For the four sets of three measured physical quantities as given below. Which of the following options is correct ? (i) A1 = 24.36, B1 = 0.0724, C1 = 256.2 (ii) A2 = 24.44, B2 = 16.082, C2 = 240.2 (iii) A3 = 25.2, B3 = 19.2812, C3 = 236.183 (iv) A4 = 25, B4 = 236.191, C4 = 19.5

A4 + B4 + C4 \gt A3 + B3 + C3 = A2 + B2 + C2 \gt A1 + B1 + C1

A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 = A4 + B4 + C4

A4 + B4 + C4 \lt A1 + B1 + C1 \lt A3 + B3 + C3 \lt A2 + B2 + C2

A4 + B4 + C4 \lt A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3

Units and Measurements PYQ — JEE Main 2020 · 9 Jan · Shift 2 · Q48 with Solution

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Correct answer: D

We calculate the sums first:

$S_1 = A_1 + B_1 + C_1 = 24.36 + 0.0724 + 256.2 = 280.6324$

$S_2 = A_2 + B_2 + C_2 = 24.44 + 16.082 + 240.2 = 280.722$

$S_3 = A_3 + B_3 + C_3 = 25.2 + 19.2812 + 236.183 = 280.6642$

$S_4 = A_4 + B_4 + C_4 = 25 + 236.191 + 19.5 = 280.691$

In addition, the result should be rounded to the least number of decimal places among the quantities being added.

For $S_1$ : decimal places are $2, 4, 1$ so keep 1 decimal place. $S_1 = 280.6324 \approx 280.6$
For $S_2$ : decimal places are $2, 3, 1$ so keep 1 decimal place. $S_2 = 280.722 \approx 280.7$
For $S_3$ : decimal places are $1, 4, 3$ so keep 1 decimal place. $S_3 = 280.6642 \approx 280.7$
For $S_4$ : decimal places are $0, 3, 1$ so keep 0 decimal places. $S_4 = 280.691 \approx 281$

Thus,

\quad S_3 = 280.7, \quad S_2 = 280.7, \quad S_1 = 280.6$$ Therefore, $$A_4+B_4+C_4 > A_3+B_3+C_3 = A_2+B_2+C_2 > A_1+B_1+C_1$$ 4. **Match with the options** This corresponds to **Option D**.