1. **Problem statement:** We have a grouped frequency distribution table with classes and frequencies given as:\n\n| Classes | Frequency (f) |\n|---|---|\n| 42 - 55 | 6 |\n| 56 - 69 | 4 |\n| 70 - 83 | 6 |\n| 84 - 97 | 15 |\n| 98 - 111 | 4 |\n| 112 - 125 | 5 |\n\nTotal frequency, $\sum f = 40$.\n\nFind the mean of the data represented by this grouped data.\n\n2. **Calculate midpoints for each class:**\nMidpoint for class $[a,b]$ is $\frac{a+b}{2}$.\n- $m_1 = \frac{42 + 55}{2} = 48.5$\n- $m_2 = \frac{56 + 69}{2} = 62.5$\n- $m_3 = \frac{70 + 83}{2} = 76.5$\n- $m_4 = \frac{84 + 97}{2} = 90.5$\n- $m_5 = \frac{98 + 111}{2} = 104.5$\n- $m_6 = \frac{112 + 125}{2} = 118.5$\n\n3. **Calculate the product of frequency and midpoint for each class:**\n- $6 \times 48.5 = 291$\n- $4 \times 62.5 = 250$\n- $6 \times 76.5 = 459$\n- $15 \times 90.5 = 1357.5$\n- $4 \times 104.5 = 418$\n- $5 \times 118.5 = 592.5$\n\n4. **Sum these products:**\n$$\sum f m = 291 + 250 + 459 + 1357.5 + 418 + 592.5 = 3368$$\n\n5. **Calculate the mean using formula:**\n$$\text{Mean} = \frac{\sum f m}{\sum f} = \frac{3368}{40} = 84.2$$\n\n**Final answer:** The mean of the grouped data is $84.2$.