correct answer is: a) 25\large\frac{5}{7}

Explanation

Here, fractions given are =1\large\frac{2}{7}\ ,\ \normalsize1\large\frac{1}{35}\ ,\large\frac{48}{49}
Or, \large\frac{9}{7}\ ,\ \large\frac{36}{35}\ ,\ \large\frac{45}{49} _{[covert mixed to improper fraction]}
We know,
The formula we use to get the LCM of two or more fractions is:
\large\frac{The\ LCM\ of\ Numerator}{The\ HCF\ of\ Denominator}

\therefore The LCM of the three fractions is:
\rightarrow LCM of Numerators 9, 36 and 45 is =180
\rightarrow HCF of Denominators 7, 35 and 49 is =7

So that, the LCM of 1\large\frac{2}{7}\normalsize,1\large\frac{1}{35},\large\frac{48}{49} is =\large\frac{180}{7} or 25\large\frac{5}{7}.
\therefore The smallest number which is exactly divisible is 25\large\frac{5}{7}.
Ans: 25\large\frac{5}{7} is the smallest number which is exactly divisible by 1\large\frac{2}{7}\ ,\ \normalsize1\large\frac{1}{35}\ ,\ \large\frac{48}{49} .

Other variants of this particular math question are:

1. What is the smallest number divisible by 1 2/7, 1 1/35, and 48/49?
2. Calculate the smallest number divisible by 1 2/7, 1 1/35, and 48/49.
3. Find the least common multiple (LCM) of 1 2/7, 1 1/35, and 48/49.
4. Find the smallest multiple of 1 2/7, 1 1/35, and 48/49.