Explanation:

This is a classic problem involving work and time, often solved using the concept of 'Man-Days' or 'Work-Rate'. The fundamental principle is that the rate of work per unit (e.g., per cat) is constant. \nLet's denote: $M_1$ = Number of cats in the first scenario = 100 $T_1$ = Time taken in the first scenario = 100 minutes $W_1$ = Work done in the first scenario (mice caught) = 100 mice $M_2$ = Number of cats in the second scenario = 7 $T_2$ = Time taken in the second scenario = ? $W_2$ = Work done in the second scenario (mice caught) = 7 mice \nThe formula relating these quantities is: $\frac{M_1 \times T_1}{W_1} = \frac{M_2 \times T_2}{W_2}$\nThis formula states that the amount of work done per worker per unit of time is constant. \nSubstitute the given values into the formula: $\frac{100 \times 100}{100} = \frac{7 \times T_2}{7}$ \nSimplify both sides of the equation: $100 = T_2$ \nSo, it will take 100 minutes for 7 cats to catch 7 mice. \nAlternatively, consider the rate of one cat. If 100 cats catch 100 mice in 100 minutes, this implies that, on average, one cat catches one mouse in 100 minutes. If we have 7 cats, each cat still takes 100 minutes to catch one mouse. If each of the 7 cats needs to catch one mouse (total 7 mice), and each cat takes 100 minutes for its task, then simultaneously, all 7 cats will complete their individual tasks of catching one mouse each in 100 minutes. \nLet's analyze the options:\nOption (A) $\frac{100}{7}$ minutes: This would be the answer if 1 cat caught 7 mice, or 7 cats caught 100 mice, which is incorrect.\nOption (B) $100$ minutes: This matches our calculated time. This is correct.\nOption (C) $\frac{49}{100}$ minutes: This is incorrect. This value is too small and doesn't logically follow from the problem.\nOption (D) $7$ minutes: This would imply a much faster rate, or a different relationship, which is incorrect. \nTherefore, the correct option is (B).