Question

Fill in the place holders with the correct symbol > or <:

Answer 1

Hint: To find the correct symbol, firstly, we will find the Least common multiple of the denominators of the given fractions. Now, to make the denominator the same for both the fractions, we will multiply the denominator and numerator of both the fractions with a number that makes the denominator equals to LCM. Now, since the denominator will become the same for both the fractions, we can simply compare the fractions by checking for its numerators.

Complete step-by-step answer:
 To find the answer, we need to compare the numbers $\dfrac{5}{{12}}$ and $\dfrac{5}{8}$ .
 Since the denominator of both the fractions is different, we will need to make it same for the comparison. We can do so by finding the least common multiple of both the denominators.
 The LCM of the numbers 12 and 8 can be expressed as:
$\begin{array}{l}
{\rm{LCM}} = 2 \times 2 \times 2 \times 3\\
{\rm{LCM}} = 24
\end{array}$
Since the LCM of the denominator is 24, we will try to make the denominator of both the fractions by multiplying them with relevant numbers. This can be expressed as
$\dfrac{5}{{12}} \times \dfrac{2}{2} = \dfrac{{10}}{{24}}$
 And
$\dfrac{5}{8} \times \dfrac{3}{3} = \dfrac{{15}}{{24}}$
  From the above expressions, we can say that,
$\dfrac{{10}}{{24}} < \dfrac{{15}}{{24}}$
 Hence,
$\dfrac{5}{{12}} < \dfrac{5}{8}$ .

Note: To solve a problem like this, we always need to make denominators the same for the fractions. We can compare any number of fractions by finding the least common multiple of their denominators and then making the denominator equal to LCM by multiplying with a relevant number. If both fractions are with negative signs, then we will follow the same steps that we did in this question, but in the last step, we will multiply both the fractions with a negative sign. There is one more case in which one of the fractions can have a negative sign, in that case, we can simply conclude that the fraction with a negative sign is less than the other.