Question : Which of the following statement(s) is/are true?
I. $\sqrt{144}\times \sqrt{36}<\sqrt[3]{125}\times \sqrt{121}$
II. $\sqrt{324}+ \sqrt{49}<\sqrt[3]{216}\times \sqrt{9}$
Option 1: Only I
Option 2: Only II
Option 3: Neither I nor II
Option 4: Both I and II
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Correct Answer: Neither I nor II
Solution :
I. $\sqrt{144}\times \sqrt{36}<\sqrt[3]{125}\times \sqrt{121}$
On the left side, $\sqrt{144} = 12$ and $\sqrt{36} = 6$, so their product is $12 \times 6 = 72$.
On the right side, $\sqrt[3]{125} = 5$ and $\sqrt{121} = 11$, so their product is $5 \times 11 = 55$.
So, the inequality $72 < 55$ is not true.
II. $\sqrt{324}+ \sqrt{49}<\sqrt[3]{216}\times \sqrt{9}$
On the left side, $\sqrt{324} = 18$ and $\sqrt{49} = 7$, so their sum is $18 + 7 = 25$.
On the right side, $\sqrt[3]{216} = 6$ and $\sqrt{9} = 3$, so their product is $6 \times 3 = 18$.
So, the inequality $25 < 18$ is not true.
Therefore, the answer is neither I nor II.
Hence, the correct answer is 'Neither I nor II'.
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