Question 128
The expression
$$\frac{1}{8} + \frac{1}{10} + \frac{1}{11} + \frac{1}{15} + \frac{1}{20} + \frac{1}{41} + \frac{1}{110} + \frac{1}{1640}$$ is equal to
Solution
Expression : $$\frac{1}{8} + \frac{1}{10} + \frac{1}{11} + \frac{1}{15} + \frac{1}{20} + \frac{1}{41} + \frac{1}{110} + \frac{1}{1640}$$
= $$(\frac{1}{41} + \frac{1}{1640}) + (\frac{1}{10} + \frac{1}{20}) + (\frac{1}{11} + \frac{1}{110}) + (\frac{1}{8} + \frac{1}{15})$$
= $$(\frac{40+1}{1640})+(\frac{2+1}{20})+(\frac{10+1}{110})+(\frac{15+8}{120})$$
= $$\frac{1}{40}+\frac{3}{20}+\frac{1}{10}+\frac{23}{120}$$
= $$\frac{3+18+12+23}{120}=\frac{56}{120}=\frac{7}{15}$$
=> Ans - (D)
