The numbers 2, 3, 4 and 5 occur (2 + 5k), (5k - 7), (2k - 3) and (k + 2) times, respectively. The average of rhe numbers is 2.85. Later on, the number 2 was replaced by 6 in all the places. What is the average of the new numbers?

The four numerals and their frequencies are:
  2 occurs $$(2+5k)$$ times, 3 occurs $$(5k-7)$$ times, 4 occurs $$(2k-3)$$ times, 5 occurs $$(k+2)$$ times.

Total number of observations
$$N = (2+5k) + (5k-7) + (2k-3) + (k+2)$$ $$N = 13k - 6$$   $$-(1)$$

Total sum of the observations
$$S = 2(2+5k) + 3(5k-7) + 4(2k-3) + 5(k+2)$$ $$S = (4+10k) + (15k-21) + (8k-12) + (5k+10)$$ $$S = 38k - 19$$   $$-(2)$$

The given average is $$2.85 = \dfrac{57}{20}$$, so
$$\frac{S}{N} = \frac{38k-19}{13k-6} = \frac{57}{20}$$

Cross-multiplying:
$$20(38k-19) = 57(13k-6)$$ $$760k - 380 = 741k - 342$$ $$19k - 38 = 0$$ $$k = 2$$

Substituting $$k = 2$$ in the frequencies:
2 appears $$2 + 5(2) = 12$$ times,
3 appears $$5(2) - 7 = 3$$ times,
4 appears $$2(2) - 3 = 1$$ time,
5 appears $$2 + 2 = 4$$ times.

Total number of terms (from $$(1)$$):
$$N = 13(2) - 6 = 20$$

After replacement, every 2 is changed to 6. Hence the new data set has:
12 copies of 6, 3 copies of 3, 1 copy of 4, and 4 copies of 5.

New sum
$$S' = 6 \times 12 + 3 \times 3 + 4 \times 1 + 5 \times 4$$ $$S' = 72 + 9 + 4 + 20 = 105$$

New average
$$\bar{x} = \frac{S'}{N} = \frac{105}{20} = 5.25$$

Hence the required average is 5.25.
Option B which is: 5.25