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CTET Paper II (Mathematics and Science) 20 Aug 2023 Solved Paper

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Question : 46 of 150

Marks: +1, -0

If

x=\sqrt{198}\times \sqrt{550}

y=\sqrt[3]{99}\times \sqrt[3]{363}

\frac{1}{x}+\frac{1}{y}

is equal to :

$0.\bar{01}$
$0.0\bar{3}$
$0\cdot \bar{33}$
$0\cdot \bar{1}$

Solution:

Concept:

Simplify each radical using prime factorization, then compute

x

,

y

, and finally

\frac{1}{x}+\frac{1}{y}

.

Explanation:

First,

x = \sqrt{198} \times \sqrt{550}

.

Factor:

198 = 2 \times 3^2 \times 11

and

550 = 2 \times 5^2 \times 11

.

So

\sqrt{198} = 3\sqrt{22}

and

\sqrt{550} = 5\sqrt{22}

.

Thus

x = 3 \times 5 \times (\sqrt{22})^2 = 15 \times 22 = 330

.

Next,

y = \sqrt[3]{99} \times \sqrt[3]{363}

.

Factor:

99 = 3^2 \times 11

and

363 = 3 \times 11^2

.

Then

y = (3^2 \cdot 11)^{1/3} \times (3 \cdot 11^2)^{1/3} = 3^{(2/3+1/3)} \times 11^{(1/3+2/3)} = 3^1 \times 11^1 = 33

.

Now

\frac{1}{x}+\frac{1}{y} = \frac{1}{330}+\frac{1}{33} = \frac{1+10}{330} = \frac{11}{330} = \frac{1}{30} = 0.0\overline{3}

.

Answer:

Option B:

0.0\overline{3}

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