Number of people in age group III of city D is ______ % more than that of age…

Concept. When a total is divided in a ratio a : b : c, the parts add up to (a + b + c), so one part equals total ÷ (a + b + c) and each age group holds its own ratio share of that one part.

Concept. The phrase "X is p% more than Y" always measures the excess against the quantity that follows the word "than". That reference quantity Y is the base, never X, so p = ((X − Y) ÷ Y) × 100.

Applying it to this table.

1. City D has 12.8 lakh people in the ratio 3 : 1 : 4. The parts add up to 3 + 1 + 4 = 8, so one part = 12.8 ÷ 8 = 1.6 lakh.

2. Age group III of city D is the 4-part share: 4 × 1.6 = 6.4 lakh.

3. City B has 16 lakh people in the ratio 2 : 3 : 3. The parts add up to 2 + 3 + 3 = 8, so one part = 16 ÷ 8 = 2 lakh.

4. Age group II of city B is the 3-part share: 3 × 2 = 6 lakh.

5. The stem reads "… more than that of age group II of city B", so city B's 6 lakh is the base and the excess is 6.4 − 6 = 0.4 lakh.

6. Required percentage = (0.4 ÷ 6) × 100 = 40 ÷ 6 = 20/3 = \(6\frac{2}{3}\%\).

Cross-check. Increase the base by that rate: 6 × 16/15 = 6.4 lakh, which is exactly the age group III figure for city D, so the rate is consistent with the table.

Where the near misses come from. Dividing the same 0.4 lakh excess by 6.4 lakh — city D's own figure — gives \(6\frac{1}{4}\%\), but that answers the reverse comparison ("by what percent is city B's group II less than city D's group III"). Rounding 20/3 up to a whole number gives 7%, and a flat 6% would mean an excess of only 0.36 lakh.

Answer. Age group III of city D exceeds age group II of city B by \(6\frac{2}{3}\%\).