Answer:
\(3\frac{1}{2} < 3\frac{19}{20} < 3\frac{39}{40}\)
Step-by-step explanation:
Given numbers,
\(3\frac{39}{40}, 3\frac{19}{20}, 3\frac{1}{2}\)
\(=\frac{159}{40}, \frac{79}{20}, \frac{7}{2}\)
Since, LCM( 40, 20, 2 ) = 40,
So, we need to make denominator 40 of all fractions for comparing them,
\(\frac{159}{40}, \frac{158}{40}, \frac{140}{40}\)
We know that,
\(\frac{140}{40} < \frac{158}{40} < \frac{159}{40}\)
\(\implies 3\frac{1}{2} < 3\frac{19}{20} < 3\frac{39}{40}\)
Which is the required order.
