The L.C.M. of $3$, $2.7$ and $0.09$ is

### Concept & Strategy Just like with H.C.F., the strategy for calculating the L.C.M. of decimals is to convert them into like decimals (same number of decimal places) by adding trailing zeros. Then, treat them as integers, find their L.C.M., and place the decimal point back in the result. ### Step-by-Step Solution * **Given:** The numbers are $3$, $2.7$, and $0.09$. * **Step 1:** The maximum number of decimal places is two (in $0.09$). Rewrite all numbers with two decimal places: $3.00$, $2.70$, and $0.09$. * **Step 2:** Remove the decimal points by multiplying each by $100$. We now need the L.C.M. of $300$, $270$, and $9$. * **Step 3:** Perform prime factorization for each number: * $300 = 2^2 \times 3 \times 5^2$ * $270 = 2 \times 3^3 \times 5$ * $9 = 3^2$ * **Step 4:** The L.C.M. is found by multiplying the highest power of every prime factor present. $$L.C.M. = 2^2 \times 3^3 \times 5^2$$ $$L.C.M. = 4 \times 27 \times 25 = 100 \times 27 = 2700$$ * **Step 5:** Adjust the decimal point by dividing by $100$ (since we multiplied by $100$ initially). $$2700 / 100 = 27$$ ### Exam Strategy & Shortcut **Multiple Check:** The L.C.M. of a set of numbers must be a multiple of the largest number in that set (or equal to it). The largest number here is $3$. The L.C.M. must therefore be $\ge 3$. Looking at the options, (a) $2.7$, (b) $0.27$, and (c) $0.027$ are all strictly less than $3$ and thus physically impossible to be multiples of $3$. By pure elimination, only $27$ can be the L.C.M. ### Common Pitfall A very common mistake is finding the L.C.M. of $3$, $27$, and $9$ (ignoring place value alignment) which gives $27$, and then guessing where to put the decimal point back (often guessing $0.27$). Always pad the numbers to equal length ($3.00$, $2.70$, $0.09$) before dropping decimals to maintain mathematical integrity. ### Final Answer **Therefore, the correct answer is 27.**