Daily sales (in ₹) at an Amritsar tea stall over several days:
120-150: 10, 150-180: 18, 180-210: 14, 210-240: 6, 240-270: f.
The mean daily sale is exactly ₹195. Using a = 195 and h = 30, find f.

Class marks 135, 165, 195, 225, 255 give $u_i=-2,-1,0,1,2$, so $\sum f_i{u}_i=-20-18+0+6+2f=2f-32$. Because the mean equals the assumed mean 195, the correction $h\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$ must be zero, so $2f-32=0$ and $f=16$ (check: $\sum f_i{x}_i=12480,\sum f_i=64$, 12480/64 = 195). "32" solves $f-32=0$, forgetting that the last class has $u_i=2$; "8" uses $u_i=4$ for it, miscounting the steps; "48" is the total of the four known frequencies.