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Significant Figures in Sum

For the four ...

Question

For the four sets of three measured physical quantities as given below, which of the following options is correct?
$(A) A_{1} = 24.36, B_{1} = 0.0724, C_{1} = 256.2$
$(B) A_{2} = 24.44, B_{2} = 16.082, C_{2} = 240.2$
$(C) A_{3} = 25.2, B_{3} = 19.2812, C_{3} = 236.1832$
$(D) A_{4} = 25, B_{4} = 236.191, C_{4} = 19.5$

A

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

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B

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

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C

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

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D

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

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E

None of the above

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Solution

The correct option is E None of the above
$Here,$
$A_{1} + B_{1} + C_{1} = 24.36 + 0.0724 + 256.2 = 280.6 A_{2} + B_{2} + C_{2} = 24.44 + 16.082 + 240.2 = 280.7 A_{3} + B_{3} + C_{3} = 25.5 + 19.2812 + 236.183 = 280.7 A_{4} + B_{4} + C_{4} = 25 + 256.191 + 19.5 = 281$

So, the correct answer should be,

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

Hence, $(D)$ is the correct answer.

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