearn · Published 1 November 2026
On a small screen, swipe across the illustration to read every label.
Write the price base at every step
A discount is a reduction from a stated price. Successive discounts are applied one after another, so the output of the first step becomes the base of the next. For a fictional ₹1,000 item, 20% off removes ₹200 and leaves ₹800. A further 10% removes ₹80, not ₹100, leaving ₹720.
The total reduction is ₹280, which is 28% of the original ₹1,000. Adding 20 + 10 would give 30%, but that treats both discounts as percentages of the same base. A step-by-step price trail makes the changing base visible.
Use retained-price multipliers
A 20% discount leaves 80% of the price, represented by 0.80. A 10% discount leaves 0.90. Multiply the retained fractions: ₹1,000 × 0.80 × 0.90 = ₹720. This is often cleaner than separately calculating two reductions.
For discounts a% and b%, the retained fraction is (1 − a/100)(1 − b/100). The equivalent single discount is 100 times one minus that product. You can derive the shortcut a + b − ab/100, but keep the multiplier explanation so the rule does not become another number to memorise.
Worked example: unequal prices and discounts
A fictional jacket marked ₹2,400 has 25% off, followed by 15% off the reduced price. The first retained price is ₹2,400 × 0.75 = ₹1,800. The second is ₹1,800 × 0.85 = ₹1,530. The overall reduction is ₹870, giving 870/2400 × 100 = 36.25%.
A flat 40% discount would instead leave ₹1,440, so it is cheaper by ₹90 when every other term is the same. Do not compare the headline numbers alone. Check whether the item, starting price, fees and conditions are genuinely comparable.
Does reversing the order change the result?
For pure percentage discounts on the changing price, with no intermediate rounding or other conditions, order does not change the final price: 0.80 × 0.90 = 0.90 × 0.80. Starting from ₹1,000, either order gives ₹720. The individual rupee reductions differ, but their total agrees.
A fixed-value coupon changes the problem. On ₹1,000, first taking ₹100 off and then 20% leaves ₹720. First taking 20% off and then ₹100 leaves ₹700. Fixed amounts do not behave like retained-price multipliers. Read the offer’s stated order rather than extending the percentage rule.
Recover the original price
If a final price of ₹720 follows 20% then 10% discounts, divide by the retained multiplier 0.72: original price = ₹720/0.72 = ₹1,000. Adding 28% to ₹720 gives ₹921.60, which is not the original because 28% would now be calculated on the smaller price.
Check the recovered value by applying both discounts forward. For ₹1,530 after 25% then 15%, divide by 0.75 × 0.85 = 0.6375 to recover ₹2,400. Keep unrounded values during the calculation unless the offer specifies rounding at each step.
Rounding, extra charges and realistic wording
If an answer involves paise, use the rounding rule supplied by the question or stated by the seller. Rounding intermediate prices can slightly change a final result. Classroom comparisons should state whether taxes, delivery and coupons are included. None of the prices here are real BananaLearn offers or recommendations to buy.
For example, a ₹720 item plus ₹60 delivery costs ₹780 in total. A different seller charging ₹750 with delivery included is cheaper on those stated terms. A larger percentage reduction does not necessarily mean the lowest total payment when starting prices or charges differ.
Check the explanation, not just the final price
Ask a learner to name the base for the second percentage and predict whether the combined discount is less than the sum. For two positive discounts below 100%, the equivalent reduction is less than their sum because the second reduction applies to a reduced base.
Use the record to compare two invented offers. Write the original price, each retained multiplier, the final item price and any extra charge. A complete solution includes a fair-comparison statement and appropriate currency units, rather than a bare percentage.
Compare offers on the same basis
Copy the record into a notebook or use it to organise your own answer. Blank spaces are for your work.
| Offer | Start | Retained multiplier | Item price | Extra charges |
|---|---|---|---|---|
| A: 20%, then 10% | ₹1,000 | 0.80 × 0.90 | ₹720 | ___ |
| B: flat 30% | ₹1,000 | 0.70 | ₹700 | ___ |
| Your offer | ___ | ___ | ___ | ___ |
12 questions with explained answers
Try the questions before opening the answers. Explain the reason, then try a changed example to check your understanding.
₹1,000 less 20%, then 10%: final price?
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₹720
1000 × 0.8 × 0.9.Equivalent discount for 20% then 10%?
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28%
1 − 0.72 = 0.28.₹800 less 25%, then 10%?
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₹540
800 × 0.75 × 0.9.Equivalent discount for 25% then 10%?
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32.5%
1 − 0.675 = 0.325.₹2,400 less 25%, then 15%?
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₹1,530
2400 × 0.75 × 0.85.Two successive 10% discounts: equivalent reduction?
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19%
1 − 0.9² = 0.19.Does changing the order of pure 20% and 10% discounts change the exact final price?
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No
Multiplication commutes, absent intermediate rounding or other conditions.₹1,000: ₹100 coupon first, then 20% off?
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₹720
(1000 − 100) × 0.8.₹1,000: 20% off first, then ₹100 coupon?
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₹700
1000 × 0.8 − 100.Final ₹720 after 20% then 10%: original?
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₹1,000
720/0.72.₹720 plus ₹60 delivery versus ₹750 delivered: cheaper?
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₹750 delivered
The other total is ₹780.Why is 20% + 10% not the equivalent discount?
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The second discount has a smaller base
It removes ₹80, not ₹100, in the ₹1,000 example.
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