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) $A_4 = 25$, $B_4 = 236.191$, $C_4 = 19.5$
Answer: (C) $A_1+B_1+C_1 < A_2+B_2+C_2 = A_3+B_3+C_3 < A_4+B_4+C_4$
In addition, the result keeps as many decimal places as the least precise term.
(i) $24.36 + 0.0724 + 256.2 = 280.6324 \to 280.6$ (one decimal place)
(ii) $24.44 + 16.082 + 240.2 = 280.722 \to 280.7$
(iii) $25.2 + 19.2812 + 236.183 = 280.6642 \to 280.7$
(iv) $25 + 236.191 + 19.5 = 280.691 \to 281$ (no decimal places)
So $280.6 < 280.7 = 280.7 < 281$.
Solution by Sreeraj P, M.Sc Physics