RRB Technician Grade 3 previous year question
If a = 0.1125, then find the value of 100[√(1 + 2(3a) + 9a²) − 4a].
Asked in RRB Technician Grade 3 (09 Mar 2026 Shift 3) · Quantitative Aptitude
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Correct answer: (B) 88.75
Since 1+6a+9a² = (1+3a)², √(1+6a+9a²)−4a = 1+3a−4a = 1−a; so 100×(1−0.1125) = 88.75.
हिंदी में प्रश्न
यदि a = 0.1125 है, तो 100[√(1 + 2(3a) + 9a²) − 4a] का मान ज्ञात कीजिए।
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सही उत्तर: (B) 88.75
चूंकि 1+6a+9a² = (1+3a)², √(1+6a+9a²)−4a = 1+3a−4a = 1−a; अतः 100×(1−0.1125) = 88.75।
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