SSC MTS 6 Feb 2026 Shift 3 Paper

Concept:Adding two fractions with conjugate surds in the denominator simplifies by using the identity $(a+\sqrt{b})(a-\sqrt{b}) = a^2 - b$.Explanation:First, write the sum as a single fraction: $\frac{1}{6+\sqrt{3}} + \frac{1}{6-\sqrt{3}}$.The common denominator is $(6+\sqrt{3})(6-\sqrt{3})$.Multiply out: $(6+\sqrt{3})(6-\sqrt{3}) = 6^2 - (\sqrt{3})^2 = 36 - 3 = 33$.Add the numerators: $(6-\sqrt{3}) + (6+\sqrt{3}) = 6+6 = 12$ (the $\sqrt{3}$ terms cancel).So the sum is $\frac{12}{33}$.Reduce the fraction by dividing numerator and denominator by $3$: $\frac{12 \div 3}{33 \div 3} = \frac{4}{11}$.Therefore, the correct answer is $\frac{4}{11}$.Answer:C. $4/11$