RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

Concept:For negative fractions, the number with the larger absolute value is smaller. Compare them by converting to decimals or using a common denominator.Explanation:Convert all fractions to a common denominator of $120$.$\frac{-7}{2} = \frac{-420}{120}$$\frac{-2}{3} = \frac{-80}{120}$$\frac{-5}{8} = \frac{-75}{120}$$\frac{-7}{10} = \frac{-84}{120}$Arranging in increasing order gives:$\frac{-420}{120} < \frac{-84}{120} < \frac{-80}{120} < \frac{-75}{120}$So, $\frac{-7}{2} < \frac{-7}{10} < \frac{-2}{3} < \frac{-5}{8}$.Option A states $\frac{-7}{2} < \frac{-2}{3} < \frac{-5}{8}$.Check each comparison:$\frac{-7}{2} < \frac{-2}{3}$ is true because $-3.5 < -0.666$.$\frac{-2}{3} < \frac{-5}{8}$ is true because $-0.666 < -0.625$.So option A is correct.Check other options:Option B: $\frac{-5}{8} < \frac{-2}{3}$ is false since $-0.625 > -0.666$.Option C: $\frac{-5}{8} < \frac{-7}{10}$ is false since $-0.625 > -0.7$.Option D: $\frac{-5}{8} < \frac{-2}{3}$ is false for the same reason.Answer:The correct inequality is option A: $\frac{-7}{2} < \frac{-2}{3} < \frac{-5}{8}$.