Question

Write 'T' for true and 'F' for false

(i) $\frac{9}{16}<\frac{13}{24}.$
(ii) Among $\frac{2}{5},\frac{16}{35}$ and $\frac{9}{14}$, the largest is $\frac{16}{35}.$
(iii) $\frac{11}{15}-\frac{9}{20}=\frac{17}{60}.$
(iv) $\frac{11}{25}$ of a litre = 440 mL.
(v) $16\frac{3}{4}\times 6\frac{2}{5}=107\frac{3}{10}.$

Answer 1

(i) F

By cross multiplication, we have:

9 $\times$ 24 = 216 and 13 $\times$ 16 = 208

However, 216 > 208

∴ $\frac{9}{16}>\frac{13}{24}$

(ii) F

The LCM of 5, 35 and 14 is 70.

Now, $\frac{2}{5}=\frac{2}{5}\times \frac{14}{14}=\frac{28}{70};\frac{16}{35}=\frac{16}{35}\times \frac{2}{2}=\frac{32}{70}\mathrm{and}\frac{9}{14}=\frac{9}{14}\times \frac{5}{5}=\frac{45}{70}$

Clearly, $\frac{28}{70}<\frac{32}{70}<\frac{45}{70}$

∴ $\frac{2}{5}<\frac{16}{35}<\frac{9}{14}$

(iii) T

The LCM of 15 and 20 = (5 $\times$ 3 $\times$ 4) = 60

∴ $\frac{11}{15}-\frac{9}{20}=\frac{44-27}{60}=\frac{17}{60}$
(iv) T

$\frac{11}{25}$ of 1 L = $\frac{11}{25}$ of 1000 ml = $\left(1000\times \frac{11}{25}\right)$ ml = (40 $\times$ 11) ml = 440 ml

(v) F

$16\frac{3}{4}\times 6\frac{2}{5}$= $\left(\frac{67}{4}\times \frac{32}{5}\right)=\left(\frac{67\times 32}{4\times 5}\right)=\left(\frac{67\times 8}{5}\right)=\frac{536}{5}=107\frac{1}{5}$