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SSC Selection Post 2019 PYQ — Question 58

SSC Selection Post - Graduation Level 2019 - 15 Oct 2019 • Quantitative Aptitude • Q58

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If x=120.21+121.22+...+129.30x=\frac{1}{20.21}+\frac{1}{21.22}+...+\frac{1}{29.30} and y=130.31+131.32+...+139.40y=\frac{1}{30.31}+\frac{1}{31.32}+...+\frac{1}{39.40}, then what is the value of xy\frac{x}{y}?

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Correct answer: B

Explanation

The correct answer is Option 4 i.e. 24\frac{2}{4}.

x=120.21+121.22+...+129.30x=\frac{1}{20.21}+\frac{1}{21.22}+...+\frac{1}{29.30}

=(120−121)+(121−122)+⋯+(129−130)= \left( \frac{1}{20} - \frac{1}{21} \right) + \left( \frac{1}{21} - \frac{1}{22} \right) + \cdots + \left( \frac{1}{29} - \frac{1}{30} \right)

=120−130=3−260=160= \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}

y=130.31+131.32+...+139.40y=\frac{1}{30.31}+\frac{1}{31.32}+...+\frac{1}{39.40}

=(130−131)+(131−132)+⋯+(139−140)= \left( \frac{1}{30} - \frac{1}{31} \right) + \left( \frac{1}{31} - \frac{1}{32} \right) + \cdots + \left( \frac{1}{39} - \frac{1}{40} \right)

=130−140=4−3120=1120= \frac{1}{30} - \frac{1}{40} = \frac{4 - 3}{120} = \frac{1}{120}

x/y = (1/60)/(1/120) = 2