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Multiplication of Fractions

Which is grea...

Question

Which is greater?

(1)

(2)

(3)

(4)

(5)

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Solution

(1)

The fraction is greater.

(2)

The fraction is greater.

(3)

Here,

Thus, is greater.

(4)

Thus, is greater.

(5)

Thus, is greater.

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Similar questions

Arrange the fractions in ascending order:
$\frac{3}{8}$ , $\frac{7}{8}$ , $\frac{6}{8}$ , $\frac{4}{8}$

8 \frac{2}{3} + 9 \frac{3}{8}

\frac{3 + \sqrt{5}}{2 \sqrt{5} + 3} = a + b \sqrt{5}

, then the numbers

a

b

- \frac{5}{9}, \frac{7}{12}, - \frac{2}{3}

\frac{11}{18}

in ascending order of their magnitudes. Identify the smallest

Q. Prove that:

(i)

\sqrt{3 \times 5^{- 3}} \div \sqrt[3]{3^{- 1}} \sqrt{5} \times \sqrt[6]{3 \times 5^{6}} = \frac{3}{5}

9^{3 / 2} - 3 \times 5^{0} - {(\frac{1}{81})}^{-}^{1 / 2} = 15

{(\frac{1}{4})}^{- 2} - 3 \times 8^{2 / 3} \times 4^{0} + {(\frac{9}{16})}^{- 1 / 2} = \frac{16}{3}

\frac{2^{1 / 2} \times 3^{1 / 3} \times 4^{1 / 4}}{10^{- 1 / 5} \times 5^{3 / 5}} \div \frac{3^{4 / 3} \times 5^{- 7 / 5}}{4^{- 3 / 5} \times 6} = 10

\sqrt{\frac{1}{4} + (0.01)^{- 1 / 2}} - (27)^{2 / 3} = \frac{3}{2}

\frac{2^{n} + 2^{n - 1}}{2^{n + 1} - 2^{n}} = \frac{3}{2}

{(\frac{64}{125})}^{- 2 / 3} + \frac{1}{{(\frac{256}{625})}^{1 / 4}} + (\frac{\sqrt{25}}{\sqrt[3]{64}}) = \frac{65}{16}

\frac{3^{- 3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times (15) -^{4 / 3} \times 3^{1 / 3}} = 28 \sqrt{2}

\frac{(0.6)^{0} - (0.1) - 1}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(- \frac{1}{3})}^{- 1}} = - \frac{3}{2}

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