Q52 / 150 · MHT CET 2009 · Mathematics

If $D_{30}$ is the set of all divisors of $30, x, y \in D_{30}$, we define $x+y=\operatorname{LCM}(x, y), x \cdot y=\operatorname{GCD}(x, y)$, $x^{\prime}=\frac{30}{x}$ and $f(x, y, z)=(x+y) \cdot\left(y^{\prime}+z\right)$, then $f(2,5,15)$ is equal to

  • A
    2
  • B
    5
  • Correct answer
    C
    10
  • D
    15

Solution

$D_{30}=\{1,2,3,5,6,10,15,30\}$ $\begin{aligned} f(2,5,15) &=(2+5) \cdot\left(5^{\prime}+15\right) \\ &=10 \cdot\left(\frac{30}{5}+15\right) \\ &=10(6+15) \\ &=10 \cdot 30 \\ &=10 \end{aligned}$

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

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