Find:(i) $\frac{7}{3}\div2$(ii) $\frac{4}{9}\div5$(iii) $\frac{6}{13}\div7$(iv) $4\frac{1}{3}\div3$(v) $3\frac{1}{2}\div4$(vi) $4\frac{3}{7}\div7$

To do:

We have to find

(i) $\frac{7}{3}\div2$

(ii) $\frac{4}{9}\div5$

(iii) $\frac{6}{13}\div7$

(iv) $4\frac{1}{3}\div3$

(v) $3\frac{1}{2}\div4$

(vi) $4\frac{3}{7}\div7$

Solution:

We know that,

$\frac{a}{b} \div \frac{c}{d}=\frac{a}{b}\times \frac{c}{d}$

Therefore,

(i) $\frac{7}{3}\div2=\frac{7}{3}\times\frac{1}{2}$

$=\frac{7\times1}{3\times2}$

$=\frac{7}{6}$

(ii) $\frac{4}{9}\div5=\frac{4}{9}\times\frac{1}{5}$

$=\frac{4\times1}{9\times5}$

$=\frac{4}{45}$

(iii) $\frac{6}{13}\div7=\frac{6}{13}\times\frac{1}{7}$

$=\frac{6\times1}{13\times7}$

$=\frac{6}{91}$

(iv) $4\frac{1}{3}\div3=\frac{4\times3+1}{3}\times\frac{1}{3}$

$=\frac{12+1}{3}\times\frac{1}{3}$

$=\frac{13}{3}\times\frac{1}{3}$

$=\frac{13\times1}{3\times3}$

$=\frac{13}{9}$

(v) $3\frac{1}{2}\div4$

$=\frac{3\times2+1}{2}\times\frac{1}{4}$

$=\frac{6+1}{2}\times\frac{1}{4}$

$=\frac{7}{2}\times\frac{1}{4}$

$=\frac{7\times1}{2\times4}$

$=\frac{7}{8}$

(vi) $4\frac{3}{7}\div7$

$=\frac{4\times7+3}{7}\times\frac{1}{7}$

$=\frac{28+3}{7}\times\frac{1}{7}$

$=\frac{31}{7}\times\frac{1}{7}$

$=\frac{31\times1}{7\times7}$

$=\frac{31}{49}$

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