Assume the population of Panchagarh in 2016 is 450000. Then the combined non-male population of cities P, Q and T is approximately what percent more/less than the combined male populations of cities R, S and T?

State total $= 450000$. Non-male share of a city $= 100\% - \text{male }\%$.

Non-male population of P, Q, T

City	City population	Non-male %	Non-male population
P	$0.20 \times 450000 = 90000$	80%	$72{,}000$
Q	$0.28 \times 450000 = 126000$	35%	$44{,}100$
T	$0.12 \times 450000 = 54000$	40%	$21{,}600$
		Total	$\mathbf{137{,}700}$

Male population of R, S, T

City	City population	Male %	Male population
R	$0.25 \times 450000 = 112500$	45%	$50{,}625$
S	$0.15 \times 450000 = 67500$	50%	$33{,}750$
T	$0.12 \times 450000 = 54000$	60%	$32{,}400$
		Total	$\mathbf{116{,}775}$

Since the first total is larger, it is more:

$\frac{137700 - 116775}{116775} \times 100 = \frac{20925}{116775} \times 100 \approx 17.9\% \approx 18\%$

Answer: 18%, More

---

Note: the value 450000 is not actually needed — it cancels. Working purely in percentage points of the state: $(16 + 9.8 + 4.8) = 30.6$ against $(11.25 + 7.5 + 7.2) = 25.95$, giving $\frac{4.65}{25.95} \times 100 \approx 18\%$.