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SSC MTS 2026 PYQ — Question 10

SSC MTS - 11 Feb 2026 - Shift 2 • Quantitative Aptitude • Q10

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Find the approximate value of:
76.91×143.977.07×12.01−3.932×195.886.93×8.031/3=?\frac{76.91 \times 143.97}{7.07 \times 12.01} - \frac{3.93^2 \times \sqrt{195.88}}{6.93 \times 8.03^{1/3}} = ?

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Correct answer: C

Explanation

76.91×143.977.07×12.01≈77×1447×12=11×12=1323.932×195.886.93×8.031/3≈42×147×2=16×1414=16⇒132−16=116\frac{76.91 \times 143.97}{7.07 \times 12.01} \approx \frac{77 \times 144}{7 \times 12} = 11 \times 12 = 132 \\ \frac{3.93^2 \times \sqrt{195.88}}{6.93 \times 8.03^{1/3}} \approx \frac{4^2 \times 14}{7 \times 2} = \frac{16 \times 14}{14} = 16 \\ \Rightarrow 132 - 16 = 116