Tag: equivalent fractions

The given numbers can be arranged in the ascending order as $\displaystyle \frac{1}{5}\, <\, \displaystyle \frac{3}{5}\, <\, \displaystyle \frac{7}{5}\, <\, \displaystyle \frac{9}{5}$
Greatest number $=\, \displaystyle \frac{9}{5}$;
Least number $=\, \displaystyle \frac{1}{5}$.
We have, $\displaystyle \frac{9}{5}\, \times\, \displaystyle \frac{x}{100}\, =\, \displaystyle \frac{1}{5}$
$x\, =\, \displaystyle \frac{100}{9}\, =\, 11\, \displaystyle \frac{1}{9}\, \%$

This question is very easy if we solve it by verification
process.

Take (A) i.e. $\displaystyle\frac{-2}{3}\,<\,\frac{4}{-9}\, \frac{-5}{12}\, <\,\frac{7}{-18}$

First take \displaystyle\frac{-2}{3},\,\frac{-4}{9}$

$-2\,\times\,9,\, 4\,\times\,3$

-18, -12

$\because\, -12\,>\,-18$

So, $\displaystyle\frac{-4}{9}, \frac{-5}{12}$

$- 4\,\times\, 12, \, - 5\,\times\, 9$

- 48, - 45

$\because\, -45\, >\, -48$

So, $\displaystyle\frac{-5}{12}\,>\, \frac{-4}{9},\,i.e.,\frac{-4}{9}\, <\, \frac{-5}{12}$

Finally, $\displaystyle\frac{-5}{12}, \frac{-7}{18}$

$5\,\times\, 18, \, -7\,\times\, 12$

- 90, - 84

$\because\, -84\, >\, -90$

So, $\displaystyle\frac{-7}{18}\, >\, \frac{-5}{12}, i.e., \frac{-5}{12}\, <\, \frac{-7}{18}$

$\therefore\, \displaystyle\frac{-2}{3}\, <\, \frac{-4}{9}\,<\,  \frac{-5}{12}\,<\, \frac{-7}{18}$

You can identify the answer by observing the question by practicing this method.

$\displaystyle \frac{4}{7},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{2}{5},\, \displaystyle \frac{5}{9}$
The above numbers in ascending order are
$\displaystyle \frac{1}{3}\, <\, \displaystyle \frac{2}{5}\, <\, \displaystyle \frac{5}{9}\, <\, \displaystyle \frac{4}{7}$
Middle two numbers are $\displaystyle \frac{2}{5}$ & $\displaystyle \frac{5}{9}$.
$\therefore$ Average = $\displaystyle \frac{\displaystyle \frac{2}{5}\, +\, \displaystyle \frac{5}{9}}{2}\, =\, \displaystyle \frac{43}{90}$.

Take any two given rational numbers
$\displaystyle \frac{-7}{13},\, \displaystyle \frac{-5}{13}$
$- 7\, \times\, 13,\, -5\, \times\, 13$
$- 91,\, -65$
$\because\, - 65\, >\, - 91$
So $\displaystyle \frac{-7}{13}$ is smaller.
Now compare this with $\displaystyle \frac{- 11}{13}$
$\displaystyle \frac{- 7}{13},\, \displaystyle \frac{- 11}{13}$
$-7\, \times\, 13,\, - 11\, \times\, 13$
$- 91\, ,\, - 143$
$\because\, - 91\, >\, - 143$
So $\displaystyle \frac{- 11}{13}$ is smaller.

By verification process, $A\,\rightarrow\, \displaystyle\frac{3}{4},\, \frac{-2}{1}, \, \frac{-11}{20},\, \frac{-4}{5}$

$3\,\times\, 1,\, -2\,\times\, 4$

3, - 8

Correct, $-2\,\times\, 20, \, -11\,\times\, 1$

- 40, - 11

Wrong.

$B\, \rightarrow\, \displaystyle \frac {3}{4},\, \frac{-11}{20}, \frac{-4}{5},\, \frac{-2}{1}$

$3\,\times\, 20, \, -11\,\times\, 4$

60, - 44

$\displaystyle\frac{3}{4}\, >\,\frac{-11}{20}$ $-11\,\times\, 5,\, -4\,\times\, 20$

- 55, - 80

$\because\, \displaystyle\frac{-11}{20}\, >\, \frac{-4}{5}$

$-4\,\times\, 1,\, -2\,\times\, 5$

- 4, - 10

$\because\, \displaystyle\frac{-4}{5}\, >\, -2$

Since $\dfrac {1}{4}=0.25$.

Given, 

Given,