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SSC MTS 2026 PYQ — Question 9

SSC MTS - 9 Feb 2026 - Shift 2 • Quantitative Aptitude • Q9

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Find the approximate value of:
270.12×215.973(2.96)×80.93−9.92+3.08168.91=?\frac{270.12 \times \sqrt[3]{215.97}}{(2.96) \times \sqrt{80.93}} - \frac{9.92 + 3.08}{\sqrt{168.91}} = ?

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Correct answer: A

Explanation

270.12×215.973(2.96)×80.93−9.92+3.08168.91215.973=6,80.93=8.996≈9,168.91=13=270.12×62.96×9−1313=1620.7226.64−1=60.84−1=59.84=59\frac{270.12 \times \sqrt[3]{215.97}}{(2.96) \times \sqrt{80.93}} - \frac{9.92 + 3.08}{\sqrt{168.91}} \\ \sqrt[3]{215.97} = 6,\quad \sqrt{80.93} = 8.996 \approx 9,\quad \sqrt{168.91} = 13 \\ = \frac{270.12 \times 6}{2.96 \times 9} - \frac{13}{13} \\ = \frac{1620.72}{26.64} - 1 \\ = 60.84 - 1 \\ = 59.84 =59