Assume missing numbers be a,band c.a/45=21/b=c/27=7/9Here, a/45=7/9a=7× 45/9=35And, 21/b=7/9b=21× 9/7=27Also, c/27=7/9c=7× 27/9=21Hence, 35/45=21/27=21/27=7/9. ...

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Byju's Answer

Standard V

Mathematics

LCD Method

/45=21/=/27=7...

Question

$\frac{}{45} = \frac{21}{} = \frac{}{27} = \frac{7}{9}$

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Solution

Assume missing numbers be $a, b$ and $c$ .
$\frac{a}{45} = \frac{21}{b} = \frac{c}{27} = \frac{7}{9}$
Here, $\frac{a}{45} = \frac{7}{9}$
$a = \frac{7 \times 45}{9} = 35$
And, $\frac{21}{b} = \frac{7}{9}$
$b = \frac{21 \times 9}{7} = 27$
Also, $\frac{c}{27} = \frac{7}{9}$
$c = \frac{7 \times 27}{9} = 21$
Hence, $\frac{35}{45} = \frac{21}{27} = \frac{21}{27} = \frac{7}{9}$ .

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Similar questions

Q. Question 27
The equivalent rational number of

\frac{7}{9}

, whose denominator is 45 is ___.

What number will replace ' $*$ ' mark?

$\frac{7}{8}$ $=$ $\frac{21}{*}$

\frac{- 45}{15} + \frac{- 3}{- 27} + \frac{8}{9} =

What is the value of the following expression?

$\frac{- 45}{15} + \frac{- 3}{- 27} + \frac{8}{9}$

Q. Question 9
Simplify the following

(i) \sqrt{45} - 3 \sqrt{20} + 4 \sqrt{5} (i i) \frac{\sqrt{24}}{8} + \frac{\sqrt{54}}{9} (i i i) 4 \sqrt{12} \times 7 \sqrt{6} (i v) 4 \sqrt{28} + 3 \sqrt{7} + \sqrt[3]{7} (v) 3 \sqrt{3} + 2 \sqrt{27} + \frac{7}{\sqrt{3}} (v i) (\sqrt{3} - \sqrt{2})^{2} (v i i) \sqrt[4]{81} - 8 \sqrt[3]{216} + 15 \sqrt[5]{32} + \sqrt{225} (v i i i) \frac{3}{\sqrt{8}} + \frac{1}{\sqrt{2}} (i x) \frac{2 \sqrt{3}}{3} - \frac{\sqrt{3}}{6}

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45=21=27=79

Assume missing numbers be a,band c.a45=21b=c27=79Here, a45=79a=7×459=35And, 21b=79b=21×97=27Also, c27=79c=7×279=21Hence, 3545=2127=2127=79.

$\frac{}{45} = \frac{21}{} = \frac{}{27} = \frac{7}{9}$