This is the boss. No calculator. Every logarithm the cascade needs is printed in the question — take $\ln 10$ as 2.3, $\ln 20$ as 3.0 and $\ln 2$ as 0.7 — and the flows have been chosen so the absorption factor lands on a whole number. Everything is sized to fit in your head.

Four lines, and every answer feeds the next: solubility gives the equilibrium slope, the slope and the two flows give the absorption factor, the factor and the recovery give the transfer units, and the transfer units times the packing's HTU give metres. Write each line down as you go, because one slip at the top rolls all the way to the quotation.

At $A = 2$ the arithmetic simplifies in a way worth seeing before you start: $1 - 1/A$ and $1/A$ are both exactly 0.5, so the bracket is $0.5\left(y_1/y_2\right) + 0.5$ and the count is simply twice the logarithm of it. That is not a coincidence of this question — it is why 2 is such a comfortable place to design an absorber.

And note what the last line actually asks. It does not want another number; it wants a decision: can this scrubber be quoted into the shell the client already owns, yes or no. The arithmetic exists only to earn you the right to make it. Watch for the trap this whole chapter has been laying — a transfer-unit count is not a length, and comparing one against a shell in metres is how a column gets sold two metres short.

Everything in the chapter rides in this one tower. 5★ under par here is the chapter crown. Clock in.