For the four sets of three measured physical quantities as given below. Which of the following options is correct?
$$(i)$$ $$A_1 = 24.36, B_1 = 0.0724, C_1 = 256.2$$
$$(ii)$$ $$A_2 = 24.44, B_2 = 16.082, C_2 = 240.2$$
$$(iii)$$ $$A_3 = 25.2, B_3 = 19.2812, C_3 = 236.183$$
$$(iv)$$ $$A_4 = 25, B_4 = 236.191, C_4 = 19.5$$

The problem asks us to compare the algebraic sums $$A_i+B_i+C_i$$ of four different data-sets and then decide which of the four statements given in the options describes their order correctly.

Whenever we add measured quantities we must obey the rule of significant figures meant for addition and subtraction.

Rule (state first): While adding or subtracting, the final result is written only up to that number of digits after the decimal point which is possessed by the addend having the least number of digits after the decimal point.

Now we evaluate every sum one by one, showing all arithmetic steps and then applying the above rule.

Set (i): $$A_1=24.36,\;B_1=0.0724,\;C_1=256.2$$

Raw sum (ordinary addition, no rounding yet):

$$24.36 + 0.0724 = 24.4324$$

$$24.4324 + 256.2 = 280.6324$$

The three addends possess respectively 2, 4 and 1 decimal places; the least is 1. So we round the result to one place after the decimal:

$$280.6324 \longrightarrow 280.6$$

Set (ii): $$A_2=24.44,\;B_2=16.082,\;C_2=240.2$$

Raw sum:

$$24.44 + 16.082 = 40.522$$

$$40.522 + 240.2 = 280.722$$

Here the addends have 2, 3 and 1 decimal places; the least is again 1. So we round to one decimal place:

$$280.722 \longrightarrow 280.7$$

Set (iii): $$A_3=25.2,\;B_3=19.2812,\;C_3=236.183$$

Raw sum:

$$25.2 + 19.2812 = 44.4812$$

$$44.4812 + 236.183 = 280.6642$$

The numbers have 1, 4 and 3 decimal places; the least is 1. Rounding to one decimal place gives:

$$280.6642 \longrightarrow 280.7$$

Set (iv): $$A_4=25,\;B_4=236.191,\;C_4=19.5$$

Raw sum:

$$25 + 236.191 = 261.191$$

$$261.191 + 19.5 = 280.691$$

Now the addends have 0, 3 and 1 decimal places; the least is 0. Hence the answer must be rounded to no decimal places at all:

$$280.691 \longrightarrow 281$$

We therefore have the four correctly rounded totals

$$\begin{aligned} A_1+B_1+C_1 &= 280.6,\\ A_2+B_2+C_2 &= 280.7,\\ A_3+B_3+C_3 &= 280.7,\\ A_4+B_4+C_4 &= 281. \end{aligned}$$

Arranging them in ascending order of magnitude gives

$$280.6 \; < \; 280.7 = 280.7 \; < \; 281,$$

or, written with the original symbols,

$$A_1+B_1+C_1 \; < \; A_2+B_2+C_2 \;=\; A_3+B_3+C_3 \; < \; A_4+B_4+C_4.$$

This statement is exactly what Option C expresses.

Hence, the correct answer is Option C.