- A. p(10⁴ – k²)² = q × 10⁸
- B. p(10⁴ – k²)² = q × 10⁴
- C. p(10⁴ – k²) = q × 10⁴
- D. p(10⁴ – k²) = q × 10⁸
Answer: A
Explanation
Let the original price be p.
1. Price after first increase of k%: p₁ = p * (1 + k/100) = p * (100+k)/100
2. Price after first decrease of k%: p₂ = p₁ * (1 – k/100) = p * (100+k)/100 * (100-k)/100
This simplifies to p * (100² – k²)/100² = p * (10000 – k²)/10000
3. Price after second increase of k%: p₃ = p₂ * (1 + k/100) = p * [(10000 – k²)/10000] * (100+k)/100
4. Price after second decrease of k% (which is q): q = p₃ * (1 – k/100) = p * [(10000 – k²)/10000] * (100+k)/100 * (100-k)/100
q = p * [(10000 – k²)/10000] * [(100² – k²)/100²]
q = p * [(10000 – k²)/10000] * [(10000 – k²)/10000]
q = p * [(10000 – k²)/10000]²
q = p * (10⁴ – k²)² / (10⁴)²
q = p * (10⁴ – k²)² / 10⁸
Rearranging to match the options:
q × 10⁸ = p × (10⁴ – k²)²
Or p(10⁴ – k²)² = q × 10⁸.
This question tests the understanding of successive percentage changes and algebraic manipulation, a fundamental concept in quantitative aptitude for CSAT.