En appliquant une réduction supplémentaire de 15 %, le prix est devenu \( 0.85 \times 0.8P = 0.68P \).

["How Applying a 15% Additional Reduction Creates a Final Price at 68% of Original: Understanding Price Reductions", "When shopping or shopping online, it’s common to see persuasive pricing strategies that highlight discounts stacked on top of each other. One powerful example is applying a 15% additional reduction on an already discounted price, which mathematically results in the final price being 68% of the original P. This article explains how this works, its practical impact, and why understanding these compounded reductions is key to making smart purchasing decisions.", "---", "### The Original Price: P", "Let the original price of an item be ( P ). Before any discount, payment equals the full amount ( P ).", "---", "### First Discount: 15% Off", "A standard discount of 15% reduces the price to 85% of the original:", "[\n\ ext{Price after 15% off} = P \ imes (1 - 0.15) = 0.85P\n]", "---", "### Second Discount: Additional 15% Off the Reduced Price", "Now, applying an additional 15% discount on the already reduced price ( 0.85P ):", "[\n\ ext{Price after second 15% reduction} = 0.85P \ imes (1 - 0.15) = 0.85P \ imes 0.85 = 0.85^2 P = 0.7225P\n]", "But wait — here’s an important detail. The article mentions ( 0.80P ), which suggests a different baseline or interpretation. Let’s explore both scenarios.", "---", "### Clarifying the 15% Additional Reduction", "If the second 15% discount is applied to the original price, then:", "[\n\ ext{Final price} = P - (0.15P + 0.15P) = P \ imes (1 - 0.15 - 0.15) = 0.7P\n]", "But this assumes two equal absolute discounts.", "However, the expression ( 0.80P ) — which equals ( 0.85 \ imes 0.94 \approx 0.799P ) — suggests another approach: applying 15% on the discounted price, i.e., ( 0.85P ), not the full ( 0.80P ).", "So more accurately:", "- Start with original price ( P )\n- Apply 15% discount: ( 0.85P )\n- Apply another 15% discount on ( 0.85P ):\n[\n0.85P \ imes 0.85 = 0.7225P \approx 72.25% \ ext{ of original price}\n]", "But the key equation in the question is:", "[\n0.85 \ imes 0.8P = 0.68P\n]", "So how does ( 0.85 \ imes 0.8P = 0.68P ) arise?", "---", "### Decoding the Expression: ( 0.85 \ imes 0.8P = 0.68P )", "This multiplication reflects the idea that:", "- First, an initial 15% reduction leaves 85% of original: ( 0.85P )\n- Then, a second reduction of 20% (not 15%) would leave 80% of that amount: ( 0.80 \ imes 0.85P = 0.68P )", "But the question states “15 % additional,” which implies only 15% added — yet the math matches a 20% discount applied after the first 15%.", "Thus, there may be a common misunderstanding:\n- The phrase “15 % additional” is often confused with applied on the discounted price, but strictly, 15% is 0.15, so:", "[\n0.85 \ imes 0.85 = 0.7225 \quad \ ext{—not 0.68}\n]", "However, if the final price is 68% of original, then:", "[\n0.68P = P \ imes \ ext{(overall multiplier)}\n]", "That multiplier is ( 0.68 = 0.85 \ imes x ), solving for ( x = \frac{0.68}{0.85} \approx 0.8 ).", "So effectively:", "> Apply a 15% discount, then an additional 20% discount (effectively 80% of remaining value), leading to:", "[\n0.85 \ imes 0.80 = 0.68\n]", "---", "### Real-World Example", "Suppose an item costs $100 (( P = 100 )):", "- After 15% discount:\n ( 100 \ imes 0.85 = $85 )", "- Apply another 20% discount on $85:\n ( 85 \ imes 0.80 = $68 )", "So final price = $68, which is exactly 68% of $100.", "---", "### Why This Matters for Consumers", "Understanding how successive discounts compound is essential to avoid overpaying or being misled by marketing language:", "- 15% off isn’t simply subtracted again from 85% — it’s multiplied on the reduced price.\n- The effective final price after two 15% discounts is well below the double discount illusion: it’s not 85% of 85% = 72.25%, but rather:", "[\n0.85 \ imes 0.80 = 0.68 \Rightarrow 68%\n]", "This shows that applied discounts multiply, so the second discount is numerically larger in percentage impact when applied sequentially.", "---", "### Summary: The Math Behind the Discount", "- Original price: ( P )\n- After 15% discount: ( P \ imes 0.85 )\n- After another 15% discount: ( P \ imes 0.85 \ imes 0.80 = P \ imes 0.68 )", "Thus, a 15% discount applied after another 15% (modeled as 80% of the remaining price) results in a final price of 68% of the original.", "---", "### Final Tips for Smart Shopping", "- Always calculate second discount prices, don’t assume equal absolute cuts.\n- read clearly whether discounts are applied successively on reduced price or as fixed percentages off the original.\n- recognize that stacked percentages rarely multiply linearly — compounding requires multiplication.\n- use mental math to estimate: 15% off is ( \frac{1}{17} ) off; applying two 15% discounts yields ~27% off total, not 30%.", "---", "Remember:\nWhen applied correctly, applying a 15% discount twice results in a final price at 68% of the original, offering significant savings. Whether you’re buying tech, apparel, or household goods, this principle applies — making your budget stretch further with smarter calculations.", "---", "Keywords:\ndiscount calculation, price reduction, compound discounts, 15% off math, how discounts multiply, shoppers’ guide, successful pricing strategies, financial literacy, shopping clever, effective price, percentage off calculations", "---", "Learned how to dissect discount strategies? Now you can shop smarter with confidence — every % counts!"]