Which of the below surds are arranged in correct ascending order?

2 \sqrt{3} > 3 \sqrt{2} < \sqrt{7} < \frac{2}{\sqrt{2}}

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\frac{2}{\sqrt{2}} < \sqrt{7} < 2 \sqrt{3} < 3 \sqrt{2}

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\frac{2}{\sqrt{2}} > \sqrt{7} > 2 \sqrt{3} < 3 \sqrt{2}

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2 \sqrt{3} > 3 \sqrt{2} > \sqrt{7} > \frac{2}{\sqrt{2}}

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Solution

The correct option is B $\frac{2}{\sqrt{2}} < \sqrt{7} < 2 \sqrt{3} < 3 \sqrt{2}$
If x>y
Then, $x^{2} > y^{2}$
So, squaring all the surds
$(2 \sqrt{3})^{2} = 4 \times 3 = 12$
$(3 \sqrt{2})^{2} = 9 \times 2 = 18$
$(\sqrt{7})^{2} = 7$
$(\frac{2}{\sqrt{2}})^{2} = \frac{4}{2} = 2$
So, the correct ascending order is given by,
2<7<12<18
That is,
$\frac{2}{\sqrt{2}} < \sqrt{7} < 2 \sqrt{3} < 3 \sqrt{2}$

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Question 22

Which of the following is an example of distributive property of multiplication over addition for rational numbers.
(a) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = [- \frac{1}{4} \times \frac{2}{3}] + [- \frac{1}{4} \times (\frac{- 4}{7})]$
(b) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = [\frac{1}{4} \times \frac{2}{3}] - (\frac{- 4}{7})$
(c) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = \frac{2}{3} + (- \frac{1}{4}) \times \frac{- 4}{7}$
(d) $- \frac{1}{4} \times {\frac{2}{3} + (\frac{- 4}{7})} = {\frac{2}{3} + (\frac{- 4}{7})} - \frac{1}{4}$

Solve:

(i) $2 - \frac{3}{5}$

(ii) $4 + \frac{7}{8}$

(iii) $\frac{3}{5} + \frac{2}{7}$

(iv) $\frac{9}{11} - \frac{4}{15}$

(v) $\frac{7}{10} + \frac{2}{5} + \frac{3}{2}$

(vi) $2 \frac{2}{3} + 3 \frac{1}{2}$

(vii) $8 \frac{1}{2} - 3 \frac{5}{8}$

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Which of the below surds are arranged in correct ascending order?

The correct option is B 2√2<√7<2√3<3√2If x>y Then, x2>y2 So, squaring all the surds (2√3)2=4×3=12 (3√2)2=9×2=18 (√7)2=7 (2√2)2=42=2 So, the correct ascending order is given by, 2<7<12<18 That is, 2√2<√7<2√3<3√2