For the four sets of three measured physical quantities as given below. Which of the following options is correct?
$\begin{array}{l} (i) A_{1} = 24.36, B_{1} = 0.0724, C_{1} = 256.2 \\ (i i) A_{2} = 24.44, B_{2} = 16.082, C_{2} = 240.2 \\ (i i i) A_{3} = 25.2, B_{3} = 19.2812, C_{3} = 236.183 \\ (i v) A_{4} = 25, B_{4} = 236.191, C_{4} = 19.5 \end{array}$

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$A_{1} + B_{1} + C_{1} < A_{2} + B_{2} + C_{2} = A_{3} + B_{3} + C_{3} < A_{4} + B_{4} + C_{4}$

$A_{4} + B_{4} + C_{4} < A_{1} + B_{1} + C_{1} < A_{3} + B_{3} + C_{3} < A_{2} + B_{2} + C_{2}$

$A_{1} + B_{1} + C_{1} = A_{2} + B_{2} + C_{2} = A_{3} + B_{3} + C_{3} = A_{4} + B_{4} + C_{4}$

$A_{4} + B_{4} + C_{4} < A_{1} + B_{1} + C_{1} = A_{2} + B_{2} + C_{2} = A_{3} + B_{3} + C_{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} I n a d d i t i o n u \sin g s i g n i f i c a n t f i g u r e s, w e n e e d t o r o u n d o f f t o l e a s t d e c i m a l p o i n t s . \\ A_{1} + B_{1} + C_{1} = 24.36 + 0.0724 + 256.2 = 280.6324 \\ = 280.6 s (U \sin g s i g n i f i c a n t f i g u r e) \\ A_{2} + B_{2} + C_{2} = 24.44 + 16.082 + 240.2 = 280.722 = 280.7 \\ A_{3} + B_{3} + C_{3} = 25.2 + 19.2812 + 236.183 = 280.6642 = 280.7 \\ A_{4} + B_{4} + C_{4} = 25 + 236.191 + 19.5 = 280.691 = 281 \end{array}$