Let total tank = $1$ unit.

Step 1 — compute full-tank times / per-minute rates for each full set (as given).

Set X: 20 X-pipes fill $0.70$ tank in 14 minutes. Rate_X_total = $0.70 / 14 = 0.05$ (tank per minute). So time to fill whole tank by 20 X-pipes = $1 / 0.05 = 20$ minutes. Equivalently, rate_20X = $1/20$ per minute.

Set Y: 10 Y-pipes fill $3/8$ tank in 6 minutes. Rate_Y_total = $(3/8) / 6 = 3/48 = 1/16$ (tank per minute). So time to fill whole tank by 10 Y-pipes = $16$ minutes. Equivalently, rate_10Y = $1/16$ per minute.

Set Z (drain): 16 Z-pipes empty $1/2$ tank in 20 minutes. Rate_Z_total = $(1/2) / 20 = 1/40$ (tank per minute) — this is draining rate. So time to empty whole tank by 16 Z-pipes = $40$ minutes. Equivalently, drain_rate_16Z = $1/40$ per minute.

Step 2 — adjust rates for "half of X closed" and "half of Y open".

Half of X closed ⇒ only 10 X-pipes working, so X rate halves: rate_X_half = $(1/20) / 2 = 1/40$ per minute (filling).

Half of Y open ⇒ only 5 Y-pipes working, so Y rate halves: rate_Y_half = $(1/16) / 2 = 1/32$ per minute (filling).

All Z open ⇒ drain_rate = $1/40$ per minute (emptying).

Step 3 — net filling rate when all three sets operate (with above adjustments).

Net rate = rate_X_half + rate_Y_half − drain_rate = $1/40 + 1/32 - 1/40$ = $1/32$ (tank per minute).

(Notice the $1/40$ from X and the $1/40$ from Z cancel.)

Step 4 — time to fill 50% ($0.5$) of tank at net rate:

Time = required_work / net_rate = $0.5 / (1/32) = 0.5 \times 32 = 16$ minutes.

Therefore the correct option is D (16 minutes).