Question

Arrange the following rational number in increasing order. [4 MARKS]

(i) $\frac{4}{3}$, $\frac{3}{2}$, $\frac{7}{9}$, $\frac{5}{4}$.

(ii) $\frac{8}{5}$, $\frac{12}{7}$, $\frac{9}{2}$, $\frac{23}{20}$.

Answer 1

Each part: 2 Marks

(i) Find the LCM of denominators, i.e. the LCM of 3, 2, 9 and 4 = 36.

Now make the denominators common to the above fraction.

So, $\frac{4\times 12}{3\times 12}$ = $\frac{48}{36}$ ; $\frac{3\times 18}{2\times 18}$ = $\frac{54}{36}$ ; $\frac{7\times 4}{9\times 4}$ = $\frac{28}{36}$ ; $\frac{5\times 9}{4\times 9}$ = $\frac{45}{36}$;

Now arrange the fractions according to their numerator, fraction having large magnitude of numerator will be of greater magnitude.

So, correct sequence is $\frac{28}{36}$, $\frac{45}{36}$, $\frac{48}{36}$, $\frac{54}{36}$.

$\Rightarrow \frac{7}{9}<\frac{5}{4}<\frac{4}{3}<\frac{3}{2}$


(ii) Find the LCM of denominators, i.e. the LCM of 5, 7, 2 and 20 = 140.

Now make the denominators common to the above fraction.

So, $\frac{8\times 28}{5\times 28}$ = $\frac{224}{140}$;
$\frac{12\times 20}{7\times 20}$ = $\frac{240}{140}$;
$\frac{9\times 70}{2\times 70}$ = $\frac{630}{140}$;
$\frac{23\times 7}{20\times 7}$ = $\frac{161}{140}$;

Now arrange the fractions according to their numerator, fraction having large magnitude of numerator will be of greater magnitude.

So, correct sequence is $\frac{161}{140}$, $\frac{224}{140}$, $\frac{240}{140}$, $\frac{630}{140}$.

$\Rightarrow \frac{23}{20}<\frac{8}{5}<\frac{12}{7}<\frac{9}{2}$