Q151 / 270 · NDA 2016 (Phase 2) · Mathematics

Let $S$ be a set of all distinct numbers of the form $\frac{\mathrm{P}}{\mathrm{Q}}$, where $\mathrm{p}, \mathrm{q} \in\{1,2,3,4,5,6\}$. What is the cardinality of the set S ?

  • A
    $21$
  • Correct answer
    B
    $23$
  • C
    $32$
  • D
    $36$

Solution

Given that P , $\mathrm{q}:(1,2,3,4,5,6$, For $\frac{P}{q}$ from, when $P=1, q=1,2,3,4,5,6$ thus, $\frac{\mathrm{P}}{\mathrm{q}}=1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ and $\frac{1}{6}$ $\left(\frac{\mathrm{P}}{\mathrm{q}}\right)=6$ When $\mathrm{P}=2, \mathrm{q}=1,3,5$ thus $\frac{\mathrm{P}}{\mathrm{q}}=2, \frac{2}{3}, \frac{2}{5}$, and $n\left(\frac{p}{q}\right)=3$ When $P=3, q=1,2,3,4,5$ thus $\frac{P}{q}=3, \frac{3}{2}, \frac{3}{4}, \frac{3}{5}$ and $n\left(\frac{p}{q}\right)=4$ When $\mathrm{P}=4, \mathrm{q}=1,3,5$ thus $\frac{\mathrm{P}}{\mathrm{q}}=4, \frac{4}{3}, \frac{4}{5}$, and $n\left(\frac{p}{q}\right)=3$ When $\mathrm{P}=5, \mathrm{q}=1,2,3,4,6$ thus $\left(\frac{\mathrm{P}}{\mathrm{q}}\right)=5, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}$, and $\frac{5}{6}$ and $n\left(\frac{\mathrm{P}}{\mathrm{q}}\right)=5$ When $\mathrm{P}=6, \mathrm{q}=1,5$ thus $\left(\frac{p}{q}\right)=6, \frac{6}{5}$ and $n\left(\frac{P}{q}\right)=2$ Hence, cordinality of the set (5) $=6+3+4+3+5+2=23$.

Hence, the correct answer is option: 'B'.

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