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Ratio

For the four ...

Question

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) $A_{2} = 24.44, B_{2} = 16.08, C_{2} = 240.2$
(iii) $A_{3} = 25.2, B_{3} = 19.2812, C_{3} = 236.183$
(iv) $A_{4} = 25, B_{4} = 236.191, C_{4} = 19.5$

A

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

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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

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

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D

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

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E

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

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Solution

The correct option is E
$A_{1} + B_{1} + C_{1} < A_{3} + B_{3} + C_{3} < A_{4} + B_{4} + C_{4} < A_{2} + B_{2} + C_{2}$

Step 1. Given data:
(i) $A_{1} = 24.36, B_{1} = 0.0724, C_{1} = 256.2$
(ii) $A_{2} = 24.44, B_{2} = 16.08, C_{2} = 240.2$
(iii) $A_{3} = 25.2, B_{3} = 19.2812, C_{3} = 236.183$
(iv) $A_{4} = 25, B_{4} = 236.191, C_{4} = 19.5$
Step 2. Finding the sums:
$A_{1} + B_{1} + C_{1} = 280.6324$
$A_{2} + B_{2} + C_{2} = 280.72$
$A_{3} + B_{3} + C_{3} = 280.6642$
$A_{4} + B_{4} + C_{4} = 280.691$
On comparing, $280.632 < 280.664 < 280.691 < 280.72$
which $\Rightarrow A_{1} + B_{1} + C_{1} < A_{3} + B_{3} + C_{3} < A_{4} + B_{4} + C_{4} < A_{2} + B_{2} + C_{2}$
Hence, option E is correct

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