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How quickly can you do this? Fill appropriate sign. ( '<', '=', '>')
(a) $ \frac{1}{2} \square \frac{1}{5} $
(b) $ \frac{2}{4} \square \frac{3}{6} $
(c) $ \frac{3}{5} \square \frac{2}{3} $
(d) $ \frac{3}{4} \square \frac{2}{8} $
(e) $ \frac{3}{5} \square \frac{6}{5} $
(f) $ \frac{7}{9} \square \frac{3}{9} $
(g) $ \frac{1}{4} \square \frac{2}{8} $
(h) $ \frac{6}{10} \square \frac{4}{5} $
(i) $ \frac{3}{4} \square \frac{7}{8} $
(j) $ \frac{6}{10} \square \frac{3}{5} $
(k) $ \frac{5}{7} \square \frac{15}{21} $
To do:
We have to write '$<$' or ' \( > \) ', '$=$' between the given pairs of fractions.
Solution:
(a) \( \frac{1}{2} \square \frac{1}{5} \)
Here, the numerators are same.
Therefore, the fraction having lesser denominator will be lesser.
This implies,
$\frac{1}{2} > \frac{1}{5}$
(b) $\frac{2}{4}=\frac{1\times2}{2\times2}$
$=\frac{1}{2}$
$\frac{3}{6}=\frac{1\times3}{2\times3}$
$=\frac{1}{2}$
This implies,
$\frac{2}{4} = \frac{3}{6}$
(c) LCM of denominators 5 and 3 is 15.
This implies,
$\frac{3}{5}\times\frac{3}{3}=\frac{3\times3}{5\times3}$
$=\frac{9}{15}$
$\frac{2}{3}\times\frac{5}{5}=\frac{2\times5}{3\times5}$
$=\frac{10}{15}$
Therefore,
$\frac{9}{15}<\frac{10}{15}$
This implies,
$\frac{3}{5} < \frac{2}{3}$
(d) $\frac{2}{8}=\frac{1\times2}{4\times2}$
$=\frac{1}{4}$
$3>1$
This implies,
$\frac{3}{4} > \frac{1}{4}$
Therefore,
$\frac{3}{4} > \frac{2}{8}$
(e) Here, the denominators are same.
Therefore, the fraction having greater numerator is greater.
This implies,
$\frac{6}{5} > \frac{3}{5}$
Therefore,
$\frac{3}{5} < \frac{6}{5}$
(f) Here, the denominators are same.
Therefore, the fraction having greater numerator is greater.
This implies,
$\frac{7}{9} > \frac{3}{9}$
(g) $\frac{2}{8}=\frac{2\times1}{2\times4}$
$=\frac{1}{4}$
Therefore,
$\frac{1}{4} = \frac{2}{8}$
(h) $\frac{6}{10}=\frac{2\times3}{2\times5}$
$=\frac{3}{5}$
Now, the denominator of $\frac{3}{5}$ and $\frac{4}{5}$ are same.
$4>3$
This implies,
$\frac{4}{5} > \frac{3}{5}$
Therefore,
$\frac{6}{10} < \frac{4}{5}$
(i) We can write $\frac{3}{4}$ as, $\frac{3}{4}=\frac{3\times2}{4\times2}$
$=\frac{6}{8}$
Now, the denominators of $\frac{6}{8}$ and $\frac{7}{8}$ are same.
$7>6$
This implies,
$\frac{7}{8} > \frac{6}{8}$
Therefore,
$\frac{3}{4} < \frac{7}{8}$
(j) $\frac{6}{10}=\frac{3\times2}{5\times2}$
$=\frac{3}{5}$
Here, numerators are same.
Therefore,
$\frac{6}{10} = \frac{3}{5}$
(k) $\frac{15}{21}=\frac{3\times5}{3\times7}$
$=\frac{5}{7}$
Here, numerators are same.
Therefore,
$\frac{5}{7} = \frac{15}{21}$