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Number systems — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSNumber systems1 Mark
Question
The value of $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}},$ is:
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Answer

  1. $\frac{4}{3}$
    Solution:
    $\sqrt{48}=\sqrt{16\times3}=4\sqrt3$
    $\sqrt{32}=\sqrt{16\times2}=4\sqrt2$
    $\sqrt{27}=\sqrt{9\times3}=3\sqrt3$
    $\sqrt{18}=\sqrt{9\times2}=3\sqrt2$
    Now, $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}=\frac{4\sqrt3+4\sqrt2}{3\sqrt3+3\sqrt2}$
    $=\frac{4\big(\sqrt{\not3}+\sqrt{\not2}\big)}{3\big(\sqrt{\not3}+\sqrt{\not2}\big)}$
    $=\frac{4}{3}$
    Hence, correct option is (a).

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