Xavier Aptitude Test 2020 Paper

Question : 60 of 100

Marks: +1, -0

When expressed in a decimal form, which of the following numbers will be non - terminating as well as non-repeating?

Solution:

Option A:

(π/2)[1/π +1]-π/2 = 1/2

Option B:

\sin^2 1° + \sin^2 2°... + \sin^2 89°= 44 +1/2 \sin^2 (89) = \cos^2 { 1}

\sin^2{1} + \cos^2{1}=1

Option C:

√2 (3√2 - 4√2) + √3 = 6-4 + √3 = 2 + √3

which is non-terminating and non repeating.
Option D:

{3√729}/3 +22/7 = 3+22/7

Option E:

(4-π) [1 + (π/4) + {π/4}^2 + {π/4}^3 + ...(\infinite \terms)]=1 {1π}/4 =4 {4}/π ⇒ (4 - π) [1 +π/4 = {π/4}^2 + {π/}^3 + ... (infinite terms)] = 4

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