If H.C.F. of $p$ and $q$ is $x$ and $q = xy$, then the L.C.M. of $p$ and $q$ is

### Concept & Algebraic Logic The fundamental rule connecting H.C.F. and L.C.M. applies not just to integers, but perfectly to algebraic variables as well. $$ \text{H.C.F.} \times \text{L.C.M.} = \text{First Number} \times \text{Second Number} $$ ### Step-by-Step Solution * **Given:** * The two numbers are $p$ and $q$. * Their H.C.F. = $x$. * We are also given a substitution identity: $q = xy$. * According to the core property: $x \times \text{L.C.M.} = p \times q$. * Isolate the L.C.M.: $\text{L.C.M.} = \frac{p \times q}{x}$. * Now, substitute the given value $q = xy$ into the equation: $\text{L.C.M.} = \frac{p \times (xy)}{x}$. * Cancel out the common term $x$ from the numerator and denominator. * $\text{L.C.M.} = p \times y = py$. ### Exam Strategy & Shortcut Since $q = xy$ and $x$ is the H.C.F., $y$ is the leftover coprime factor of $q$. The L.C.M. must contain all prime factors at their highest powers. It will simply be the first number $p$ multiplied by the leftover coprime factor of the second number $y$. Hence, $p \times y = py$. ### Common Pitfall Students often get intimidated by algebraic variables and try to plug in random numbers to solve the problem. While plugging in numbers (e.g., $p=4, q=6$) can work, it takes significantly longer than simply trusting the algebraic formula and substituting the variables directly. ### Final Answer **Therefore, the correct answer is $py$.**